Theory of Ordinary Differential Equations

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 24 by Jishan Hu and Wei-Ping Li

Contents 1 Existence and Uniqueness 1 11 Some Basics 1 12 Uniqueness Theorem 6 13 Continuity 8 14 Existence Theorem 13 15 Local Existence Theorem and The Peano Theorem 19 151 Local Existence Theorem 19 152 The Peano Theorem 21 16 Linear Systems 23 17 Continuation of Solutions 27 18 Miscellaneous Problems 29 2 Plane Autonomous Systems 33 21 Plane Autonomous Systems 33 22 Linear Autonomous Systems 39 23 Complete Classification for Linear Autonomous Systems 44 24 Liapunov Direct Method 52 25 A Test For Instability 61 26 Nonlinear Oscillations 65 261 Undamped Oscillations 65 iii

iv CONTENTS 262 Damped Oscillations 66 27 Miscellaneous Problems 67

1 Existence and Uniqueness 11 SOME BASICS A normal system of first order ordinary differential equations (ODEs is 1 n = X 1 (x 1,, x n ; t, = X n (x 1,, x n ; t (11 Many varieties of ODEs can be reduced to this form For example, consider an n-th order ODE d n x n = F ( t, x,,, dn 1 x n 1 Let x 1 = x, x 2 =,, x n = dn 1 x Then the ODE can be changed to n 1 1 n 1 n = x 2, = x n, = F (t, x 1,, x n 1

2 1 EXISTENCE AND UNIQUENESS Let us review some notations and facts on vectors and vector-valued functions The normal system (11 can be written as its vector form: = X(x, t For a vector x = (x 1,, x n, define x = x 2 1 + + x2 n The inner product is defined to be x y = x 1 y 1 + + x n y n where y = (y 1,, y n We have the triangle inequality and the Schwarz inequality x + y x + y, x y x y For a vector valued function x(t = (x 1 (t,, x n (t, we have x (t = (x 1(t,, x n(t, and b ( b x(t = x 1 (t,, b a a a x n (t We have the following useful inequality b x(t a b a x(t A vector field X(x = ( X 1 (x 1,, x n,, X n (x 1,, x n is aid to be continuous if each X i is a continuous function of x 1,, x n 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 stu whether the solution is unique, subject some additional initial conditions Example 111 Show that the differential equation = 1 2 y 1 does not have solution satisfying y( = for t > Solution Acturally, the general solution of this differential equation is y 2 = t + C, where C is an arbitrary constant The initial condition implies C = Thus, we have y 2 = t, that shows there exists no solution for t >

11 SOME BASICS 3 Example 112 Show that the differential equation x = x 2/3 has infinitely many solutions satisfying x( = on every interval [, b] Solution Define, if t < c; x c(t = (t c 3, if c t b 27 It is easy to check for any c, the function x c satisfies the differential equation and x c( = Definition 111 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 whenever (x, t R and (y, t R 1 X(x, t X(y, t L x y, (12 Lemma 112 If X(x, t has continuous partial derivatives X i x j R, then it satisfies a Lipschitz condition in R 2 on a bounded closed convex domain Proof Denote M = sup X i/ x j x R 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 R and s 1, (1 sx + sy is a line segment connecting x and y Thus it is in the domain R since R is convex For each component X i, we have, for fixed x, y, t, regarding X i((1 sx + sy as a function of the variable s, n d Xi((1 sx + sy, t = ds Using the mean value theorem, we have X i(y, t X i(x, t = n k=1 k=1 X i x k ((1 sx + sy, t(y k x k X i x k ((1 σ ix + σ iy, t(y k x k, 1 It is worth to point out the following: (i The left hand side of the inequality (12 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 2 Here 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 ( (ii That X(x, t = X 1 (x 1,, x n, t,, X n(x 1,, x n, t has continuous partial derivatives means that X i x j and X i are continuous for all i, j Sometimes we use X(x, t C 1 (D for this t (iii A convex domain D means that for any two points x and y in D, (1 tx + ty D for t 1

4 1 EXISTENCE AND UNIQUENESS for some σ i between and 1 The Schwarz inequality gives n X i ((1 σ ix + σ iy, t(y k x k x k k=1 ( n Xi 2 1/2 ( n 1/2 ((1 σ ix + σ iy, t x k y k x k 2 k=1 k=1 ( n 1/2 M 2 y x = nm y x k=1 Thus, ( n 1/2 X(y, t X(x, t = X i(y, t X i(x, t 2 i=1 ( n 1/2 ( 2 nm y x = nm y x The Lipschitz condition follows, with the Lipschitz constant nm i=1 Example 113 Determine whether the function X(x, t = x2 + 1 x satisfies a Lipschitz condition in the domains: (1 R 1 = [1, 2] [, 1]; (2 R 2 = (1, 2 [, 1]; (3 R 3 = [1, 2] [, + ; (4 R 4 = [1, + [, T ]; (5 R 5 = (, 1 [, 1] Solution (1 Since the function X(x, t is continuously differentiable in the bounded closed convex domain R = [1, 2] [, 1], by Lemma 112, we know that the function satisfies a Lipschitz conditio in this region (2 Since the function X(x, t satisfis a Lipschitz condition in R 1, and R 2 R 1, we know that the function satisfis the same Lipschitz inequality in R 2 Hence the function X(x, t satisfis a Lipschitz condition in R 2 (3 Since R 3 = [1, 2] [, + is not a bounded region, we cannot apply Lemma 112 in this case Since X(x, t X(y, t = x y xy 1 xy t > t, as t +, there exists no contant L, independent of x, y and t, such that t X(x, t X(y, t L x y Hence, the function X is not a Lipschitz function in R 3 (4 Again, R 4 = [1, + [, T ] is not bounded and Lemma 112 does not apply in this case For (x, t, (y, t R 4, X(x, t X(y, t = xy 1 xy t x y ( 1 1 T x y xy T x y Thus, the function X is a Lipschitz function in R 4, with L = T

11 SOME BASICS 5 (5 Since R 5 = (, 1 [, 1] is not a closed region, Lemma 112 does not apply in this case In fact, the function X is continuously differentiable in R, with X(x, t x = x2 1 x 2 t X(x, t But the derivative is not bounded when t, since as x + when t x Actually, we can show that the function X does not satisfy any Lipschitz condition in R 5 To see that, we take x = 1 n, y = 2 and t = 1, and have n X(x, t X(y, t = n2 2 1, for n 2 Obviously, there exists no constant L such that n2 1 L for all n Therefore, the function X 2 does not satisfy any Lipschitz conditions in the domain R 5 Example 114 Reduce the ODE d3 x 3 + x2 = 1 to an equivalent first-order system and determine in which domain or domains the resulting system satisfies a Lipschitz condition Solution Let x 1 = x, x 2 = x, x 3 = x, we get the equivalent first order system 1 = x 2, 2 = x 3, 3 = 1 x 2 1 Denote x = (x 1, x 2, x 3, y = (y 1, y 2, y 3 and X(x, t = (x 2, x 3, 1 x 2 1 Then, X(x, t X(y, t = (x 2 y 2, x 3 y 3, x 2 1 + y 2 1, which implies X(x, t X(y, t = (x 2 y 2 2 + (x 3 y 3 2 + (x 1 y 1 2 (x 1 + y 1 2 We can see that whenever x 1 and y 1 are bounded, to be precise, x 1 + y 1 is bounded, the Lipschitz condition holds In fact, consider the domain { } R = (x 1, x 2, x 3, t x 1 M for some constant M Take L = max{4m 2, 1}, then whenever (x, t, (y, t R, we have X(x, t X(y, t (x 2 y 2 2 + (x 3 y 3 2 + 4M 2 (x 1 y 1 2 L 2 (x 2 y 2 2 + L 2 (x 3 y 3 2 + L 2 (x 1 y 1 2 = L x y Therefore X(x, t satisfies the Lipschitz condition with the Lipschitz constant L = max{4m 2, 1} in the domain R Exercise 11 1 Show that the function X(x, t = x satisfies a Lipschitz condition everywhere, but X x does not exist at x =

