# Green's Function for Differential Equation $t\dfrac{d^2}{dt^2} - \dfrac{d}{dt}$

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

$$$$G_a(t) = \begin{cases}At^2 + B &ta\end{cases}$$$$

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.

• 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. – MeowBlingBling Mar 21 '19 at 2:53
• Do note that $t=0$ is a singular point of the equation, so you can't easily enforce continuity on $(-1,1) – Dylan Mar 21 '19 at 3:44 • With$a=0\$, after enforcing the boundary conditions, are there any other tricks to solve for the actual value of the variables? – 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 $$u(1) = -\frac{8}{3}\not= 0$$).