2
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

Hints:

  • 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, $$ |u(x)| \leq \max_{ 0 \leq t \leq x} | \phi(x)| (e^{x} -1).$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you explain how did you find $ u_{ \max} $...? $\endgroup$ – passenger Jan 7 '13 at 18:40
  • $\begingroup$ I hope it is now better understandable? $\endgroup$ – Fabian Jan 7 '13 at 18:41
  • $\begingroup$ Thank you very much for your time! Nice solution! Every step in now clear to me! $\endgroup$ – passenger Jan 7 '13 at 18:49
  • $\begingroup$ 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)...? $\endgroup$ – passenger Jan 7 '13 at 18:52
  • $\begingroup$ Yes. See above :-) (write down an equation/solution for $y-e^x$ and estimate) $\endgroup$ – Fabian Jan 7 '13 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.