2
$\begingroup$

I am trying to numerically evaluate a Greens function for this equation:

$$ \left[\frac{\partial^2}{\partial x^2} + f(x) \right] G(x) = \delta (x-x_0) $$

With Neumann boundary conditions. Here, the function $f(x)$ is known. The difficulty in solving this lies in the numerical approximation of the $\delta$ function. Rather than approximate $\delta \sim \frac{1}{x}$ I attempted to resolve the $\delta$ by integrating both sides in a small interval around $x_0$, but this leaves me with conditions on the derivative of $G$ at $x_0$ that I don't know how to implement in a finite elements method.

I would really appreciate a reference for solving such equations (preferably with a finite elements method), or if someone could lend a guiding hand for solving such an equation.

$\endgroup$
  • $\begingroup$ What is your domain? Is it the whole real line? If that's the case, why do you need an $x_0$? $\endgroup$ – nicoguaro Feb 2 '17 at 21:11
  • $\begingroup$ The domain is bounded: $x\in [0,L]$ with $0<x_0<L$. The conditions of $G$ on the boundary are Neumann: $G'(0)=C_1$ and $G'(L)=C_2$ where both $C_1$ and $C_2$ are known. $\endgroup$ – alexvas Feb 2 '17 at 21:17
  • $\begingroup$ I suppose that depending on $f(x)$ you can write your operator as a Sturm-Liouville operator, and find the eigenvectors, then express $G(x)$ as an eigenvector expansion. $\endgroup$ – nicoguaro Feb 2 '17 at 21:29
  • $\begingroup$ I'm not sure I understand. To turn this into an eigenfunction equation I need to have $G$ on both sides. I could move $f(x)G(x)$ over to the RHS, but what do I do with the $\delta$? I don't see how to rewrite this as a SL equation. $\endgroup$ – alexvas Feb 2 '17 at 21:56
  • $\begingroup$ No, check this. And also this. $\endgroup$ – nicoguaro Feb 2 '17 at 21:59
2
$\begingroup$

Why approximate the delta function? In the finite element method, you need to evaluate the weak form of the equation, for which the right hand side will look like this: $$ \int \varphi_i(x) \delta(x-x_0), $$ where $\varphi_i$ is the $i$th shape function. But you know what this integral is: $$ \int \varphi_i(x) \delta(x-x_0) = \varphi_i(x_0). $$ Of course, most shape functions will be zero at $x_0$, but the shape functions defined on the cell that contains $x_0$ will not, and those are going to be the only nonzero entries in your right hand side vector.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.