I am studying the following non-local ODE $$\dot p(x) \nu_{\varepsilon, \alpha}(x) + \int_{x}^{2x_0}\frac{\dot p(s)}{s + \varepsilon} ds = c \quad \text{for } x \in [0,2x_0].$$ The number $x_0$ can be chosen arbitrarily close to 0 and we set $0 < \varepsilon \ll x_0$ as small as we want. Moreover, the function $\nu_{\varepsilon, \alpha}$ is defined by $$\nu_{\varepsilon, \alpha}(x) = \log\left(\frac{(x + \epsilon)^{1 + \alpha}}{|p(x)|^2}\right),$$ for some arbitrarily small constant $\alpha > 0 $. Finding an explicit solution to such an ODE seems quite hard as we have this $\log$ that introduces the non-linearity and makes everything much harder. To get rid of this integral making this equation non-local, I took the derivative on both sides of the equation so that we reach $$\ddot p(x) \nu_{\varepsilon, \alpha}(x) + \left(\dot \nu_{\varepsilon, \alpha}(x) - \frac{1}{x + \varepsilon}\right) \dot p(x) = 0,$$ with the derivative of $\nu_{\varepsilon, \alpha}$ given by $$\dot \nu_{\varepsilon, \alpha}(x) = \frac{1 + \alpha}{x + \varepsilon} - \frac{p(x) \dot p(x)}{|p(x)|^3}.$$ Therefore, the equation becomes $$\ddot p(x) \nu_{\varepsilon, \alpha}(x) + \left(\frac{\alpha}{x + \varepsilon} + \frac{p(x) \dot p(x)}{|p(x)|^3}\right)\dot p(x) = 0.$$ Solving the latter amounts to solve the desired non-local equation up to a constant. But as we are working with a non-linear ODE, I am not sure at all the we can even find an explicit solution to this equation. However, I would be interested in studying the property of the solution, such as the smoothness and see if the solution blows-up. My knowledge in ODEs being rather shallow, I was wondering if anyone here could provide me more information on how study that kind of problem.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Cross-posted on MSE: math.stackexchange.com/q/4783577/42969. $\endgroup$– Martin RCommented Oct 9, 2023 at 12:28
-
1$\begingroup$ See meta.mathoverflow.net/a/2638/116247 for some guidelines about cross-posting (wait some days, provide links, ...) $\endgroup$– Martin RCommented Oct 9, 2023 at 12:28
Add a comment
|