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Consider this problem

$$ \begin{cases} -\Delta u=10 \hspace{6mm} \mbox{in} \hspace{6mm} \Omega \\ u=0 \hspace{6mm}\mbox{on}\hspace{6mm}\Gamma_d \\ \frac{\partial u}{\partial n}=-\sqrt{4x^2+64y^2} \hspace{3mm} \mbox{on} \hspace{3mm} \Gamma_N \end{cases} $$ $$ \Omega=\{(x,y)\mid x^2+4y^2<16\}\\ \Gamma_d=\{(x,y)\in \partial \Omega\mid y\le 0\} \\ \Gamma_N=\{(x,y)\in \partial \Omega\mid y> 0\} $$

I want to convert above poisson equation to laplace equation. I know that the exact solution of the above equation is $$ u_e(x,y)=16-x^2-4y^2 $$ My idea for converting it would be to consider $u=u_0 + \widetilde{u}$ and $u_0=\frac{-5}{2}x^2-\frac{5}{2}y^2$. But I am not exactly sure of side effects of substituting this into the above equation. Will it change the boundary conditions? How so?
FreeMind
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    Yes, of course. You need to see what your correction term must now do on the boundary so that the combination will behave appropriately for the original problem. To do that you just subtract. – Ian Jul 17 '21 at 12:59

1 Answers1

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So first off your $\tilde{u}$ will in fact satisfy the Laplace equation if $u$ satisfies your Poisson equation.

As for the boundary conditions, recall that the goal is to have $u=0$ on $\Gamma_d$ and $\frac{\partial u}{\partial n}=-\sqrt{4x^2+64y^2}$ on $\Gamma_N$. But $\tilde{u}=u-u_0$.

So you find that $\tilde{u}$ needs to be equal to $0-u_0=\frac{5}{2} x^2+\frac{5}{2} y^2$ on $\Gamma_d$ and that $\frac{\partial \tilde{u}}{\partial n}$ needs to be equal to $-\sqrt{4x^2+64y^2}-\frac{\partial}{\partial n} \left ( -\frac{5}{2} x^2 - \frac{5}{2} y^2 \right )$ on $\Gamma_N$. Now $n=\left \langle \frac{2x}{\sqrt{4x^2+64y^2}},\frac{8y}{\sqrt{4x^2+64y^2}} \right \rangle$, so $\frac{\partial \tilde{u}}{\partial n}=-\sqrt{4x^2+64y^2}-\frac{2x}{\sqrt{4x^2+64y^2}}(-5x)-\frac{8y}{\sqrt{4x^2+64y^2}}(-5y)=\frac{6x^2-24y^2}{\sqrt{4x^2+64y^2}}$.

Personally, even if I didn't already have the exact solution in hand, I would probably have tried $u_0$ being an appropriate multiple of $x^2+4y^2$ (which ends up being $-(x^2+4y^2)$ in this case) so that the Dirichlet BC for $\tilde{u}$ would end up just being $\tilde{u}=c$ for some $c$. But that does end up sorta solving the entire question "by inspection", because if you go this way and fix up the Dirichlet BC then the Neumann BC "happens" to line up perfectly as well.

Ian
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    @FreeMind Your old version would've worked just as well to be honest. But I will change it. – Ian Jul 17 '21 at 15:21
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    @FreeMind Just take the gradient of $x^2+4y^2$ (the function whose level set is the boundary) and normalize it. That gives you a normal (maybe inward maybe outward) then you check from the geometry that it points outward. – Ian Jul 17 '21 at 15:31