6 1 EXISTENCE AND UNIQUENESS 2 For f(x, t = x 1 3 t, does it satisfy a Lipschitz condition over the interval [1, 2] [, T ]? Does it satisfy a Lipschitz condition over the interval [ 1, 1] [, T ]? Here T is a constant 3 Does the function f(x, t = x 3 satisfy a Lipschitz condition over [, + [, +? 4 Does the function f(x, t = xt x 2 + t 2 satisfy a Lipschitz condition over [, + [, T ]? Here T is a constant + 1 12 UNIQUENESS THEOREM In this section, we prove the following uniqueness theorem Theorem 121 (Uniqueness Theorem If the vector field X(x, t satisfies a Lipschitz condition in a domain R, then there is at most one solution x(t of the differential system that satisfies a given initial condition x(a = c in R Proof Assume that X(x, t satisfies the Lipschitz condition = X(x, t, (13 X(x, t X(y, t L x y, for any (x, t, (y, t R Let x(t = (x 1(t,, x n(t and y(t = (y 1(t,, y n(t be two solutions satisfying the same initial condition x(a = c = y(a Denote σ(t = x(t y(t 2 = σ (t = = By the Schwarz inequality, we have The last inequality implies n k=1 [x k (t y k (t] 2 Then, n 2 [ x k(t y k(t ] [x k (t y k (t] k=1 n 2 [X k (x(t, t X k (y(t, t] [x k (t y k (t] k=1 = 2 [X(x(t, t X(y(t, t] [x(t y(t] σ (t σ (t = 2 (X(x, t X(y, t (x y 2 X(x, t X(y, t x y 2L x y 2 = 2Lσ(t d ( σ(te 2Lt = ( σ (t 2Lσ(t e 2Lt Hence σ(te 2Lt is a decreasing function Therefore for t > a, σ(te 2Lt σ(ae 2La = Since σ(t, we have σ(t = for t a, ie, x(t = y(t for t a To argue for t < a, as above we again have σ (t σ (t 2Lσ(t,

12 UNIQUENESS THEOREM 7 which implies d ( σ(te +2Lt = ( σ (t + 2Lσ(t e +2Lt So σ(te +2Lt is an increasing function Therefore for t < a, σ(te +2Lt σ(ae +2La = Again we have σ(t = for t a, ie, x(t = y(t for t a Example 121 Find a region where the differential equation x = x + 3x 1/3 has a unique solution, ie, find (x, t such that the solution x(t of the differential equation with x(t = x is unique in a neighborhood of (x, t Solution We will show the following (1 the differential equation has a unique solution with x(t = x, x ; (2 there are more than one solution satisfying x(t = (1 For any given (x, t, with x, we choose a small δ > such that [x δ, x + δ] By Lemma 112, we know that the function X(x, t = x + 3x 1/3 satisfies a Lipschitz condition in the region { } R = (x, t x x δ, t t T = [x δ, x + δ] [t T, t + T ], where T > is any fixed constant 3 By Theorem 121, we conclude that the differential equation has a unique solution with x(t = x, x (2 It is easy to see that x(t is one solution of the differential equation with x(t = We only need to show that there exists another solution which also satisfies x(t = Consider the improper integral x du u + 3u 1/3 For any c >, we know that c du c < u + 3u < du 1/3 3u = 1 1/3 2 c2/3 Hence the improper integral converges for c > This allows us to define an implicit function x(t by x du = t t u + 3u1/3 We should have x(t =, since the last equation becomes an identity when setting t = t Obviously, this function x(t, otherwise we will have t t, a contradiction This function x(t certainly satisfies the differential equation, which can be seen easily by differentiating both sides of the last equation 3 We can make the following argument Let (x, t, (y, t R = [x δ, x + δ] [t T, t + T ] Then, by the mean value theorem, we have X(x, t X(y, t = X (ξ, t 1 x x y = ξ 2/3 x y, where ξ is between x and y This implies ξ [x δ, x + δ] The function X x (x, t = 1 + x 2/3 is continuous in x in the compact set [x δ, x + δ], since [x δ, x + δ] This indicates that there is a number L (might depend on x and δ such that 1 ξ 2/3 L Thus, the function X(x, t satisfies a Lipschitz condition in [x δ, x + δ] R

8 1 EXISTENCE AND UNIQUENESS Exercise 12 1 Show that for < α < 1, the initial value problem has at least two solutions = xα, x( =, 2 Show that the initial value problem has a unique solution = sin t sin x, x( = c, 3 Show that the initial value problem: has at most one solution = 1 + x2 1 + t 2, x( = 1, 4 Show that the initial value problem has at most one solution 5 Show that the initial value problem has a unique solution = t sin x, x( = 1, du = tu + sin u, u( = 1, 6 State and prove a uniqueness theorem for the differential equation y = F (t, y, y 13 CONTINUITY Theorem 131 (Continuity Theorem Let x(t and y(t be any two solutions of the differential equation (13 in T 1 t T 2, where X(x, t is continuous and satisfies the Lipschitz condition (12 in some region R that contains the region where x(t and y(t are defined Then for any a, t [T 1, T 2 ] x(t y(t e L t a x(a y(a, Proof Let us first assume that t a Then, for the function σ(t = x(t y(t 2, as in the proof of Uniqueness Theorem 121, we have σ (t 2Lσ(t, which implies d ( σ(te 2Lt

