# Green's function

1. Apr 26, 2008

### Semo727

Hello! I have problem with my homework, but what I'm going to ask you is not homework problem so I hope it is OK I'm writing it here :)

I need to find Green's function for differential operator
$$L=a\frac{d^2}{dx^2}+b\frac{d}{dx}+c$$
i.e. find solution for differential equation equation
$$LG(x,y)=\delta(x-y)$$
I'm looking for solution in interval $$[0,1]$$ and boundary conditions: $$G(0,y)=0, \frac{dG(1,y)}{dx}=0$$. Well, my problem is, that I don't know what conditions the derivative $$\frac{dG(x,y)}{dx}$$ must satisfy for $$x\rightarrow y$$ from the left and from the right. I just know, that $$G(x,y)$$ must be continuous function (when we look at it as function in x argument) also in the point y.

On wikipedia there is said, that there should be derivative "jump". i.e.
$$\frac{dG(x,y)}{dx}\vert_{y^+}-\frac{dG(x,y)}{dx}\vert_{y^-}=1/p(y)$$, where $$p(y)$$ comes from the Sturm-Liouville differential operator which they consider, i.e. they just examine the case
$$L=\frac{d}{dx}(p(x)\frac{d}{dx})+q(x)$$

So my question is: what are the conditions for the derivative jump in case of linear operator above? Or if there are none, how can I find Green function satisfying $$LG(x,y)=\delta(x-y)$$ as I have just three equations (comming from boundary problems and continuity) for four constants and these equations don't consider the delta function on the right side of the equation?

Thanks for your soon reply and I apologize for my very non-mathematical approach to that problem while formulating my questions, but this is homework problem from my physics classes and they didn't formulated theory of Green's functions very mathematicaly correctly so my all understanding of this problem is only intuitive.

2. Apr 26, 2008

### HallsofIvy

Staff Emeritus
Then put it in that (self adjoint) form! Remember finding "integrating factors" for linear first order equations? Do the same thing here. Imagine multiplying the entire equation by u(x) so you have $au(x)d^2x/dt^2+ bu(x)dx/dt+ cu(x)u= 0$. In order for that to be of the form $d(p(t)dx/dt)/dt+ q(x)= p(t)d^2x/dt+ p'(t)dx/dt+ q(x)= au(x)d^2x/dt^2+ bu(x)dx/dt+ cu(x)= 0$ you must have p(t)= a u(x) and p'(t)= au'= bu. Integrating that equation, $u(t)= e^{(b/a)t}$. That is, you can write your equation as
$$\frac{d(e^{(b/a)t}\frac{dx}{dt}}{dt}+ ce^{(b/a)t}x= 0$$

3. Apr 26, 2008

### Semo727

Oh, yes, thanks. OK, so we can transfer the linear operator to desired form, but they are not the same. I mean, that
$$L=a\frac{d^2}{dx}+b\frac{d}{dx}+c\neq L'=\frac{d}{dx}(e^{(b/a)x}\frac{d}{dx})+ce^{(b/a)x}=Le^{(b/a)x}$$
So I think that also right side of the equation: $$\delta(x-y)$$ should be multiplied by that factor, so we have
$$L'G(x,y)=e^{(b/a)x}\delta(x-y)$$
So what doest it imply for derivative jump? That it should be $$e^{(b/a)y}/e^{(b/a)y}=1$$?

Last edited: Apr 26, 2008
4. Apr 26, 2008

### Semo727

oh, there is a mistake in mine (and also HallsofIvy's) post. $$L'$$ should be
$$L'=\frac{d}{dx}(\frac{ae^{(b/a)x}d}{dx})+ce^{(b/a)x}$$ and so the derivative "jump" should be probably
$$\frac{e^{(b/a)x}}{ae^{(b/a)x}}=1/a$$ which is also what one can guess also from the original linear operator
$$a\frac{d^2}{dx^2}+b\frac{d}{dx}+c$$
but what is interesting is, that the fact, that first derivative doesn't exist has no effect for this jump. I mean, that similar linear operator
$$a\frac{d^2}{dx^2}+c$$
would give the same jump $$1/a$$.
But I'm not sure at all that my thoughts about the jump $$1/a$$ are O.K., because they are all ONLY intuitive, i.e. unfortunately I don't know anything what is mathematically correct about distributions and functional analysis - that means anything about delta function.

Last edited: Apr 26, 2008