Let T > 0 and L greaterthanorequalto 0. Consider C [0, T], the space of all cont

Question

Let T > 0 and L greaterthanorequalto 0. Consider C [0, T], the space of all continuous real valued functions on [0, T], with the metric rho defined by rho (x, y) = sup_0 lessthanorequalto t lessthanorequalto T e^-Lt |x (t) - y (t)|. (a) Prove that (C[0, T], rho) is a complete metric space. (b) Consider the initial value problem (*) {x' (t) = f(t, x(t)) for t > 0, x(0) = x_0 where f: R times R rightarrow R is continuous and globally Lipschitz continuous with respect to x, ie, there exists L greaterthanorequalto 0 such that |f (t, x) - f (t, y)| lessthanorequalto L | x - y | for all t, x. y elementof R Construct an integral operator as in the proof of Theorem 13.9.2 in the notes and show that this operator is a contraction on (C[0, T], rho). Hence deduce that (*) has a unique solution in C^1 [0, T]1 (c) Deduce that (*) has a unique solution in C^1 [0, infinity).

Explanation / Answer

In (b) you have proven that (*) has unique solution in C1[0,T]. Now since that is true for arbitrary T. Consider that is not true for C1[0, infinity) then there is a largest such T such that the statement is true. So it contradicts with arbitrary T (non-existence of maximum T) . Hence done.

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