13 CONTINUITY 9 Integrating the last inequality from a to t gives σ(te 2Lt σ(ae 2La, which yields the desired inequality In the case t a, again as in the proof of Uniqueness Theorem 121, we obtain σ (t 2Lσ(t, which implies d ( σ(te +2Lt Now we integrate the last inequality from t to a to have σ(te 2Lt σ(ae 2La This implies the desired inequality in the case t a Corollary 132 Let x(t be the solution of the differential equation (13 satisfying the initial condition x(a, c = c Let the hypotheses of Continuity Theorem 133 be satisfied, and let the function x(t, c be defined for c c K and t a T Then (1 x(t, c is a continuous function of both variables; (2 if c c, then x(t, c x(t, c uniformly for t a T Proof For (a, it is obvious that x(t, c is continuous in t since x(t, c is a solution of the differential equation (13, which is differentiable To see that x(t, c is continuous in c, we take c 1 and c 2, with c 1 c K and c 2 c K Since x(t, c 1 and x(t, c 2 are solutions of (13, by Continuity Theorem 133, we get x(t, c 1 x(t, c 2 e L t a x(a, c 1 x(a, c 2 = e L t a c 1 c 2 ɛ Let t be fixed Hence, for any given ɛ >, take δ = e, whenever c 1 c 2 < δ, we have L t a x(t, c 1 x(t, c 2 < e L t a δ = ɛ Therefore, x(t, c is continuous in c For (b, in above, we can take δ = { } 1 Since min t [a T,a+T ] e L t a continuous in c uniformly for t a T { } 1 min ɛ t [a T,a+T ] e L t a > is a finite number, hence, δ > is independent of t Therefore, x(t, c is Theorem 133 (Strong Continuity Theorem Assume that (1 x(t and y(t satisfy the differential equations = X(x, t and = Y(y, t,

1 1 EXISTENCE AND UNIQUENESS respectively, on T 1 t T 2 (2 The functions X and Y be defined and continuous in a common domain R, and satisfy X(z, t Y(z, t ɛ, for any (z, t R, with T 1 t T 2 (3 X(x, t satisfies the Lipschitz condition (12 in R 4 Then x(t y(t x(a y(a e L t a + ɛ L ( e L t a 1, for any a, t [T 1, T 2 ] Proof Assume that a, t [T 1, T 2] Let us first consider the case t a Define σ(t = x(t y(t 2 = n [x k (t y k (t] 2 We have k=1 σ (t = 2 [ x (t y (t ] [x(t y(t] = 2 [X(x(t, t Y(y(t, t] [x(t y(t] = 2 [X(x(t, t X(y(t, t] [x(t y(t] + 2 [X(y(t, t Y(y(t, t] [x(t y(t] The Schwarz inequality implies σ (t σ (t 2 X(x(t, t X(y(t, t x(t y(t + 2 X(y(t, t Y(y(t, t x(t y(t 2L x(t y(t 2 + 2ɛ x(t y(t = 2Lσ(t + 2ɛ σ(t (1 Let us assume that σ(a > Consider the initial value problem du = 2Lu + 2ɛ u, t a, u(a = σ(a Formally, we can introduce the substitution v(t = u(t This gives the equivalent differential equation 2v dv = 2Lv2 + 2ɛv v(t = for all t is not a solution since v(a = u(a Thus we can divide both sides of the equation by v(t to obtain a linear differential equation for v This leads to the following initial value problem dv Lv = ɛ, t a, v(a = u(a 4 The function Y(y, t may not satisfy a Lipschitz condition

13 CONTINUITY 11 The solution of the initial value problem is u(t = v(t = u(ae L(t a + ɛ ( e L(t a 1 L Claim: σ(t u(t Suppose not, there exists t 1 a such that σ(t 1 > u(t 1 Let t = sup{t a t < t 1, σ(t u(t} Take a sequence t n t, t n t such that σ(t n u(t n By the continuity of σ(t and u(t, we must have σ(t = lim n n t n t with t n > t, we must have σ(t n < u(t n Taking limits, we have σ(t u(t Therefore we σ(t = u(t Since u(t >, we have, for t t t 1, [σ(t u(t] 2L[σ(t u(t] + 2ɛ[ σ(t u(t] = 2L[σ(t u(t] + 2ɛ σ(t u(t σ(t + u(t 2L[σ(t u(t] + 2ɛ σ(t u(t, u(t which implies [ { ( [σ(t u(t] exp 2L + 2ɛ/ u(t t}] Integrating the last inequality from t to t, we have { ( [σ(t u(t] exp 2L + 2ɛ/ } { ( u(t t [σ(t u(t ] exp 2L + 2ɛ/ } u(a a = Therefore σ(t u(t for t t t 1 which is a contradiction to σ(t 1 > u(t 1 The Claim is proved Thus, σ(t u(t, which implies x(t y(t = σ(t u(t = u(ae L(t a + ɛ ( e L(t a 1 L = σ(ae L(t a + ɛ ( e L(t a 1 L = x(a y(a e L(t a + ɛ ( e L(t a 1 L (2 For σ(a =, we have to modify the discussion above For each positive integer n, we consider the following initial value problem du = 2Lu + 2ɛ u, t a, u(a = 1/n The discussion above can be applied to this problem, and we have the solution [ u n(t = n 1/2 e L t a + ɛ ( 2 e L t a 1] L If we can show that σ(t u n(t for t a, then, after taking n, we obtain the desired inequality in the case σ(a = The last inequality can be proved by contradiction In fact, if σ(t 1 > u n(t 1,

12 1 EXISTENCE AND UNIQUENESS for some t 1 > a, then there exists t to be the largest t in the interval a < t t 1 such that σ(t u n(t Obviously, σ(t = u n(t > and σ(t > u n(t for t < t t 1 But this is impossible according to the discussion in the case σ(a > by replacing a by t We can use the inequality σ (t σ (t 2Lσ(t + 2ɛ σ(t and a similar argument to prove the following inequality x(t y(t x(a y(a e L(a t + ɛ L ( e L(a t 1 Exercise 13 1 Assume that ϕ(t, ψ(t, w(t are continuous on [a, b], with w(t > If the inequality ϕ(t ψ(t + w(sϕ(sds a holds on [a, b], then { } ϕ(t ψ(t + w(sψ(s exp w(udu ds a s Further, if ψ(t is non-negative and monotonically increasing, then { } ϕ(t ψ(t exp w(sds a 2 Prove Strong Continuity Theorem in detail for t a 3 Let x(t and y(t be the solutions of the initial value problems x + (1 + t 2 x =, < t < 1, x( = 1, x ( = and respectively Estimate y + (1 + t 2 y = δ, < t < 1, δ = const, y( = 1, y ( =, max x(t y(t t [,1] 4 Let X(x, t, s be continuous for x c K, t a T, and s s S, and let it satisfy X(x, t, s X(y, t, s L x y Show that the solution x(t, s of = X(x, t, s satisfying x(a, s = c is a continuous function of s

