Alexander Viktorovich Abrosimov Date of birth: November 16, 1948 (1948 11 16) Place of birth: Kuibyshev Date of death ... Wikipedia

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Makarskaya E. V. In the book: Days of student science. Spring - 2011. M.: Moscow State University of Economics, Statistics and Informatics, 2011. P. 135-139.

The authors consider the practical application of the theory of linear differential equations for the study of economic systems. The paper provides an analysis of the dynamic models of Keynes and Samuelson Hicks with the finding of equilibrium states of economic systems.

Ivanov A. I., Isakov I., Demin A.V. et al. 5. M.: Word, 2012.

The manual examined the quantitative methods for the study of oxygen consumption by man during tests with dosage exercise performed in the GSC of the Russian Academy of Sciences-AMBP RAS. The manual is intended for scientists, physiologists and doctors working in the field of aerospace, underwater and sports medicine.

Mikheev A. V. SPb.: Department of Operational Printing NSU HSE - St. Petersburg, 2012.

This collection contains tasks at the rate of differential equations, readable by the author at the Faculty of Economics Economics HSE - St. Petersburg. At the beginning of each topic, a brief summary of the main theoretical facts is given and examples of solutions of typical tasks are dealt. For students and listeners of higher professional professional programs.

Konakov V.D. STI. WP BRP. Publishing the Board of Trustees of the Mechanics and Mathematics Faculty of Moscow State University, 2012. 2012.

At the heart of this training manual, there is a special course on the selection of a student, read by the author on the Mechanically - Mathematical Faculty of Moscow State University. M.V. Lomonosov in 2010-2012 academic years. The manual introduces the reader with the parameterix method and its discrete analogue developed in the most lately The author of the benefit and his colleagues co-authors. It combines together the material that was previously kept only in a number of magazine articles. Not striving for the maximum generality of the presentation, the author set the goal to demonstrate the possibilities of the method in the proof of local limit theorems on the convergence of Markov chains to the diffusion process and upon receipt of bilateral assessments of the ARONSON type for some degenerate diffusion.

ISS. 20. NY: Springer, 2012.

This publication is a collection of individual articles "Third International Conference on Information Systems", which took place at the University of Florida, February 16-18, 2011. The purpose of this conference was to collect together scientists and engineers from industry, governments and scientific circles So that they can exchange new discoveries and results in matters related to theory and practice of the dynamics of information systems. Dynamics of information systems: Mathematical discovery is a modern study and is intended to students - graduate students and researchers who are interested in the latest discoveries in informational theory and dynamic Systems. Scientists of other disciplines can also benefit from the application of new developments in their areas of research.

Palvelev R., Sergeyev A. G. Proceedings of the Mathematical Institute. V.A. Steklov wounds. 2012. T. 277. P. 199-214.

The adiabatic limit is studied in Landau Ginzburg hyperbolic equations. Using the specified limit, the correspondence is established between the solutions of the Ginzburg-Landau equations and adiabatic trajectories in the space of static solutions, called the vortices. Menthon proposed the heuristic adiabatic principle that postulates that any solution of the Ginzburg-Landau equations with a sufficiently low kinetic energy can be obtained as a perturbation of some adiabatic trajectory. Strict proof of this fact found recently the first author

We Give An Explicit Formula for A Quasi-ISomorphism Between The Operads Hycomm (The Homology of The Homotopy Quitient of Batalin-Vilkovisky Operad by The BV-operator). In Oter Words We Derive An Equivalence of Hycomm-Algebras and BV-Algebras Enhanced with a Homotopy That Trivializes The BV-Operator. These Formulas Are Given in Terms of the Givental Graphs, And Are Proved in Two Different Ways. One Proof Use The Givental Group Action, And The Other Proof Goes Through A Chain of Explicit Formulas On Resolutions of Hycomm and BV. The Second Approach Gives, in Particular, A Homological Explanation of the Givental Group Action on Hycomm-Algebras.

Under scientific Edited: A. Mikhailov. 14. M.: Sociological Faculty of Moscow State University, 2012.

Articles of this collection are written on the basis of reports made in 2011 at the Sociological Faculty of Moscow State University. M.V. Lomonosov at a meeting of the XIV interdisciplinary annual scientific seminar "Mathematical modeling of social processes" them. Hero of Socialist Labor Academician A.A. Samara.

The publication is intended for researchers, teachers, students of universities and scientific institutions of wounds interested in problems, development and implementation of the methodology for mathematical modeling of social processes.

