$\Delta u(x,y) = x^2 $
$x^2 + y^2 <9$
On the boundary $x^2 + y^2 = 9$, $\frac {\delta u}{\delta n} = y$.
I've seen similar problems solved in texts (Strauss, Bleecker, Stavroulakis), but where $\Delta u(x,y) = 0 $. How can I solve it for the given case?
$\endgroup$1 Answer
$\begingroup$One way you can do this is by reducing this to the case of the Laplacian being equal to $0$.
Let $B_3$ be the ball $$\left\{(x,y)|x^2+y^2<9\right\}.$$Consider a special solution to $\Delta v=x^2$; for example, $v(x)=\frac{x^4}{12}$, and set $w=u-v$. Then, $$\Delta w=\Delta u-\Delta v=x^2-x^2=0,$$ and $$\frac{\partial w}{\partial\nu}=\frac{\partial u}{\partial\nu}-\frac{\partial v}{\partial\nu}=y-\nabla v\cdot\nu=y-\left(\frac{x^3}{3},0\right)\cdot\left(\frac{x}{3},\frac{y}{3}\right)=y-\frac{x^4}{9}.$$ So, $w$ is a harmonic function, and it solves the equation $$\left\{\begin{array}{c l}\Delta w=0,&{\rm in}\,\,B_3\\ \frac{\partial w}{\partial\nu}=y-\frac{x^4}{9},&{\rm on}\,\,\partial B_3\end{array}\right..$$
If you know how to find $w$, then you can find $u$ using the relation $$u=v+w=\frac{x^4}{12}+w.$$
There is also a more general way, using the representation of solutions via Green's function.
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