Theory of Ordinary Differential Equations Existence, Uniqueness and Stability Jishan Hu and Wei-Ping Li Department of Mathematics The Hong Kong University of Science and Technology ii Copyright c 2004 by Jishan Hu and Wei-Ping Li Contents 1 Existence and Uniqueness 1 1.1 Some Basics 1 1.2 Uniqueness Theorem 6 1.3 Continuity 8 1.4 Existence Theorem 13 1.5 Local Existence Theorem and The Peano Theorem 19 1.5.1 Local Existence Theorem 19 1.5.2 The Peano Theorem 21 1.6 Linear Systems 23 1.7 Continuation of Solutions 27 1.8 Miscellaneous Problems 29 2 Plane Autonomous Systems 33 2.1 Plane Autonomous Systems 33 2.2 Linear Autonomous Systems 39 2.3 Complete Classification for Linear Autonomous Systems 44 2.4 Liapunov Direct Method 52 2.5 A Test For Instability 61 2.6 Nonlinear Oscillations 65 2.6.1 Undamped Oscillations 65 iii iv CONTENTS 2.6.2 Damped Oscillations 66 2.7 Miscellaneous Problems 67 1 Existence and Uniqueness 1.1 SOME BASICS A normal system of first order ordinary differential equations (ODEs) is 8 dx1 > = X1(x1; : : : ; xn; t); > dt <> . (1.1) . > > dx :> n = X (x ; : : : ; x ; t): dt n 1 n Many varieties of ODEs can be reduced to this form. For example, consider an n-th order ODE dnx dx dn−1x = F t; x; ;:::; : dtn dt dtn−1 dx dn−1x Let x = x; x = ; : : : ; x = . Then the ODE can be changed to 1 2 dt n dtn−1 8 dx1 > = x2; > dt > > . <> . dx > n−1 = x ; > dt n > > dx :> n = F (t; x ; : : : ; x ): dt 1 n 1 2 1 EXISTENCE AND UNIQUENESS Let us review some notations and facts on vectors and vector-valued functions. The normal sys- tem (1.1) can be written as its vector form: dx = X(x; t): dt p 2 2 For a vector x = (x1; : : : ; xn), define kxk = x1 + ··· + xn. The inner product is defined to be x · y = x1y1 + ··· + xnyn where y = (y1; : : : ; yn). We have the triangle inequality kx + yk ≤ kxk + kyk; and the Schwarz inequality jx · yj ≤ kxk · kyk: For a vector valued function x(t) = (x1(t); : : : ; xn(t)), we have 0 0 0 x (t) = (x1(t); : : : ; xn(t)) ; and Z b Z b Z b ! x(t)dt = x1(t)dt; : : : ; xn(t)dt : a a a We have the following useful inequality Z b Z b x(t)dt ≤ kx(t)k dt: a a A vector field X(x) = X1(x1; : : : ; xn);:::;Xn(x1; : : : ; xn) is aid to be continuous if each Xi is a continuous function of x1; : : : ; xn. Since only a few simple types of differential equations can be solved explicitly in terms of known elementary function, in this chapter, we are going to explore the conditions on the function X such that the differential system has a solution. We also study whether the solution is unique, subject some additional initial conditions. Example 1.1.1 Show that the differential equation dy 1 = − y−1 dt 2 does not have solution satisfying y(0) = 0 for t > 0. Solution Acturally, the general solution of this differential equation is y2 = −t + C; where C is an arbitrary constant. The initial condition implies C = 0. Thus, we have y2 = −t, that shows there exists no solution for t > 0. 2 1.1 SOME BASICS 3 Example 1.1.2 Show that the differential equation x0 = x2=3 has infinitely many solutions satisfying x(0) = 0 on every interval [0; b]. Solution Define 8 0; if 0 ≤ t < c; < xc(t) = (t − c)3 : ; if c ≤ t ≤ b. 27 It is easy to check for any c, the function xc satisfies the differential equation and xc(0) = 0. 2 Definition 1.1.1 A vector-valued function X(x; t) is said to satisfy a Lipschitz condition in a region R in (x; t)-space if, for some constant L (called the Lipschitz constant), we have kX(x; t) − X(y; t)k ≤ Lkx − yk; (1.2) whenever (x; t) 2 R and (y; t) 2 R.1 @X Lemma 1.1.2 If X(x; t) has continuous partial derivatives i on a bounded closed convex domain @xj R, then it satisfies a Lipschitz condition in R.2 Proof Denote M = sup j@Xi=@xj j : x2R 1≤i;j≤n Since X has continuous partial derivatives on the bounded closed region R, we know that M is finite. For x; y 2 R and 0 ≤ s ≤ 1, (1 − s)x + sy is a line segment connecting x and y. Thus