Study of the properties of a non-local ODE

Question

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 $$\dot \nu_{\varepsilon, \alpha}(x) = \frac{1 + \alpha}{x + \varepsilon} - \frac{p(x) \dot p(x)}{|p(x)|^2}.$$ Therefore, the equation becomes $$\ddot p(x) \nu_{\varepsilon, \alpha}(x) + \left(\frac{\alpha}{x + \varepsilon} + \frac{p(x) \dot p(x)}{|p(x)|^2}\right)\dot p(x) = 0.$$ 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.

Answer 1

Partial answer. After toying around for a little bit I came up with a pretty promising substituation. If we define $$ p(x)=\exp\left(\frac{f(y)+(1+\alpha)y}{2B}\right)\\ y=A+B\ln(x+\epsilon) $$

your second order ODE can be significantly reduced to the following autonomos version that on top no longer involves $\epsilon$. $$ \Big(-A\alpha-A-2B-f(y)\Big)f'(y)^2+\Big(-2\alpha f(y)-2A\alpha^2+(-2A-2B)\alpha-4B\Big)f'(y)-2B\Big(A\alpha+A+f(y)\Big)f''(y)-(1+\alpha)((\alpha-1)f(y)+A\alpha^2-A+2B)=0 $$

Applying the standard procedure for autonomous equations $g(f(y))=f'(y)\Rightarrow g'g=f''$ we can integrate once and get

$$ f'(y) = -1-\alpha+\frac{(A\alpha+A+f(y))^\alpha e^{f(y)/B}}{\int_0^{f(y)}((\alpha+1)A+s)^\alpha e^{s/B}\left(\frac{(1+\alpha)A+2B+s}{2B((1+\alpha)A+s)}\right)ds-C} $$

From here on I couldnt really get much further. Maybe a clever choice of $A,B$ simpliefies it? The obvious choice $A=0,B=1$ leads to:

$$ f'(y) = -1-\alpha-\frac{ 2e^{f(y)}}{ (-f(y))^{-\alpha} \left(2 \Gamma(\alpha, -f(y)) - \Gamma(1 + \alpha, -f(y))\right)-2C} $$

which is seperable but obviously not easily solved in a closed form. The only thing i could manage is the case $\alpha=1$ which involves the dilog operator.

Are there any specific initial conditions you are interested in?