Prove that the solution exists for all time.

Question

Suppose that $x^{\prime}=-x+f(x)$ where $f(x)$ is Lipschitz continuous and $0<f(x)<1$ for all $x \in R$. Prove that the solution exists for all time.

I am reading from Teschl ODEFrom the notation of page 37-38( I am basically using Picard Lindeloff theorem). If I choose any $T,\delta >>0$ then $M=max_{(t,x)\in V} |f(t,x)|\leq \delta+1$ and then $\frac{\delta}{M}=\frac{\delta}{\delta+1}$(approx) and then $T_0=\min\{T,\frac{\delta}{M}\}\leq 1$. Hence, I am stuck about how to show that the solution exists for all time.

I am confused about the fallacy here as the function $-x+f(x)$ is globally Lipschitz the solution should exist for all $t$. Please help me or give me a proof.

Answer 1

The Picard Lindeloff theorem guarantees you existence of a local solution, what you need is to use it to show that every solution can be extended unless it is already defined on whole $\mathbb R$.

Let me clarify one detail. Since you talk about the solution, then you need some initial condition $x(t_0)=x_0$, so we can talk about the initial value problem (IVP).

Let $\bar t$ be the supremum of all all $\hat t$ such that the IVP has a (unique) solution on $[t_0,\hat t]$. Our aim is to show that $\bar t = \infty$. Suppose to the contrary that $\bar t$ is finite. Then the IVP has solution on the interval $[t_0,\bar t - \varepsilon]$ for any $\varepsilon >0$. Applying the Picard Lindeloff theorem at the point $t_1 = \bar t - \varepsilon$ you show that the solution can be extended. All you need to check is that it is possible to choose $\varepsilon >0$ small enough so that Picard Lindeloff theorem guarantees you existence of solution beyond $\bar t$. Essentially it follows from the fact that $-x+f(x)$ is locally bounded, but I will skip the technical details. Once you show that the solution can be extended, you obtain a contradiction with that $\bar t<\infty$ was indeed the supremum of all $\hat t$.

Extending the solution "backwards-in time" is analogous.

Answer 2

I solved it in another way later. Suppose $f$ is globally Lipschitz with constant $C$ then, if we consider cor 2.6 page no 41 of the Teschl(I am referring the cor because I referred the book in the question)

$U=\Bbb R$ and $$L(t)=\sup_{x\neq y \in [-M,M]}\frac{|f(t,x)-f(t,y)|}{|x-y|} \leq 1+C$$

and hence, $$\int_{-T}^TL(t) \leq \int_{-T}^T(1+C)dt=(1+C)2T<\infty.$$

From the corollary $x(t)$ exists for all time $t$.