Existence and uniqueness: Picard s theorem First-order equations Consider the equation y = f(x, y) (not necessarily linear). The equation dictates a value of y at each point (x, y), so one would expect there to be a unique solution curve through a given point. We will prove the following theorem: THEOREM 1. Let I be a real interval (possibly R itself), and let f be a continuous real function on I R, satisfying the Lipschitz condition f(x, y 2 ) f(x, y 1 ) K y 2 y 1 on this set. Let x I. Then there is a unique function y on I such that y (x) = f[x, y(x)] for x I and y(x ) = y. Note. By the mean-value theorem, the Lipschitz condition will be satisfied if we have f(x, y) K on I R. y LEMMA 1. It is enough to prove the result for the case x = y =. Proof. Suppose the result is known for this case and x, y are given. Let g(x, y) = f(x + x, y + y ). By assumption, there is a unique function z such that z (x) = g[x, z(x)] when x + x I and z() =. Let y(x) = z(x x ) + y. Then y(x ) = y and for x I, y (x) = z (x x ) = = = The next step is to show that the required properties of y can be equally expressed by an integral. LEMMA 2. Let I be an interval containing. For a differentiable function on I, the following statements are equivalent: 1
(i) y (x) = f[x, y(x)] for x I and y() = ; (ii) y(x) = Proof. (i) (ii). f[t, y(t)] dt for x I. Let C(I) be the space of all continuous functions on I. We define a mapping A from C(I) to itself as follows: given y C(I), (Ay)(x) = f[t, y(t)] dt. By Lemma 2, y is the required solution if and only if A(y) = y. We take y =, y 1 = A(y ), y 2 = A(y 1 ),..., y n = A(y n 1 ) = A n (y ). We will show that (y n ) converges (uniformly on I) to a limit function y. One would then expect that A(y) = lim n A(y n ) = lim n y n+1 = y, as required. LEMMA 3. Suppose that u(x) v(x) C for x I. Then for each n 1, (A n u)(x) (A n v)(x) C (K x )n n! for x I. Proof. By the Lipschitz condition, f[t, u(t)] f[t, v(t)] CK for t I, so (Au)(x) (Av)(x) = [f(t, u(t)) f(t, v(t))] dt CK x, i.e. the statement holds for n = 1. Now suppose it holds for a certain n. Then, by the Lipschitz condition, f [t, (A n u)(t)] f [t, (A n v)(t)] KC (K t )n. n! So (for x > ), (A n+1 u)(x) (A n+1 v)(x) x = {f [t, (A n u)(t)] f [t, (A n v)(t)]} dt 2
(Similarly for x <, with x instead of x.) Proof of the Theorem. For the moment, assume that I is a bounded, closed interval. Then x R (say) for x I, and continuous functions are bounded on I, so there exists M such that f(t, ) M for t I. Uniqueness. Suppose that Au = u and Av = v. Then A n u = u and A n v = v for all n. Also, there exists C such that u(x) v(x) C for x I. By Lemma 3, u(x) v(x) C (K x )n n! for all n 1. But for each x, (K x ) n /n! as n. Hence u(x) = v(x). Existence. Let y =, y 1 = A(y ), y 2 = A(y 1 ),..., y n = A n (y ),.... Then y 1 (x) y (x) = y 1 (x) = f(t, ) dt M x C for x I, where C = MR. By Lemma 3, y n+1 (x) y n (x) for x I. Now n=1 (KR)n /n! is convergent (to e KR ), so by the M-test, n=1 (y n+1 y n ) is uniformly convergent on I (say to y). But (y 1 y ) + (y 2 y 1 ) + + (y n y n 1 ) = y n y = y n, so this says that y n y uniformly on I. In other words, y n (x) y(x) δ n for x I, where δ n as n. By the case n = 1 in Lemma 3, y n+1 (x) (Ay)(x) = (Ay n )(x) Ay(x) K x δ n for x I, so for x I, as required. (Ay)(x) = lim n y n+1 (x) = y(x) 3
Finally, suppose that I is an open or unbounded interval. Then we can express it as n=1i n, where I 1 I 2 I 3... are bounded, closed intervals with I 1. By the above, there is for each n a unique solution y (n) on I n with y (n) () =. By this uniqueness, y (n+1) agrees with y (n) on I n, so it is consistent to define y on I by: y(x) = y (n) (x) for x I n. Clearly, y satisfies the equation on I. We now give two quite simple examples to show that both parts of the theorem can fail if the Lipschitz condition is not satisfied. Example 1. Consider the equation y = y 2, with y(1) = 1, on the interval [ 1, 1]. Solving by elementary methods, we have 1 y 2 y = 1, 1 y = x + c. The condition y(1) = 1 selects the solution y = 1/x. This is clearly the unique solution on (, 1]. It cannot be extended to a differentiable function at, so there is no solution on [ 1, 1]. To see that the Lipschitz condition fails, note that f(x, y) = y 2. We have f(x, y + 1) f(x, y) = (y + 1) 2 + y 2 = 2y 1, which is unbounded on R. Example 2. Consider y = 3y 2/3 with y() =, on I = R. Clearly, is one solution. Elementary methods give 1 3y 2/3 y = 1, y 1/3 = x + c, y = (x + c) 3. Hence x 3 is another solution with y() =. There are actually infinitely many solutions: for each c >, a solution is y c (x) = { (x c) 3 for x c for x < c. (Then y c is differentiable at c, with derivative ). The Lipschitz condition fails because y 2/3 y = 1 y 1/3 as y +. 4