Ministry of Education and Science of the Russian Federation National Research Nuclear University "MIFI" T. I. Bukharova, V. L. Kamynin, A. B. Kostin, D. S. Tkachenko Lectures on ordinary differential equations Recommended by UMO Nuclear Physics and Technologies quality of textbook for students of higher educational institutions Moscow 2011 UDC 517.9 BBK 22.161.6 B94 Bukharova T.I., Kamynin V.L., Kostin A.B., Tkachenko D.S. Common lecture course differential equations : Tutorial. - M.: Niau Mafi, 2011. - 228 p. The training manual was created on the basis of the course of lectures read by the authors in the Moscow Engineering and Physical Institute over the years. It is intended for students by Niya mythi of all faculties, as well as for students of universities with increased mathematical preparation. The manual was prepared in the framework of the program for the creation and development of Niya Mafi. Reviewer: Dr. Fiz.-Mat. Sciences N.A. Kudryashov. ISBN 978-5-7262-1400-9 © National Research Nuclear University "MIII", 2011 Table of contents Preface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 I. Introduction to the theory of ordinary differential equations Basic concepts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cauchy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 6 11 II. The existence and uniqueness of the solution of the Cauchy problem for the 1st order equation is theorem of the uniqueness for the first order. . . . . . . . . . . . . . . . . . . Solving the solution of the Cauchy problem for the first order. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continue the solution for the first order. . . . . . . . . . . . . . . . . . . . III. Cauchy's task for normal N-th order system basic concepts and some auxiliary properties of vector function. . . . Identity of solving the Cauchy problem for normal system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; . The concept of metric space. Priniype of squeezing displays. . . . . . Theores of the problem and the uniqueness of the solution of the Cauchy problem for normal systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 14 23 34 38 34 43 44 48 IV. Some classes of ordinary differential equations solved in quadratures Equation with separating variables. . . . . . . . . . . . . . . . . . . . . . . Linear first order. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniform equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equation Áernlli. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equation in complete differentiated dealers. . . . . . . . . . . . . . . . . . . . . . . . . . 55 55 58 63 64 65 V. 67 The first-order equations that are not permitted relative to the derivative of the problem of the problem and the uniqueness of the solution of the OÄU, not permitted relative to the derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Special solution. ÄChrimnant curve. Enveling. . . . . . . . . . . . . . . . The parameter administration parameter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lagrana equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Clairo equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vi. Linear ODE systems basic concepts. Theorem of the problem and the uniqueness of solving the problem homogeneous systems of linear OÄU. . . . . . . . . . . . . . . . . . . . . . . . The determinant is ârnoy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Complex solutions for a homogeneous system. Transition to essential CCR. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The systemic systems of linear Oäu. ÌTode Variaii constant. . . . . Uniform systems of linear OÄU with constant cobfinger. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Indicative function from the matrian. . . . . . . . . . . . . . . . . . . . . . . . 3 67 70 77 79 81 85 Cauchy 85. . . 87. . . 91. . . . . . 96 97. . . 100 . . . 111 The systemic systems of linear OÄU with constant cobfinger. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 VII. Linear high-order ADU is reduced to system linear OÄU. Theorem of the problem and the uniqueness of the solution of the Cauchy problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniform linear high order. . . . . . . . . . . . . . . . . . . . . . Properties of complex solutions of a homogeneous linear high order. Transition from the complex CPC to the essential. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Foundation linear high-order lines. ÌTode Variaii constant. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniform linear high-order solutions with constant cobfinger. . . . . . . . . . . . . . . . . . . . . . . . . . . This is a high-order linear linear with constant co-effect. . . . . . . . . . . . . . . . . . . . . . . . . . . 126 VIII. Theory of stability The basic concepts and definitions relating to sustainability. . . . . . . . . . . . . . . . . . . . Stability of the solutions of the linear system. . . . . . Oregious Lyapunov on stability. . . . . . . . . . Stability according to the first approximation. . . . . . . The behavior of phase trajectories near the rest point 162. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 128 136 139 142 150 162 168 172 182 187 IX. The first integrals of the ODU 198 systems 198 The first integrals of autonomous systems of ordinary different delicual equations198 The autonomous systems of the OÄU. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 Symmetric record systems OÄU. . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 X. Equations in private derivatives of the first order Uniform linear equations in the private derivatives of the first order of the Cauchy ç for a linear equation in the private derivatives of the first order. . . . . . . . . . . . . . . . . . . . . . . Quasilinear equations in private first-order derivatives. . . . Cauchy has a quasilinear equation in the private derivatives of the first order. . . . . . . . . . . . . . . . . . . . . . . Bibliography. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . -4- 210. . . . . 210. . . . . 212. . . . . 216. . . . . 223. . . . . 227 Preface When preparing the book, the authors were aimed at assessing in one place and state information on most issues related to the theory of ordinary differential equations in an affordable form. Therefore, in addition to the material included in the mandatory program of the course of ordinary differential equations reading in NIII (and in other universities), the allowance includes additional questions for which, as a rule, there is not enough time at lectures, but which will be useful for a better understanding. subjects and will come up with current students in their further professional activities. All statements of proposed benefits are mathematically strict evidence. These evidence are usually not original, but all are redesigned in accordance with the style of presentation of mathematical courses in the MEPh. According to widespread among teachers and scientists, mathematical disciplines should be studied with full and detailed evidence, moving gradually from simple to complex. The authors of this manual adhere to the same opinion. The theoretical information cited in the book is supported by the analysis of a sufficient number of examples that, as we hope, simplify the reader to study the material. The manual is addressed to students of universities with increased mathematical preparation, first of all, students of Niya MEPI. At the same time, it will also be useful to everyone who is interested in the theory of differential equations and uses this section of mathematics in its work. -5- Chapter I. Introduction to the theory of ordinary differential equations 1. 1. Basic concepts everywhere in the manual through HA, Bi will be denoted by any of the sets (A, B) ,, (a, b], we obtain X0 2 ZX LN 4C + 3 U (T) V (T) DT5 ZX V (T) DT. LN C 6 x0 x0 After the potentiation of the last inequality and use (2.3), we have 2 x 3 zx z u (x) 6 C + U (T) V (T) DT 6 C EXP 4 V (T) DT5 x0 x0 at all x 2 [1, 1]. We estimate the difference JF (x, y2) f (x, y1) j \u003d sin x y1 y2 6 at all (x , y) 2 g. Thus, f satisfies the Lipschitz condition with L \u003d 1 in fact, even with L \u003d SIN 1 along y. However, the FY0 derivative at the points (x, 0) 6 \u003d (0, 0) does not even exist. The following theorem, an interesting in itself, will allow to prove the uniqueness of the solution of the Cauchy problem. Theorem 2. 1 (on the assessment of the difference between two solutions). Let G region 2 in R, and f (x, y) 2 Cg and satisfies in the G Lipschitz condition Y with constant L. If Y1, Y2 two solutions of the equation y 0 \u003d F (x, y) on the segment, then it is fair to inequality (evaluation): JY2 (X) Y1 (X) J 6 JY2 (X0) Y1 (x0) j exp L (x x0) 6 y1 at all x 2. -19- y2 proof. By definition 2. 2 solutions of equation (2.1) we obtain that 8 x 2 points x, y1 (x) and x, y2 (x) 2 g. For all T 2, we have faithful equalities Y10 (T) \u003d FT, Y1 (T ) and y20 (t) \u003d ft, y2 (t), which integrate by t on the segment, where x 2. Integration is legal, as the right and left parts are continuous on functions. We obtain the system of equalities ZX Y1 (X) Y1 (x0) \u003d x0 zx y2 (x) y2 (x0) \u003d f t, y1 (t) dt, f t, y2 (t) dt. X0 subtracting one of the other, we have JY1 (X) Y2 (X) J \u003d Y1 (X0) Y2 (X0) + ZX HFT, Y1 (T) IFT, Y2 (T) DT 6 X0 ZX 6 Y1 (X0) Y2 ( x0) + ft, y1 (t) ft, y2 (t) dt 6 x0 zx 6 y1 (x0) y2 (x0) + l y1 (t) y2 (t) dt. X0 Denote by C \u003d Y1 (x0) Y2 (x0)\u003e 0, V (T) \u003d L\u003e 0, U (T) \u003d Y1 (T) Then, in the inequality of Hronolla-áellman, we obtain a rating: JY2 (X) Y1 (X) J 6 JY2 (X0) Y1 (x0) j exp L (x x0) Y2 (T)\u003e 0. For all x 2. Theorem is proved. As a result of a proven theorem, we obtain the theorem of the uniqueness of the solution of the Cauchy problem (2.1), (2.2). Alternate 1. Let the function f (x, y) 2 C G and satisfies the Lipschitz condition in Y, and the functions of Y1 (X) and Y2 (X), the two solutions of equation (2.1) on the same segment, and X0 2. If y1 (x0) \u003d y2 (x0), then y1 (x) y2 (x) on. Evidence. Consider two cases. -20- 1. Let x\u003e x0, then, from Theorem 2. 1 it follows that H i i.e. Y1 (x) Y1 (x) Y2 (x) 6 0 exp L (x x0), y2 (x) at x\u003e x0. 2. Let x 6 x0 let the replacement t \u003d x, then yi (x) \u003d yi (t) y ~ i (t) at i \u003d 1, 2. Since x 2, T 2 [x0, x1] and the equality y ~ 1 (x0) \u003d y ~ 2 (x0) is made. We find out what equation is satisfying y ~ i (t). The next chain of equalities is true: d y ~ i (t) \u003d dt d ~ yi (x) \u003d dx f x, yi (x) \u003d f (t, y ~ i (t)). Here we took advantage of the differentiation of the complex function and the fact that Yi (x) is solutions of equation (2.1). Since the function f ~ (t, y) f (t, y) is continuous and satisfies the Lipschitz condition for Y, then by Theorem 2. 1 we have that y ~ 1 (t) y ~ 2 (t) on [x0, x1 ], i.e. y1 (x) y2 (x) on. Combining both cases considered, we obtain the approval of the investigation. Alternatively 2. (On continuous dependence on the initial data) let the function f (x, y) 2 Cg and satisfies the Lipschitz condition along Y with the constant L, and the functions y1 (x) and y2 (x) are solutions of equation (2.1) defined on. We correspond to l \u003d x1 x0 and δ \u003d y1 (x0) y2 (x0). If at 8 x 2, the inequality y1 (x) y2 (x) 6 Δ el l is true. The proof should be immediately from Theorem 2. 1. Inequality from Corollary 2 is called the assessment of the sustainability of the decision on initial data. Its meaning is that if at x \u003d x0, the solutions are "close", then on the final segment they are also close. Theorem 2. 1 gives an important estimate of the difference module of two solutions for applications, and a consequence of 1 is the uniqueness of the solution of the Cauchy problem (2.1), (2.2). There are also other sufficient conditions for uniqueness, one of which we now give. As noted above, the geometrically the uniqueness of the solution of the Cauchy problem means that through a point (x0, y0) region G can pass no more than one integral curve of equation (2.1). Theorem 2. 2 (Osguda about uniqueness). Suppose that the function f (x, y) 2 cg and for 8 (x, y1), (x, y2) 2 g is performed inequality f (x, y1) f (x, y2) 6 6 φ jy1 y2 j, where φ ( U)\u003e 0 at u 2 (0, β], φ (u) is continuous, and zβ du! +1, when ε! 0+. If the point (x0, y0), the region φ (u) ε g is not more One integral curve (2.1). -21- Proof. Suppose there are two solutions y1 (x) and y2 (x) of equation (2.1), such that y1 (x0) \u003d y2 (x0) \u003d y0, denotes z (x) \u003d y2 (x) y1 (x). DYI as \u003d f (x, yi), at i \u003d 1, 2, then for z (x) the equality dx dz \u003d f (x, y2) f (x, y1 ). dx dz \u003d f (x, y2) f (x, y1) jzj 6 φ jzj jzj, i.e. right then z dx 1 D inequality jzj2 6 φ jzj jzj, from which it follows with JZJ 6 \u003d 0 DX Dual inequality: zjz2 j zx2 dx 6 x1 2 D jzj 6 2 jzjφ jzj zx2 dx, (2.5) x1 jz1 j where integration is carried out according to any segment on which z (x)\u003e 0, and zi \u003d z (xi), i \u003d 1, 2. By assumption, z (x) 6 0 and, moreover, it is continuous, so such a segment is found, we will choose it and fix it. Consider the sets n o x1 \u003d x x< x1 и z(x) = 0 , n o X2 = x x > x2 and z (x) \u003d 0. If one of these sets is not empty, since z (x0) \u003d 0 and x0 62. Let, for example, x1 6 \u003d ∅, it is limited from above, therefore 9 α \u003d sup x1. Note that z (α) \u003d 0, i.e. α 2 x1, since it is assumed that z (α)\u003e 0, by virtue of continuity, we will have z (x)\u003e 0 at some interval α δ1, α + δ1, and this is contrary to the definition α \u003d sup x1. From the condition Z (α) \u003d 0 it follows that α< x1 . По построению z(x) > 0 for all x 2 (α, x2], and by virtue of the continuity z (x)! 0+ at x! Α + 0. Repeat the arguments in pin (2.5), integrating on the segment [α + δ, x2], where x2 It was chosen above and fixed, and δ 2 (0, x2 α) - arbitrary, we obtain the inequality: zjz2 j zx2 dx 6 α + δ d jzj2 6 2 jzjφ jzj jz (α + δ) j zx2 dx. α + Δ in this double I will define Δ! 0+, then z (α + δ)! z (α) \u003d 0, from zjz2 jd jzj2! +1, according to the continuity of z (x), and then the integral 2 jzjφ jzj theorem. jz (α + Δ) j -22- right part of the inequality RX2 DX \u003d X2 α δ 6 x2 α is limited α + Δ from above the final value, which is simultaneously impossible. The resulting contradiction proves theorem. 2. 2. Essential the solution of the Cauchy problem for the first order What is the task of Cauchy (2.1), (2.2) it is understood as the following task of finding the function y (x): 0 y \u003d f (x, y), (x, y) 2 g, y (x0) \u003d y0, (x0, y0 ) 2 g, where f (x, y) 2 cg and (x0, y0) 2 g; g is a region in R2. Lemma 2. 2. Let F (X, Y) 2 CG. If the following statements take place: 1 ) Everything re φ (x) equation (2.1) on the range of HA, Bi, satisfying (2.2) x0 2 ha, Bi, is a solution on HA, Bi integral equation ZX y (x) \u003d y0 + f τ, y (τ) dτ ; (2.6) x0 2) if φ (x) 2 C ha, Bi solution of the integral equation (2.6) on HA, Bi, 1 where X0 2 HA, Bi, then φ (x) 2 C ha, Bi and is a solution (2.1 ), (2.2). Evidence. 1. Let φ (x), the decision (2.1), (2.2) on HA, BI. Then, according to the remark 2.2 φ (x) 2 C ha, Bi and 8 τ 2 ha, Bi, we have the equality φ 0 (τ) \u003d f τ, φ (τ), integrating which from x0 to x, we get (at any X 2 HA , Bi) rx φ (x) φ (x0) \u003d f τ, φ (τ) dτ, with φ (x0) \u003d y0, i.e. φ (x) - Solution (2.6). x0 2. Let y \u003d φ (x) 2 C ha, Bi - solution (2.6). Since f x, φ (x) is continuous on HA, Bi by condition, then zx φ (x) y0 + f τ, φ (τ) dτ 2 C 1 HA, Bi x0 as an integral with a variable upper limit from the continuous function. Differentiating the last equality of x, we obtain φ 0 (x) \u003d f x, φ (x) 8 x 2 ha, bi and, obviously, φ (x0) \u003d y0, i.e. φ (x) is the solution of the Cauchy problem (2.1), (2.2). (As usual, under the derivative at the end of the segment it is understood as the corresponding one-sided derivative.) -23 "Review 2. 6. Lemma 2. 2 is called the lemma about the Cauchy problem (2.1), (2.2) of the integral equation (2.6). If we prove that the solution of equation (2.6) exists, we obtain the solvability and objectives of the Cauchy (2.1), (2.2). This plan is implemented in the following theorem. Theorem 2. 3 (local existence theorem). Let the rectangle p \u003d (x, y) 2 R2: JX X0 J 6 α, JY Y0 J 6 β lies entirely in G of the function of determining the function f (x, y). The F (X, Y) 2 C G and satisfies the Lipschitz condition for N y OV G with constant L. The corresponding β m \u003d max f (x, y), h \u003d min α, m. If there is a solution of the task of êshoshi (2.1), (2.2). Evidence. On the cut, we establish the existence of a solution of the integral equation (2.6). To do this, consider the following sequence of functions: zx y0 (x) \u003d y0, y1 (x) \u003d y0 + f τ, y0 (τ) dτ, ... x0 zx yn (x) \u003d y0 + f τ, yn 1 (τ ) Dτ, etc. x0 1. We show that 8 n 2 n functions yn (successive approximations) are defined, i.e. We show that at 8 x 2, the inequality yn (x) y0 6 β is performed for all n \u003d 1, 2 ,. . . We use the method of mathematical induction (MMI): a) the induction basis: n \u003d 1. zx y1 (x) y0 \u003d f τ, y0 (τ) dτ 6 m0 x0 6 m h 6 β, x0 where m0 \u003d max f (x , y0) at jx x 0 j 6 α, m0 6 m; b) assumption and induction step. Let the inequality are true for yn 1 (x), we prove it for yn (x): zx yn (x) y0 \u003d f τ, yn 1 (τ) dτ 6 m x x0 So, if jx x0 j 6 h, then yn ( x) Y0 6 β 8 N 2 N. -224 - X0 6 M H 6 β. Our goal will be the proof of convergence follower of the nearest 1st yk (x) k \u003d 0, for this it is convenient to represent it in the form: yn \u003d y0 + n x yk 1 (x) \u003d y0 + y1 yk (x) y0 + y2 y1 +. . . + yn yn 1, k \u003d 1 i.e. Sequences of partial sums of the functional series. 2. We estimate the members of this series by proving the following inequalities 8 N 2 N and 8 x 2: X0 jn yn (x) yn 1 6 m0 l 6 m0 ln n! Apply the method of mathematical induction: JX N 1 1 HN. N! (2.7) a) induction base: n \u003d 1. y1 (x) x y 0 6 m0 x0 6 m0 h, proved above; b) assumption and induction step. Let the inequality are true for n each for n: zx yn (x) yn 1 f τ, yn 1 (τ) \u003d f τ, yn 2 (τ) 1, dτ 6 x0 zx i yn 6 for Lipsshitz 6 L H Yn 1 2 Dτ 6 x0 H ZX i 6 by the assumption of induction 6 L n 2 m0 l jτ x0 jn 1 dτ \u003d (n 1)! x0 m0 ln 1 \u003d (n 1)! ZX Jτ N 1 x0 j m0 ln 1 jx x0 jn m0 l n 6 dτ \u003d (n 1)! N N! 1 x0 Rx Here we used the fact that the integral I \u003d Jτ x0 at x\u003e x0 with x< x0 Rx I = (τ x0 Rx I = (x0 n 1 x0) τ)n 1 dτ = dτ = x0 (x (x0 x)n n Таким образом, неравенства (2.7) доказаны. -25- x0)n и n = jx x0 jn . n x0 jn 1 dτ : hn . 3. Рассмотрим тождество yn = y0 + ним функциональный ряд: y0 + 1 P n P yk (x) yk 1 (x) и связанный с k=1 yk 1 (x) . Частичные суммы это- yk (x) k=1 го ряда равны yn (x), поэтому, доказав его сходимость, получим сходимость 1 последовательности yk (x) k=0 . В силу неравенств (2.7) функциональный ряд мажорируется на отрезке k 1 P k 1 h числовым рядом M0 L . Этот числовой ряд сходится k! k=1 по признаку Даламбера, так как M0 Lk hk+1 k! ak+1 = ak (k + 1)! M0 L k 1 hk = h L ! 0, k+1 k ! 1. Тогда по признаку Вейерштрасса о равномерной сходимости функциональный 1 P ряд y0 + yk (x) yk 1 (x) сходится абсолютно и равномерно на отрезk=1 ке , следовательно и функциональная последовательность 1 yk (x) k=0 сходится равномерно на отрезке к некоторой функ- ции ϕ(x), а так как yn (x) 2 C , то и ϕ(x) 2 C как предел равномерно сходящейся последовательности непрерывных функций. 4. Рассмотрим определение yn (x): Zx yn (x) = y0 + f τ, yn 1 (τ) dτ (2.8) x0 – это верное равенство при всех n 2 N и x 2 . У левой части равенства (2.8) существует предел при n ! 1, так как yn (x) ⇒ ϕ(x) на , поэтому и у правой части (2.8) тоже существует Rx предел (тот же самый). Покажем, что он равен функции y0 + f τ, ϕ(τ) dτ , x0 используя для этого следующий критерий равномерной сходимости функциональной последовательности: X yn (x) ⇒ ϕ(x) при n ! 1 () sup yn (x) y(x) ! 0 при n ! 1 . x2X X Напомним, что обозначение yn (x) ⇒ ϕ(x) при n ! 1 принято использовать для равномерной на множестве X сходимости функциональной последователь 1 ности yk (x) k=0 к функции ϕ(x). -26- Покажем, что y0 + Rx X f τ, yn 1 (τ) dτ ⇒ y0 + x0 здесь X = . Для этого рассмотрим f τ, yn 1 (τ) f τ, ϕ(τ) dτ 6 x2X x0 Zx h i yn 1 (τ) 6 по условию Липшица 6 sup L ϕ(τ) dτ 6 x2X x0 6 L h sup yn 1 (τ) ϕ(τ) ! 0 при n ! 1 τ 2X X в силу равномерной при n ! 1 сходимости yn (x) ⇒ ϕ(x). Таким образом, переходя к пределу в (2.8), получим при всех x 2 верное равенство Zx ϕ(x) = y0 + f τ, ϕ(τ) dτ, x0 в котором ϕ(x) 2 C . По доказанной выше лемме 2. 2 ϕ(x) – решение задачи Коши (2.1), (2.2). Теорема доказана. Замечание 2. 7. В теореме 2. 3 установлено существование решения на отрезке . По следствию 1 из теоремы 2. 1 это решение единственно в том смысле, что любое другое решение задачи Коши (2.1), (2.2), определенное на совпадает с ним на этом отрезке. Замечание 2. 8. Представим прямоугольник P в виде объединения двух (пересекающихся) прямоугольников P = P [ P + , (рис. 2. 3) где n P = (x, y) n P = (x, y) + x 2 ; x 2 ; -27- jy jy o y0 j 6 β , o y0 j 6 β . Рис. 2. 3. Объединение прямоугольников Обозначим f (x, y . M − = max − f (x, y , M + = max + P P Повторяя,с очевидными изменениями, доказательство теоремы 2. 3 отдель но для P + или P − , получим существование (и единственность) решения на отрезке n o β + + , где h = min α, M + или, соответственно, на , n o β − . Отметим, что при этом, вообще говоря, h+ 6= h− , а h h = min α, M − из теоремы 2. 3 есть минимум из h+ и h− . Замечание 2. 9. Существование решения задачи (2.1), (2.2) теоремой 2. 3 гарантируется лишь на некотором отрезке . В таком случае говорят, что теорема является локальной. Возникает вопрос: не является ли локальный характер теоремы 2. 3 следствием примененного метода ее доказательства? Может быть, используя другой метод доказательства, можно установить существование решения на всем отрезке , т.е. глобально, как это было со свойством единственности решения задачи Коши (2.1), (2.2)? Следующий пример показывает, что локальный характер теоремы 2. 3 связан с «существом» задачи, а не с методом ее доказательства. Пример 2. 1. Рассмотрим задачу Коши 0 y = −y 2 , (2.9) y(0) = 1 n o в прямоугольнике P = (x, y) jxj 6 2, jy − 1j 6 1 . Функция f (x, y) = −y 2 непрерывна в P и fy0 = −2y 2 C P , поэтому все условия тео1 β , α = и ремы 2. 3 выполнены, а M = max f (x, y) = 4. Тогда h = min P P M 4 -28- теорема 2. 3 гарантирует существование решения на отрезке 1 1 . Решим − , 4 4 эту задачу Коши, используя «разделение переменных»: − dy = dx y2 () y(x) = 1 . x+C 1 – решение задачи Коши (2.9). x+1 График решения представлен на рис. (2.4), из которого видно, что решение 1 при x < x = − покидает прямоугольник P , а при x 6 −1 даже не 2 существует. Подставляя x = 0, найдем C = 1 и y(x) = Рис. 2. 4. Локальный характер разрешимости задачи Коши В связи с этим возникает вопрос об условиях, обеспечивающих существование решения на всем отрезке . На приведенном примере мы видим, что решение покидает прямоугольник P , пересекая его «верхнее» основание, поэтому можно попробовать вместо прямоугольника P в теореме 2. 3 взять полосу: o n 2 A 6 x 6 B − 1 < y < +1 , A, B 2 R. Q = (x, y) 2 R Оказывается, что при этом решение существует на всем отрезке A, B , если f (x, y) удовлетворяет условию Липшица по переменной y в Q. А именно, имеет место следующая важная для приложений теорема. Теорема 2. 4. Пусть функция f (x, y) определена, непрерывна и удовлетворяет условию Липшица по y с константой L в полосе Q = (x, y) 2 R2: A 6 x 6 B, y 2 R , -29- где A, B 2 R. Òогда при любых начальных данных x0 2 , y0 2 R т.е. (x0 y0) 2 Q существует и притом единственное решение задачи Êоши (2.1), (2.2), определенное на всем . Доказательство. Áудем считать, что x0 2 (A, B). Проведем рассуждения по схеме теоремы 2. 3 отдельно для полосы o n + 2 x 2 y 2 R и Q = (x, y) 2 R n Q = (x, y) 2 R2 o x 2 y 2 R . + Если x0 = A или x0 = B, то один из этапов рассуждений (для Q или, соответственно, для Q) отсутствует. + Возьмем полосу Q и построим последовательные приближения yn+ (x), + как в теореме 2. 3. Поскольку Q не содержит ограничений на размер по y, то пункт 1) доказательства теоремы 2. 3 не проверяем. Далее, как в предыдущей теореме, от последовательности переходим к ряду с частичными суммами yn+ (x) = y0 + n X yk+ (x) yk+ 1 (x) , где x 2 . k=1 Повторяя рассуждения, доказываем оценку вида (2.7) x0 jn x0)n n 1 (B 6 M0 L 6 M0 L (2.10) n! n! при всех x 2 ; здесь M0 = max f (x, y0) при x 2 , откуда yn+ (x) yn+ 1 n 1 jx yn+ (x) как и выше в теореме 2. 3 получим, что ⇒ ϕ+ (x), n ! 1, причем ϕ+ (x) 2 C , ϕ+ (x) – решение интегрального уравнения (2.6) на . Возьмем полосу Q и построим последовательность yn (x). Действуя ана логично, получим, что 9 ϕ (x) 2 C , ϕ (x) – решение интегрального уравнения (2.6) на . Определим функцию ϕ(x) как «сшивку» по непрерывности ϕ+ и ϕ , т.е. + ϕ (x), при x 2 , ϕ(x) = ϕ (x), при x 2 . Отметим, что ϕ+ (x0) = ϕ (x0) = y0 и потому ϕ(x) 2 C . Функции ϕ (x) по построению удовлетворяют интегральному уравнению (2.6), т.е. Zx ϕ (x) = y0 + f τ, ϕ (τ) dτ, x0 -30- где x 2 для ϕ+ (x) и x 2 для ϕ (x), соответственно. Следовательно, при любом x 2 функция ϕ(x) удовлетворяет инте 1 гральному уравнению (2.6). Тогда по лемме 2. 2, ϕ(x) 2 C и является решением задачи Коши (2.1), (2.2). Теорема доказана. Из доказанной теоремы 2. 4 нетрудно получить следствие для интервала (A, B) (открытой полосы). Ñледствие. Пусть функция f (x, y) определена, непрерывна в открытой полосе Q = (x, y) 2 R2: x 2 (A, B), y 2 R , причем A и B 2 R могут быть символами 1 и +1 соответственно. Предположим, что f (x, y) удовлетворяет в полосе Q условию: 9 L(x) 2 C(A, B), такая, что 8 x 2 (A, B) и 8 y1 , y2 2 R выполняется неравенство f (x, y2) f (x, y1) 6 L(x) jy2 y1 j. Òогда при любых начальных данных x0 2 (A, B), y0 2 R т.е. (x0 y0) 2 Q существует и притом единственное решение задачи Êоши (2.1), (2.2), определенное на всем (A, B). Доказательство. Для любой полосы Q1 = (x, y) 2 R2: x 2 , y2R , где A1 > A, B1.< B, лежащей строго внутри Q и содержащей (x0 , y0), справедлива теорема 2. 4, так как при доказательстве оценок вида (2.10), необходимых для обоснования равномерной на сходимости последовательности yn (x) , используются постоянные M0 = max f (x, y0) при x 2 и L = max L(x) x 2 . Эти постоянные не убывают при расширении (A, B). Возьмем последовательность расширяющихся отрезков , удовлетворяющих условиям B > Bk + 1\u003e bk for all k 2 n; 1) A.< Ak+1 < Ak , 2) x0 2 при всех k 2 N; 3) Ak ! A, Bk ! B при k ! 