14 EXISTENCE THEOREM 13 14 EXISTENCE THEOREM In this section, we stu existence of the differential equation (13 The idea is to establish an equivalent integral equation for any given initial value problem Then we show that the iteration of the integral operator converges to a solution Theorem 141 Let X(x, t be a continuous function Then a function x(t is a solution of the initial value problem = X(x, t, (14 x(a = c if and only if it is a solution of the integral equation x(t = c + a X(x(s, sds (15 Proof Let us assume that x(t is a solution of the initial value problem (14 The Fundamental Theorem of Calculus implies that x k (t = x k (a + a x k(sds Using (14, we have the integral equation (15 Conversely, if x(t is a solution of the integral equation (15, then x(a = c and, by the Fundamental Theorem of Calculus, we have x k(t = X k (x(t, t, k = 1,, n These imply that x(t satisfies = X(x, t For a given X(x, t, if it is defined for all x in t a T, and is continuous, then we can define an operator U by U(x = c + X(x(s, sds (16 a For this operator, its domain is { x(t x(t is continuous in the interval t a T } and its range is { y(t y(t is continuously differentiable in the interval t a T and y(a = c } Thus, a solution of the integral equation (15 is a fixed point of the operator U: x = U(x Theorem 141 can be re-stated as the following: Theorem 142 Let X(x, t be a continuous function Then a function x(t is a solution of the initial value problem (14 if and only if the operator U defined in (16 has a fixed point in C[a T, a + T ]

14 1 EXISTENCE AND UNIQUENESS Now we can use the operator U to generate a sequence of functions {x n } from the given initial data by the successive iteration: x (t c, x n = U(x n 1 = U n (x, n = 1, 2, (17 This is call a Picard iteration 5 Now we show that, under some conditions, a Picard iteration converges to a solution of the initial value problem (14 Lemma 143 Assume that X(x, t is continuous and satisfies the Lipschitz condition (12 on the interval t a T for all x, y Then the Picard iteration (17 converges uniformly for t a T Proof Let M = sup t a T X(c, t < + Without loss of generality we can assume that a = and t a In other words, we prove the lemma on the interval t T The proof for the general a and t < a can be deduced from this case by the substitutions t t + a and t a t We first prove by induction that In fact, for n = 1, For n = 2, x n (t x n 1 (t (M/L(Lt n, n = 1, 2, (18 n! x 1 (t x (t t = X(x (s, s ds x 2 (t x 1 (t = X(x (s, s ds M L L ds = Mt = (M/L(Lt1 1! [ X(x 1 (s, s X(x (s, s ] ds X(x 1 (s, s X(x (s, s ds Assume that the desired inequality holds for n = k: x 1 (s x (s ds Ms ds = LMt2 2 x k (t x k 1 (t (M/L(Ltk k! = (M/L(Lt2 2! 5 The Picard iteration is started from x = c, given by the initial data

14 EXISTENCE THEOREM 15 Then, x k+1 (t x k (t = [ ] X(x k (s, s X(x k 1 (s, s ds X(x k (s, s X(x k 1 (s, s ds L L x k (s x k 1 (s ds (M/L(Ls k k! ds = (M/L(Ltk+1 (k + 1! Hence, the estimates (18 hold Next, we show that the sequence {x n (t} converges uniformly for t T Indeed, for t T, since and the positive series (M/L(Lt k k! (M/L(LT k, k! (M/L(LT k is convergent to (M/L ( e LT 1, by the Comparison Test, the series x (t + k=1 k=1 k! [ ] x k (t x k 1 (t is also convergent Actually, the convergence is uniform for t T, since t The n-th partial sum, x (t + n k=1 [ ] x k (t x k 1 (t = x n (t Therefore, the sequence of functions {x n (t} converges uniformly (M/L(LT k k! is independent of Theorem 144 (Existence Theorem Assume that X(x, t is continuous and satisfies the Lipschitz condition (12 on the interval t a T for all x, y Then the initial value problem (14 has a unique solution on the interval t a T Proof The uniqueness is a direct consequence of Uniqueness Theorem 121 We only need to prove the existence By Lemma 143, the sequence {x n (t} defined by the Picard iteration with x (t c is uniformly convergent Denote x (t the limit function We show that x (t is a solution of the integral equation (15 By the definition, x n+1 (t = c + X(x n (s, s ds The left hand side is uniformly convergent to x (t By the Lipschitz condition, X(x m (s, s X(x n (s, s L x m (s x n (s, and so the integral on the right hand side is also uniformly convergent Since X(x, s is continuous, we know that X(x n (s, s X(x (s, s

16 1 EXISTENCE AND UNIQUENESS Hence, we obtain x (t = c + X(x (s, s ds Finally, by Theorem 141, we conclude that the function x (t is a solution of the initial value problem (14 Example 141 Solve the initial value problem by the Picard iteration = x, x( = 1 Solution For this initial value problem, the integral operator U is defined as Thus, the Picard iteration gives U(x(t = 1 + x(s ds x (t = 1, x 1(t = U(x (t = 1 + x 2(t = U(x 1(t = 1 + = 1 + t + t2 2, 1 ds = 1 + t, (1 + s ds x n(t = 1 + t + t2 2 + + tn n!, x n+1(t = U(x n(t = 1 + (1 + s + s2 2 + + sn n! = 1 + t + t2 2 + + tn n! + tn+1 (n + 1! We see that the sequence {x n(t} converges uniformly to the function e t Hence we get the solution x(t = e t by the Picard iteration Example 142 Verify the Taylor series for sin t and cos t by applying the Picard iteration to the first order system corresponding to the second order initial value problem x = x, x( =, x ( = 1 ds Solution The associated first order system is 1 = x 2, 2 = x 1,

14 EXISTENCE THEOREM 17 with the initial condition x( = (x 1(, x 2( = (, 1 The corresponding X(x, t = (x 2, x 1, and the initial value c = (, 1 The Picard iteration yields x (t = (, 1, x 1 (t = (, 1 + (1, ds = (, 1 + (t, = (t, 1, x 2 (t = (, 1 + = (, 1 + (1, s ds = ( t, t2 2 ( t, 1 t2, 2 x 3 (t = (, 1 + (1 s2 2, s ds = (, 1 + (t t3 3!, t2 2 = (t t3 3!, 1 t2, 2 x 4 (t = (, 1 + (1 s2 s3, s + ds 2 3! = (, 1 + (t t3 3!, t2 2 + t4 4! = (t t3 3!, 1 t2 2! + t4 4! We claim that for n =, 1, 2,, x 2n+1 (t = (t + + ( 1 n t 2n+1 (2n + 1!, 1 + + t 2n ( 1n, (2n! x 2n+2 (t = (t + + ( 1 n t 2n+1 (2n + 1!, 1 + + t 2n+2 ( 1n+1 (2n + 2! We prove this Claim by mathematical induction When n =, they hold due to the calculations above Suppose they hold when n = k Then we have x 2k+3 (t = (, 1 + (1 + + ( 1 k+1 s 2k+2 (2n + 2!, s s 2k+1 ( 1k ds (2k + 1! = (t + + ( 1 k+1 t 2k+3 (2k + 3!, 1 + + t 2k+2 ( 1k+1, (2k + 2! x 2k+4 (t = (, 1 + (1 + + ( 1 k+1 s 2k+2 (2k + 2!, s s 2k+3 ( 1k+1 ds (2k + 3! = (t + + ( 1 k+1 t 2k+3 (2k + 3!, 1 + + t 2k+4 ( 1k+2 (2k + 4!

