Let $L$ be an operator and suppose the Green's function exists. That is there exist a function $G$ such that $LG=\delta$ where $\delta$ is the Dirac delta function. If $L$ is linear, one can represent the solution $u$ of the Poisson equation $Lu=f$ by $u(x)=\int_{\Omega}G(x,y)f(y)dy$ as discussed in the following link :https://en.wikipedia.org/wiki/Green%27s_function
My question is can we get the same representation of $u$ if $L$ becomes nonlinear?
Thank you very much in advance.
Suppose you have a function $G$ such that $LG=\delta$, with boundary condition $G=0$, and define $$ u(x)=\int_\Omega G(x,y)f(y)dy. $$
Then apply the operator $L$ to $u$ and get: $$ Lu(x)=L \int_\Omega G(x,y)f(y)dy. $$ If its true that $L \int_\Omega G(x,y)f(y)dy = \int_\Omega LG(x,y)f(y)dy,$ then $$ Lu(x)=\int_\Omega \delta(x,y)f(y)dy=f(x) $$ and $u$ solves the equation $Lu=f$ in $\Omega$. The question now is when it is possible to bring the operator $L$ inside the integral. If the operator is linear it can be easily done. If $L$ is nonlinear, it cannot be done in many cases.