MA18 ODE: Picard s Theorem Preeti Raman IIT Bombay MA18
Existence and Uniqueness The IVP s that we have considered usually have unique solutions. This need not always be the case. MA18
Example Example: Consider the IVP dy dx = y 1 3 ; y() =. This is a separable first order DE. Hence, separating the variables, we get: y 1 3 dy = dx; i.e., 3 2 y 2 3 = x +c, or [ ]3 2 2 y = (x +c). 3 The initial condition y() = gives c =. Hence [ ]3 2x 2 y = φ(x) = ; x 3 is a solution of the given IVP. MA18
Example continued [ ]3 2x 2 y = φ(x) = ; x 3 is a solution of the IVP dy dx = y 1 3 ; y() =. Check: [ ]3 2x 2 y = φ(x) = ; x 3 is also a solution of the IVP. Check: y = ψ(x) is also a solution of the IVP. Check: For any a >, { if x [,a) y = φ a (x) = ± [ 2 3 (x a)]3 2 if x a is continuous, differentiable, and gives a solution of the given IVP. MA18
Existence and Uniqueness That is, we get infinitely many solutions of the given IVP. MA18
Example Example: Consider the IVP dy dx = 3x2 +4x +2 ; y() = 1. 2y 2 This is a separable non-linear first order DE. Hence, separating the variables, we get: Integrating this equation, we get (2y 2)dy = (3x 2 +4x +2)dx. y 2 2y = x 3 +2x 2 +2x +c. The initial condition y() = 1 implies that c = 1. Hence i.e., (y 1) 2 = x 3 +2x 2 +2x, y = 1± x 3 +2x 2 +2x. This equation gives us two functions satisfying the given IVP. MA18
Existence and Uniqueness Consider the IVP y 1 = f(t,y); y(t ) = y. We study the existence and uniqueness of solutions for the above IVP. By a linear translation of coordinates, we may assume that the initial condition given is y() =. Theorem Consider the IVP Suppose f and f y y 1 = f(t,y); y() =. are continuous on the rectangle a t a, b y b. Then there is an interval I = [ h,h] [ a,a] and a function y = φ(t) : I R which is the unique solution of the IVP. MA18
Existence and Uniqueness Note: Compare the example given earlier with the hypotheses of the above theorem. MA18
Example Example: Solve the IVP: y 1 = y 2 ; y() = 1 and find the interval in which the solution exists. As f(t,y) = y 2 & f y = 2y are continuous everywhere, the IVP has a unique solution on an interval containing t = by the above theorem. The given DE is separable and we have dy y 2 = dt. Thus, 1 y = t +c; i.e., y = 1 t +c. MA18
Example continued The initial condition y() = 1 gives c = 1. Hence y(t) = 1 1 t is the solution to the given DE. This function is unbounded as t 1. Hence, y : (,1) R defined by is the solution to the given IVP. y(t) = 1 1 t MA18
Corollary Consider the IVP y 1 +p(t)y = g(t); y(t ) = y, where p and g are continuous functions on an interval I with t I. Then there is a unique solution on I of the given IVP. Proof. Since y 1 = p(t)y g(t), it follows that f(t,y) = p(t)y g(t) and f y are both continuous on I R. By the existence and uniqueness theorem, the given IVP has a unique solution on a subinterval J I with t J. MA18
Picard s Iteration Method Picard s iteration method gives us a rough idea on how to construct solutions to IVP s. Consider the IVP y 1 = f(t,y); y() =. Suppose y = φ(t) is a solution to the IVP. Then, dφ dt = f(t,φ(t)), φ() =. That is, φ(t) = f(s,φ(s))ds; φ() =. The above equation is called an integral equation in the unknown function φ. MA18
Picard s Iteration Method Conversely, if the integral equation holds i.e., φ(t) = f(s,φ(s))ds; φ() =, then by the Fundamental Theorem of Calculus, dφ dt = f(t,φ(t)), so that y = φ(t) is a solution to the IVP y 1 = f(t,y); y() =. Thus, solving the integral equation is equivalent to solving the IVP. MA18
Picard s Iteration Method Picard s iteration describes a way to look for solutions of the integral equation φ(t) = f(s,φ(s))ds. We define iteratively a sequence of functions φ n (t) for every integer n as follows: Let φ (t) φ 1 (t) = φ 2 (t) = φ n+1 (t) = f(s,φ (s))ds f(s,φ 1 (s))ds f(s,φ n (s))ds. MA18
Picard s Iteration Method Note: Each φ n satisfies the initial condition φ n () =. None of the φ n may satisfy y 1 = f(t,y). Suppose for some n, φ n+1 = φ n. Then, φ n+1 = φ n = f(s,φ n (s))ds, and this implies d dt (φ n(t)) = f(t,φ n (t)) is a solution of the given IVP. In general, the sequence {φ n } may not terminate. In fact, all the φ n may not even be defined outside a small region in the domain. However, it is possible to show that, under the hypotheses of the above theorem, the sequence converges to a function φ(t) = lim n φ n(t) which is the unique solution to the given IVP. MA18
Example Example: Solve the IVP: y 1 = 2t(1+y); y() =. The corresponding integral equation is φ(t) = Let φ (t). Then, 2s(1+φ(s))ds. φ 1 (t) = φ 2 (t) = φ 3 (t) = 2sds = t 2, 2s(1+s 2 )ds = t 2 + t4 2, 2s(1+s 2 + s4 2 )ds = t2 + t4 2 + t6 6. MA18
Example continued We claim: φ n (t) = t 2 + t4 2 + t6 6 Use induction to prove this: +...+ t2n n!. φ n+1 (t) = = 2s(1+φ n (s))ds 2s (1+s 2 + s4 2 = t 2 + t4 2 + t6 6 ) s2n +...+ ds n! +...+ t2n n! + t2n+2 (n+1)!. Hence φ n (t) is the n-th partial sum of the series k=1 t 2k k!. MA18
Example continued Recall that φ n (t) is the n-th partial sum of the series Applying the ratio test, we get: t 2k+2 (k +1)! k! t 2k = t2 k +1 for all t as k. Thus, lim φ n(t) = n k=1 t 2k k! = et2 1. Hence, y(t) = e t2 1 is a solution of the IVP. k=1 t 2k k!. MA18
Example continued Uniqueness of solution: Suppose the IVP y 1 = 2t(1+y),y() = has two solutions φ and ψ.thus, both these satisfy the integral equation as well, i.e., φ(t) = It follows that 2s ( 1+φ(s) ) ds and ψ(t) = φ(t) ψ(t) 2s ( φ(s) ψ(s) ) ds. 2s ( 1+ψ(s) ) ds. Suppose we restrict the solutions to the interval t A/2, for an arbitrary constant A. Then φ(t) ψ(t) 2s φ(s) ψ(s) ds A φ(s) ψ(s) ds. MA18
Example continued Let φ(t) ψ(t) A U(t) = φ(s) ψ(s) ds. φ(s) ψ(s) ds. Then U() =, U(t) for all t and U (t) = φ(t) ψ(t).so, Thus: or equivalently, that U (t) AU(t). e At U (t) e At AU(t) [e At U(t)]. Integrating from to, t we get d ( ) e At U(s) ds ds which implies that e At U(t) for all t. Hence U(t) and U(t), and so φ(t) ψ(t) for all t. MA18