it is in the domain R since R is convex. For each component Xi, we have, for fixed x; y; t, regarding Xi((1 − s)x + sy) as a function of the variable s, n d X @Xi Xi((1 − s)x + sy; t) = ((1 − s)x + sy; t)(yk − xk): ds @xk k=1 Using the mean value theorem, we have n X @Xi Xi(y; t) − Xi(x; t) = ((1 − σi)x + σiy; t)(yk − xk); @xk k=1 1It is worth to point out the following: (i) The left hand side of the inequality (1.2) is evaluated at (x; t) and (y; t), with the same t for these two points. (ii) The constant L is independent of x, y and t. However, it depends on R. In other words, for a given function X(x; t), its Lipschitz constant may change if the domain R is different. In fact, for the same function X(x; t), it can be a Lipschitz function in some regions, but not a Lipschitz function in some other regions. 2Here are some explanation of some concepts and terminologies in the Lemma. (i) A bounded closed domain is also called compact. It has the following property that any continuous function on a compact set has an absolute maximum and absolute minimum. @Xi (ii) That X(x; t) = X1(x1; : : : ; xn; t);:::;Xn(x1; : : : ; xn; t) has continuous partial derivatives means that @xj @X and i are continuous for all i; j. Sometimes we use X(x; t) 2 C1(D) for this. @t (iii) A convex domain D means that for any two points x and y in D, (1 − t)x + ty 2 D for 0 ≤ t ≤ 1. 4 1 EXISTENCE AND UNIQUENESS for some σi between 0 and 1. The Schwarz inequality gives n n 2!1=2 n !1=2 X @Xi X @Xi X 2 ((1 − σi)x + σiy; t)(yk − xk) ≤ ((1 − σi)x + σiy; t) jyk − xkj @xk @xk k=1 k=1 k=1 n !1=2 X 2 p ≤ M · ky − xk = nM · ky − xk: k=1 Thus, n !1=2 X 2 kX(y; t) − X(x; t)k = jXi(y; t) − Xi(x; t)j i=1 n !1=2 X p 2 ≤ nM · ky − xk = nM · ky − xk: i=1 The Lipschitz condition follows, with the Lipschitz constant nM. 2 Example 1.1.3 Determine whether the function x2 + 1 X(x; t) = · t x satisfies a Lipschitz condition in the domains: (1) R1 = [1; 2] × [0; 1]; (2) R2 = (1; 2) × [0; 1]; (3) R3 = [1; 2] × [0; +1); (4) R4 = [1; +1) × [0;T ]; (5) R5 = (0; 1) × [0; 1]. Solution (1) Since the function X(x; t) is continuously differentiable in the bounded closed convex domain R = [1; 2] × [0; 1], by Lemma 1.1.2, we know that the function satisfies a Lipschitz conditio in this region. (2) Since the function X(x; t) satisfis a Lipschitz condition in R1, and R2 ⊂ R1, we know that the function satisfis the same Lipschitz inequality in R2. Hence the function X(x; t) satisfis a Lipschitz condition in R2. (3) Since R3 = [1; 2] × [0; +1) is not a bounded region, we cannot apply Lemma 1.1.2 in this case. Since jX(x; t) − X(y; t)j xy − 1 = · jtj > jtj ! 1; jx − yj xy as t ! +1, there exists no contant L, independent of x; y and t, such that jX(x; t) − X(y; t)j ≤ L jx − yj : Hence, the function X is not a Lipschitz function in R3. (4) Again, R4 = [1; +1) × [0;T ] is not bounded and Lemma 1.1.2 does not apply in this case. For (x; t); (y; t) 2 R4, xy − 1 jX(x; t) − X(y; t)j = · jtj · jx − yj xy 1 ≤ 1 − · T · jx − yj xy ≤ T jx − yj: Thus, the function X is a Lipschitz function in R4, with L = T . 1.1 SOME BASICS 5 (5) Since R5 = (0; 1) × [0; 1] is not a closed region, Lemma 1.1.2 does not apply in this case. In fact, the function X is continuously differentiable in R, with @X(x; t) x2 − 1 = t: @x x2 @X(x; t) But the derivative is not bounded when t 6= 0, since ! 1 as x ! 0+ when t 6= 0. @x Actually, we can show that the function X does not satisfy any Lipschitz condition in R5. To see that, 1 2 we take x = , y = and t = 1, and have n n n2 jX(x; t) − X(y; t)j = − 1; 2 n2 for n ≥ 2. Obviously, there exists no constant L such that − 1 ≤ L for all n. Therefore, the function X 2 does not satisfy any Lipschitz conditions in the domain R5. 2 d3x Example 1.1.4 Reduce the ODE + x2 = 1 to an equivalent first-order system and determine dt3 in which domain or domains the resulting system satisfies a Lipschitz condition.