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Let $ \phi : \mathbb R \to \mathbb R $ a continuous function. Prove that the initial value problem

$$ y' =y + \phi (x) e^{ -y^{2}} ; \quad y(0)=1$$

has a unique solution on the real line.

(ii) If $ y$ is the unique solution prove that

$$ | y(x) -e^{x} | \leq ( e^x -1) \max_{ 0 \leq t \leq x} | \phi (t) | \quad \forall x>0 .$$

I have prove that there is a unique solution but I can't prove the inequality in (ii).

Any help?

Thanks in advance!

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up vote 3 down vote accepted


  • Write down a differential equation for $u(x) = y(x) - e^x$.

  • For $0\leq t \leq x$, $$\left|\phi(t) e^{-y^2}\right| \leq \max_{ 0 \leq t \leq x} | \phi(x)|. $$

Spoiler below:

$$u' = u + \phi(x) e^{-y^2}, u(0)=0,$$ You can get the estimates for $u(x)$ from the fact that $u'=u + f(x)$ has the solution $$ u(x) =e^{x} \int_0^x e^{-t} f(t) dt $$ here $f(x) = \phi(x) e^{-y^2}$. We have $$\left|\int_0^x e^{-t} f(t) dt\right| \leq \max_{ 0 \leq t \leq x} | \phi(x)| \int_0^x e^{-t} dt = \max_{ 0 \leq t \leq x} | \phi(x)| (1- e^{-x}).$$ Thus, $$ |u(x)| \leq \max_{ 0 \leq t \leq x} | \phi(x)| (e^{x} -1).$$

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Can you explain how did you find $ u_{ \max} $...? – passenger Jan 7 '13 at 18:40
I hope it is now better understandable? – Fabian Jan 7 '13 at 18:41
Thank you very much for your time! Nice solution! Every step in now clear to me! – passenger Jan 7 '13 at 18:49
Just one small question: Is this a general way to attack problems like this one, where we need to estimate the quantity $ | y(x) - e^x|$ ( where $y$ is a solution to a initial valu problem)...? – passenger Jan 7 '13 at 18:52
Yes. See above :-) (write down an equation/solution for $y-e^x$ and estimate) – Fabian Jan 7 '13 at 18:57

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