18 1 EXISTENCE AND UNIQUENESS That is, the claim is also true for n = k + 1 It can be easily shown that the associated initial value problem has a unique solution x(t = (sin t, cos t The uniqueness can be verified by checking that the conditions of Uniqueness Theorem 121 hold for the associated problem Since {x n (t} converges to the unique solution by Existence Theorem 144, we have sin t = t t3 3! + + ( 1n 1 t 2n 1 (2n 1! +, cos t = 1 t2 2! + + ( 1n t 2n (2n! + Example 143 Let f(x, t be a function satisfying the Lipschitz condition f(x, t f(y, t L x y for all t T and all x and y Suppose f(x, t is continuous and bounded Let M = f(x, t Let x(t be a solution of x = f(x, t with initial value x( = c and x k (t sup <x< t T be the k-th term in the Picard s approximation method Prove for t x(t x k (t MLk (k + 1! tk+1, Solution: Use mathematical induction When k =, we have x(t x (t = x(t x( = x (ξ for some < ξ < t by the Mean Value Theorem Since x (ξ = f(ξ, t M, the desired inequality holds for k = Suppose the inequality holds for k = n Since x(t x n+1 (t = f(x(s, sds f(x n (s, sds f(x(s, s f(x n (s, s ds L x(s x n (s ds L MLn (n + 1! sn+1 ds = ML n+1 t n+2 (n + 2!, we know that the inequality holds when k = n + 1 By mathematical induction, the inequality holds for all k

15 LOCAL EXISTENCE THEOREM AND THE PEANO THEOREM 19 Exercise 14 1 For the initial value problem x = tx, x( = 1, obtain the n-th approximation of the Picard iteration Use mathematical induction to justify the formula 2 For the initial value problem x = t + x, x( =, obtain x (t, x 1 (t, x 2 (t and the n-th term of the sequence of the Picard approximation Use mathematical induction to justify the formula 3 Assume that X(x, t is continuous and satisfies the Lipschitz condition (12 on the interval t a T for all x, y For any function f(t, continuous on t a T, define the sequence {x n (t} by x (t = f(t, x n (t = c + X(x n 1 (s, sds, n = 1, 2, a Show that {x n (t} is uniformaly convergent to the unique solution of the initial value problem (14 4 Show that the initial value problem y + y 2 1 =, y( = y, y ( = y 1 is equivalent to the integral equation y(t = y + y 1 t (t s [ y 2 (s 1 ] ds 5 Solve the integral equation y(t = cos t + sin(τ t y(τdτ 6 (1 Show that the initial value problem { y + (1 + t 2 y =, t >, y( = 1, y ( = is equivalent to the integral equation y(t = cos t + sin(τ t τ 2 y(τdτ (2 Show that the successive approximations {φ n(t} defined by φ (t =, φ n(t = cos t + sin(τ t τ 2 φ n 1 (τdτ, n = 1, 2, converges uniformly to the solution in (1 for t < T Here T is any fixed constant 15 LOCAL EXISTENCE THEOREM AND THE PEANO THEOREM 151 Local Existence Theorem In the Existence Theorem 144, the function X(x, t satisfies a Lipschitz condition (12 for all x This condition is quite strong and many functions may fail this condition

2 1 EXISTENCE AND UNIQUENESS Example 151 Show that the conclusion of Theorem 144 fails for the initial value problem = ex, x( = c Solution The solution of the initial value problem is given implicitly by e c e x = t It is obvious that the function is defined only in < t < e c Hence, there is no ɛ > such that the differential equation has a solution defined on all of t < ɛ for every initial value, since we can always take a sufficient large c > such that e c < ɛ Thus, the conclusion of Theorem 144 fails for this initial value problem The cause of this failure is that the function X(x = e x does not satisfy a Lipschitz condition for all x In fact, is unbounded for large values of x X(x, t X(, t x = ex 1 x However, if the function X(x, t satisfies a Lipschitz condition in a bounded domain, then a solution exists in a limited region Theorem 151 (Local Existence Theorem Assume that X(x, t is continuous and satisfies the Lipschitz condition (12 in the closed domain x c K, t a T Then the initial value problem (14 has a unique solution in the interval t a min{t, K/M}, where M = sup X(x, t x c K t a T Proof The existence can be proved as in the proof of Theorem 144, except that we have to modify the estimates for the sequence {x n (t} by showing that if x(t is defined and continuous on t a min{t, K/M} satisfying (a x(t is defined and continuous on t a min{t, K/M}; (b x(a = c; (c x(t c K on t a min{t, K/M}, then one iteration y = U(x still satisfies these three conditions Indeed, (a and (b are obvious To show (c, we have the following y(t c = X(x(s, s ds a X(x(s, s ds a M t a M K M = K

15 LOCAL EXISTENCE THEOREM AND THE PEANO THEOREM 21 152 The Peano Theorem The Lipschitz condition plays an important role in the proof of the Local Existence Theorem However, if we drop it, we are still able to prove the existence, by a more sophisticated argument Theorem 152 (Peano Existence Theorem Assume that X(x, t is continuous in the closed domain x c K, t a T Then the initial value problem (14 has at least one solution in the interval t a min{t, K/M}, where M = sup X(x, t x c K t a T To prove this theorem, we need the following definition and the famous Arzelá-Ascoli Theorem Definition 153 A family of functions F is said to be equicontinuous on [a, b] if for any given ɛ >, there exists a number δ > such that x(t x(s < ɛ whenever t s < δ for every function x F and t, s [a, b] Arzelá-Ascoli Theorem Assume that the sequence {x n (t} is bounded and equicontinuous on [a, b] Then there exists a subsequence {x ni (t} that is uniformly convergent on [a, b] Now we are rea to prove Peano Existence Theorem 152 Proof Denote T 1 = min{t, K/M} As argued at the beginning of the proof of Lemma 143, we only need to prove the theorem on t T 1, with a = We first construct a sequence of bounded equicontinuous functions {x n (t} on [, T 1] For each n, define c, for t T 1/n, x n (t = T1 /n c + X (x n (s, s ds, for T 1/n < t T 1 The above formula defines the value of x n (t recursively in terms of the previous values of x n (t We can use mathematical induction to show that x n (t c K on [, T 1] Indeed, on [, T 1/n], it is trivial since x n (t = c If we assume that the inequality holds on [, k T 1/n] ( k < n, then on [k T 1/n, (k + 1 T 1/n], T1 x n /n (t c = X (x n (s, s ds M t T 1/n M T 1 K Hence, the sequence {x n (t} is uniformly bounded on [, T 1]: x n (t c + K