1. Заметим сразу, что S = (A, B) и, более того, для любого x 2 (A, B) k найдется номер x 2 . N (x) 2 N, такой, что при всех -31- k > N is performed by proveing \u200b\u200bthis auxiliary statement for the case of A, B 2 R (i.e., a and b are finite; if a \u003d 1 or b \u003d + 1, then in the same way). Take x a b x, arbitrary x 2 (a, b) and δ (x) \u003d min, δ (x)\u003e 0. 2 2 2 of the number Δ from the convergence of AK! A and BK! B We obtain that 9 n1 (δ) 2 n: 8 k\u003e n1, a< Ak < A + δ < x, 9 N2 (δ) 2 N: 8 k > N2, X.< B δ < Bk < B. Тогда для N = max N1 , N2 справедливо доказываемое свойство. Построим последовательность решений задачи Коши (2.1), (2.2) Yk (x), применяя теорему 2. 4 к соответствующему отрезку . Любые два из этих решений совпадают на общей области определения по следствию 1 из теоремы 2.1. Таким образом, два последовательных решения Yk (x) и Yk+1 (x) совпадают на , но Yk+1 (x) определено на более широком отрезке . Построим решение на всем (A, B). Возьмем и построим ϕ(x) – решение задачи (2.1), (2.2) на всем (по теореме 2. 4). Затем продолжим это решение на , . . . , . . . Получим, что решение ϕ(x) определено на всем (A, B). Докажем его единственность. Предположим, что существует решение ψ(x) задачи Коши (2.1), (2.2), также определенное на всем (A, B). Докажем, что ϕ(x) ψ(x) при любом x 2 (A, B). Пусть x – произвольная точка (A, B), найдется номер N (x) 2 N, такой, что x 2 при всех k > N. Applying a consequence of 1 p. 2.1 (that is, the uniqueness theorem), we obtain that φ (t) ψ (t) at all T 2 and, in particular, at t \u003d x. Since X is an arbitrary point (A, B), the uniqueness of the solution, and with it, the consequence is proved. Note 2. 10. In the proven investigation, we first met with the concept of continuing the decision on a wider set. In the next paragraph, we will study it in more detail. We give a few examples. p Example 2. 2. For equation y 0 \u003d EJXJ X2 + Y 2, find out if its solution exists on everything (a, b) \u003d (1, +1). Consider this equation in the "strip" Q \u003d R2, the function p jxj f (x, y) \u003d e x2 + y 2 ∂f y \u003d ejxj p, fy0 6 ejxj \u003d l (x). ∂Y x2 + y 2 according to claim 2. 1 of paragraph 2.1 The function f (x, y) satisfies the condition of Lipschitz on Y with the "constant" L \u003d L (x), X is fixed. Then all the conditions of the consequence are performed, and with any initial data (X0, Y0) 2 R2, the solution of the Cauchy problem exists and moreover is the only one on (1, +1). Note that the equation itself in quadratures is not solved, but approximate solutions can be built numerically. It is determined and continuous in Q, -32- Example 2. 3. For the Y 0 \u003d EX Y 2 equation, find out whether its solutions defined on R. If we again consider this equation in the "strip" Q \u003d R2, where the function ∂ FF (X, Y) \u003d EX Y 2 is defined and continuous, A \u003d 2YEX, we can notice, ∂Y that the condition of the investigation is violated, namely, there is no such continuous function L (x) that f (x, y2) f (x, y1) 6 l (x) JY2 Y1 J with all y1, y2 2 r. In fact, f (x, y2) f (x, y1) \u003d ex JY2 + Y1 J JY2 Y1 J, and the expression JY2 + Y1 J is not limited to Y1, Y2 2 R. Thus, the consequence is not applicable. I decide this equation "separation of variables", we obtain a general solution: "y (x) \u003d 0, y (x) \u003d 1. EX + C Take for definiteness x0 \u003d 0, Y0 2 R. If Y0 \u003d 0, then y (x ) 0 - the solution of the Cauchy problem on R. 1 - the solution of the Cauchy problem. At y0 2 [1, 0) ex it is defined at all x 2 R, and at Y0 2 (1, 1) [(0, +1) Y0 + 1 can be continued through the point x \u003d ln. More precisely, if x\u003e 0, then y0 1 solution y (x) \u003d y0 +1 is defined at x 2 (1, x), and if x< 0 x e y0 y0 < 1 , то решение определено при x 2 (x , +1). В первом случае lim y(x) = +1, а во втором – lim y(x) = 1. Если y0 6= 0, то y(x) = x!x 0 y0 +1 y0 x!x +0 -33- Для наглядности нарисуем интегральные кривые при соответствующих значениях y0 (рис. 2. 5). Рис. 2. 5. Интегральные кривые уравнения y 0 = ex y 2 Таким образом, для задачи Коши 0 y = ex y 2 , y(0) = y0 имеем следующее: 1) если y0 2 [ 1, 0], то решение существует при всех x 2 R; y0 + 1 2) если y0 < 1, то решение существует лишь при x 2 ln ; +1 ; y0 y0 + 1 . 3) если y0 > 0, then the solution exists only at x 2 1; LN Y0 This example shows that the limit on the growth of the function f (x, y) in the current consequence of Theorem 2. 4 is essential to continue the solution for all (A, B). Similarly, examples with the function f (x, y) \u003d f1 (x) y 1 + ε with any ε\u003e 0, in the given example, ε \u003d 1 only for the convenience of presentation is taken. 2. 3. Continuation of the solution for the first order of the first order. Definition 2. 5. Consider the equation y 0 \u003d f (x, y) and let y (x) - its solution on HA, BI, and Y (X) - its solution on HA , Bi, with HA, BI is contained in HA, Bi and y (x) \u003d y (x) on HA, BI. Then y (x) is called the continuation of the solution y (x) on HA, Bi, and about y (x) they say that it is continued on HA, Bi. -34- in paragraph 2.2, we proved the local theorem of the existence of the Cauchy problem (2.1), (2.2). Under what conditions does this decision be continued on a wider gap? This issue is devoted to this issue. The main result is as follows. Theorem 2. 5 (on the continuation of the solution in a limited closed area). Let the function f (x, y) 2 Cg and satisfies the Lipschitz condition along Y in R2, A (X0, Y0) the inner point of the limited closed region G G. The solution of the equation y 0 \u003d f (x , y), continued up to ∂G border of the region G, i.e. It can be continued on such a segment that points a, y (a) and b, y (b) lie on ∂G. ∂f (x, y) is continuous in limited by ot, closed, convex by y region G, then the function f (x, y) satisfies the Lipschitz condition in the variable y. See the consequence of Approval 2. 1 ∂F from paragraph 2.1. Therefore, this theorem will be valid if it is continuous in ∂y G. Remark 2. 11. Recall that if proof. Since (x0, y0) is an internal point G, then there is a closed rectangle NO 2 p \u003d (x, y) 2 R x0 6 α, y y0 6 β, the whole lying in G. Then by Theorem 2. 3 of . 2.2 There is H\u003e 0 such that there is a solution on the segment (and the only one) solution y \u003d φ (x) of the equation y 0 \u003d f (x, y). I will first continue this decision to right up to the border of the G of the region, breaking the proof to individual steps. 1. Consider the set ER: NO E \u003d α\u003e 0 The solution y \u003d φ (x) continually on the solution y \u003d φ1 (x) of the equation y 0 \u003d f (x, y), satisfying the Cauchy conditions φ1 ~ b \u003d φ ~ b . Thus, φ (x) and φ1 (x) are solutions on the segment of ~ b H1, ~ B one equations that match the point x \u003d ~ B, so they coincide on the whole segment ~ b H1, ~ B and, therefore, φ1 (x) is a continuation of the solution φ (x) from the segment ~ B H1, ~ B to ~ B H1, ~ B + H1. Consider the function ψ (x): φ (x), x 2 x0, ψ (x) \u003d φ1 (x), x 2 ~ b ~ b, h1, ~ b + h1 ~ b h1, x0 + α0 + h1, which It is a solution to the equation y 0 \u003d F (x, y) and satisfies the condition of the Cauchy ψ (x0) \u003d y0. Then the number α0 + H1 2 E, and this is contrary to the definition of α0 \u003d sup e. Therefore, case 2 is impossible. Similarly, the solution φ (x) continues to the left, on the segment, where the point A, φ (a) 2 ∂G. Theorem is completely proved. -37- Chapter III. Cauchy's task for the normal system of N-th order 3. 1. Basic concepts and some auxiliary properties of vector functions in this chapter will consider the normal N-th order system of the form 8\u003e T, Y ,. . . , y y _ \u003d f 1 n 1 1\u003e,< y_ 2 = f2 t, y1 , . . . , yn , (3.1) . . . > \u003e: y_ \u003d f t, y ,. . . , Y, N n 1 n where unknown (the desired) are the functions Y1 (T) ,. . . , yn (t), and FI functions are known, i \u003d 1, n, the point over the function denotes the derivative of T. It is assumed that all fi are defined in the region G Rn + 1. It is convenient to record the system (3.1) in vector form: y_ \u003d f (t, y), where y (t) y1 (t). . . , yn (t), f (t, y) f1 (t, y). . . , fn (t, y); Arrogors in the designation of vectors will not write for brevity. Such a record will also be denoted (3.1). Let the point T0, Y10 ,. . . , Yn0 lies in G. The Cauchy problem for (3.1) is to find a solution φ (t) of the system (3.1) satisfying the condition: φ1 (t0) \u003d y10, φ2 (t0) \u003d y20, ..., φn (t0) \u003d yn0, (3.2) or in vector form φ (t0) \u003d y 0. As noted in Chapter 1, under the solution of the system (3.1), the vector function φ (t) \u003d φ1 (t) is understood as the system of HA, Bi. . . , φn (t), satisfying conditions: 1) 8 T 2 HA, Bi point T, φ (t) lies in G; 2) 8 t 2 ha, bi 9 d dt φ (t); 38 3) 8 T 2 HA, Bi φ (t) satisfies (3.1). If such a decision additionally satisfies (3.2), where T0 2 HA, BI, then it is called the solution of the Cauchy problem. Conditions (3.2) are called on the forces or Cauch conditions, and the number T0, Y10 ,. . . , YN0 - Cauchy data (initial data). In the particular case, when the vector function f (t, y) (n + 1) variable depends on Y1 ,. . . , yn linearly, i.e. It has the form: f (t, y) \u003d a (t) y + g (t), where a (t) \u003d aij (t) - n n matrix, system (3.1) is called linear. In the future, we will need the properties of the vector functions that we here are here for the convenience of links. Rules of addition and multiplication by the number for vectors are known from the course of linear algebra, these basic operations are fully implemented. n If in R to introduce a scalar product x, y \u003d x1 y1 +. . . + XN Yn, we obtain the Euclidean space, which will also be denoted by Rn, with a length S q n p vector jxj \u003d x, x \u003d x2k (or by the Euclidean norm). For scalar k \u003d 1, the works and lengths are fair two main inequalities: 1) 8 x, y 2 rn 2) 8 x, y 2 rn). x + y 6 x + y x, y 6 x (triangle inequality); Y (Cauchy's inequality belongs - from the course of the mathematical analysis of the second semester, it is known that the convergence of the sequence of points (vectors) in the Euclidean space (finite-dimensional) is equivalent to the convergence of sequences of the coordinate of these vectors, they say is equivalent to the coordination convergence. It is easily followed from inequalities: QP Max x 6 x21 +... Here are some inequalities for vector functions used in the future. 1. For any vector function y (t) \u003d y1 (t) ,. . . , yn (t) integrable (for example, continuous) on, fairly inequality zb zb y (t) dt 6 ay (t) dt a -39- (3.3) or in coordinate form 0 zb zb y1 (t) dt, @ Y2 (T) DT ,. . . , A 1 ZB A ZB Q Yn (T) DT A 6 Y12 (T) +. . . YN2 (T) DT. A A proof. Note first, that in the inequality does not exclude the case of< a, для этого случая в правой части присутствует знак внешнего модуля. По определению, интеграл от вектор-функции – это предел интегральn P ных сумм στ (y) = y(ξk) tk при характеристике («мелкости») разбиения k=1 λ(τ) = max tk стремящейся к нулю. По условию στ ! k=1, N Rb y(t) dt , а по a неравенству треугольника получим στ 6 n X Zb y(ξk) tk ! k=1 при λ(τ) ! 0 y(t) dt, a (здесь мы для определенности считаем a < b). По теореме о переходе к пределу в неравенстве получим доказываемое. Случай b < a сводится к изученному, Rb Ra так как = . a b Аналоги теорем Ролля и Лагранжа отсутствуют для вектор-функций, однако можно получить оценку, напоминающую теорему Лагранжа. 2. Для любой вектор-функции x(t), непрерывно дифференцируемой на , имеет место оценка ¾приращения¿: x(b) x(a) 6 max x 0 (t) b a. (3.4) Доказательство. Неравенство (3.4) сразу получается из (3.3) при y(t) = x 0 (t). При доказательстве теоремы разрешимости для линейных систем нам понадобятся оценки с n n матрицами, которые мы сейчас и рассмотрим. 3. Пусть A(t) = aij (t) n n матрица, обозначим произведение Ax через y. Как оценить y через матрицу A и x ? Оказывается, справедливо неравенство Ax 6 A -40- 2 x, (3.5) где x = p jx1 j2 + . . . + jxn j2 , A 2 = n P ! 12 a2ij , а элементы матрицы i,j=1 A и координаты вектора x могут быть комплексными. Доказательство. Для любого i = 1, n, ai – i-я строка матрицы A, тогда: 2 2 2 yi = ai1 x1 + ai2 x2 + . . . + ain xn = ai , x 6 h i 2 6 по неравенству Коши-Áуняковского 6 jai j2 x = ! ! n n X X 2 2 aik xl = , k=1 суммируя эти неравенства по i = 1, n, имеем: 0 1 n X 2 2 2 aik A x = A y [Email Protected] 2 2 L \u003d 1 2 x, k, i \u003d 1 From where it follows (3.5). Definition 3. 1. To say that the vector function F (T, Y) satisfies the Lipschitz condition on the vector variable Y on the MNA 1 g of variables (T, Y), if 9 l\u003e 0 is such that with any t, y , 2 t, y 2 g is performed inequality ft, y 2 ft, y 1 6 l y 2 y 1. As in the case of the function of two variables (see approval 2.1), a sufficient condition for lipshese in the "convex on y" region G is the limited partial derivatives. Let's give an exact definition. Definition 3. 2. The range G of variables (T, Y) is called convex 1 2 by y, if for any two points T, Y and T, Y lying in G, it belongs entirely to it and the segment connecting these two points, t. e. The set n o t, y y \u003d y 1 + τ y 2 y 1, where τ 2. Approval 3. 1. If the range G of variables (T, Y) is convex to y, and ∂fi private derivatives are continuous and limited to the constant L in G with ∂yj all i, j \u003d 1, n, then the FT vector function satisfies In G, the Lipschitz condition for y with a constant L \u003d n l. 1 2 Proof. Consider arbitrary points T, Y and T, Y from G and 1 2 segments, connecting them, i.e. The set T, Y, where y \u003d y + τ y y1, T is fixed, and τ 2. -41- We introduce the vector function of a single scalar argument G (τ) \u003d ft, y (τ), 2 1 then g (1) g (0) \u003d ft, yft, y, and on the other hand - z1 g (1) g (0) \u003d dg (τ) dτ \u003d dτ z1 a (τ) dy (τ) dτ \u003d dτ 0 0 H \u003d by virtue y \u003d y 1 + τ y 2 yi 1 z1 \u003d a (τ) y 2 y 1 dτ 0 where a (τ) is a matrix with elements ∂fi, and ∂yj y2 y 1 is the corresponding column. Here we took advantage of the differentiation rule of the complex function, namely, at all i \u003d 1, n, t - fixed, we have: gi0 (τ) \u003d ∂fi ∂Y1 ∂Fi ∂Y2 ∂Fi ∂YN d Fi T, Y (τ) \u003d + + ... + \u003d dτ ∂y1 ∂τ ∂y2 ∂τ ∂Yn ∂τ ∂fi ∂fi, ..., y2 y1. \u003d ∂Y1 ∂YN Recalling this in matrix form, we obtain: 0 2 1 G (τ) \u003d a (τ) y y with n n matrix A (τ) \u003d aij (τ) ∂fi ∂yj. Using the estimate of the integral (3.3) and inequality (3.5), after the substitution, we obtain: ft, y 2 ft, y 1 z1 \u003d g 0 (τ) dτ \u003d 0 z1 6 a (τ) y 2 z1 y1 a (τ) y 2 0 z1 dτ 6 0 a (τ) a (τ) dτ y2 y1 6 y2 y1 6 nl 0 6 max a (τ) Since 2 y 1 dτ 6 2 2 n p ∂fi \u003d i, j \u003d 1 ∂Yj 2 y2 Y1, 2 6 N2 L2 at 8 τ 2. The statement is proved. -42- 3. 2. The pericity of the solution of the Cauchy problem for the normal system of Theorem 3. 1 (on the assessment of the difference of two solutions). Let G be some region Rn + 1, and the vector function f (x, y) is continuous in G and satisfies the Lipschitz condition on a vector variable y on the set G with constant L. if Y 1, Y 2 two solutions of the normal system (3.1) y_ \u003d F (x, y) on the segment, then the rating Y 2 (T) Y 1 (T) 6 Y 2 (T0) Y 1 (T0) EXP L (T T0) is valid for all T 2. The proof of literally, taking into account the obvious reissues, repeats the proof of Theorem 2.1 from paragraph 2.1. 2 From here it is easy to obtain the theorem of the uniqueness and stability of the decision on initial data. Alternately 3.1. Let the vector function f (t, y) continuous in the region G and satisfies in the G Lipschitz condition along y, and the functions y 1 (t) and y 2 (t) two solutions of the normal system (3.1) on the same segment, Moreover, T0 2. If y 1 (t0) \u003d y 2 (t0), then y 1 (t) y 2 (t) on. Alternately 3.2. (on continuous dependence on the initial data). Let the vector function f (t, y) be continuous in the region G and satisfies in the G Lipschitz condition along Y with constant L\u003e 0, and the vector functions Y 1 (T) and Y 2 (T) are solutions of the normal system (3.1) defined on. If at 8 t 2, the inequality y 1 (t) is true where δ \u003d y 1 (T0) Y 2 (T0), and L \u003d T1 y 2 (T) 6 Δ EL L, T0. The proof of the consequences of literally, taking into account the apparent reissues, repeats the proof of the consequences 2.1 and 2.2. 2 Study of the solvability of the Cauchy problem (3.1), (3.2), as in the one-dimensional case, is reduced to the solvability of the integral equation (vector). Lemma 3. 1. Let F (T, Y) 2 C G; Rn 1. The following statements take place: 1) Any solution φ (t) equation (3.1) on the gap HA, Bi, satisfying (3.2) T0 2 HA, Bi, is a continuous solution on HA, Bi 1 through C G; H is adopted to denote the set of all the functions continuous in the G region with the values \u200b\u200bin the space H. For example, F (T, Y) 2 C G; Rn components) defined on the set G. - the set of all continuous vector functions (from the N -43 integral equation y (t) \u003d y 0 + zt f τ, y (τ) dτ; (3.6) T0 2) if vector -Function φ (T) 2 C HA, Bi is a continuous solution of the integral equation (3.6) on HA, Bi, where T0 2 HA, Bi, then φ (t) has a continuous derivative on HA, Bi and is a solution (3.1), (3.2). Evidence. 1. Let 8 τ 2 HA, Bi are performed by the equality dφ (τ) \u003d f τ, φ (τ). Then integrating from T0 to T, taking into account (3.2), semiDτ rt 0 chim, that φ (t) \u003d y + f τ, φ (τ) dτ, i.e. φ (t) satisfies equation (3.6). T0 2. Let the continuous vector function φ (t) satisfy the equation (3.6) on HA, Bi, then ft, φ (t) is continuous on HA, Bi by the continuity theorem of the complex function, and therefore the right-hand side (3.6) ( And, therefore, the left part) has a continuous derivative of T to HA, BI. At t \u003d t0 from (3.6) φ (t0) \u003d y 0, i.e. φ (t) is the solution of the Cauchy problem (3.1), (3.2). Note that as usual, under the derivative at the end of the segment (if it belongs to it) is understood to be one-sided derivative function. Lemma is proved. Remark 3. 1. Using an analogy with a one-dimensional case (see Chapter 2) and the above approvals, it is possible to prove the terrhem on the existence and continuing to solve the Cauchy problem, constructing an iterative sequence converges to solving the integral equation (3.6) on a certain segment T0 H, T0 + H. Here we present another proof of the existence theorem (and uniqueness) solutions based on the principle of compressive mappings. We do this for dating the reader with more modern methods of the theory, which will be applied in the future, in courses of integral equations and equations of mathematical physics. To implement our plan, you will need a number of new concepts and auxiliary assertions, which we will proceed. 3. 3. The concept of metric space. The principle of compressing mappings The most important concept of limit in mathematics is based on the concept of "proximity" of points, i.e. The opportunity to find the distance between them. On the numeric axis, the distance is the module of the two numbers, on the plane is a well-known formula of Euclidean distance, etc. Many of the analysis facts do not use the algebraic properties of the elements, and they are based on the concept of the distance with honey. The development of this approach, i.e. The allocation of the "creature" relating to the concept of limit leads to the concept of metric space. -44- Definition 3. 3. Let X be a plurality of arbitrary nature, and ρ (x, y) - the actual function of two variables x, y 2 x, satisfying the three axioms: 1) ρ (x, y)\u003e 0 8 x, y 2 x, and ρ (x, y) \u003d 0 only at x \u003d y; 2) ρ (x, y) \u003d ρ (y, x) (symmetry axiom); 3) ρ (x, z) 6 ρ (x, y) + ρ (y, z) (triangle inequality). In this case, the set x with a given function ρ (x, y) is called the metric space (ìP), and the function ρ (x, y): X x 7! R, satisfying 1) - 3), - metric or distance. We present some examples of metric spaces. Example 3. Let x \u003d R with a distance ρ (x, y) \u003d x y, we obtain the MP R. N O n xi 2 R, i \u003d 1, n is example 3. 2. Let x \u003d r \u003d x1 ,. . . , XN set of ordered sets from n valid numbers S n 2 p x \u003d x1 ,. . . , Xn with a distance ρ (x, y) \u003d xk yk, we obtain N1 k \u003d 1 N dimensional Euclidean space R. n Example 3. 3. Let x \u003d c a, b; R is the set of all continuous on a, b functions with values \u200b\u200bin Rn, i.e. continuous vector functions, with a distance ρ (f, g) \u003d max f (t) g (t), where f \u003d f (t) \u003d f1 (t) ,. . . , Fn (T), T2 S N 2 P G \u003d G (T) G1 (T) ,. . . , Gn (T), F G \u003d FK (T) GK (T). k \u003d 1 for examples 3. 1 -3. 3 MP axioms are checked directly, leave it as an exercise for a conscientious reader. As usual, if each natural n is put in accordance with the electron of Xn 2 x, it is said that the sequence of points Xn ìP X is given. Definition 3. 4. The sequence of points XN MP X is called the x 2 x point, if Lim ρ xn, x \u003d 0. N! 1 Definition 3. 5. The XN sequence is called fundamental, if for any ε\u003e 0 there is such a natural number N (ε) that for all N\u003e N and M\u003e N inequality ρ xn, xm< ε. Определение 3. 6. МП X называется полным (ПÌП), если любая его фундаментальная последовательность сходится к элементу этого пространства. -45- Полнота пространств из примеров 3. 1 и 3. 2 доказана в курсе математиче ского анализа. Докажем полноту пространства X = C a, b ; Rn из примера 3. 3. Пусть последовательность вектор-функций fn (t) фундаментальна в X. Это означает, что 8 ε > 0 9 n (ε) 2 n: 8m, n\u003e n \u003d) max fm (t) fn (t)< ε. Поэтому выполнены условия критерия Коши равномерной на a, b сходи мости функциональной последовательности, т.е. fn (t) ⇒ f (t) при n ! 1. Как известно, предел f (t) в этом случае – непрерывная функция. Докажем, что f (t) – это предел fn (t) в метрике пространства C a, b ; Rn . Из равномерной сходимости получим, что для любого ε > 0 There is a number N (ε) such that for all N\u003e n and for all T 2 A, B is performed inequality Fn (T) F (T)< ε, а так как в левой части неравенства стоит непрерывная функция, то и max fn (t) f (t) < ε. Это и есть сходимость в C a, b ; Rn , следовательно, полнота установлена. В заключение приведем пример МП, не являющегося полным. Пример 3. 4. Пусть X = Q – множество рациональных чисел, а расстояние ρ(x, y) = x y – модуль разностиpдвух чисел. Если взять последовательность десятичных приближений числа 2 , т.е. x1 = 1; x2 = 1, 4; x3 = 1, 41; . . ., p то, как известно, lim xn = 2 62 Q. При этом данная последовательность n!1 сходится в R, значит она фундаментальна в R, а следовательно, она фундаментальна и в Q. Итак, последовательность фундаментальна в Q, но предела, лежащего в Q, не имеет. Пространство не является полным. Определение 3. 7. Пусть X – метрическое пространство. Отображение A: X 7! X называется сжимающим отображением или сжатием, если 9 α < 1 такое, что для любых двух точек x, y 2 X выполняется неравенство: ρ Ax, Ay 6 α ρ(x, y). (3.7) Определение 3. 8. Точка x 2 X называется неподвижной точкой отображения A: X 7! X, если Ax = x . Замечание 3. 2. Всякое сжимающее отображение является непрерывным, т.е. любую сходящуюся последовательность xn ! x, n ! 1, переводит в сходящуюся последовательность Axn ! Ax, n ! 1, а предел последовательности – в предел ее образа. Действительно, если A – сжимающий оператор, то положив в (3.7) X X y = xn ! x, n ! 1, получим, что Axn ! Ax, n ! 1. Теорема 3. 2 (Принцип сжимающих отображений). Пусть X полное метрическое пространство, а отображение A: X 7! X является сжатием. Òогда A имеет и притом единственную неподвижную точку. Доказательство этого фундаментального факта см. , . -46- Приведем обобщение теоремы 3. 2, часто встречающееся в приложениях. Теорема 3. 3 (Принцип сжимающих отображений). Пусть X полное метрическое пространство, а отображение A: X 7! X таково, что оператор B = Am с некоторым m 2 N является сжатием. Òогда A имеет и притом единственную неподвижную точку. Доказательство. При m = 1 получаем теорему 3. 2. Пусть m > 1. Consider B \u003d AM, B: X 7! X, B - compression. By Theorem 3. 2, the operator B has a single fixed point x. Since A and B are permutable AB \u003d BA and since BX \u003d X, we have B Ax \u003d A BX \u003d AX, i.e. Y \u003d AX is also a fixed point B, and since such a point by Theorem 3. 2 is unique, then y \u003d x or ax \u003d x. From here X is a fixed point of the operator A. We will prove the uniqueness. Suppose that x ~ 2 x and a ~ x \u003d x ~, then m m 1 b x ~ \u003d a x ~ \u003d a x ~ \u003d. . . \u003d x ~, i.e. x ~ - also a fixed point for b, where x ~ \u003d x. Theorem is proved. A special occasion of the metric space is a linear normalized space. We give the exact definition. Definition 3. 9. Let X be a linear space (real or complex), which defines the numerical function X acting from x to R and satisfying the axioms: 1) 8 x 2 x, x\u003e 0, and x \u003d 0 only with x \u003d θ; 2) 8 x 2 x and for 8 λ 2 R (or c) 3) 8 x, Y 2 x is performed). x + y 6 x + y λx \u003d jλj x; (The inequality of the triangular - then X is called the normalized space, X: X 7! R, satisfying 1) - 3), - the norm. And the function in the normalized suction can be introduced the distance between the elements by the formula ρ x, y \u003d x y. Performing an axiom of MP is easily checked. If the obtained metric space is fully, then the corresponding normalized space is called bana space. Often enter the norm on the same linear space in different ways. In this regard, such a concept arises. Definition 3. 10. Let X be a linear space, and - two 1 2 standards entered on it. The norms are called 1 2 norms, if 9 c1\u003e 0 and c2\u003e 0: 8 x 2 x c1 x 1 6 x 2 6 c2 x 1. Remark 3. 3. If both are two equivalent norms on x, and 1 2 space X is complete one of them, then it is fully and on another norm. This easily follows from the fact that the Xn X sequence, fundamental software, fundamental also in, and converges to 1 2 of the same element x 2 X. -47- Review 3. 4. Often Theorem 3. 2 (or 3. 3) It is used when a closed ball of this space O br (a) \u003d x 2 x ρ x, a 6 r, where R\u003e 0 and A 2 x is fixed as a complete n space. Note that a closed ball in PMP himself is a PMP with the same distance. Proof of this fact Leave the reader as an exercise. Note 3. 5. The completeness of the space was established above. 