Picard s iteration. The proof of Picard s theorem provides a way of constructing successive approximations to the solution. With the initial condition y(x ) = y, this means we define y (x) = y and y n (x) = (Ay n 1 )(x) = y + x f[t, y n 1 (t)] dt. Example. Consider the equation y = y with y() = 1. Of course, the solution is e x. Picard s iteration gives the following: f(x, y) = y, hence y (x) = 1, y 1 (x) = 1 + f(t, 1) dt = Example. The equation y = 2x + y 2, with y() =. Start with y (x) =. Then Systems of equations and higher order equations Now consider a system of first-order equations y j(x) = f j [x, y 1 (x),..., y n (x)] (1 j n). In vector notation, y(x) = [y 1 (x),..., y n (x)], and the system becomes y (x) = f[x, y(x)]. By a vector adaptation of the proof, we shall obtain a generalization of Theorem 1 to systems of equations. As we have seen before, higher order equations can be seen as a special case of a system. For y = (y 1, y 2,..., y n ) R n, write y = max y j. This is a norm on R n, giving x y as a notion of generalized distance between x and y. Note that x y M simply means that x j y j M for each j. 5
Remark. Another norm is y 2 = ( n j=1 y2 j ) 1/2. The two norms are equivalent in the sense that y y 2 n y for all y. This means that either can be used to determine convergence. Though 2 is arguably the true measure of distance, it is more complicated to use in proofs like those that follow. Our y is often denoted by y. Our theorem is: THEOREM 2. Suppose that each f j is continuous on I R n and f j (x, y) f j (x, z) K y z on I R n. Suppose that x I and real numbers α 1,..., α n are given. Then there is a unique vector function y(x) such that y (x) = f[x, y(x)] for x I and y j (x ) = α j for each j. As before, it is sufficient to consider the case where x = α j = for all j. Exactly as in Lemma 2, the required conditions are equivalent to or in vector notation y j (x) = f j [t, y(t)] dt y(x) = f[t, y(t)] dt. (j = 1,..., n), Given y = (y 1,..., y n ), where each y j is in C(I), define Ay to be z, where We require y such that Ay = y. z(x) = f[t, y(t)] dt. LEMMA 3. Let u, v be vector functions such that u(x) v(x) C for x I. Then for each k 1, (A k u)(x) (A k v)(x) C (K x )k k! for x I. Proof. Just like Lemma 3, with the notation adjusted. We describe the induction step. Assume the statement holds for a certain k. Then, by the Lipschitz condition, for each j: [ fj t, (A k u)(t) ] [ f j t, (A k v)(t) ] (K t ) k KC. k! Write A k+1 u = (g 1,..., g n ) and A k+1 v = (h 1,..., h n ). This means that g j (x) = f j [t, (A k u)(t)] dt 6
and similarly for h j (x) with v instead of u. Exactly as in Lemma 3, we deduce that for x I, which is equivalent to g j (x) h j (x) C (K x )k+1 (k + 1)! (j = 1,..., n), (A k+1 u)(x) (A k+1 v)(x) C (K x )k+1 (k + 1)!. Theorem 2 now follows exactly as in the case n = 1. Let y () = and y (k) = (y (k) 1,..., y n (k) ) = A k (). (Here the notation y (k) is being used temporarily for the kth function of the sequence, not the kth derivative.) As before, we see that for each j, the functions y (k) j tend to a limit function y j, uniformly on I, as k (in other words, y (k) y uniformly on I), and Ay = y. THEOREM 3. Suppose that f is continuous on I R n and f(x, y) f(x, z) K y z on I R n. Suppose that x I and real numbers α 1,..., α n are given. Then there is a unique function y on I such that y (n) (x) = f[x, y(x), y (x),..., y (n 1) (x)] for x I and y(x ) = α, y (j) (x ) = α j for 1 j n 1. Proof. The equation is equivalent to the system y j = y j+1 for j n 2, y n 1 = f(x, y, y 1,..., y n 1 ). In the notation of Theorem 2 (with j now ranging from to n 1), we have f n 1 = f, while for j n 2, f j (x, y, y 1,..., y n 1 ) = y j+1. We only need to verify that the functions f j satisfy the Lipschitz condition. This is trivial for j n 2, and ensured by our hypothesis for j = n 1. Finally, we specialize this to linear equations, showing that the Lipschitz condition is then automatically satisfied. 7
THEOREM 4. Suppose that the functions p, p 1,..., p n 1 and g are continuous on an interval I. Let x I and real numbers α 1,..., α n be given. Then there is a unique function y such that y (n) (x) = p (x)y(x) + p 1 (x)y (x) + + p n 1 (x)y (n 1) (x) + g(x) for x I and y(x ) = α, y (j) (x ) = α j for 1 j n 1. Proof. First assume that I is a bounded, closed interval (then extend to open and unbounded intervals as before). Then each p j is bounded on I: say p j (x) K for x I. Apply Theorem 3 with n 1 f(x, y, y 1,..., y n 1 ) = p j (x)y j + g(x). Since it is a sum of products of continuous functions of single variables, f is continuous on I R n. The Lipschitz condition is satisfied, because: n 1 n 1 f(x, y) f(x, z) = p j (x)(y j z j ) K y j z j nk y z. j= j= j= 8