22 1 EXISTENCE AND UNIQUENESS The equicontinuity of the sequence {x n (t} on [, T 1] can be proven by the following estimates: for any t 1, t 2 [, T 1],, if t 1, t 2 [, T 1/n], 2 T 1/n X(x n (s, sds, if t1 [, T1/n] and t2 (T1/n, T1], x n (t 1 x n (t 2 = 1 T 1/n X(x n (s, sds, if t2 [, T1/n] and t1 (T1/n, T1], 2 T 1/n X(x n (s, sds, if t1, t2 (T1/n, T1] t 1 T 1/n M t 1 t 2 By Arzelá-Ascoli Theorem, we know that there exists a uniformly convergent subsequence {x n i (t} that converges to a continuous function x (t on [, T 1] as n i We can show that the function x (t is actually a solution of the initial value problem (14 Indeed, for any fixed t (, T 1], we take n i sufficiently large such that T 1/n i < t Thus, by the definition of {x n (t}, we have x n i (t = c + X (x n i (s, s ds X (x n i (s, s ds t T 1 /n i As n i, since X (x, t is uniformly continuous, we have X (x n i (s, s ds second integral of the last equation tends to zero, since X (x n i (s, s ds t T 1 /n i Mds = M T1 t T 1 /n i n i Hence, we know that the function x (t satisfies the integral equation x (t = c + X (x (s, s ds X (x (s, s ds; the Without the Lipschitz condition, it is known that solution of initial value problem is not necessarily unique For instance, it can be earily see that the initial value problem is not unique, with the following two solutions = 3 2 x1/3, t, x( =, x 1 (t =, x 2 (t = t 3/2 Exercise 15

16 LINEAR SYSTEMS 23 1 Determine the existence region of the solution to the initial value problem = t2 + x 2, x( =, where the equation is defined in the region R = {(x, t x 1, t 1} Find the first four approximations x (t, x 1 (t, x 2 (t, x 3 (t of the Picard iteration 2 Determine the existence region of the solution to the initial value problem = t2 x 2, x( =, where the equation is defined in the region R = {(x, t x 1, t 1} Find the first four approximations x (t, x 1 (t, x 2 (t, x 3 (t of the Picard iteration 3 Assume that the functions F (x 1, x 2,, x n, t is continuous and satisfies a Lipschitz condition in t a T, x c K Then the initial value problem { u (n = F (u, u, u,, u (n 1, t, u (i (a = c i, i n 1, has a unique solution on t a min{t, K/M}, where M = ( sup x 2 2 + + x 2 n + F 2 (x 1, x 2,, x n, t 1/2 x c K t a T 16 LINEAR SYSTEMS In this section, we apply the theorems obtained in the previous sections to linear differential systems of the form n i = a ij (tx j (t + b i (t, 1 i n, (19 Or, in a matrix form, we write j=1 where A(t = (a ij (t and b(t = (b 1 (t,, b n (t = A(tx + b(t, (11 Theorem 161 Assume that the functions a ij (t and b i (t are continuous for t a T, 1 i, j n Then the initial value problem (19 with x(a = c = (c 1,, c n has a unique solution on t a T Proof The initial value problem can be re-written in terms of the vector form (14, with ( n n X(x(t, t = a 1j(tx j(t + b 1(t,, a nj(tx j(t + b n(t, j=1 j=1 and x(a = (c 1,, c n

24 1 EXISTENCE AND UNIQUENESS The function X(x, t is continuous and satisfies the Lipschitz condition (12 on the interval t a T for all x, y Indeed, by the Schwarz inequality, we have = = X(x, t X(y, t 2 ( n 2 ( n a 1j(t(x j y j + + a nj(t(x j y j j=1 j=1 ( n ( n ( n ( n a 1j(t 2 x j y j 2 + + a nj(t 2 x j y j 2 j=1 j=1 ( a ij(t 2 x y 2 i,j ( sup a ij(t 2 x y 2 t a T i,j j=1 2 j=1 Hence the function X(x, t satisfies the conditions in Existence Theorem 144 This gives the existence of the solution The uniqueness follows from Uniqueness Theorem 121 If b i (t, 1 i n, the linear differential system (19 is called homogeneous, for which we can construct its general solution Assume that x 1,, x n are n solutions of a homogeneous linear differential system Obiously, the matrix Ψ(t = ( x 1 (t,, x n (t = x 1 1(t x n 1 (t x 1 n(t x n n(t, (111 satisfies dψ(t = A(tΨ(t If these n solutions are linearly independent at every point t, Ψ(t is called a fundamental matrix of the homogeneous linear differential system Theorem 162 Assume that the functions a ij (t are continuous for t a T, 1 i, j n If x 1,, x k are k solutions of the homogeneous linear differential system, then the constant vectors x 1 (t,, x k (t are linearly independent for some t if and only if they are linearly independent for every t on t a T Proof Assume that x 1 (t,, x k (t are linearly independent for some t Let t be another point on t a T other than t We need to show that if α 1x 1 ( t + + α nx k ( t =, then, α 1 = = α k = In fact, since x 1,, x k are solutions, we have two solutions of the homogeneous system, and α 1x 1 + + α nx k, both vanishing at t By the uniqueness of Theorem 161, we conclude that α 1x 1 (t + + α nx k (t =

16 LINEAR SYSTEMS 25 for every t on t a T In particular, we have α 1x 1 (t + + α nx k (t = The linear independence of x 1 (t,, x k (t implies that all α i =, 1 i k The other direction is obvious Theorem 163 Assume that the functions a ij (t are continuous for t a T, 1 i, j n Let x i (t, 1 i n, be n solutions of the homogeneous linear differential system (19 and assume that they are linearly independent at some t Define Ψ(t = ( x 1 (t,, x n (t Then the solution of the homogeneous linear differential system satisfying the initial condition x(a = c = (c 1,, c n is given by x(t = Ψ(tΨ 1 (ac In particular, if x i (t be the solution of the homogeneous linear differential system (19 that satisfies the initial condition x i k (a =, k i, xi i (a = 1 The the solution satisfying the initial condition x(a = c = (c 1,, c n is given by x(t = c 1 x 1 (t + + c n x n (t Proof Since x 1 (t,, x n (t are linearly independent, by Theorem 162, we know that the matrix Ψ(t = ( x 1 (t,, x n (t is non-singular for every t on t a T and hence, Ψ(t is a fundamental matrix Let x(t be the unique solution of the homogeneous system satisfying the initial condition x(a = c Its existence is guaranteed by Theorem 161 Consider the function x(t Ψ(tΨ 1 (ac Since Ψ(t is a fundamental matrix, we know this function is again a solution of the homogeneous system Now we have two solutions of the homogeneous system, and x(t Ψ(tΨ 1 (ac, both vanishing at t = a By the uniqueness of Theorem 161, we have x(t Ψ(tΨ 1 (ac =, That is, x(t = Ψ(tΨ 1 (ac In particular, if x i (t satisfies the initial condition x i k(a =, i, x i i(a = 1, then Ψ(a is the identity matrix Therefore x(t = Ψ(tc = c 1x 1 (t + + c nx n (t Theorem 164 Let A(t = (a ij (t and b(t be continuous on t a T The solution of the non-homogeneous linear system (11, with initial condition x(t = x, is given by x(t = Φ(tΦ 1 (t x + Φ(t t Φ 1 (sb(sds where Φ(t is any fundamental matrix of the corresponding homogeneous system In particular, if A is a constant matrix, then x(t = Φ(tΦ 1 (t x + t Φ(t s + t Φ 1 (t b(sds