3. Note that in the linear space X \u003d C 0, T, R, you can enter the rate kxk \u003d max x (t) so that the normalized value will be Banach. On the same set of continuous on space 0, T vector functions, you can enter the equivalent norm by the formula KxKα \u003d max e αt x (t) with any α 2 R. When α\u003e 0, the equivalence follows from the inequalities E αT X (T) 6 E αt x (t) 6 x (t) at all T 2 0, T, from where E αT KXK 6 KXKα 6 KXK. We will use this property of equivalent norms in the proof of the theorem on the unambiguous solvability of the Cauchy problem for linear (normal) systems. 3. 4. The theorems of the existence and uniqueness of the solution of the Cauchy problem for normal systems Consider the Cauchy problem (3.1) - (3.2), where the initial data T0, Y 0 2 G, G Rn + 1 is the field of determining the vector function F (T, Y ). In this paragraph, we assume that G has a certain n appearance G \u003d A, B O, where the region Rn, and BR (Y 0) \u003d takes place the theorem. Y 2 Rn y Y0 6 R is entirely in. Theorem 3. 4. Let the vector function f (t, y) 2 C g; Rn, with 9 m\u003e 0 and L\u003e 0 such that conditions 1) 8 (t, y) 2 g \u003d a, b f (t, y) 6 m are performed; 2) 8 (t, y 1), (t, y 2) 2 g f t, y 2 f t, y 1 6 l y 2 y 1. Fix the number Δ 2 (0, 1) and let T0 2 (A, B). R 1 δ 9 h \u003d min; ; T0 A; B T0\u003e 0 ML such that exists and more than the solution of the problem of êshoshi (3.1), (3.2) y (t) on the section JH \u003d T0 H, T0 + H, and Y (T) Y 0 6 R with all T 2 JH. -48- Proof. In Lemma 3. 1, the Cauchy problem (3.1), (3.2) is equivalent to an integral equation (3.6) on the segment, and therefore, on JH, where H is chosen above. Consider the Banach space x \u003d c (jh; rn) - a plurality of continuous vector-functions X (T) with the norm kxk \u003d max x (t) and we introduce a closed set: T2JH SR Y 0 N 8 T 2 JH \u003d y (t) 2 x y (t) n \u003d y (t) 2 x yy (t) O 0 6R \u003d O 0 y 6R Closed ball in X. Operator A, determined by rule: Ay \u003d Y 0 + ZT F τ , Y (τ) Dτ, T 2 JH, T0 translates SR Y 0 to itself, since y 0 \u003d max ay zt t2jh f τ, y (τ) dτ 6 h \u200b\u200bm 6 r t0 for condition 1 of the theorem and the definition H. We prove that A is on the SR compression operator. Take an arbitrary 0 1 2 and we estimate the value: y (t), y (t) 2 Sr y A 2 Ay 1 \u003d max zt h t2jh f τ, y 2 (τ) if τ, y 1 (τ) dτ 6 t0 ZT 6 MAX T2JH F τ, Y 2 (τ) f τ, y 1 (τ) dτ 6 t0 6h l y2 y1 \u003d q y2 y1, where q \u003d h l 6 1 δ< 1 по условию теоремы. Отметим (см. замечание 3.4), что замкнутый шар SR y 0 в банаховом пространстве X является ПМП. Поэтому применим принцип сжимающих отображений (теорема 3. 2), по которому существует единственное решение y(t) 2 X интегрального уравнения (3.6) на отрезке Jh = t0 h, t0 + h . Теорема доказана. Замечание 3. 6. Если t0 = a или t0 = b, то утверждение теоремы сохраняется с небольшими изменениями в формуле для h и отрезка Jh . Приведем эти изменения для случая t0 = a. В этом случае число h > 0 Select according to R formula H \u003d min m; 1L Δ; b a, and as a segment JH, it is necessary to take -49- jh \u003d t0, t0 + h \u003d a, a + h. All other conditions of the theorem do not change, its proof, taking into account the reissued, is achieved. For the case of T0 \u003d B, similarly, h \u003d min m; 1L Δ; b a, and jh \u003d b h, b. n Remark 3. 7. In Theorem 3. 4, the condition F (T, Y) 2 C G; R, where G \u003d A, B D can be weakened by replacing the requirement of continuity f (t, y) by variable T each time Y 2, with the preservation of conditions 1 and 2. The proof will not change. Note 3. 8. It is enough that conditions 1 and 2 of Theorems 3. 4 were performed 0 at all T, Y 2 A, B Br y, and the constant M and L depend on, 0 Generally speaking, from Y and R. with more hard The limitations on the FT, Y vector function, similar to Theorem 2.4, is valid for the existence theorem and the uniqueness of the solution of the Cauchy problem (3.1), (3.2) on the entire segment A, b. N Theorem 3. 5. Let the FX vector function FX, Y 2 CG, R, where G \u003d A, B Rn, exists L\u003e 0, such that the condition 8 T, Y 1, T, Y 2 2 G ft , Y 2 ft, y 1 6 l y 2 y 1. If any T0 2 and Y 0 2 RN, on A, B exists and moreover, the only solution of the task of the êshoshi (3.1), (3.2). Evidence. Take arbitrary T0 2 and Y 0 2 Rn and fix them. The set G \u003d A, B Rn is present in the form: G \u003d G [G +, where Rn, and G + \u003d T0, B Rn, believing that T0 2 A, B, otherwise one G \u003d A, T0 from the evidence stages will be absent. Conduct reasoning for the G + strip. On the segment T0, b, the Cauchy problem (3.1), (3.2) is equivalent to equation (3.6). We introduce the operator integral n A: X 7! X, where x \u003d c t0, b; R, according to the formula AY \u003d Y 0 + ZT F τ, Y (τ) Dτ. T0 Then the integral equation (3.6) can be written in the form of an operator equation AY \u003d Y. (3.8) If we prove that the operator equation (3.8) has a solution in the PMP X, then we obtain the solvability of the Cauchy problem on T0, B or on A, T0 for G. If this solution is the only one, due to equivalence, the only solution will be the solution of the Cauchy problem. We give two evidence of the unambiguous solvability of equation (3.8). Proof 1. Consider arbitrary vector functions 1 2 N y, y 2 x \u003d c t0, b; R, then values \u200b\u200bare valid for any -50- T 2 T0, b AY 2: AY 1 ZT HF τ, Y 2 (τ) \u003d 1 f τ, y (τ) i dτ 6 t0 zt y 2 (τ) 6l y 1 (τ) Dτ 6 L T T0 MAX Y 2 (τ) Y 1 (τ) 6 τ 2 T0 6L T T0 Y2 Y1. Recall that in x the norm is entered as follows: kxk \u003d max x (τ). From the resulting inequality we will have: 2 2 AY 2 1 AY ZT HF τ, AY 2 (τ) \u003d 1 I τ T0 Dτ F τ, AY (τ) Dτ 6 T0 6 L2 ZT AY 2 (τ) AY 1 (τ ) Dτ 6 L2 T0 ZT Y2 Y1 6 T0 6 L2 T T0 2! 2 y2 y1. Continuing this process, it can be proved by induction that 8 K 2 N AK Y 2 AK Y 1 6 L T T0 K! K y2 y1. Hence, finally, we obtain an estimate of AK Y 2 AK Y 1 \u003d Max AK Y 2 L B T0 AK Y 1 6 L B T0 K! K y2 y1. K Since α (k) \u003d! 0 for k! 1, then there is a k0 that, k! that α (k0)< 1. Применим теорему 3. 3 с m = k0 , получим, что A имеет в X неподвижную точку, причем единственную. Доказательство 2. В банаховом пространстве X = C t0 , b ; Rn введем семейство эквивалентных норм, при α > 0 (See Remark 3. 5) by formula: X α \u003d max e αt x (t). -51- We will show that you can choose α so that the operator A in the space x with the norm at α\u003e l will be compressive. Indeed, α ay 2 ay 1 α zt hf τ, y 2 (τ) αt \u003d max e 1 f τ, y (τ) i dτ 6 t0 6 max e αt zt y 2 (τ) l y 1 (τ) dτ \u003d T0 \u003d L max e zt αt e ατ y 2 (τ) Eατ dτ 6 y 1 (τ) t0 6 l max e αt zt eατ dτ max e ατ y 2 (τ) y 1 (τ) \u003d y2 α t0 \u003d L MAX E αT Since α\u003e l, then q \u003d l α 1 1 αt e α e eαt0 l \u003d α α b t0 y 2 y1 y 1 α \u003d 1 E α b T0.< 1 и оператор A – сжимающий (например, с α = L). Таким образом, доказано, что существует и притом единственная вектор + функция ϕ (t) – решение Коши (3.1), (3.2) на t0 , b . задачи Rn задачу Коши сведем к предыдущей при Для полосы G = a, t0 помощи линейной замены τ = 2t0 t. В самом деле, для вектор-функция y(t) = y 2t0 τ = y~(t), задача Коши (3.1), (3.2) запишется в виде: y~(τ) = f (2t0 τ, y~(τ)) f~ (τ, y~(τ)) , y~(t0) = y 0 на отрезке τ 2 t0 , 2t0 a . Поэтому можно применить предыдущие рассуждения, взяв b = 2t0 a. Итак, существует и притом единственное решение задачи Коши y~(τ) на всем отрезке τ 2 t0 , 2t0 a и,следовательно, ϕ (t) = y~ 2t0 t – решение задачи Коши (3.1), (3.2) на a, t0 . Возьмем «сшивку» вектор-функций ϕ (t) и ϕ+ (t), т.е. вектор-функцию ϕ (t), при t 2 a, t0 ; ϕ(t) = ϕ+ (t), при t 2 t0 , b . d dτ Как при доказательстве теоремы 2.4, устанавливаем, что ϕ(t) – это решение задачи Коши (3.1), (3.2) на a, b . Единственность его следует из следствия 3.1. Теорема доказана. -52- Замечание 3.9. Утверждение 3. 1 дает достаточное условие того, что векторфункция f t, y в выпуклой по y области G удовлетворяет условию Лип∂fi шица. А именно, для этого достаточно, чтобы все частные производные ∂yj были непрерывны и ограничены некоторой константой в G. Аналогично следствию из теоремы 2.4 получаем такое утверждение для нормальных систем. Ñледствие 3.3. Пусть вектор-функция f (t, y) определена, непрерывна в открытой полосе o n n Q = (t, y) t 2 (A, B), y 2 R , причем A и B могут быть символами 1 и +1 соответственно. Предположим, что вектор-функция f (t, y) удовлетворяет в полосе Q условию: 9 L(t) 2 C(A, B) такая, что 8 t 2 (A, B) и 8 y 1 , y 2 2 Rn выполняется неравенство f t, y 2 f t, y 1 6 L(t) y 2 y 1 . Òогда при любых начальных данных t0 2 (A, B), y 0 2 Rn существует и притом единственное решение задачи Êоши (3.1), (3.2) на всем интервале (A, B). Доказательство проводится повторением соответствующих рассуждений из п. 2.2, оставляем его добросовестному читателю. В качестве других следствий из доказанной теоремы 3. 5 получим теорему о существовании и единственности решения задачи Коши для линейной системы. Речь идет о задаче нахождения вектор-функции y(t) = (y1 (t), . . . , yn (t)) из условий: d y(t) = A(t)y(t) + f 0 (t), t 2 a, b , (3.9) dt y(t0) = y 0 , (3.10) где A(t) = aij (t) – n n матрица, f 0 (t) – вектор-функция переменной t, t0 2 a, b , y 0 2 Rn – заданы. n 0 Теорема 3. 6. Пусть a (t) 2 C a, b , f (t) 2 C a, b ; R , ij t0 2 a, b , y 0 2 Rn заданы. Òогда существует и притом единственное решение задачи Êоши (3.9), (3.10) на всем отрезке a, b . Доказательство. Проверим, что для функции f t, y = A(t)y + f 0 (t) выполнены теоремы 3. 5. Во-первых, f t, y 2 C G; Rn , где условия G = a, b Rn , как сумма двух непрерывных функций. Во-вторых, (см. неравенство (3.5)): Ay 2 Ay 1 = A(t) y 2 y 1 6 A 2 y 2 y 1 6 L y 2 y 1 , -53- поскольку A n P 2 ! 21 aij (t) 2 – непрерывная на a, b функция. Тогда i,j=1 по теореме 3. 5 получим доказываемое утверждение. Теорема 3. 7. Пусть aij (t) 2 C (R), f 0 (t) 2 C (R; Rn) заданы. Òогда при любых начальных данных t0 2 R, y 0 2 Rn существует и притом единственное решение задачи Êоши (3.9), (3.10) на всей числовой прямой. Доказательство. Проверим, что выполнены все условия следствия из теоре мы 3. 5 с A = 1, B = +1. Вектор-функция f t, y = A(t)y + f 0 (t) непрерывна в полосе Q = R Rn как функция (n + 1) переменной. Кроме того, L(t) y 2 y 1 , f t, y 2 f t, y 1 6 A(t) 2 y 2 y 1 где L(t) – непрерывная по условию теоремы на A, B = 1, +1 функция. Таким образом, все условия следствия выполнены, и теорема доказана. -54- Глава IV. Некоторые классы обыкновенных дифференциальных уравнений, решаемых в квадратурах В ряде случаев дифференциальное уравнение может быть решено в квадратурах, т.е. для его решения может быть получена явная формула. В таких случаях методика решения, как правило, следующая. 1. Предполагая, что решение существует, находят формулу, по которой решение выражается. 2. Существование решения затем доказывается непосредственной проверкой, т.е. подстановкой найденной формулы в исходное уравнение. 3. Используя дополнительные данные, (например, задавая начальные данные Коши) выделяют конкретное решение. 4. 1. Уравнение с разделяющимися переменными В данном параграфе применим уже использовавшуюся выше методику для решения уравнений с разделяющимися переменными, т.е. уравнений вида y 0 (x) = f1 (x) f2 (y), Áудем предполагать, что f1 (x) 2 C (ha, bi) , x 2 ha, bi, f2 (y) 2 C (hc, di) , y 2 hc, di. (4.1) f2 (y) 6= 0 на hc, di а следовательно, в силу непрерывности функции f2 (y), она сохраняет знак на hc, di . Итак, предположим, что в окрестности U(x0) точки x0 2 ha, bi существует решение y = ϕ(x) уравнения (4.1). Тогда имеем тождество dy = f1 (x) f2 (y), dx y = ϕ(x), 55 x 2 U(x0). Но тогда равны дифференциалы dy = f1 (x) dx f2 (y) мы учли, что f2 (y) 6= 0 . Из равенства дифференциалов вытекает равенство первообразных с точностью до постоянного слагаемого: Z Z dy = f1 (x) dx + C. (4.2) f2 (y) После введения обозначений Z F2 (y) = Z dy , f2 (y) F1 (x) = f1 (x) dx, получаем равенство F2 (y) = F1 (x) + C. (4.3) Заметим, что F20 (y) = 1/f2 (y) 6= 0, поэтому к соотношению (4.3) можно применить теорему об обратной функции, в силу которой равенство (4.3) можно разрешить относительно y и получить формулу y(x) = F2 1 F1 (x) + C , (4.4) справедливую в окрестности точки x0 . Покажем, что равенство (4.4) дает решение уравнения (4.1) в окрестности точки x0 . Действительно, используя теорему о дифференцировании обратной функции и учитывая соотношение F10 (x) = f1 (x), получим y 0 (x) = dF2 1 (z) dz z=F1 (x)+C F10 (x) = 1 F20 (y) y=y(x) F10 (x) = f2 y(x) f1 (x), откуда следует, что функция y(x) из (4.4) является решением уравнения (4.1). Рассмотрим теперь задачу Коши для уравнения (4.1) с начальным условием y(x0) = y0 . (4.5) Формулу (4.2) можно записать в виде Zy dξ = f2 (ξ) Zx f1 (x) dx + C. x0 y0 Подставляя сюда начальное условие (4.5), находим, что C = 0, т.е. решение задачи Коши определяется из соотношения Zy y0 dξ = f2 (ξ) Zx f1 (x) dx. x0 -56- (4.6) Очевидно, оно определяется однозначно. Итак, общее решение уравнения (4.1) задается формулой (4.4), а решение задачи Коши (4.4), (4.5) находится из соотношения (4.6). Замечание 4. 1. Если f2 (y) = 0 при некоторых y = yj , (j = 1, 2, . . . , s), то, очевидно, решениями уравнения (4.1) являются также функции y(x) yj , j = 1, 2, . . . , s, что доказывается непосредственной подстановкой этих функций в уравнение (4.1). Замечание 4. 2. Для уравнения (4.1) общее решение определяем из соотношения F2 (y) F1 (x) = C. (4.7) Таким образом, левая часть соотношения (4.7) постоянна на каждом решении уравнения (4.1). Соотношения типа (4.7) можно записать и при решении других ОДУ. Такие соотношения принято называть интегралами (общими интегралами) соответствующего ОДУ. Дадим точное определение. Определение 4. 1. Рассмотрим уравнение y 0 (x) = f (x, y). (4.8) Соотношение (x, y) = C, (4.9) где (x, y) – функция класса C 1 , называется общим интегралом уравнения (4.8), если это соотношение не выполняется тождественно, но выполняется на каждом решении уравнения (4.8). При каждом конкретном значении C 2 R мы получаем частный интеграл. Общее решение уравнения (4.8) получается из общего интеграла (4.9) с использованием теоремы о неявной функции. Пример 4. 1. Рассмотрим уравнение x (4.10) y 0 (x) = y и начальное условие y(2) = 4. (4.11) Применяя для решения уравнения (4.10) описанный выше метод разделения переменныõ, получаем y dy = x dx, откуда находим общий интеграл для уравнения (4.10) y 2 x2 = C. Общее решение уравнения (4.10) запишется по формуле p y= C + x2 , а решение задачи Коши (4.10), (4.11) – по формуле p y = 12 + x2 . -57- 4. 2. Линейные ОДУ первого порядка Линейным ОДУ первого порядка называется уравнение y 0 (x) + p(x)y(x) = q(x), Если q(x) 6 Если q(x) x 2 ha, bi. (4.12) 0, то уравнение называется неоднородным. 0, то уравнение называется однородным: y 0 (x) + p(x)y(x) = 0. (4.120) Теорема 4. 1. 1) Если y1 (x), y2 (x) решения однородного уравнения (4.120), α, β произвольные числа, то функция y (x) αy1 (x) + βy2 (x) также является решением уравнения (4.120). 2) Для общего решения неоднородного уравнения (4.12) имеет место формула yон = yоо + yчн; (4.13) здесь yон общее решение неоднородного уравнения (4.12), yчн частное решение неоднородного уравнения (4.12), yоо общее решение однородного уравнения (4.120). Доказательство. Первое утверждение теоремы доказывается непосредственной проверкой: имеем y 0 αy10 + βy20 = αp(x)y1 βp(x)y2 = p(x) αy1 + βy2 = p(x)y . Докажем второе утверждение. Пусть y0 – произвольное решение уравнения (4.120), тогда y00 = p(x)y0 . C другой стороны, 0 yчн = p(x)yчн + q(x). Следовательно, 0 y0 + yчн = p(x) y0 + yчн + q(x), а значит y y0 + yчн является решением уравнения (4.12). Таким образом, формула (4.13) дает решение неоднородного уравнения (4.12). Покажем, что по этой формуле могут быть получены все решения уравнения (4.12). Действительно, пусть y^(x) – решение уравнения (4.12). Положим y~(x) = y^(x) yчн. Имеем y~ 0 (x) = y^ 0 (x) 0 yчн (x) = p(x)^ y (x) + q(x) + p(x)yчн (x) = p(x) y^(x) q(x) = yчн (x) = p(x)~ y (x). Таким образом, y~(x) – решение однородного уравнения (4.120), и мы имеем y^(x) = y~(x) + yчн, что соответствует формуле (4.13). Теорема доказана. -58- Ниже будем рассматривать задачи Коши для уравнений (4.12) и (4.120) с начальным условием y(x0) = y0 , x0 2 ha, bi. (4.14) Относительно функций p(x) и q(x) из (4.12) будем предполагать, что p(x), q(x) 2 C (ha, bi). Замечание 4. 3. Положим F (x, y) = p(x)y + q(x). Тогда в силу наложенных выше условий на p(x) и q(x) имеем F (x, y), ∂F (x, y) 2 C G , ∂y G = ha, bi R1 , а следовательно, для задачи Коши (4.12), (4.14) справедливы теоремы существования и единственности решения, доказанные в главе 2. В доказанных ниже теоремах 4. 2, 4. 3 будут получены явные формулы для решений уравнений (4.120) и (4.12) и будет показано, что эти решения существуют на всем промежутке ha, bi. Рассмотрим сначала однородное уравнение (4.120). Теорема 4. 2. утверждения: Пусть p(x) 2 C (ha, bi). Òогда справедливы следующие 1) любое решение уравнения (4.120) определено на всем промежутке ha, bi; 2) общее решение однородного уравнения (4.120) задается формулой y(x) = C e где C R p(x) dx , (4.15) произвольная константа; 3) решение задачи Êоши (4.120), (4.14) задается формулой Rx y(x) = y0 e x0 p(ξ) dξ . (4.16) Доказательство. Выведем формулу (4.15) в соответствии с данной в начале главы методикой. Прежде всего заметим, что функция y 0 является решением уравнения (4.120). Пусть y(x) – решение уравнения (4.120), причем y 6 0 на ha, bi. Тогда 9 x1 2 ha, bi такая, что y(x1) = y0 6= 0. Рассмотрим уравнение (4.120) в окрестности точки x1 . Это уравнение с разделяющимися переменными, причем y(x) 6= 0 в некоторой окрестности точки x1 . Тогда, следуя результатам предыдущего параграфа, получим явную формулу для решения Z dy = p(x) dx, ln y = p(x) dx + C, y -59- откуда R y(x) = C e p(x) dx , c 6= 0, что соответствует формуле (4.15). Áолее того, решение y 0 также задается формулой (4.15) при C = 0. Непосредственной подстановкой в уравнение (4.120) убеждаемся, что функция y(x), задаваемая по формуле (4.15) при любом C, является решением уравнения (4.120), причем на всем промежутке ha, bi. Покажем, что формула (4.15) задает общее решение уравнения (4.120). Действительно, пусть y^(x) – произвольное решение уравнения (4.120). Если y^(x) 6= 0 на ha, bi, то повторяя предыдущие рассуждения, получим, что эта функция задается формулой (4.15) при некотором C: именно, если y^(x0) = y^0 , то Rx p(ξ) dξ . y^(x) = y^0 e x0 Если же 9x1 2 ha, bi такая, что y^(x1) = 0, то задача Коши для уравнения (4.120) с начальным условием y(x1) = 0 имеет два решения y^(x) и y(x) 0. В силу замечания 4. 3 решение задачи Коши единственно, поэтому y^(x) 0, а следовательно, задается формулой (4.15) при C = 0. Итак, доказано, что общее решение уравнения (4.120) определено на всем ha, bi и задается формулой (4.15). Формула (4.16), очевидно, является частным случаем формулы (4.15), поэтому задаваемая ею функция y(x) является решением уравнения (4.120). Кроме того, x R0 p(ξ) dξ y(x0) = y0 e x0 = y0 , поэтому формула (4.16) действительно задает решение задачи Коши (4.120), (4.14). Теорема 4. 2 доказана. Рассмотрим теперь неоднородное уравнение (4.12). Теорема 4. 3. Пусть p(x), q(x) 2 C (ha, bi). Òогда справедливы следующие утверждения: 1) любое решение уравнения (4.12) определено на всем промежутке ha, bi; 2) общее решение неоднородного уравнения (4.12) задается формулой Z R R R p(x) dx p(x) dx q(x)e p(x) dx dx, (4.17) y(x) = Ce +e где C произвольная константа; 3) решение задачи Êоши (4.12), (4.14) задается формулой Rx y(x) = y0 e x0 Zx p(ξ) dξ + q(ξ)e x0 -60- Rx ξ p(θ) dθ dξ. (4.18) Доказательство. В соответствии с теоремой 4. 1 и формулой (4.13) yон = yоо + yчн требуется найти частное решение уравнения (4.12). Для его нахождения применим так называемый метод вариации произвольной постоянной. Суть этого метода заключается в следующем: берем формулу (4.15), заменяем в ней константу C на неизвестную функцию C(x) и ищем частное решение уравнения (4.12) в виде yчн (x) = C(x) e R p(x) dx . (4.19) Подставим yчн (x) из (4.19) в уравнение (4.12) и найдем C(x) так, чтобы это уравнение удовлетворялось. Имеем R R 0 yчн (x) = C 0 (x) e p(x) dx + C(x) e p(x) dx p(x) . Подставляя в (4.12), получим C 0 (x) e R p(x) dx + C(x) e R p(x) dx p(x) + C(x)p(x) e R p(x) dx = q(x), откуда R C 0 (x) = q(x) e p(x) dx . Интегрируя последнее соотношение и подставляя найденное C(x) в формулу (4.19), получим, что Z R R p(x) dx yчн (x) = e q(x) e p(x) dx dx. Кроме того, в силу теоремы 4. 2 R yоо = C e p(x) dx . Поэтому используя формулу (4.13) из теоремы 4. 1, получаем, что Z R R R p(x) dx p(x) dx y(x) = yоо + yчн = Ce +e q(x)e p(x) dx dx, что совпадает с формулой (4.17). Очевидно, что формула (4.17) задает решение на всем промежутке ha, bi. Наконец, решение задачи Коши (4.12), (4.14) задается формулой Rx y(x) = y0 e Rx p(ξ) dξ x0 +e p(θ) dθ Zx Rξ p(θ) dθ q(ξ)ex0 x0 dξ. (4.20) x0 Действительно, формула (4.20) является частным случаем формулы (4.17) при C = y0 , поэтому она задает решение уравнения (4.12). Кроме того, x R0 y(x0) = y0 e x0 x R0 p(ξ) dξ +e p(θ) dθ Zx0 Rξ q(ξ)e x0 x0 x0 -61- p(θ) dθ dξ = y0 , поэтому удовлетворяются начальные данные (4.14). Приведем формулу (4.20) к виду (4.18). Действительно, из (4.20) имеем Rx y(x) = y0 e Zx p(ξ) dξ + x0 Rξ q(ξ)e x p(θ) dθ Rx dξ = y0 e Zx p(ξ) dξ + x0 x0 Rx q(ξ)e p(θ) dθ dξ, ξ x0 что совпадает с формулой (4.18). Теорема 4. 3 доказана. Ñледствие(об оценке решения задачи Коши для линейной системы). x0 2 ha, bi, p(x), q(x) 2 C (ha, bi), причем p(x) 6 K, q(x) 6 M Пусть 8 x 2 ha, bi. Òогда для решения задачи Êоши (4.12), (4.14) справедлива оценка M Kjx x0 j Kjx x0 j y(x) 6 y0 e + e 1 . K (4.21) Доказательство. Пусть сначала x > x0. By virtue of (4.18), we have Rx zx k dξ y (x) 6 y0 ex0 rx k dθ m eξ + dξ \u003d y0 ek (x x0) zx + m x0 \u003d y0 e k (x x0) ek (x ξ) dξ \u003d x0 m + k e k (x ξ) ξ \u003d x ξ \u003d x0 \u003d y0 e kjx x0 j m kjx + e k x0 j 1. Now let X.< x0 . Тогда, аналогично, получаем x R0 y(x) 6 y0 e x K dξ Zx0 + Rξ M ex K dθ dξ = y0 eK(x0 x) Zx0 +M x = y0 e K(x0 eK(ξ x) dξ = x M x) eK(ξ + K x) ξ=x0 ξ=x M h K(x0 x) = y0 e + e K M Kjx Kjx x0 j e = y0 e + K i 1 = K(x0 x) x0 j Таким образом, оценка (4.21) справедлива 8 x 2 ha, bi. Пример 4. 2. Решим уравнение y = x2 . x Решаем сначала однородное уравнение: y0 y0 y = 0, x dy dx = , y x ln jyj = ln jxj + C, -62- y = C x. 1 . Решение неоднородного уравнения ищем методом вариации произвольной постоянной: y чн = C(x) x, Cx = x2 , x 0 C x+C 0 C = x, x2 C(x) = , 2 откуда x3 , 2 y чн = а общее решение исходного уравнения y =Cx+ x3 . 2 4. 3. Однородные уравнения Однородным уравнением называется уравнение вида y 0 y (x) = f , (x, y) 2 G, x (4.22) G – некоторая область в R2 . Áудем предполагать, что f (t) – непрерывная функция, x 6= 0 при (x, y) 2 G. Однородное уравнение заменой y = xz, где z(x) – новая искомая функция, сводится к уравнению с разделяющимися переменными. В силу данной замены имеем y 0 = xz 0 + z. Подставляя в уравнение (4.22), получим xz 0 + z = f (z), откуда z 0 (x) = 1 x f (z) z . (4.23) Уравнение (4.23) представляет собой частный случай уравнения с разделяющимися переменными, рассмотренного в п. 4.1. Пусть z = ϕ(x) – решение уравнения (4.23). Тогда функция y = xϕ(x) является решением исходного уравнения (4.22). Действительно, y 0 = xϕ 0 (x) + ϕ(x) = x 1 x f (ϕ(x)) ϕ(x) + ϕ(x) = xϕ(x) y(x) = f ϕ(x) = f =f . x x Пример 4. 3. Ðешим уравнение y0 = y x -63- ey/x . Положим y = zx. Тогда xz 0 + z = z откуда y(x) = 1 z e, x z0 = ez , dz dx = , e z = ln jzj + C, z e x z = ln ln Cx , c 6= 0, x ln ln Cx , c 6= 0. 4. 4. Уравнение Áернулли Уравнением Áернулли называется уравнение вида y 0 = a(x)y + b(x)y α , α 6= 0, α 6= 1 . x 2 ha, bi (4.24) При α = 0 или α = 1 получаем линейное уравнение, которое было рассмотрено в п. 4.2. Áудем предполагать, что a(x), b(x) 2 C (ha, bi). Замечание 4. 4. Если α > 0, then, obviously, the function y (x) 0 is a solution of equation (4.24). To solve the Equation of the Bernilli (4.24) α 6 \u003d 0, α 6 \u003d 1, we divide both parts of the equation on Y α. At α\u003e 0, it is necessary to take into account that by virtue of comments 4. 4, the function y (x) 0 is the solution of equation (4.24), which will be lost with this division. Consequently, in the future it will need to add to the general decision. After division, we obtain the ratio y α y 0 \u003d a (x) y 1 α + b (x). Introducts a new desired function z \u003d y 1 α, then z 0 \u003d (1 therefore, we arrive at the equation relative to z z 0 \u003d (1 α) a (x) z + (1 α) y α) b (x). α y 0, A (4.25) Equation (4.25) is a linear equation. Such equations are considered in clause 4.2, where the formula of a general solution was obtained, by virtue of which the solution z (x) of equation (4.25) is written in the form Z (x) \u003d Ce R (α 1) A (x) DX + + (1 α ) E R (α 1) A (x) DX 1 Z b (x) E R (α 1) A (x) DX DX. (4.26) Then the function y (x) \u003d z 1 α (x), where Z (x) is defined in (4.26), is a solution of the Bernilli equation (4.24). -64- In addition, as described above, at α\u003e 0, the solution also is the function y (x) 0. Example 4. 4. 4. The equation y 0 + 2y \u003d y 2 ex. (4.27) We divide equation (4.27) to Y 2 and we will replace Z \u003d we obtain a linear inhomogeneous equation 1 y. As a result, z 0 + 2z \u003d ex. (4.28) We first decide the homogeneous equation: z 0 + 2z \u003d 0, dz \u003d 2dx, z ln jzj \u003d 2x + c, z \u003d Ce2x, C 2 R1. Solution of the inhomogeneous equation (4.28) We are looking for a variation of an arbitrary constant: Zn \u003d C (x) E2x, C 0 E2X 2CE2X + 2CE2X \u003d EX, C 0 \u003d EX, C (X) \u003d EX, from where Zhen \u003d EX, and the general solution of the equation (4.28) z (x) \u003d Ce2x + EX. Consequently, the solution of the Bernilli equation (4.24) is recorded as y (x) \u003d 1. EX + CE2X In addition, the solution of equation (4.24) is also the function y (x) this solution we lost when this equation is divided into Y 2. 0. 4. 5. The equation in full differentials Consider the equation in differentials M (x, y) dx + n (x, y) dx \u003d 0, (x, y) 2 g, (4.29) G is some region in R2. This equation is called the equation in full differentials if there is a function f (x, y) 2 C 1 (g), called the potential, such that Df (x, y) \u003d m (x, y) dx + n (x, y ) DY, (X, Y) 2 G. We will assume that m (x, y), n (x, y) 2 C 1 (g), and the area G is a single-connected. In these assumptions, aware of mathematical analysis (see, for example), it is proved that the potential F (X, Y) for equation (4.29) exists (i.e. (4.29) - the equation in complete differentials) if and only when My (x, y) \u003d nx (x, y) -65- 8 (x, y) 2 g. At the same time (x, z y) f (x, y) \u003d m (x, y) dx + n (x, y) dy, (4.30) (x0, y0) where the point (x0, y0) is some fixed point From G, (x, y) - the current point in G, and the curvilinear integral is taken along any curve connecting the points (x0, y0) and (x, y) and the entire lying in the region G. If equation (4.29) is equation

