Existence of unique solution of initial value problem

Question

Let us take initial value problem $y'+\frac{{2}}{{t}}y=4t$, $y(1)=2$ where $p(t)=\frac{{2}}{{t}}$ and $q(t)=4t$. Here $q(t)$ is continuous for all $t$ while $p(t)$ is continuous only for $t<0$ or $t>0$. The interval $t>0$ contains the initial point; consequently theorem guarantees that the initial value problem has unique solution on the interval $0<t<\infty$, but if the initial condition $y(1)=2$ is changed to $y(-1)=2$ the existence of solution will be on the interval $-\infty<t<0$ containing the initial point.

Now my question is that is it necessary to look upon the interval which necessarily must contain initial point for the existence of solution of initial value problem? Is it valid or does there exist solution on those interval which doesn't contain initial point but those coefficients $p(t)$ and $q(t)$ are continuous for all the points contained in those interval?

Answer 1

You can have a unique solution in an interval different from the one over which $p(t)$ and $q(t)$ are continuous from the initial condition.

Let's look at your example. Using for instance an integrating factor of $t^2$ and integrating the resulting exact differential leads to the general solution $y=t^2+C/t^2$, and then $C=1$ from your initial condition. So you know that $y(-1)=2$ even though the initial condition is at $t=1$ and the singularity at $t=0$ is in the way.

Dance around the maypole

This case is better understood by considering the problem in complex variables. Not always, but often enough, when you go to complex variables the function is well-defined everywhere around the singularity. That is the case here. Starting at $t=1$ you can integrate along any path that dodges $t=0$ by going through some other point on the imaginary axis, arrive at $t=-1$ and the value of your function, in this case, is uniquely $y=2$. In effect, $t=-1$ is in the right interval after all, because you can go around $t=0$ in the complex plane without anything else blocking you.

This does not always work, though, because sometimes when you go to the complex domain the singularity may not stay isolated, so your $y$ value outside the "right" interval depends on your integration path. If you had had $y'+(2/t)y=4/t^3$ instead of the actual $4t$ on the right, you would run into this problem. You learn about this when you study complex variables.