Consider the problem:
- $u_{t}=Ku_{xx}$
- $u_{x}(0,t)=g_{0}(t)$
- $u_{x}(L,t)=g_{1}(t)$
The question is "Transform this problem in other where the frontier conditions are homogeneous, suposing $g_{0}$ and $g_{1}$ differentiable."
What I was thinking is introducing the function $g(x,t)$ = $g_{0}x + \frac{(g_{1} - g_{0})}{L}\frac{x^{2}}{2}$ beacuse it satisfies the last two equations! So $u(x,t)$ is solution of this problem iff $w(x,t)=u(x,t)+g(x,t)$ is solution of:
- $w_{t}=Kw_{xx} + l(x,t)$ (where $l$ is a function dependent of $x$ and $t$)
- $w_{x}(0,t)=0$
- $w_{x}(L,t)=0$
Is this the best we can do? It would be amazing to find some special function $g$ such $u(x,t)$ is solution of this problem iff $w(x,t)=u(x,t)+g(x,t)$ is solution of:
- $w_{t}=Kw_{xx}$
- $w_{x}(0,t)=0$
- $w_{x}(L,t)=0$
Is this possible? Or the thing I did first is the best we can do in this situation? Thank you in advance.