This lecture course is read more than 10 years for students of theoretical and applied mathematics in the Far Eastern State University. Complies with the standard II generation but these specialties. Recommended students and undergraduates of mathematical specialties.

Cauchy Theorem on the existence and uniqueness of the solution of the Cauchy problem of the first order equation.
In this paragraph, we, imposing certain limitations on the right-hand part of the first-order differential equation, we prove the existence and uniqueness of the solution determined by the initial data (X0, U0). The first proof of the existence of the solution of differential equations belongs to Cauchy; The proof below is given by the Picar; It is performed using the method of consecutive approximations.

TABLE OF CONTENTS
1. First order equations
1.0. Introduction
1.1. Equations with separating variables
1.2. Uniform equations
1.3. Generalized homogeneous equations
1.4. Linear equations of the first order and leading to them
1.5. Bernoulli equation
1.6. Riccati equation
1.7. Equation in full differentials
1.8. Integrating multiplier. The simplest cases of finding an integrating multiplier
1.9. Equations that are not resolved relative to the derivative
1.10. Cauchy Theorem on the existence and uniqueness of the solution of the Cauchy challenge of the first order
1.11. Special Points
1.12. Special solutions
2. Equations of higher orders
2.1. Basic concepts and definitions
2.2. Types of N-order equations that are resolved in quadratures
2.3. Intermediate integrals. Equations that allow decrease in order
3. Linear differential equations on order
3.1. Basic concepts
3.2. Linear homogeneous differential equations on order
3.3. Lowering the order of a linear homogeneous equation
3.4. Heterogeneous linear equations
3.5. Decrease of order in a linear inhomogeneous equation
4. Linear equations with constant coefficients
4.1. Uniform linear equation with constant coefficients
4.2. Inhomogeneous linear equations with constant coefficients
4.3. Linear equations of second order with fluctuating solutions
4.4. Integration by power series
5. Linear systems
5.1. Inhomogeneous and homogeneous systems. Some properties of solutions of linear systems
5.2. Required and sufficient conditions for linear independence to solutions of a linear homogeneous system
5.3. The existence of a fundamental matrix. Building a general solution of a linear homogeneous system
5.4. Construction of the whole set of fundamental matrices of a linear homogeneous system
5.5. Inhomogeneous systems. Construction of a general solution by the method of variation of arbitrary constant
5.6. Linear homogeneous systems with constant coefficients
5.7. Some information from the theory of functions from matrices
5.8. Constructing a fundamental matrix of a system of linear homogeneous equations with constant coefficients in the general case
5.9. Theorem of the existence and theorem on the functional properties of solutions of normal systems of the first-order differential equations
6. Elements of the theory of stability
6.1
6.2. The simplest types of rest points
7. Equations in partial derivatives of the 1st order
7.1. Linear homogeneous equation in partial derivatives of the 1st order
7.2. Inhomogeneous linear equation in private derivatives of 1 order
7.3. System of two partial differential equations with 1 unknown function
7.4. Pfaffa equation
8. Options for test tasks
8.1. Test №1
8.2. Examination number 2.
8.3. Examination number 3.
8.4. Examination number 4.
8.5. Examination number 5.
8.6. Examination number 6.
8.7. Examination number 7.
8.8. Examination number 8.

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"Lectures on ordinary differential equations Part 1. The elements of the general theory in the textbook are presented the provisions that make up the basis for the theory of ordinary differential equations: ..."

-- [ Page 1 ] --

A. E. Mamontov

Lectures on ordinary

Differential equations

Elements of general theory

The textbook sets out the provisions constituting

the basis of the theory of ordinary differential equations: the concept of solutions, their existence, uniqueness,

dependence on parameters. Also (in § 3), a certain attention is paid to the "explicit" solution to some classes of equations. The manual is intended for in-depth study Course "Differential Equations" by students studying at the Mathematics Faculty of the Novosibirsk State Pedagogical University.

UDC 517.91 BBK B161.61 Preface Tutorial Designed for students of the Mathematical Faculty of Novosibirsk State Pedagogical University, who want to explore the obligatory course "Differential equations" in an extended volume. The readers are offered the main concepts and results that make up the foundation of the theory of ordinary differential equations: the concept of decisions, theorems on their existence, uniqueness, depending on the parameters. The described material is presented in the form of a logical inseparable text in §§ 1, 2, 4, 5. Also (in § 3, standing by several mansion and temporarily interrupting the main thread of the course) briefly consider the most demanded techniques of "explicit" finding solutions of some classes of equations. When first reading § 3 you can skip without significant damage for the logical structure of the course.

Exercises play an important role, in large quantities included in the text. The reader is strongly recommended to sharpen their "hot pixels", which guarantees the assimilation of the material and serve as a test. Moreover, often these exercises fill a logical tissue, that is, without solving them, not all provisions will be strictly proven.

In square brackets in the middle of the text made comments that bear the role of comments (extended or side explanations). Lexically, these fragments are interrupting the main text (i.e., for a coherent reading, they need to "not notice"), but still they are needed as an explanation. In other words, these fragments need to be perceived as if they were put on the fields.

The text there are separately categorized "comments for the teacher" - they can be omitted when reading students, but useful for the teacher who will use the manual, for example, when reading lectures - they help better understand the logic of the course and indicate the direction of possible improvements (extensions) of the course . However, the development of these comments to students can only be welcomed.

A similar role is played by "justifications for the teacher" - they are in extremely compressed form give the proof of some provisions offered to the reader as exercises.

The most common (key) terms are used in the form of abbreviations, the list of which is given for convenience at the end. There also provides a list of mathematical designations encountered in the text, but not related to the most common (and / or unknown in the literature unambiguous).

The symbol means the end of the proof, the formulation of approval, comments, etc. (where it is necessary to avoid confusion).

The numbering of formulas is independently conducted in each paragraph. As referenced to a part of the formula, indices are used, for example (2) 3 means the 3rd part of the formula (2) (fragments, separated typographic spaces, and from logical positions - a ligament "and").

This manual cannot completely replace the deep study of the subject, which requires independent exercises and reading additional literature, for example, the list of which is given at the end of the manual. However, the author tried to set out the main provisions of the theory in a sufficiently compressed form suitable for the lecture course. In this regard, it should be noted that when reading the lecture course, about 10 lectures takes place on this manual.

An edition of another 2 parts (volumes), which continues this manual and the current cycle of lectures on the subject of "ordinary differential equations" are planned: Part 2 (linear equations), part 3 (further theory of nonlinear equations, equations in partial derivatives of the first order).

