How to offset a known solution to the poisson equation?

Question

Given the (2D) poisson equation $\nabla^2u=x+C$ and given $x$ and a known solution $u$ and constant $C$, what is the simplest function $v=f(u)$, such that $\nabla^2v=x$ ?

In my concrete case everything is discretized, so the laplace operator uses a 2D 5-point stencil.

Answer 1

so are you asking if $u=u(x,y)$ and if $u_{xx}+u_{yy}=x+C$ then what $f$ satisfies $f_{xx}(u)+f_{yy}(u)=x$?

you can expand this by saying: $$f_{xx}(u)=\frac{\partial^2f}{\partial x^2}=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right]=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}\right]=\frac{\partial f}{\partial u}\frac{\partial^2u}{\partial x^2}+\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial u}\right]\frac{\partial u}{\partial x}=f_uu_{xx}+(f_u)_x$$ and doing the same for $f_{yy}$ we get: $$f_u(u_{xx}+u_{yy})+(f_u)_x+(f_u)_y=x$$ $$f_u(x+C)+(f_u)_x+(f_u)_y=x$$ now we have an equation of the form: $$F.g(x)+F_x+F_y=x$$

Answer 2

Since $\nabla^2(C(x^2+y^2)/4)=C$ we can simply offset $u$ by $C(x^2+y^2)/4$.