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AFAIK, a Green's function $G(x,\epsilon)$ has a singularity for $x=\epsilon$. This is clear in many analytical expressions for $G(x,\epsilon)$.

I do not understand, however, what happens to the singularity when $G(x, \epsilon)$ is expanded in terms of the linear operator eigenfunctions:

$G(x, \epsilon) = \sum_i \frac{u_i(x)u_i(\epsilon)}{\lambda_i}$.

For example, for rectangular-like domains and the dirichlet laplacian, eigenfunctions $u(x)$ are sines. How can I show that the series of $G(x, \epsilon)$ is still singular at $x=\epsilon$?

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  • $\begingroup$ The singularity can arise from the condition that the set of eigenfunctions is complete. The complete set is defined by the condition $\sum_{i}u_i(x)u_{i}(y)=\delta(x-y)$ - delta -function. If we have an operator $P$, (full) set of its eigenfunctions $u_i(x)$ and eigenvalues $\lambda_i$, we can build a function $G(x,y)=\sum_{i}\frac{1}{\lambda_i}u_i(x)u_{i}(y)$ such that $P\dot{G(x,y)}=\sum_i\lambda_i\frac{1}{\lambda_i}u_i(x)u_{i}(y)=\sum_{i} u_i(x)u_{i}(y)=\delta(x-y)$ $\endgroup$
    – Svyatoslav
    Commented Feb 1, 2021 at 15:02
  • $\begingroup$ Thanks. I see the singularity in $\sum_i u_i(x)u_i(y) = \delta(x-y)$, but how does that imply that $\sum_i \frac{u_i(x)u_i(y)}{\lambda_i}$ is also singular for $x=y$? $\endgroup$
    – zooond
    Commented Feb 1, 2021 at 16:38

1 Answer 1

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I found an answer in Courant & Hilbert's book pgs. 351-2.

The Green's function itself may be continuous and well-defined at $x=\epsilon$. However, the Green's function is singular in the sense that its first derivative does not exist at $x=\epsilon$.

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