I have a Second Order Differential Equation $tu'' - u' = 1-t^2$. How do you find the Green's function $G_a(t)$ for any $a\in(-1,1)$? We define the Green's function $G_a(t)$ as a solution to $tG_a'' - G_a' = \delta(t-a)$

My Attempt

I found the Homogeneous solution $u(t) = At^2 + B$ for constants $A,B$. For a fixed $a$ the Green's function $G_a(t)$ should obey the Homogeneous solution on either side of $a$. So

\begin{equation} G_a(t) = \begin{cases}At^2 + B &t<a \\Ct^2 + D &t>a\end{cases} \end{equation}

Integrating the equation from $a-\epsilon$ to $a+\epsilon$ as $\epsilon\to 0$ lead to $D-B = \dfrac{1}{2}$. At this point I've tried to enforce continuity of the Green's function and the Jump in Derivative but no luck.

Any help is appreciated.

  • $\begingroup$ I was interested in boundary conditions that are $u(-1)=u(1)=0$. All four of these conditions can’t hold in this case. Also @Winther when a is 0 The continuity won’t hold at all. $\endgroup$ – MeowBlingBling Mar 21 '19 at 2:53
  • 2
    $\begingroup$ Do note that $t=0$ is a singular point of the equation, so you can't easily enforce continuity on $(-1,1) $\endgroup$ – Dylan Mar 21 '19 at 3:44
  • $\begingroup$ With $a=0$, after enforcing the boundary conditions, are there any other tricks to solve for the actual value of the variables? $\endgroup$ – MeowBlingBling Mar 21 '19 at 12:08

The boundary value problem $tu''(t) - u'(t) = f(t)$ with $u(-1) = u(1) = 0$ does not have a unique solution (or a solution at all) so your problem is ill defined and we can't find a Green's function for this problem.

To see more clearly why, note that we can find a general solution to the equation by using an integrating factor. We write the ODE as $\frac{d}{dt}\left(\frac{u'(t)}{t}\right) = \frac{f(t)}{t^2}$ which after integration, multiplication by $t$ and a final integration leads to

$$u(t) = D + \frac{t^2-1}{2}C + \int_{-1}^{t}t'\int_{-1}^{t'}\frac{f(t'')}{t''^2}\,{\rm d}t''\,{\rm d}t'$$

where $C$ and $D = u(-1)$ are integration constants and importantly we see that $C$ cannot be constrained by specifying $u(\pm 1)$. Imposing $u(-1) = u(1) = 0$ we see that a solution exists only if $f$ has the property that $$\int_{-1}^{t}t'\int_{-1}^{t'}\frac{f(t'')}{t''^2}\,{\rm d}t''\,{\rm d}t' = 0$$ and in that case any choice of $C$ is valid. In your case where $f(t) = 1-t^2$ then the condition above is not satisfied and there is no solution (if you try to solve the ODE with $u(-1) = 0$ then you will find $u(1) = -\frac{8}{3}\not= 0$).


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