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?

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    $\begingroup$ 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

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).

  • $\begingroup$ 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

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