# Showing Global Lipschitz $\implies$ Global Solution ODE

Let $$f$$ satisfy a global Lipschitz condition in the second argument. $$\|f(x,y)-f(x,y^*)\|_2\le L\|y-y^*\|_2$$ Show every initial value problem has exactly one global solution. (Show the maximum existence interval is $$\mathbb{R}$$).

I'm trying to show this by showing that the operator $$(Ay)(x):=y_0+\int_{x_0}^xf(t,y(t))dt$$ is a contraction and therefore must have a fixed point, which would be our solution. With rather elementary transformations I got: $$\|(Au)(x)-(Av)(x)\|_2\le (x-x_0)L\|u-v\|_2$$ Heres my problem: If I limit x such that $$(x-x_0)\le L^{-1}$$ I have a contraction and therefore I have shown there is a solution. But know I've limited my maximum existence interval to $$[x_0,x_0+L^{-1}]$$.

My two questions:

1. Can I stich together smaller existence intervals? How do I do it? My idea would be to start a "new" initial value problem at the edge of the last existence interval, is that formally possible?
2. If I can stich them together to show there is a global solution, what about the solution on the interval $$(-\infty,x_0)$$ to the left of $$x_0$$?

First, you have established that for any $$x_0,y_0$$ that there is a unique solution $$y$$ defined on an interval $$(x_0-{1 \over L} , x_0+{1 \over L})$$ that satisfied the ODE and $$y(x_0) = y_0$$. Note that this interval includes points for $$x.
If you choose the collection of points $$t_k = k \delta$$ where $$0<\delta < { 1\over L}$$ then at least one of those points will lie in the above interval and you can repeatedly extend the solution (in both directions) to define a solution $$y$$ on $$\mathbb{R}$$. Any other solution $$\tilde{y}$$ passing through $$x_0,y_0$$ must equal $$y$$ on the first interval, then the neighbouring intervals, etc, etc. and hence $$\tilde{y} = y$$.
Another approach is to note that if $$A^n$$ is a contraction, then $$A$$ has a unique fixed point, and it is not hard to show that $$A^n$$ has a Lipschitz rank of no more than $${(x-x_0)^n \over n!}L$$, so we can show that there is a unique solution on any bounded interval containing $$x_0$$. This unique solution must also be the unique solution on any larger interval, hence this defines a solution on all $$\mathbb{R}$$.
Finally, even if you only define a solution for $$x\ge x_0$$, note that the same analysis applies to the backwards equation $$z'=-f(x,z)$$, and the resulting solution, when run backwards satisfies the original equation. So you can extend the solution to $$x in this way.
• To anyone else reading: by swapping the bounds of integration you can swap $x$ and $x_0$ in the second inequality of the question, thereby showing that the interval extends to the left of $x_0$ Commented Nov 23, 2019 at 20:27