3
$\begingroup$

Suppose that we have a singular BVP of the form $$\dot{x}=\frac{f(x,t)}{t^n}, \qquad x(0)=x_0$$ where $x\in \mathbb{R}^n$, $n\in \mathbb{N}$.

There is of course no more uniqueness and the existence is not obvious as well. But what I would like to do is to consider instead a regularized version of the problem: $$\dot{x}=\frac{f(x,t+\varepsilon)}{(t+\varepsilon)^n}, \qquad x(0)=x_0$$ and to give enough information to describe the pointwise limit $\lim_{\varepsilon \to 0+}x(\varepsilon,t)$.

For example, I have found that for a one dimensional BVP problem of the form $$ \dot{x} = c - \frac{x^2}{(t+\varepsilon)^2}, \qquad x(0)=x_0, $$ $x(\varepsilon,t)$ will converge pointwise for $t>0$ to a unique solution of the above equation with $\varepsilon =0$ and boundary conditions $x(0) = 0$, $\dot{x}(0) = -(1+\sqrt{1+4c})/2$. The problem is that I was able to do this one as a very-very special case. For example, I am completely stuck if we add two extra equations like $$ \dot{x} = c - \frac{x^2}{(t+\varepsilon)^2}, $$ $$ \dot{y} = \frac{xy}{(t+\varepsilon)^2}, $$ $$ \dot{z} = d - \frac{y^2}{(t+\varepsilon)^2}. $$

So my question is: has anyone seen some similar problems and techniques to solve them in the literature? Any ideas or references would be very helpful.

$\endgroup$
  • $\begingroup$ Do they really converge even in the first case? Take $x_0=-1$, for instance. Then you have $\dot x=c-\frac{x^2}{(t+\varepsilon)^2}\le c-\frac{x^2}{\delta^2}\le -\frac 12\frac{x^2}{\delta^2}$ on $[0,\delta]$ if $\delta>0$ is small enough. So $(x^{-1})'\ge \frac{1}{2\delta^2}$ on $[0,\delta]$, but you have no room between $-1$ and $0$ to move that fast for so long time. So it just blows up like crazy with this initial value. What am I missing? $\endgroup$ – fedja Nov 30 '17 at 3:05
  • $\begingroup$ Oh, yes, sorry. You are totally right. We will have convergence if we compactify the real line and consider this equation as an ode on the projective line. Then it will converge if $1+4c\geq 0$, or else it just starts oscillating faster and faster on $\mathbb{P}^1$ and there is no limit. If we look only at $\mathbb{R}$, then solutions that start with $x_0< 0$ will blow up to minus infinity, as you have seen. The same story with the second system. It should be seen as an equation on a Grassmanian, but if we assume that $x_0>0$ then mathematica suggests that everything should be ok even in R^3. $\endgroup$ – Ivan Nov 30 '17 at 10:49

Your Answer

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

Browse other questions tagged or ask your own question.