Consider a differential equation such as the eigenvalue problem
$$D f(x,y)=E f(x,y)$$
where $D$ is a second order differential operator in $x$ and $y$, (take for instance $D=a\frac{d^2}{dx^2}+b\frac{d^2}{dy^2}+c$ with constants $a,b,c$), and $E$ is an eigenvalue constant.
I am interested in finding solutions to this eigenvalue problem of the form
$$f(x,y)=\delta(x-y)\tilde f(x,y)$$
where $\delta(x)$ is the Dirac Delta. An issue that I notice is that the action of $D$ on such $f$ produces derivatives of the Dirac Delta, which are not present on the right hand side. This is a bit confusing. Have such problems been studied in mathematics? Any hint how to tackle this problem?
I think the following solves it.
Expand the Dirac Delta in its integral representation
$$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty dz \, e^{i z x}$$
then each derivative acting on $\delta$ in $\delta(x-y)f(x,y)$ effectively pulls down a factor of $z$. Drop the $z$ integration and the common factor $e^{i z (x-y)}$ for the time being. Since $f(x,y)$ does not depend on $z$, coefficients of all different powers of $z$ must vanish independently.
Another way to say this is:
Coefficients of all different derivatives of Dirac Delta must vanish independently, so
$$\delta (x-y) \left(a f^{(2,0)}(x,y)+b f^{(0,2)}(x,y)+(c-E) f(x,y)\right)=0\\ \delta '(x-y) \left(2 a f^{(1,0)}(x,y)-2 b f^{(0,1)}(x,y)\right)=0\\ \delta ''(x-y) (a f(x,y)+b f(x,y))=0$$
Third equation is satisfied only if $b=-a$, then second equation tells us that $f(x,y)=f(x-y)$, and finally first equation implies that $f(x-y)=0$, or in case of $c=E$ the function $f(x-y)$ can be arbitrary.
What do you think about the above? Can you see anything wrong with this discussion?