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$\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?

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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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  • $\begingroup$ Could you please explain what $B_3$ is? $\endgroup$ – Dr. John A Zoidberg Mar 17 '16 at 10:14
  • $\begingroup$ Also how does one obtain $u$ from $w$? $\endgroup$ – Dr. John A Zoidberg Mar 17 '16 at 10:42
  • $\begingroup$ Please check the edits. $\endgroup$ – detnvvp Mar 17 '16 at 11:08
  • $\begingroup$ Thanks! could you give me some advice on solving this math.stackexchange.com/questions/1701501/… too? It's a similar problem, I tried the same approach that you gave but it didn't work. Thanks! $\endgroup$ – Dr. John A Zoidberg Mar 17 '16 at 11:31

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