# Singularity of Green's function and eigenfunction expansion

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 $$x=\epsilon$$?

• 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 $P\dot{G(x,y)}=\sum_i\lambda_i\frac{1}{\lambda_i}u_i(x)u_{i}(y)=\sum_{i} u_i(x)u_{i}(y)=\delta(x-y)$ Commented Feb 1, 2021 at 15:02
• 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 $x=y$? Commented Feb 1, 2021 at 16:38

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 $$x=\epsilon$$.