26 1 EXISTENCE AND UNIQUENESS Furthermore, if Ψ(t is the fundamental matrix satisfying Ψ(t = I, then x(t = Ψ(tx + t Ψ(t s + t b(sds Proof For any given fundamental matrix Φ(t of the homogeneous system, we postulate the solution x(t of the nonhomogeneous system satisfying the initial condition x(t = x to be in the form x(t = Φ(tΦ 1 (t {x + φ(t} Then, φ(t satisfies the initial condition φ(t = To find the equation satisfied by φ(t, we substitute the expression into the nonhomogeneous system to have Φ (tφ 1 (t {x + φ(t} + Φ(tΦ 1 (t φ (t = A(tΦ(tΦ 1 (t {x + φ(t} + b(t Since Φ (t = A(tΦ(t, the last equation gives whose solution satisfying φ(t = is Φ(tΦ 1 (t φ (t = b(t, φ(t = Φ(t Φ 1 (sb(sds t Thus x(t = Φ(tΦ 1 (t x + Φ(t Φ 1 (sb(sds t If A is a constant matrix, to show it is sufficient to show that x(t = Φ(tΦ 1 (t x + t Φ(t s + t Φ 1 (t b(sds, Φ(tΦ 1 (s = Φ(t s + t Φ 1 (t, for any t, s and t Indeed, let U(t = Φ(tΦ 1 (s and V (t = Φ(t s + t Φ 1 (t, with s and t being two parameters It is obvious that U(s = V (s = I We know that they also satisfy the same differential system, since U (t = Φ (tφ 1 (s = AΦ(tΦ 1 (s = AU(t, V (t = Φ (t s + t Φ 1 (t = AΦ(t s + t Φ 1 (t = AV (t By the uniqueness of Theorem 161, the corresponding columns of U and V are identical since they satisfy the same equation with the same initial conditions Exercise 16 1 Assume that the functions a ij (t, 1 i, j n, are continuous on t a T Let Ψ(t be a fundamental matrix defined by n solutions x 1 (t,, x n (t of a homogeneous system = A(tx, with A(t = (a ij(t (a Show that the Wronskian W (t, defined to be det Ψ(t, satisfies the differential equation ( n dw (t = a ii (t W (t i=1

17 CONTINUATION OF SOLUTIONS 27 (b Prove that the Wronskian is either always zero or nowhere zero on t a T 2 Assume that the functions a ij (t, 1 i, j n, are continuous on t a T Let Φ(t and Ψ(t be two fundamental matrices of the homogeneous system = A(tx, with A(t = (a ij(t Prove that Φ(tΨ 1 (t is a constant matrix 3 Let A(t = (a ij (t and b(t be continuous on t a T Prove that there exist n+1 solutions x 1, x 2,, x n+1 of the non-homogeneous system = A(tx+b(t such that every solution x of this equation can be expressed as x(t = α 1 x 1 (t + α 2 x 2 (t + + α n+1 x n+1 (t, for some constants α 1, α 2,, α n+1 with α 1 + α 2 + + α n+1 = 1 4 Assume that the matrix A(t is continuous for t (a, b and R(x, t is continous for all x and t (a, b Prove that the initial value problem = A(tx + R(x, t, x(t = x, is equivalent to the integral equation x(t = Φ(tΦ 1 (t x + Φ(t s + t Φ 1 (t R(x(s, sds, t where Φ(t is any fundamental matrix of the homogeneous system = A(tx 5 Consider the initial value problem where A is a constant matrix = Ax, x( = x, (a Show that the initial value problem is equivalent to the integral equation x(t = x + Ax(sds (b Show that the Picard iteration defined by the integral equation gives x n (t = (I + At + A 2 t2 tn + + An x 2! n! (c Show that the solution of the initial value problem is (I + At + A 2 t2 2! + x 17 CONTINUATION OF SOLUTIONS Theorem 171 Assume that X(x, t is continuously differentiable in an open region R of (x, t- space For any point (c, a R, the initial value problem (14 has a unique solution x(t defined

28 1 EXISTENCE AND UNIQUENESS over an interval a t < b (b is a finite number or infinite such that if b < +, either x(t is unbounded as t b or (x(t, t approaches the boundary of the region R 6 Proof By the Local Existence Theorem 151, we know that there exist solutions of the initial value problem (14 in some interval [a, T Given two solutions x(t and y(t, defined in [a, T 1 and [a, T 2 respectively, we define a new function z(t to be either x(t or y(t wherever either is defined Then z(t is again a solution of (14 defined on [a, max{t 1, T 2} Thus, we can denote { } b = sup T the initial value problem (14 has a solution in [a, T We can define a single solution, denoted again by x(t, called the maximal solution, defined in [a, b The above construction indicates the existence of the maximal solution It is also unique, by Uniqueness Theorem 121 Let us consider the limiting behavior of x(t as t b Then there are only possibilities: Case 1: b = + Case 2: b < + and x(t is unbounded as t b Case 3: b < + and x(t is bounded as t b We only need to show that in the third case, x(t approaches the boundary of the region R as t b Indeed, let {t n} be any sequence such that t n b Since the sequence of points {(x(t n, t n} is bounded, there exists at least one limit point, say (d, b We now show that the point (d, b is on the boundary of R In fact, if it is an interior point, then there exists a closed neighborhood D : x d ɛ, t b ɛ of (d, b also in R Let M = max D X(x(t, t Take δ < min{ɛ, ɛ/2m} and let G D be the open set { } G = (x,, t x d < ɛ, t b < δ We can take k large enough such that (x(t k, t k G Applying Local Existence Theorem 151 to the equation = X(y(t, t, we know that there exists a unique solution in the interval t b < δ satisfying y(t k = x(t k If we define { x(t, a t < b, z(t = y(t, b t < b + δ, then clearly z(t is a solution of the initial value problem (14 over the interval [a, b + δ, contradicting to the maximality of b Example 171 Find the maximal solutions for the following initial value problems: (1 (3 = x, x( = c; = 1 t 2 cos 1 t, x(t = c, t (2 = x2, x( = c; 6 The conclusion of Theorem 171 indicates that there are only three possible outcomes 1 b = + or 2 b < + and x(t is unbounded as t b or 3 b < + and (x(t, t approaches the boundary of R

