1
$\begingroup$

Problem:

Let $p$ and $q$ be two integers, and $q > p>0$. Does the following ODE have a general solution on some finite time interval $[0,T]$? If yes, how can I obtain the solution?

$$ \dot x(t) = (x(t))^{p/q} + u(t), $$ where $x(0)=x_0>0$, and $0<u(t)<b<+\infty$ for all $t$.

Based on the problem formulation, $x(t)$ for $t\in[0, T]$ should be bounded. If it is not possible to find a general solution, can we find a tight bound of $|x(t)|$ in terms of $|u(t)|$ and $x_0$?


Here is my thinking:

As the right hand side of ODE satisfy the global Lipschitz condition ($x_0 > 0$) in $x$, so the ODE has a unique solution.

Denote $f(x)=x^{p/q}$. For any constant $\epsilon>0$, $|f(x)|\leq L(\epsilon)|x|+\epsilon$, where $L$ is a constant which depends on $\epsilon$. Therefore, $$ |\dot x(t)| \leq L|x(t)| + \epsilon + b, $$ which implies that $|x(t)|\leq e^{Lt}x_0+\frac{1}{L}(e^{Lt}-1)(\epsilon+b)$. In this way, the bound of $x$ depends on $\epsilon$ and $L$, which are not related to the problem. But I wish to find a more accurate bound that only depends on $x_0$ and $u$. I am not sure if it is possible to do so. Anyone can help?

$\endgroup$
  • 1
    $\begingroup$ Your $L$ also depends on $x_0$ (try $x_0:=0$). $\endgroup$ – Loïc Teyssier Nov 17 '13 at 13:14
  • $\begingroup$ @LoïcTeyssier Thanks for pointing out this mistake in my analysis. Indeed, $L$ depends on the initial condition $x_0$. $\endgroup$ – Zhang Changhe Nov 18 '13 at 2:22
4
$\begingroup$

I understand that your ODE is one-dimensional: in that case you can avoid Lipschitz continuity and replace it by transversality: The autonomous equation $$ \dot x =f(x),\quad x(0)=x_0, $$ has a unique local solution provided $f$ is continuous and $f(x_0)\not=0$. Peano's theorem provides existence whereas uniqueness follows follows from separability of the equation, which can be written as $$ \frac{dx}{ f(x)} =dt. $$ I know that your problem is not autonomous, but the same direct integration could be used: existence and uniqueness of a positive $C^1$ solution on $[0,T)$ follows from the classical Cauchy-Lipschitz theorem. We have $$ x(t)-x_0=\int_0^t (x(s)^{p/q}+u(s))ds=R(t), \tag 1 $$ so that $ \dot R=x^{p/q}+u\le \bigl(R+x_0\bigr)^{p/q}+b. $ The Gronwall reasoning shows that the solutions of this differential inequality are smaller that the unique solution of the ODE $$ \dot R=\bigl(R+x_0\bigr)^{p/q}+b, R(0)=0.\tag 2 $$ (2) has a unique solution since $x_0>0$ and is actually separable so can be integrated explicitly. As a result, a bound for $R$ provides a bound for $x$, thanks to (1).

$\endgroup$
  • $\begingroup$ Thanks so much for your clear and detailed explanation. I get your point. Your method renders a closed bound for $x(t)$. $\endgroup$ – Zhang Changhe Nov 18 '13 at 2:24

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.