§ 1. Introduction The differential equation (DB) is the ratio of the form U1 U1 Un, the highest derivatives f y, u (y), ..., \u003d 0, y1 y2 yk (1) where y \u003d (y1, ..., yk) Rk - independent variables, and u \u003d u (y) - unknown functions1, u \u003d (u1, ..., un). Thus, in (1) there are n unknown, so that N equations are required, i.e. f \u003d (f1, ..., fn), so (1) is, generally speaking, the system from N equations. If an unknown function is one (n \u003d 1), then equation (1) is scalar (one equation).

So, the function (s) f is set (s), and u is searched. If k \u003d 1, then (1) is called ODU, and otherwise - uch. The second case is the subject of the Special Course of UMF, set out in the series of textbooks. In this series of benefits (consisting of 3 parts-volumes), we will study only the ODU, with the exception of the last paragraph of the last part (volume), in which we will start learning some particular cases of uch.

2U U Example. 2 \u003d 0 - This is an urgent.

y1 y Unknown values \u200b\u200bU can be real or complex, which is insignificant, since this moment only applies to the form of recording equations: any comprehensive record can be turned into a real, separating the real and imaginary parts (but, of course, doubles the number of equations And unknown), and vice versa, in some cases it is convenient to move to an integrated record.

du d2v dv · 2 \u003d uv; U3 \u003d 2. This is a system of 2 ODE example.

dY DY DY for 2 unknown functions from an independent alternating y.

If k \u003d 1 (ODU), then the "direct" icon d / dy is used.

u (y) du Example. EXP (SIN Z) DZ is an ORD, since it has an example. \u003d U (u (y)) at n \u003d 1 is not a remote control, but the function is a developing differential equation.

This is not a remote control, but an integro-differential equation, we will not study such equations. However, the equation (2) is easily reduced to the ODU:

The exercise. To reduce (2) to the ODU.

But in general, integral equations are a more complex object (it is partially studied in a course of functional analysis), although, as we will see below, it is with their help that some results are obtained for ODU.

Du arise from both intramathematics needs (for example, in differential geometry) and in applications (historically for the first time, and now mainly in physics). The simplest Du is the "basic task of differential calculus" on the restoration of the function according to its derivative: \u003d H (y). As is known from the analysis, its decision has the form u (y) \u003d + h (s) DS. More general Du for its solutions require special methods. However, as we will see, almost all methods for solving ODU "explicitly" are essentially reduced to the specified trivial case.

In applications most often, the Code occurs when describing the processes developing in time, so that the role of an independent variable is usually playing time t.

thus, the meaning of the ODU in such applications is to describe the change in the parameters of the system over time, so convenient when building general Theory ODE denotes an independent variable through T (and called it with time with all the consequences of the terminological consequences), and the unknown function (AI) - through x \u003d (x1, ..., xn). Thus, the general view of ODU (Systems ODU) is as follows:

where F \u003d (F1, ..., Fn) is that e. This is a system of N ODU for n functions x, and if n \u003d 1, then one ODE for 1 function x.

In this case, x \u003d x (t), T R, and x in generally speaking is complex (this is for convenience, since then some systems are more compactly recorded).

It is said that the system (3) has order M on the XM function.

The derivatives are called elders, and the rest (including XM \u003d) - younger. If all M \u003d, then they simply say that the order of the system is equal.

True, often the order of the system is called the number M, which is also natural, as it becomes clear further.

The question of the need to study the ODU and their applications, we will be considered sufficiently reasonable other disciplines (differential geometry, mathematical analysis, theoretical mechanics, etc.), and it is partially covered in practical training in solving problems (for example, from the task). In the present course, we will be engaged exclusively to mathematical study of the systems of the form (3), which implies the answer to the following questions:

1. What does it mean to "decide" the equation (system) (3);

2. How to do this;

3. What properties have these decisions, how to explore them.

Question 1 is not so obvious as it seems - see below. Immediately note that any system (3) can be reduced to the first order system, denoting the younger derivatives as new unknown functions. The easiest thing is to explain this procedure using the example:

of 5 equations for 5 unknowns. It is easy to understand that (4) and (5) are equivalent in the sense that the solution of one of them (after the corresponding reissued) is the solution of the other. At the same time, it should only be aware of the question of the smoothness of the decisions - this will be done further when we encounter a higher order (i.e. not the first one).

But now it is clear that it is enough to study only one of the first order, and others may be required only for the convenience of designations (this situation will sometimes occur).

And now we restrict ourselves of the first order:

dimx \u003d dimf \u003d n.

The study of the equation (system) (6) is inconvenient due to the fact that it is not resolved relative to DX / DT derivatives. As is known from the analysis (from the theorem on the implicit function), under certain conditions on F, the equation (6) can be resolved relative to the DX / DT and write it in the form where F: Rn + 1 Rn is set, and X: Rn is desired. It is said that (7) there is an ODU, permitted relative to derivatives (ORD of a normal form). During the transition from (6) to (7), naturally, difficulties may arise:

Example. EXP (X) \u003d 0 equation cannot be recorded in the form (7), and does not have solutions at all, that is, the EXP does not have zeros even in the complex plane.

Example. The equation x 2 + x2 \u003d 1 is written in the form of two normal ODE x \u003d ± 1 x2. Each of them should be solved and then interpret the result.

Comment. For information (3) to (6), difficulty may occur if (3) has 0 order for some function or part of functions (i.e. it is a functional differential equation). But then these functions must be excluded by the theorem on the implicit function.

Example. x \u003d y, xy \u003d 1 x \u003d 1 / x. It is necessary to find X from the obtained ODU, and then Y from the functional equation.

But in any case, the problem of transition from (6) to (7) relates rather to the field of mathematical analysis than to the Du, and we will not do it. However, when solving the ODE (6), interesting moments may occur from the point of view, so this question is appropriate to study when solving tasks (as this is done for example B) and it will be slightly affected in § 3. But in the rest of the course we will deal only with normal systems and equations. So, consider the ODU (System ODU) (7). We write it 1 time in the pumidation form:

The concept of "solving (7)" (and in general, any do) for a long time was understood as the search for "explicit formula" to solve (i.e., in the form of elementary functions, their primary, or special functions, etc.), without Accent on the smoothness of the solution and the interval of its definition. However, the current state of the theory of ODU and other sections of mathematics (and in general sciences) shows that such an approach is unsatisfactory - at least because the share of ODUs that giveaway to such "explicit integration" is extremely small (even for the simplest ODE X \u003d F (T) It is known that the solution in elementary functions is rare, although there is a "obvious formula").

Example. Equation X \u003d T2 + X2, despite its extreme simplicity, does not have solutions in elementary functions (and there is even "no formula").

And although you know the classes of ODU, for which the "explicit" solution is possible (in the same way as it is useful to be able to "count the integrals", when it is possible, although it is possible extremely rarely), in this connection, the terms are characterized: "InitEngrilators ODU "," Integral ODU "(outdated analogues of modern concepts" to solve ODU "," ODU decision "), which reflect the former concepts about the decision. How to understand the modern terms, we will state now.

and this issue will be considered in § 3 (and also traditionally much attention is paid to him when solving problems in practical classes), but one should not expect any versatility from this approach. As a rule, under the decision process (7) we will understand completely different steps.

It should be clarified which function x \u003d x (t) may be referred to as a solution (7).

First of all, we note that the clear wording of the concept of solving is impossible without specifying the set on which it is defined, if only because the solution is a function, and any function (according to school definition) is a law that matches any element of a certain set (called the definition region This function) some element of another set (function values). Thus, it is absurd by definition without specifying the area of \u200b\u200bits definition. Analytical functions (more widely elementary) serve here "exception" (misleading) under the reasons listed below (and some others), but in the case of Du such liberty is unacceptable.

and in general, without specifying the sets for the definition of all functions involved in (7). It will be clear from further, it is advisable to rigidly link the concept of solving with a multitude of its definition, and consider solutions to different, if the sets of their definition are different, even if the crosses of these sets coincide.

Most often, in specific situations, this means that if solutions are constructed in the form of elementary functions, so that 2 solutions have a "identical formula", then it is necessary to check whether the sets on which these formulas are written are coincided. The confusion, for a long time reigned in this matter, was an exclusion, while decisions were considered in the form of elementary functions, since the analytical functions clearly continue on wider intervals.

Example. X1 (T) \u003d ET on (0.2) and x2 (t) \u003d et on (1,3) - different solutions of the equation x \u003d x.

At the same time, naturally, in the quality of the definition of any solution to take an open interval (maybe endless), since this set should be:

1. Open so that at any point it makes sense to talk about the derivative (bilateral);

2. Svyaznoy, so that the solution does not fit into incoherent pieces (in this case it is more convenient to talk about several solutions) - see the previous example.

Thus, the solution (7) is a pair (, (a, b)), where A B + is defined on (A, B).

Comment for the teacher. In some textbooks, it is allowed to include the end of the segment in the field of determination of the decision, but it is inappropriate due to the fact that only complicates the presentation, and the real generalization does not give (see § 4).

To make it easier to understand further reasoning, it is useful to use a geometric interpretation (7). In the space Rn + 1 \u003d ((t, x)) at each point (T, x), where F is defined, the vector f (t, x) can be considered. If you construct a solution graph (7) in this space (it is called an integral curve of the system (7)), then it consists of points of the form (T, X (T)). When the T (a, b) is changed, this point moves along the IR. The tangent to the IR at the point (T, X (T)) has the form (1, x (t)) \u003d (1, f (t, x (t))). Thus, IR is those and only those curves in the RN + 1 space, which in each of its own point (T, X) have a tangential, parallel vector (1, f (t, x)). At this idea, it was built so called. The isocline method for the approximate construction of IR, which is used as graphs of solutions of specific ODUs (see

eg ). For example, at n \u003d 1, our construction means the following: at each point IR, its slope to the T axis has the property TG \u003d F (T, X). It is natural to assume that by taking any point from a set of definition F, we can spend an IR through it. This idea will be strictly justified further. As long as we lack the strict formulation of smoothness of solutions - this will be done below.

Now it is necessary to clarify the set B, which defines f. This is a lot of naturally take:

1. open (so that the IR can be built in the neighborhood of any point from b), 2. connected (otherwise all connected pieces can be considered - anyway, IR (as a chart of a continuous function) can not jump out of one piece in another, so that Community search for solutions will not affect).

We will consider only classical solutions (7), i.e. such that the X and its X is continuous on (A, B). Then naturally require f c (b). Next, this requirement will be implied by us. So, we finally get a definition. Let B Rn + 1 be the area, F C (B).

A pair (, (a, b)), AB +, is defined by (a, b), is called the solution (7) if C (A, B), each time T (A, B), (T, (T) ) B and exists (T), with (T) \u003d F (T, (T)) (then automatically C 1 (A, B)).

It is geometrically clear that (7) will have a lot of solutions (which is easy to understand graphically), since to carry out IR, starting with points of the form (T0, X0), where T0 is fixed, then we will receive different IR. In addition, the change in the decision definition interval will give another solution, according to our definition.

Example. x \u003d 0. Solution: X \u003d \u003d const Rn. However, if you select some T0 and fix the value x0 of the solution at the point T0: X (T0) \u003d X0, the value is defined uniquely: \u003d x0, that is, the solution is unique with an accuracy of the interval selection (A, B) T0.

The presence of a "faceless" set of solutions is inconvenient to work with Nimi2 - it is more convenient to "import" them as follows: add to (7) additional conditions so as to highlight the only (in a certain sense) the decision, and then, overgoing these conditions, work with each The solution is separate (the geometrically solution may be one (IR), and there are many pieces - with this inconvenience will look later).

Definition. The task for (7) is (7) with additional conditions.

We substantially invented the simplest task - this is the task of Cauchy: (7) with the conditions of the form (Cauchy data, initial data):

C Point of view of applications This task is natural: for example, if (7) describes the change in some parameters x with time t, then (8) means that in some (initial) moment of time the value of the parameters is known. It is necessary to learn other tasks, we will talk about it later, but for now we will focus on the Cauchy task. Naturally, this task makes sense at (T0, x0) B. Accordingly, the solution of problem (7), (8) is called a solution (7) (in the sense of the definition given above) such that T0 (A, B), and is executed (eight).

Our nearest task is to prove the existence of the solution of the Cauchy problem (7), (8), and with certain additional example - a square equation, it is better to write x1 \u003d ..., x2 \u003d ... than x \u003d b / 2 ± ...

assumptions on F - and its uniqueness in a certain sense.

Comment. We need to clarify the concept of the norm of the vector and the matrix (although the matrices will only be needed in part 2). Due to the fact that in the finite-dimensional space all the rules are equivalent, the choice of a particular norm does not matter if we are interested only as estimates, and not accurate values. For example, for vectors you can apply | x | p \u003d (| xi | p) 1 / p, p - peano segment (pickara). Consider the cone k \u003d (| x x0 | f | t t0 |) and its truncated part k1 \u003d k (t ip). It is clear that just k1 C.

Theorem. (Peano). Let the requirements for F in the problem (1) specified in determining the solution, i.e.:

f C (b), where b is an area in Rn + 1. Then, at all (T0, x0) b, there is a solution of task (1).

Evidence. Set arbitrarily (0, T0] and construct the so-called. Loars of Euler with a step, namely: it is a broken in Rn + 1, in which each link has a projection on the axis t long, the first link to the right begins at the point (T0, X0) and This is that it is DX / DT \u003d F (T0, X0); the right end of this link (T1, X1) serves as the left end for the second, on which DX / DT \u003d F (T1, X1), and so on, And similarly to the left. The resulting broken defines a piecewise linear function x \u003d (t). While T IP, the broken remains in K1 (and even more so in C, and therefore in b), so the construction is correct - for this actually and was done Auxiliary construction before the theorem.

In fact, everywhere except the points of the break there exists, and then (s) (t) \u003d (z) dz, where there are arbitrary derivative values \u200b\u200bat the break points.

At the same time (moving along the induction lifriction) in particular, | (t) x0 | F | T T0 |.

Thus, on IP functions:

2. Equatively continuous, since Lipshtsy:

Here, if necessary, refresh my knowledge about such concepts and results as: equitable continuity, uniform convergence, the theorem of Arzel Ascol, etc.

By the arzel-ascol theorem there is a sequence k 0 such that K on IP, where C (IP). By construction, (t0) \u003d x0, so it remains to verify that we will prove it for s t.

The exercise. Similar to consider S t.

Set 0 and we find 0 so that for all (T1, x1), (T2, x2) C, this can be done due to the uniform continuity of F on Compact C. We will find M n so that you fix T int (IP) and take any s INT (IP) is that TST +. Then for all Z we have | k (z) k (t) | F, so in mind (4) | k (z) (t) | 2F.

Note that k (z) \u003d k (z) \u003d f (z, k (z)), where Z is the abscissa of the left end of the broken cut containing the point (z, k (z)). But the point (z, k (z)) enters the cylinder with parameters (2f), built at the point (T, (T)) (in fact, even in a truncated cone - see fig. But it does not matter now), So in view of (3) we obtain | k (z) f (t, (t)) |. For broken, we have, as mentioned above, the formula for k it will give (2).

Comment. Let F C 1 (B) be. Then the solution defined on (A, B) will be class C 2 (A, B). In fact, on (a, b) we have: There are f (t, x (t)) \u003d ft (t, x (t)) + (t, x (t)) x (t) (here - the matrix of Jacobi ) - Continuous function. Cheat, there is also 2 C (a, b). You can continue to increase the smoothness of the solution if F smooth. If F is analytic, then you can prove the existence and uniqueness of the analytical solution (this is so called. Cauchy theorem), although it should not be in the previous reasoning!

Here it is necessary to remember what an analytical function is. Not to be confused with the function, representable by the power number (it is only a representation of the analytical function on, generally speaking, parts of its definition)!

Comment. At specified (T0, X0), it is possible, varying T and R, try to maximize T0. However, this is usually not so important, since to study the maximum interval of the existence of the solution there are special methods (see § 4).

In Perano Theorem, nothing stated about the uniqueness of the decision. With our understanding of the solution, it is always not the only one, since if there is some solution, then its narrowings will be other solutions for narrower intervals. This moment we will consider in more detail later (in § 4), and so far, under uniqueness, we understand the coincidence of any two solutions at the intersection of the intervals of their definition. Even in this sense, the Peaano Theorem does not say anything about the uniqueness that it is not by chance, since it is impossible to guarantee uniqueness.

Example. n \u003d 1, f (x) \u003d 2 | x |. The Cauchy problem has a trivial solution: x1 0, and besides X2 (T) \u003d T | T |. Of these two solutions, a whole 2-parametric family family can be compiled:

where + (endless values \u200b\u200bmean the absence of the corresponding branch). If you count for the area of \u200b\u200bdefinition of all these solutions all R, then they are still infinitely a lot.

We note that if you apply the proof of the Peano theorem through the broken Euler, then only the zero solution will be. On the other hand, if in the process of building a broken Euler to allow a small error at each step, even after the desire of the error parameter to zero, all solutions will remain. Thus, Peano theorem and broken Euler are natural as a method for building solutions and closely related to numerical methods.

The trouble observed in the example is due to the fact that the function F is non-deployed by x. It turns out that if you impose additional requirements for the regularity F by x, then the unity can be ensured, and this step is in a certain sense necessary (see below).

Recall some concepts from the analysis. The function (scalar or vector) G is called a Helder with an indicator (0, 1] on the set, if the Lipshitz condition is correct. At 1, it is possible only for permanent functions. The function specified on the segment (where choice is irrelevant) is called the continuity module, If they say that G satisfies in the generalized Helder condition with a module if in this case is called the continuity module G c.

It can be shown that any continuity module is a continuity module of some continuous function.

The reverse fact is important to us, namely: any continuous function on the compact has its own continuity module, i.e. satisfies (5) with some. We prove it. Recall that if it is a compact, and G c (), then it is uniformly continuous in, i.e.

\u003d (): | x Y | \u003d | G (x) G (y) |. It turns out that this is equivalent to condition (5) with some. In fact, if there is, it is enough to build a continuity module such that (()), and then at | x y | \u003d \u003d () We obtain (and) arbitrarily, x and y can be any.

And vice versa, if (5) is true, it is enough to find such that (()), and then at | x y | \u003d () We get it remains to substantiate logical transitions:

For monotonous and enough to take inverse functions, and in general, it is necessary to use so called. Generalized feedback. Their existence requires a separate proof that we will not lead, but let's just say an idea (it is useful to accompany reading drawings):

for any f, we define f (x) \u003d min f (y), f (x) \u003d max f (y) are monotone functions, and they have inverse. VVIF is easy to verify that x x f (f (x)), (f) 1 (f (x)) x, f ((f) 1 (x)) x.

The best continuity module is linear (Lipschits condition). These are "almost differentiable" functions. To give strict sense, the latest statement requires certain efforts, and we will limit ourselves to two comments:

1. Strictly speaking, not all Lipschitz function differentiating, as the example G (x) \u003d | x | on R;

2. But from differentiability follows Lipshetsevity, as the following statement shows. Any function G, which has all M on a convex set, satisfies the Lipschitz condition on it.

[While for brevity, consider scalar functions g.] Proof. For all X, Y we have clearly that this statement is true for vector functions.

Comment. If f \u003d f (t, x) (generally speaking, vector illustration), then you can enter the concept of "F Lipschitsev by X", i.e. | f (t, x) f (t, y) | C | XY |, and also prove that if D is convex in x for all T, then for the lipshetsevity f by x in D, it is enough to have derivatives F by x, limited in the statement we received an estimate | G (x) G (Y) | through | x y |. With n \u003d 1, it is usually done using a finite increment formula: G (x) g (y) \u003d G (z) (xy) (if G is a vector function, then z is its own for each component). For n 1 it is convenient to use the following analogue of this formula:

Lemma. (Adamara). Let F c (d) be (generally speaking, the vector function), where D (T \u003d T) convex at any t, and f (t, x) f (t, y) \u003d a (t, x, y) · (XY), where a is a continuous rectangular matrix.

Evidence. With any fixed T, we apply the calculation from the proof of the approval for \u003d D (T \u003d T), G \u003d Fk. We obtain the desired representation with a (t, x, y) \u003d a is in fact continuous.

Let's return to the question of the uniqueness of the solution of the problem (1).

We will put the question like this: what should be the continuity module F by x, so that the solution (1) is the only thing in the sense that 2 solutions defined in the same interval coincide? The answer is given to the following theorem:

Theorem. (Osgood). Let under the conditions of the PEANO the continuity of the continuity of the X in B, i.e. the function in inequality satisfies the condition (C). Then the problem (1) cannot have two different solutions defined in the same form interval (T0 A, T0 + B).