18 MISCELLANEOUS PROBLEMS 29 Solution (1 The function X(x, t = x is continuously differentiable at any point (x, t (, + (, + The differential system has a unique solution x(t = ce t, which is the maximal solution defined for t [, + (2 The function X(x, t = x 2 is continuously differentiable at any point (x, t (, + (, + If c =, the maximal solution is x(t =, defined for t [, + If c, the maximal solution is x(t = 1, which is defined for t [, + if c < ; or t [, c if c > Obviously, in the later case, x(t c t becomes unbounded as t c (3 The function X(x, t = 1 t cos 1 is continuously differentiable at any point (x, t (, + 2 t (, + with t The maximal solution is x(t = c sin 1 t + sin 1, which is defined for t [t, if t t < ; or for t [t, + if t > In the former case, the function x(t is still bounded, but t = is the boundary of { } R = (x, t t Exercise 17 1 Let f(x, y be a function defined on the whole plane with bounded continuous partial derivatives Suppose f(x, = for all x Let y = ϕ(x be a nontrivial solution of the differential equation = f(x, y and lim ϕ(x = Then a must be ± x a 2 Give an example to demonstrate that if the maximal interval (, b of existence of the solution for an initial value problem = f(x, t, x( = c, is finite, and if the solution x(t is unbounded as t approaches b, then b varies as c does 18 MISCELLANEOUS PROBLEMS 1 Show that the following initial value problem = x2 + y cos s, = y2 + sin s, x( = s + 1, y( = s 2 /(s 2 1 has a unique solution (x s(t, y s(t in t T, s 1/2 for some T Prove Solution lim (xs(t, ys(t = (1, s For any K 1 > and T 1 >, in the closed bounded region { } R = (x, y, t, s x2 + y 2 K 1, t T 1, s 1/2 the function ( x 2 +y cos s, y 2 +sin s is continuously differentiable for all its variables (x, y, t, s Thus, we know that it is a Lipschitz function in this region By the Local Existence Theorem, for any fixed

3 1 EXISTENCE AND UNIQUENESS s with s 1/2, the given initial value problem has a unque solution for t T s = min{t 1, K 1/M s}, where M s = sup [x2 + y cos s] 2 + [y 2 + sin s] 2 It is obvious that x 2 +y 2 K 1 t T 1 M = max M s < + s 1/2 Thus, for any s with s 1/2, the given initial value problem has a unque solution for t T = min{t 1, K 1/M} ( Denote the Lipschitz constant of the function x 2 + y + t cos s, y 2 + sin s in R to be L It is easy to see that (x 2 + y cos s, y 2 + sin s (x 2 + y 1, y 2 = (1 cos s, sin s By the Strong Continuity Theorem, the solutions (x s(t, y s(t and (x (t, y (t = (1,, of the following initial value problems = x2 + y cos s, = y2 + sin s, x( = s + 1, y( = s 2 /(s 2 1, and respectively, satisfy = x2 + y 1, = y2, x( = 1, y( =, (x s(t, y s(t (1, (s + 1, s 2 /(s 2 1 (1, e L t + 1 (e L (1 cos s, sin s L t 1 Let s, since the right hand side of the last inequality approaches zero, we have lim(xs(t, ys(t = (1, s 2 Use the Local Existence Theorem to show that the initial value problem = 1 + x2, x( =, has a unique solution for t 1/2 Determine the region where the true solution is defined by solving this initial value problem What is the limiting behavior of the true solution as t approaches the end point(s of the maximal interval of existence? Solution Consider the given initial value problem in the region { } (x, t x 1, t 1

18 MISCELLANEOUS PROBLEMS 31 The function X(x, t = 1 + x 2 satisfies the following Lipschitz condition X(x, t X(y, t = x + y x y 2 x y for x 1 and y 1 Hence, by the Local Existence Theorem, there exists a unique solution of the given initial value problem for t min{1, 1/M}, where ( M = sup X(x, t = sup 1 + x 2 = 2 x 1 x 1 t 1 t 1 That is, there exists a unique solution for t 1/2 The unique true solution of the given initial value problem is x(t = tan t, whose maximal interval of existence is ( π/2, π/2 As t ±π/2, x(t becomes unbounded 3 Assume that the function f(x, t is continuously differentiable in R = (, + (a, b, and satisfies the inequality f(x, t A(t x + B(t, for some non-negative continuous functions A(t and B(t Show that any solution of has a maximal interval of existence (a, b = f(x, t, x(t = c, t (a, b, Proof Let x = x(t be a solution of the initial value problem We only show that it can be extend to the interval [t, b The continuation of the solution to (a, x ] can be proved similarly We prove it by contradiction Suppose that the maximal interval of existence is [t, β, with β < b Select t 1 and t 2 such that t < t 1 < β < t 2 < b and t 2 t 1 < t 1 t Denote T = t 2 t 1 > Let A M and B M be positive upper-bounds of A(t and B(t in the interval [t, t 2], respectively Thus, by the condition, we have f(x, t A M x + B M, for (x, t (, + [t, t 2] We assume A M is large enough such that T < 1 A M Now we will see that the solution x = x(t can be extended to the interval [t, t 2, a contradiction In fact, since t 1 (t, β and the solution x = x(t exists on [t, β, for any positive number K, the region { } R 1 = (x, t x x(t 1 K, t t 1 T

32 1 EXISTENCE AND UNIQUENESS is a bounded closed subset of R In R 1, since f(x, t A M x + B M A M ( x(t 1 + K + B M = M, by the Local Existence Theorem, the solution curve (x(t, t exists and remains in the region { } R 2 = (x, t x x(t 1 K, t t 1 h, where h = min {T, K/M} Since R 2 is a bounded closed region, by Theorem 171, the solution curve (x(t, t can be extended to the boundary of R 2 That is, the solution exists in [t, t 1 + h Since we know that for sufficient large K, K lim K + M = 1 > T, A M h = min {T, K/M} = T Thus, the solution x = x(t can be extended to [t, t 1 + T = [t, t 2 This contradiction implies β = b

2 Plane Autonomous Systems 21 PLANE AUTONOMOUS SYSTEMS Recall a normal form of a system of first-order ordinary differential equations: 1 = X 1 (x 1,, x n ; t, n = X n (x 1,, x n ; t If n = 2 and the functions X i are independent of t, we have a so-called plane autonomous system In this case, we write the autonomous system as = X(x, y, = Y (x, y (21 Let (x(t, y(t be a solution of system (21, we get a curve in R 2, the phase plane, which is called the solution curve of the system The differential system gives the tangent vector of the solution curve, (X(x(t, y(t, Y (x(t, y(t A R 2 plane together with all solution curves is called the phase plane of the differential system The orientation of a solution curve is the direction of the movement of points on the curve when t increases For example, the solution curve ( e 2t, e t of the system = 2x, = y, 33

34 2 PLANE AUTONOMOUS SYSTEMS is a parabola in Fig 21(a The solution curve ( e 2t, e t of the system = 2x, = y, is the same parabola in Fig 21(b, with the opposite orientation y y x x (a (b Fig 21 Solution curves with orientations Example 211 Consider the differential system = 2x, = y, Sketch its oriented solutions curves in (x, y-plane Solution The general solution is given by (x(t, y(t = (c 1e 2t, c 2e t, where c 1 and c 2 are arbitrary constants For different values of c 1 and c 2, we have one solution curve They are all shown in Fig 22 In Example 211, the point (, is a solution curve of the differential system even though it is not a curve in the ordinary sense In Fig 22, the point (, is a special point, at which the direction is indeterminate Such points are of particular importance for the stu of autonomous systems Definition 211 A point (x, y is called a critical point of the autonomous system (21, if X(x, y = y(x, y = If a point is not a critical point, we call it an ordinary point Locally, the solution curves near an ordinary point are a family of parallel lines However, a local structure near a critical point can be very complicated If (x 1, y 1 is a neighboring point of a critical point (x, y, the solution curve