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Show that the non-linear integral equation

$v(x)=\cos^2(x)+\int_0^x e^{-v^2(s)}ds, \ x\in [0,\infty)$

has a solution in $C^1([0,\infty))$.

In previous questions of this sort, we have been able to use the contraction mapping theorem, but are having difficulty using it here due to the infinite domain and haven't been able to find a contraction yet.

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    $\begingroup$ You definitely can use a fixed point argument, but you'd really be reproving an existence/uniqueness theorem for your differential equation. Let $B=C^1([0,\infty))$ and your norm be $|| \cdot || = \max_{[0,\infty)}{| \cdot |}$, you should be able to show that $||Tv - T\bar{v}|| < L||v-\bar{v}||$ with $L<1$. Another way to show this is to turn your integral equation into a initial value problem and show that your right hand side, $f(x,v)$ will be lipschitz in this interval in $v$. $\endgroup$
    – DaveNine
    Aug 12, 2015 at 17:05
  • $\begingroup$ @DaveNine I'd rather say "convert integral equation to differential and show that solution doesn't blow up in finite time", but mostly I agree with you. $\endgroup$
    – Evgeny
    Aug 12, 2015 at 17:07
  • $\begingroup$ This is an inactive question, but I was working on essentially the same problem and couldn't get it. I don't know if contraction mapping will work since $C^1([0,\infty))$ is not complete w.r.t. sup norm . . . Could you explain further how this f from an initial value problem Lipschitz in v? I am stuck. $\endgroup$
    – user288742
    Jul 21, 2016 at 7:42

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I would try to argue that you can solve over an interval $[0,a]$ by choosing $a$ small enough. Then continue by solving $$ v(x) = \cos^{2}x + \int_{0}^{a}e^{-v^{2}(s)}ds+\int_{a}^{x}e^{-v^{2}(s)}ds \\ = \cos^{2}x+A+\int_{a}^{x}e^{-v^{2}(s)}ds. $$

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  • $\begingroup$ Yes, I think something like this might work, though I don't have it rigorously yet. A key step in solving it on $C([0,1])$ was showing $e^{-x^2}$ is Lipshitz with constant $\dfrac{1}{\sqrt{2}}$ on $x \in [0,\infty)$ $\endgroup$
    – NaiveHalmos
    Aug 13, 2015 at 2:00

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