Compare with an example of an interconnection above.

Lemma. If z C 1 (,), then all (,):

1. At points where Z \u003d 0, exists | z |, and || z | | | z |;

2. At points, where Z \u003d 0, there are unilateral derivatives | z | ±, and || z | ± | \u003d | z | (In particular, if z \u003d 0, then exists | z | \u003d 0).

Example. n \u003d 1, z (t) \u003d t. At point T \u003d 0 derivative from | z | There is no, but there are one-sided derivatives.

Evidence. (Lemmas). At those points where z \u003d 0, havez · z it: exists | z | \u003d, and || z | | | z |. At those points T, where z (t) \u003d 0, we have:

Case 1: Z (T) \u003d 0. Then we get existence | z | (T) \u003d 0.

Case 2: Z (T) \u003d 0. Then, at +0 or 0 Ochiz (T +) | | z (t) | The module of which is equal | z (t) |.

By condition, F C 1 (0,), F 0, F, F (+0) \u003d +. Let z1,2 be two solutions (1) defined by (T0, T0 +). Denote z \u003d z1 z2. We have:

Suppose it is T1 (for definitely T1 T0) such that z (T1) \u003d 0. The set A \u003d (T T1 | Z (T) \u003d 0) is not empty (T0 A) and limited from above. It means that it has an upper face T1. By construction, z \u003d 0 on (, T1), and due to the continuity of Z, we have z () \u003d 0.

By Lemma | z | C 1 (, T1), and at this interval right | z | | z | (| Z |), so that the integration of software (T, T1) (where T (, T1)) gives f (| z (t) |) f (| z (t1) |) T1 T. With T + 0, we get a contradiction.

Corollary 1. If, under the conditions of Peano F theorem, Lipshitsev according to X in B, then the problem (1) has a single solution in the sense described in the Osgood theorem, since in this case () \u003d c satisfies (7).

Corollary 2. If under the conditions of Peaano C (B) theorem, then the solution (1) defined on int (IP), the only one.

Lemma. Any decision (1), defined on IP, is required to satisfy the rating | x | \u003d | F (T, X) | F, and its schedule - lie in K1, and even more so in C.

Evidence. Suppose, there will be T1 IP such that (T, X (T)) C. For definiteness, let T1 T0. Then there is T2 (T0, T1] such that | x (t) x0 | \u003d R. Similarly, the arguments in the proof of the Osca's theorem can be considered that T2 is the most left such point, and on we have (t, x (t)) C, so | f (t, x (t)) | f, and therefore (t, x (t)) k1, which contradicts | x (t2) x0 | \u003d R. So, (T, X (T) ) C on all IP, and then (repeating the calculations) (t, x (t)) k1.

Evidence. (Corollary 2). C is a compact MNOF that F Lipszytsev software in C, where the graphs lie all solutions in view of the lemma. By investigation 1 we get the desired.

Comment. Condition (7) means that the Lipschitz condition for F cannot be substantially weakened. For example, the condition of the Helder C 1 is no longer suitable. Only the continuity modules are close to linear - such as bonding "the bad":

The exercise. (complicated enough). Prove that if satisfies (7), then there is 1, satisfying (7) such that 1 / in zero.

In general, it is not necessary to require something from the continuity module F by x for uniqueness - various kinds of special occasions are possible, for example:

Statement. If, in the conditions of the peano theorem, any 2 solutions (1), defined on (9), can be seen that X C 1 (A, B), and then differentiation (9) gives (1) 1, and (1) 2 obvious .

In contrast to (1), for (9), it is natural to build a solution on a closed segment.

Picar suggested for solving (1) \u003d (9) the following method of consecutive approximations. Denote x0 (t) x0, and then on the induction of the theorem. (Cauchy Picara). Let under the conditions of the Peano theorem function F Lipshitsev according to x in any convex of X compact K from the region B, i.e.

Then for any (T0, x0) b, the Cauchy task (1) (it (9)) has a single solution on int (IP), and XK x on IP, where XK is defined in (10).

Comment. It is clear that the theorem remains strength if the condition (11) is replaced by C (B), because of this condition, follows (11).

Comment for the teacher. In fact, it is not necessary for all convex software by X compacts, but only cylinders, but the wording is done exactly that, because in § 5 more general compacts will be required, and more precisely, it is precisely this wording.

Evidence. Choose arbitrarily (t0, x0) b and we will make the same auxiliary construction as before Peano theorem. We prove by induction that all xk is defined and continuous on the IP, and the graphs are lying in K1, and even more so in C. for x0 it is obvious. If this is true for XK1, then from (10) it is clear that XK is defined and continuous on the IP, and this is belonging to the K1.

Now we prove by induction an assessment on IP:

(C is convex by X CD in B, and L (C) is defined for it). At k \u003d 0, this is the proven estimate (T, X1 (T)) K1. If (12) is true for k: \u003d k 1, then from (10) we have what was required. Thus, a number of majorized on IP convergent numerics and therefore (this is called Weierstrass theorem) evenly on IP converges to some function X C (IP). But it means XK X on IP. Then in (10) on the IP go to the limit and get (9) on IP, and therefore (1) on Int (IP).

The uniqueness is immediately obtained by a consequence of 1 of the Osgood theorem, but it is useful to prove it and in another way using equation (9). Suppose that there are 2 solutions x1.2 tasks (1) (i.e. (9)) on int (IP). As mentioned above, then the graphs are necessarily lying in K1, and even more so in C. Let T i1 \u003d (T0, T0 +), where - some positive number. Then \u003d 1 / (2L (C)). Then \u003d 0. Thus, x1 \u003d x2 on i1.

Comment for the teacher. There is still proof of the uniqueness with the help of the Lemma of Hronole, it is even more natural, since it takes immediately globally, but while the lemma of Hronulla is not very convenient, since it is difficult to adequately perceive it.

Comment. The last proof of uniqueness is instructive in that once again shows in a different light, as a local uniqueness leads to a global (which is incorrect to existence).

The exercise. Prove the uniqueness immediately on all IP, arguing from the opponent as in the proof of the Osgood theorem.

An important private case (1) is linear ODU, i.e., in which the value f (t, x) is linear by x:

In this case, in order to enter the conditions of general theory, it should be required in this way, in this case, the string protrudes, and the condition of lindsery (and even differentiability) according to X is automatically performed: at all T (a, b), x, y rn we have | f (t, x) f (t, y) | \u003d | A (T) (X Y) | | A (T) | · | (X y) |.

If you temporarily allocate a compact (a, b), then it will be obtained | f (t, x) f (t, y) | L | (x y) |, where L \u003d max | A |.

From the Peano and Osgood Theorems or Cauchy-Picar, the unambiguous solvability of problem (13) on some interval (peano-picking) containing T0 should be unambiguous. Moreover, the solution on this interval is the limit of consistent approximations of the pickara.

The exercise. Find this interval.

But it turns out that in this case all these results can be proved immediately globally, i.e., on everything (A, B):

Theorem. Let it be true (14). Then the problem (13) has a single solution on (a, b), and the sequential approximations of the picking converge evenly on any compact (A, B).

Evidence. Again, as in TK-P, we build a solution to the integral equation (9) with the help of consecutive approximations by formula (10). But now we do not need to check the condition to enter the cone and cylinder, since.

f is defined at all x, while T (A, B). It is only necessary to verify that all XK are defined and continuous on (A, B), which is obvious to induction.

Instead of (12), now we will now show a similar estimate of the species where N is a number depending on the choice. The first induction step for this assessment of another (T. K. is not associated with K1): for k \u003d 0 | x1 (t) x0 | N due to continuity x1, and the following steps are similar (12).

You can not paint it, because it is obvious, but we can again notice xk x on, and X is the solution of the corresponding (10) on. But thereby we built a solution on everything (A, B), because the choice of compact is arbitrary. The uniqueness follows from the Osgood's theorem or the coschi-pickara (and reasoning above about global uniqueness).

Comment. As mentioned above, the TC-P is formally imprisonment due to the presence of Peano and Osgood theorems, but it is useful for 3 reasons - she:

1. Allows you to associate the Cauchy problem for ODU with an integral equation;

2. Invites the constructive method of consecutive approximations;

3. It makes it easy to prove the global existence for linear ODUs.

[Although the latter can be derived from the reasoning § 4.] Next, we will most often refer to it.

Example. x \u003d x, x (0) \u003d 1. Sequential approximation means x (t) \u003d E - the solution of the initial task throughout R.

Most often a number will be obtained, but a certain design remains. You can also evaluate the error of X XK (see).

Comment. Of Peano, Osguda, theorem and Cauchy Picar, it is easy to obtain the corresponding theorems for the highest order.

The exercise. Formulate the concepts of the Cauchy problem, solutions of the system and the Cauchy tasks, all theorems for the highest order, using the minimization of first-order systems set out in § 1.

Several violating the logic of the course, but with the goal of better assimilation and substantiation of methods for solving problems in practical classes, we temporarily interrupt the presentation of the general theory and we will deal with the technical problem of the "explicit decision of the ODU".

§ 3. Some integration techniques So, consider the scalar equation \u003d f (t, x). Prodt the steps of a private case that learned to integrate is so called. URP, i.e. equation in which f (t, x) \u003d a (t) b (x). The formal method of integrating the URP is to "divide" the variables T and X (hence the name): \u003d a (t) dt, and then take the integral:

chim X \u003d B (A (T)). Such a formal reasoning contains several points requiring justification.

1. Decision on B (x). We believe that F is continuous, so A c (,), b c (,), i.e., the rectangle protrudes () (,)(Generally speaking, endless). Sets (b (x) 0) and (b (x) 0) open and therefore are finite or countable sets of intervals. Between these intervals there are points or segments, where b \u003d 0. If B (x0) \u003d 0, then the Cauchy task has a solution x x0. Perhaps this solution is not the only one, then in its definition area there are intervals, where b (x (t)) \u003d 0, but then can be divided into b (x (t)). We simultaneously notice that at these intervals the function B monotonne and therefore can be taken b 1. If b (x0) \u003d 0, then in the neighborhood of T0 knowingly b (x (t) \u003d 0, and the procedure is legal. Thus, the described procedure should, generally speaking, apply when dividing the area for determining the solution on the part.

2. Integrating the left and right parts according to different variables.

Method I. Let we want to find the solution of the KDD (T) task (1) x \u003d (t). We have: \u003d a (t) b ((t)), from where they got the same formula strictly.

Method II. The equation is so called. Symmetric record of the original ODU, i.e., this in which it is not specified, which variable is independent, and which is dependent. This form makes sense just in the case of one of the first-order equation under consideration due to the theorem on the invariance of the form of the first differential.

It is appropriate to understand more detail with the concept of differential, illustrating it with an example of a plane (((T, X)), curves on it, emerging links, degrees of freedom, parameter on the curve.

Thus, equation (2) binds the differentials T and x along the desired IR. Then the integration of equation (2) in the method shown at the beginning is completely legal - it means, if you want, integrating on any variable selected as an independent.

In the method I, we showed this by choosing as an independent variable t. Now we will show this by choosing the parameter S as an independent variable along the IR (since this is more clearly shown by the equality T and X). Let the value S \u003d S0 correspond to the point (T0, X0).

Then we have: \u003d a (t (s)) T (s) DS, that after it gives it to make an emphasis on the versatility of a symmetric recording, an example: a circle is not recorded either x (t), nor as t (x), but as x (s), t (s).

Some other ODU of the first order are reduced to URPs, which can be seen when solving tasks (for example, in a task).

Another important case is a linear Code:

Method I. Variation to constant.

this is a special case of a more general approach, which will be considered in Part 2. The meaning is that the search for solutions in a special form reduces the order of the equation.

I must first be so called. Uniform equation:

By virtue of uniqueness either x 0, either everywhere x \u003d 0. In the latter case (let it give that (4), it gives all solutions (3) 0 (including zero and negative).

In formula (4) there is an arbitrary constant C1.

The variation method is constant consisting in the fact that the solution (3) C1 (T) \u003d C0 + is visible (as for algebraic linear systems) the structure of Orna \u003d Chrn + Orou (about it in more detail in part 2).

If we want to solve the Cauchy task X (T0) \u003d X0, then you need to find C0 from the Cauchy data - easily obtaining C0 \u003d X0.

Method II. We will find them, i.e., such a function V, to which you need to multiply (3) (recorded so that all unknown are assembled in the left side: xa (t) x \u003d b (t)) so that the derivative is derived from some Comfortable combination.

We have: VX Vax \u003d (VX) if V \u003d AV, i.e. (such equation, (3) is equivalent to an equation that is already easily solved and gives (5). If the Cauchy task is solved, then in (6) is convenient immediately Take a specific integral to linear ODU (3) Some others are reduced, as can be seen when solving problems (for example, in a task). More important case of linear ODUs (immediately for any n) will be considered in part 2.

Both of the considered situations are a special case so called. PED. Consider the first order ODU (with n \u003d 1) in a symmetrical form:

As mentioned, (7) sets the IR in the plane (T, X) without clarifying which variable is considered independent.

If multiply (7) on an arbitrary function M (T, X), the equivalent form of recording of the same equation will be obtained:

Thus, the same thing has many symmetrical records. Among them, a special role is played so called. Entries in full differentials, the name of the UPD is unsuccessful, since this property is not equation, but the forms of its record, i.e. such that the left part (7) is equal to DF (T, X) with some F.

It is clear that (7) there is a UPD then and only if a \u003d ft, b \u003d fx with some F. As is known from the analysis, it is necessary for the latter, and we do not justify strictly technical moments, for example, the smoothness of all functions. The fact is that § plays a minor role - it is generally not needed for other parts of the course, and I would not want to spend excessive efforts on its deployed presentation.

Thus, if (9) is performed, then there is such f (it is unique with an accuracy to the additive constant), which (7) rewrite in the form of DF (T, X) \u003d 0 (along IC), i.e.

F (t, x) \u003d const along IR, i.e. IR essence of the level line of the function F. We obtain that the integration of the UPD is a trivial task, since the search f by a and b satisfying (9) is not difficult. If (9) is not fulfilled, it should be found so called. It is M (T, X) such that (8) is the UPD, for which it is necessary and enough to perform analogue (9), which takes the form:

As follows from the theory of uch of first order (which we consider in part 3), the equation (10) always has a solution, so that they exist. Thus, any form equation (7) has an entry in the form of the UPD and therefore allows "explicit" integration. But these arguments do not give a constructive method in the general case, since to solve (10) Generally speaking, it is required to find a solution (7), which we are looking for. Nevertheless, there are a number of search techniques to them, which are traditionally considered in practical training (see for example).

Note that the above techniques for the decision of the URP and linear ODU are a special case of ideology.

In fact, URP DX / DT \u003d A (T) B (X), recorded in a symmetric form Dx \u003d A (T) B (x) DT, is solved by multiplying to it 1 / b (x), since after This turns into DX / B (X) \u003d A (T) DT, etc. DB (x) \u003d Da (T). Linear equation dx / dt \u003d a (t) x + b (t) recorded in a symmetric form DX A (T) XDT B (T) DT is solved by multiplying to them, almost all methods of solving ODU "explicitly"

(With the exception of a large block associated with linear systems), it is that with the help of special methods for lowering the order and replacement variables, they are reduced to the first order, which are then reduced to the UPD, and they are solved by the use of the main differential calculus theorem: df \u003d 0 F \u003d const. The question of a decrease in order is traditionally included in the course of practical training (see for example).

Let's say a few words about the first order of the first order, not permitted relative to the derivative:

As mentioned in § 1, you can try to resolve (11) relative to X and get a normal form, but it is not always appropriate. Often it is more convenient to solve (11) directly.

Consider the space ((t, x, p)), where p \u003d x is temporarily considered as an independent variable. Then (11) defines the surface in this space (f (t, x, p) \u003d 0), which can be written parametrically:

It is useful to remember what it means, for example, with the help of a sphere in R3.

The desired solutions will correspond to the curves on this surface: T \u003d S, x \u003d x (s), p \u003d x (s) - one degree of freedom is lost because there is a connection DX \u003d PDT on solutions. We write this link in terms of parameters on the surface (12): GU DU + GV DV \u003d H (FUDU + FV DV), i.e.

Thus, the desired solutions correspond to the curves on the surface (12), in which the parameters are connected by equation (13). The latter is ODE in a symmetric form that can be solved.

Case I. If in some area (GU HFU) \u003d 0, then (12) then T \u003d F ((V), V), X \u003d G ((V), V) gives a parametric recording of the desired curves in the plane ( (t, x)) (i.e. we project on this plane, since we do not need).

Case II. Similarly, if (GV HFV) \u003d 0.

Case III. At some points at the same time, GU HFU \u003d GV HFV \u003d 0. It requires a separate analysis, whether it is a lot of solutions (they are then called special).

Example. Equation Clero X \u003d TX + X 2. We have:

x \u003d TP + P2. Parametrize this surface: T \u003d U, P \u003d V, X \u003d UV + V 2. Equation (13) will take the form (U + 2V) DV \u003d 0.

Case I. Not implemented.

Case II. U + 2V \u003d 0, then DV \u003d 0, i.e. v \u003d c \u003d const.

So, T \u003d U, x \u003d Cu + C 2 - parametric IR recording.

Easily burn it in a clear form x \u003d Ct + C 2.

Case III. U + 2V \u003d 0, i.e. v \u003d u / 2. So, T \u003d U, X \u003d U2 / 4 - the parametric record of the "Candidate in IR".

To check whether it is an IR, we write it explicitly x \u003d T2 / 4. It turned out that this is (special) decision.

The exercise. Prove that a special solution concerns all the others.

This is a common fact - the schedule of any special solution is the envelope of the family of all other solutions. This is based on another definition of a special solution precisely as envelope (see).

The exercise. To prove that for a more general Clero equation x \u003d tx (x) with a convex function, a particular solution is viewed x \u003d (t), where - the conversion of the current from, i.e. \u003d () 1, or (t) \u003d max (TV (v)). Similarly, for the equation x \u003d tx + (x).

Comment. Read more and carefully content § 3 is set out in the textbook.

Comment for the teacher. When reading the course of lectures, it can be useful to expand § 3, giving it a more rigorous form.

Now back to the main canvas of the course, continuing the presentation started in §§§§.

§ 4. Global solvability of the Cauchy problem in § 2 We have proved the local existence of the solution of the Cauchy problem, i.e. only at a certain interval containing a point T0.

With some additional assumptions on F, we also proved the uniqueness of the solution, understanding it as a coincidence of two solutions defined in the same interval. If F is linear by x, a global existence is obtained, i.e., throughout the interval, where the coefficients of the equation (system) are determined and continuous. However, as an attempt to use to the linear system of general theory shows, the peano-picking interval is generally less than on which the solution can be constructed. Natural questions arise:

1. How to determine the maximum interval on which the solution is to be approved (1)?

2. Is this interval always coincide with the maximum on which the right part (1) 1 does it makes sense?

3. How to neatly formulate the concept of the uniqueness of the solution without reservations about the interval of its definition?

That the answer to question 2 generally speaking negative (or rather, requires great accuracy), says next example. x \u003d x2, x (0) \u003d x0. If x0 \u003d 0, then x 0 - there are no other solutions on the Osgood theorem. If X0 \u003d 0, then we decide to make a picture). The interval of the existence of the solution cannot be greater than (1 / x0) or (1 / x0, +), respectively, with x0 0 and x0 0 (the second branch of hyperboles is not related to the solution! - This is a typical student error). At first glance, nothing in the initial task "did not foresee such an outcome." In § 4 We will find an explanation for this phenomenon.

On the example of the equation X \u003d T2 + X2, a typical error of students on the interval of the solution is manifested. Here the fact that the "equation is everywhere is defined" is not at all entails the continuity of the decision on the entire direct. This is clear even from a purely everyday point of view, for example, in connection with legal laws and processes developing under them: Eneciling the law is clearly not prescribed the termination of the existence of a company in 2015, it does not mean that this company will not go rules for this year. For internal reasons (although in force within the law).

In order to answer questions 1-3 (and even to clearly formulate them), the concept of a short solution is necessary. We will (as we agreed above) consider solutions of equation (1) 1 as a pair (, (TL (), TR ())).

Definition. Solution (, (TL (), TR ())) there is a continuation of the solution (, (TL (), TR ())), if (TL (), TR ()) (TL (), TR ()), and | (TL (), TR ()) \u003d.

Definition. Solution (, (TL (), TR ())) - short, if it does not have non-trivial (i.e. other than it) continuation. (See Example above).

