Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From the texts I've used, the Green's function is of a problem is $G(x,y)$ such that $LG(x,y) = \delta(x-y)$. The fundamental solution is u(x) such that $Lu(x)=\delta(x)$. They seem to be used for the same purpose (solving inhomogenous odes/pdes), but I'm curious how they are related. For example, there is a general methodology for finding Green's functions for simple types of ODEs (i.e. regular Sturm-Liouville problems), but I'm having trouble finding such a methodology for finding fundamental solutions. Also, are the Green's functions and fundamental solutions directly related in some way? I was talking to a friend and his idea was that the Green's function had the convolution built into it (since you use it as a kernel), whereas the fundamental solution is kind of more basic than that since you convolve it with the inhomogenous part.

share|improve this question
    
Mk, well I'm thinking that if you have the fundamental solution, you can just substitute $x \rightarrow x-y$ to get the Green's function, $G(x,y) = u(x-y)$. But going the reverse direction I'm not sure about since Green's functions don't always take the form of $h(x-y)$. –  Hanmyo Dec 19 '13 at 9:22

1 Answer 1

The difference shows up in operators which are not invariant under translation. For an invariant operator such as the Laplacian, the solution of $\nabla^{2}L(x)=\delta(x)$ can be translated to give a solution of $\nabla^{2}L_{y}(x)=\delta(x-y)$ by setting $L_{y}(x)=L(x-y)$. But as soon as you add a potential such as $1/|x|$ to $-\nabla^{2}$ (such as for an atomic Hamiltonian) then the fundamental solution does not generate the Green function. Closed-form Green functions are not going to be so easy to find in general, even though asymptotics may be realistic.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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