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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 $x_0$?
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There are a few tricks you can use here.

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 $x<x_0$.

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 $x<x_0$ in this way.

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  • $\begingroup$ 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 $x_0$ $\endgroup$ Commented Nov 23, 2019 at 20:27

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