It is clear that it is NR that is particular value, and in their terms it is necessary to prove existence and uniqueness. There is a natural question - can I build HP, based on a local decision, or on the Cauchy task? It turns out yes. To understand this, we introduce the concepts:

Definition. A set of solutions ((, (TL (), TR ())))) is consistent if any 2 solutions from this set coincide at the intersection of the intervals of their definition.

Definition. A consistent set of solutions is called Maximum, if one cannot add another solution to it so that the new set is consistent and contained new points in combining solutions definition areas.

It is clear that the construction of the MNN is equivalent to building HP, namely:

1. If there is HP, then any MNN, which is containing, can only be a set of its narrowings.

The exercise. Check.

2. If there is a MNN, then HP (, (T, T +)) is built as:

put (t) \u003d (t), where - any ENN element defined at this point. Obviously, such a function will be uniquely defined on everything (T, T +) (unambiguity follows from the consistency of the set), and it coincides at each point with all the ENN elements defined at this point. For any T (T, T +) there is some kind of defined in it, and therefore in its surrounding, and so on. In this neighborhood there is a solution (1) 1, then - too. Thus, there is a solution (1) 1 on everything (T, T +). It is short, since, otherwise, nontrivial continuation could be added to the MNN contrary to its maximumness.

Constructing the MNH problem (1) in the general case (under the conditions of Peano theorem), when there are no local uniqueness, it is possible (see,), but quite cumbersome - it is based on step-by-step use of the Peano theorem with an estimate of the bottom of the interval-continuation. Thus, HP always exists. We justify this only in the case when there is a local uniqueness, then the construction of the MNN (and therefore HP) is trivial. For example, for definiteness, we will act within the framework of the TC-P.

Theorem. Let the TK-P conditions in the region B Rn + 1 are performed. Then for any (T0, x0) b, the problem (1) has a single HP.

Evidence. Consider the set of all solutions of problem (1) (it is not empty by TK-P). It forms the MNN - consistent due to local uniqueness, and the maximum due to the fact that this is a lot of all all solutions of the Cauchy problem. So HP exists. It is the only one for local uniqueness.

If you want to build HP based on the existing local decision (1) 1 (and not the task of the Cauchy), then this problem in the case of local uniqueness is reduced to the Cauchy task: you need to choose any point on the available IR and consider the corresponding Cauchy task. HP of this task will be a continuation of the initial solution due to uniqueness. If there is no uniqueness, the continuation of a given solution is carried out by the procedure specified above.

Comment. HP cannot be devoted at the ends of the interval of its existence (regardless of the condition of the uniqueness) so that it is a solution and in terminal points. To justify, it is necessary to clarify that understand under the decision of the ODU at the ends of the segment:

1. Approach 1. Let under decision (1) 1 on the segment it is understood as a function that satisfies the equation in the ends in the sense of one-sided derivative. Then the possibility of the specified daunting of some solution, for example, at the right end of the interval of its existence (T, T +] means that IR has an end point inside B, and C 1 (T, T +]. But then deciding the Cauchy X (T +) task. \u003d (T +) for (1) and finding its solution, we obtain, for the right end T + (at point T + both unilateral derivatives exist and are equal to F (T +, (T +)), which means there is a common derivative), i.e. not It was HP.

2. Approach 2. If under the solution (1) 1 on the segment it is understood as a function, only continuous at the ends, but such that the ends of the IR are lying in B (even if the equation is not required) - it turns out the same reasoning Only in terms of the corresponding integral equation (see detail).

Thus, immediately limited to open intervals as sets of definition of solutions, we did not violate the community (and only avoided the unnecessary vene with one-sided derivatives, etc.).

As a result, we answered the question 3, delivered at the beginning of § 4: when performing the condition of uniqueness (for example, Osgood or Cauchy-Picara), the uniqueness of HP solving the Cauchy problem takes place. If the condition of uniqueness is broken, then there may be a lot of HP task Cauchy, each with its interval of existence. Any solution (1) (or simply (1) 1) can be continued to HP.

To answer questions 1.2, it is necessary to consider not the variable T separately, but the behavior of the IC in the space Rn + 1. On the question of how IR "near the ends" behaves, we will note that the existence interval has ends, and IR may not have them (the end of IR in B always does not exist - see the remark above, but may not exist B - see below).

Theorem. (about leaving the compact).

we formulate it in conditions of local uniqueness, but this is not necessarily - see, there TPK is formulated as the criterion of HP.

In the conditions of the TC-p, a chart of any HP equation (1) 1 leaves any compact k b, that is, k b (t, t +): (t, (t)) k with t.

Example. K \u003d (((t, x) b | ((t, x), b)).

Comment. Thus, the IR NR near T ± is approaching B: ((t, (t)), b) 0 at t t ± - the process of continuing the solution cannot break through strictly inside B.

positively, here as an exercise it is useful to prove the positivity between the non-destroyed closed multiple sets, one of which is a compact.

Evidence. Fix K B. Take any 0 (0, (k, b)). If B \u003d Rn + 1, then by definition we consider (k, b) \u003d +. The set k1 \u003d (((t, x) | (((t, x), k) 0/2) There is also a compact in B, so there is a F \u003d max | f |. Choose the numbers T and R DOK are constantly small so that any cylinder of the form for example, it is enough to take T 2 + R2 2/4. Then the task of the Cauchy type has a solution on the TC-P solution on the interval not already than (T T0, T + T0), where T0 \u003d min (T, R / F) for all (T, X) K.

Now as a desired segment can be taken \u003d. In fact, it is necessary to show that if (t, (t)) k, then T + T0 T T + T0. We will show, for example, the second inequality. The solution of the Cauchy problem (2) with X \u003d (T) exists to the right at least to the point T + T0, but the HP of the same problem, which in view of the uniqueness there is a continuation, therefore T + T0 T +.

Thus, the HP chart always "reaches B", so that the interval of the existence of HP depends on the geometry of IR.

For example:

Statement. Let b \u003d (a, b) Rn (the interval finite or infinite), f satisfies the conditions of the TK-P in B, is HP problem (1) with T0 (A, B). Then either t + \u003d b, or | (t) | + at t t + (and similar to T).

Evidence. So, let T + B, then T + +.

Consider the compact k \u003d b B. With any R + according to the TPK, there is (R) T + such that at T (((R), T +) point (T, (T)) K. But since t +, then it is possible only for account | (t) | R. But it means | (t) | + at t t +.

In this particular case, we see that if F is defined "with all x", the interval of the NR existence may be less than the maximum possible (A, B) only due to the desire of HP by the approach to the ends of the interval (T, T +) (in general case - to the border B).

The exercise. To summarize the last statement in case when B \u003d (A, B), where Rn is an arbitrary area.

Comment. It is necessary to understand that | (t) | + does not mean any k (t).

Thus, we answered question 2 (cf. Example at the beginning § 4): IR comes to b, but its projection on the axis T may not reach the ends of the projection B on the T axis. The question remains 1 - Are there any signs for which, without solving ODU, can be judged about the possibility of continuing the decision on the "wide range of interval"? We know that for linear ODUs, this continuation is always possible, and in the example at the beginning of § 4 it is impossible.

Consider first to illustrate the special case of URP at n \u003d 1:

the convergence of the incompatible integral H (S) DS (incompatible in view \u003d + or due to the features H at the point) does not depend on the choice (,). Therefore, let's just write H (S) DS when it comes to convergence or divergence of this integral.

this could be done already in the Osguda theorem and in the associated allegations.

Statement. Let A C (,), B C (, +) be positive at their intervals. Let the Cauchy problem (where T0 (,), x0) has HP x \u003d x (t) on the interval (T, T +) (,). Then:

Corollary. If a \u003d 1, \u003d +, then T + \u003d + proof. (Approval). Note that x monotonically increases.

The exercise. Prove.

Therefore, there are X (T +) \u003d Lim X (T) +. We have a case 1. T +, X (T +) + is impossible on TPK, since X is HP.

Both integral either finite or endless.

The exercise. Finish proof.

Justification for the teacher. In the end, we obtain that in the case of 3: a (s) DS +, and in the case of 4 (if it is generally implemented) the same.

Thus, for the simplest ODUs at n \u003d 1 of the species x \u003d f (x), the continuity of solutions is determined in more detail about the structure of solutions such (so-called.

autonomous) Equations see Part 3.

Example. For f (x) \u003d x, 1 (in particular, the linear case \u003d 1), and f (x) \u003d x ln x, you can guarantee the continuity of (positive) solutions to +. For f (x) \u003d x and f (x) \u003d x ln x, with 1 solutions, "destroyed by the final time".

In general, the situation is determined by many factors and is not so simple, but the importance of "growth rate f by x remains the importance. For n 1, formulate the criteria for the continuation is difficult, but sufficient conditions exist. As a rule, they substantiate with the help of the so-called. a priori evaluations of solutions.

Definition. Let H c (,), H 0. It is said that for solutions of some ODD, there is an AO | X (T) | h (t) on (,), if any solution of this ODE satisfies this estimate on the part of the interval (,), where it is determined (i.e. it is not assumed that the solutions are necessarily defined throughout the interval (,)).

But it turns out that the availability of AO guarantees that solutions will still be defined on all (,) (and therefore satisfy the assessment throughout the interval), so the a priori assessment turns into a posteriori:

Theorem. Let the Cauchy problem (1) satisfy the conditions of TK-P, and for its solutions there is an AO on the interval () with some H C (,), and the curvilinear cylinder (| x | h (t), t (,)) b . Then Hp (1) is defined on all (,) (and therefore satisfies the JSC).

Evidence. We prove that T + (T is similar). Suppose T +. Consider the compact k \u003d (| x | h (t), t) b. According to TPK at T T +, the point of the graph (T, X (T)) leaves K, which is impossible due to the JSC.

Thus, to proof the continuity of the solution for some interval, it is enough to formally appreciate the solution on the entire required interval.

Analogy: The measurability of the leb function and the formal assessment of the integral entix the actual existence of the integral.

Let's give some examples of situations as this logic works. Let's start with the illustration of the above-mentioned thesis on "growth f by x enough slow."

Statement. Let B \u003d (,) Rn, F satisfies the conditions of the TK-P in B, | F (T, X) | A (T) B (| x |), where A and B satisfy the conditions of the previous approval C \u003d 0, and \u003d +. Then the HP problem (1) exists on (,) at all T0 (,), x0 Rn.

Lemma. If continuous, (t0) (t0); With T T proof. Note that in the surroundings (T0, T0 +): if (t0) (t0), then this is immediately obvious, and otherwise (if (t0) \u003d (t0) \u003d 0) have (t0) \u003d G (t0, 0) (T0), which gives the required again.

Suppose now that there is a T1 T0 such that (T1). Obvious reasoning can be found (T1) T2 (T0, T1] such that (T2) \u003d (T2), and on (T0, T2). But then at the point T2 we have \u003d, - contradiction.

g any, and really you need only, C, and everywhere where \u003d, there. But in order not to score head, consider as in Lemma. Here is strictly inequality, but it is non-linear ODU, and there is still so called.

Comment for the teacher. Inequalities of this kind as in Lemma is called Chaplygin inequalities (LF). It is easy to see that in the lemma there was no need to uniqueness, so that such a "strict LF" is true and within the framework of Peano theorem. "Non-Strong LF" is obviously wrong without uniqueness, since equality is a special case of non-strict inequality. Finally, the "non-stroke" as part of the condition of uniqueness is true, but it is possible to prove it only locally - with the help of it.

Evidence. (Approval). We prove that T + \u003d (T \u003d is similar). Suppose T +, then according to the approval above | x (t) | + at t t +, so that can be considered x \u003d 0 to. If we prove JSC | x | H for) (a ball for convenience is closed).

The Cauchy task X (0) \u003d 0 has the only HP x \u003d 0 on R.

We indicate a sufficient condition on F, in which the existence of HP on R + can be guaranteed at all sufficiently small x0 \u003d x (0). To do this, suppose that (4) has the so-called. Function Lyapunova, i.e. such a function V, which:

1. v C 1 (B (0, R));

2. SGNV (X) \u003d SGN | X |;

Check the fulfillment of the conditions A and B:

A. Consider the Cauchy task where | x1 | R / 2. We construct the cylinder B \u003d R b (0, R) is the field of determining the function F, where it is limited and class C 1, so that there is F \u003d max | f |. According to TC-P, there is a solution (5), determined on the interval (T1 T0, T1 + T0), where T0 \u003d min (T, R / (2F)). The choice of sufficiently large t can be achieved T0 \u003d R / (2F). It is important that T0 does not depend on the choice (T1, X1), just | x1 | R / 2.

B. While the solution (5) is determined and remains in a ball B (0, R), we can carry out the following reasoning. We have:

V (x (t)) \u003d f (x (t)) · v (x (t)) 0, i.e. v (x (t)) V (x1) m (r) \u003d max v (y) . It is clear that M and M do not decrease, incredia | R frivans in zero, m (0) \u003d m (0) \u003d 0, and outside zero they are positive. Therefore, there is R 0 such that M (R) M (R / 2). If | x1 | R, then V (x (t)) V (x1) M (R) M (R / 2), from where | x (t) | R / 2. Note that R r / 2.

Now we can formulate theorem that from PP. A, B displays the global existence of solutions (4):

Theorem. If (4) has a Lyapunov function in b (0, R), then at all x0 b (0, r) (where R is defined above) Hp Cauchy task X (T0) \u003d X0 for system (4) (with any T0) Defined to +.

Evidence. By virtue of P. A, the solution can be constructed on, where T1 \u003d T0 + T0 / 2. This solution lies in B (0, r) and to it use paragraph b, so that | x (t1) | R / 2. We again apply to n. A and we obtain a solution on, where T2 \u003d T1 + T0 / 2, i.e., now the solution is built on. To this solution, apply paragraph b and get | x (t2) | R / 2, etc. For the countable number of steps, we obtain a solution to § 5. The dependence of the ODU decisions from Consider the Cauchy task where Rk. If at some, t0 (), x0 () this Cauchy task has HP, then it is x (t,). The question arises: how to study dependence X from? This issue is important for various applications (and it will arise especially in part 3), one of which (although it is not the most important thing) is an approximate decision of the ODU.

Example. Consider the task of Cauchy. Its HP exists and is the only one, as follows from TC-P, but it is impossible to express it in elementary functions. How then to explore its properties? One way to: Note that (2) "close" to the problem y \u003d y, y (0) \u003d 1, the solution of which is easily located: y (t) \u003d et. It can be assumed that x (t) y (t) \u003d et. This idea is accounted for: Consider the problem at \u003d 1/100 this (2), and at \u003d 0 is a task for y. If we prove that x \u003d x (t,) is continuous by (in a certain sense), we obtain that x (t,) y (t) at 0, and this means x (t, 1/100) y ( T) \u003d ET.

True, it remains unclear how close X to Y, but the proof of continuity of the X software is the first necessary step without which it is impossible to promote further.

Similarly, it is useful and study depending on the parameters in the initial data. As we will see later, this dependence easily reduces to the parameter on the parameter in the right part of the equation, so that we still restrict ourselves to the task of the form be F C (D), where D is the region in Rn + K + 1; F Lipszytseva by x in any convex on x compact from D (for example, sufficient C (D)). Fix (t0, x0). Denote M \u003d Rk | (T0, X0,) D is a plurality of permissible (in which task (4) makes sense). Note that M is open. We will assume that (t0, x0) is chosen so that M \u003d. According to TK-P, for all m, there is a single NR problem (4) - the function x \u003d (t,), determined on the interval T (T (), T + ()).

Strictly speaking, because it depends on many variables, you need to record (4) so:

where (5) 1 is made on the set G \u003d ((t,) | m, t (t (), t + ())). However, the difference between the D / DT and / T icons is purely psychological (their use depends on the same psychological concept of "fixing"). Thus, the set G is a natural maximum set definition of the function, and the question of continuity should be investigated on G.

We will need an auxiliary result:

Lemma. (Holonolla). Let the function C, 0, satisfies the estimate for all T. Then, at all, the remark is true for the teacher. When reading a lecture, you can not remember this formula in advance, but leave a place, and after the output to enter.

But then keep this formula in sight, since it will be necessary in ton.

h \u003d A + B AH + B, from where you get the desired.

Meaning of this lemma: Differential equation and inequality, the relationship between them, integral equation and inequality, the relationship between them all, the differential and integral lemma of the gronaolel and the relationship between them.

Comment. You can prove this lemma and with more general assumptions about, a and b, but it is not necessary for us yet, and it will be done in the UMF course (so it is easy to see that we did not use continuity a and b, etc.).

Now we are ready to clearly formulate the result:

Theorem. (TONS) with the assumptions of f and in the notation of the introduced above it can be argued that G is open, and C (G).

Comment. It is clear that the set M is generally not connected, so that G may not be connected.

Comment for the teacher. However, if we were included (T0, X0) in the number of parameters, the connectivity would be so done in.

Evidence. Let (t,) G. need to prove that:

Suppose for definiteness t t0. We have: M, so that (t,) is defined by (t (), T + ()) T, T0, and therefore on some segment such that T dot (T, (T,),) Running compact curve D (parallel hyperplane (\u003d 0)). So, a lot of type definition needs to keep before your eyes constantly!

also there is a compact in D with sufficiently small A and B (convex by x), so that the function f Lipshitsev is on x:

[This assessment must be kept before the eyes constantly! ] and evenly continuous on all variables, and even more so | f (t, x, 1) f (t, x, 2) | (| 12 |), (t, x, 1), (t, x, 2).

[This assessment must be kept before the eyes constantly! ] Consider arbitrary 1 such that | 1 | Bi the corresponding solution (T, 1). The set (\u003d 1) is a compact in D (\u003d 1), and at T \u003d T0, the point (T, (T, 1), 1) \u003d (T0, X0, 1) \u003d (T0, (T0,), 1) (\u003d 1), and according to TPK at T T + (1), the point (T, (T, 1), 1) leaves (\u003d 1). Let T2 T0 be (T2 T + (1)) be the very first value at which the said point comes on.

By construction, T2 (T0, T1]. Our task will show that T2 \u003d T1 with additional restrictions on. Let now T3. Have (with all such T3, all values \u200b\u200bare used further defined by construction):

(T3, 1) (T3,) \u003d F (T, (T, 1), 1) F (T, (T,),) DT, let's try to prove that this value is less than a.

where is the integrated function assessed as follows:

± f (t, (t,),), and not ± f (t, (t,),), because on the difference | (t, 1) (t,) | There is no rating for now, so (t, (t, 1),) is unclear, but for | 1 | There are, and (t, (t,), 1) is known.

so in the end | (T3, 1) (T3,) | K | (t, 1) (t,) | + (| 1 |) dt.

Thus, the function (T3) \u003d | (T3, 1) (T3,) | (This is a continuous function) satisfies the conditions of the gronaole lemma with a (s) k 0, b (s) (| 1 |), T \u003d T2, \u003d 0, so we get along this lemma [this assessment you need to keep before your eyes! ] If you take | 1 | 1 (T1). We assume that 1 (T1) b. All our arguments are true for all T3.

Thus, with such a choice 1, when T3 \u003d T2, though | (T2, 1) (T2,) | A, as well as | 1 | b. So, (T2, (T2, 1), 1) is possible only due to the fact that T2 \u003d T1. But this in particular means that (T, 1) is defined on the whole segment, i.e. T1 T + (1), and all points of the form (T, 1) G, if T, | 1 | 1 (T1).

That is, although T + depends on, but the segment remains the left T + () with, closely close to the. The figure is similar to T t0, the existence of numbers T4 T0 and 2 (T4) is shown. If T T0, then the point (t,) b (, 1) g is similar to the T T0, and if t \u003d t0, then both cases are applicable, so (t0,) b (, 3) g, where 3 \u003d min (12). It is important that with a fixed (t,) one can find T1 (T,) so that T1 T 0 (or respectively T4), and 1 (T1) \u003d 1 (T,) 0 (or respectively 2), so choosing 0 \u003d 0 (t,) is clear (t. To. In the resulting cylindrical neighborhood you can enter the ball).

in fact, a more subtle property was proved: if Hp is determined on a certain segment, then it defines all HP with sufficiently close parameters (i.e.

all little indignant HP). However, on the contrary, this property follows from the openness G, as will be shown below, so these are equivalent wording.

Thus, we proved to be 1.

If we are in the specified cylinder in space, then the rating is true | 1 | 4 (, t,). At the same time | (T3,) (T,) | at | T3 T | 5 (, t,) due to continuity by t. As a result, at (t3, 1) b ((t,),) we have | (t3, 1) (t,) |, where \u003d min (4, 5). This is paragraph 2.

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