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Hello,

as I'm not an analyst, I'm having difficulties with the following, certainly well-known problem: one is given the PDE $\Delta u(x,y)=\sqrt{x^2+y^2}$ in the "region" $x^2+y^2\leq1$ with the boundary coundition $u(x,y)=0$ whenever $x^2+y^2=1$. The most obvious "answer" would be $u(x,y)=\sqrt{x^2+y^2}$, but the partial derivatives of $u(x,y)$ are not defined at $(x,y)=(0,0)$ (the singularity w.r.t. polar coordinates). Am I overlooking something, i.e. is there a well-behaved solution ?

Any help would be greatly appreciated !

Kind regards,

Stephan.

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The usual way to approach this would be to first search for radial solutions, i.e. look for functions $v$ such that $v(|(x,y)|)=u(x,y)$ and $v$ solves the equation. This should reduce the problem to an ODE in this special case, from which it is easy to find a solution.

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  • $\begingroup$ First off, don't ask me what made me think $u(x,r)=r:=\sqrt{x^2+y^2}$ was a solution; next, I did try looking for a radial solution $\hat u(r)$, i.e. tried solving $\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\hat u}{\partial r})=r$, but must have miscalculated ... What bothers me is the assumption made in most books of the existence of a radially symmetric solution, simply because of the corresponding symmetry of the (domain of the) equation and the boundary condition ... $\endgroup$ – Stephan F. Kroneck Apr 19 '11 at 12:53
  • $\begingroup$ ... Isn't some sort of symmetry-breaking imaginable ? Or is this discluded due to uniqueness considerations ? Anyway, thanks for your comment ! Kind regards, St. $\endgroup$ – Stephan F. Kroneck Apr 19 '11 at 12:56
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The solution is $\frac19(r^3-1)$.

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  • $\begingroup$ Thank you very much ! I recalculated (see my comments to the aforegoing answer) and arrived at your solution. Kind regards, St. $\endgroup$ – Stephan F. Kroneck Apr 19 '11 at 12:45
  • $\begingroup$ With the exception of my musings concerning symmetry-breaking, I feel that my original (obviously elementary) question has been answered; how can one close this question ? Stephan. $\endgroup$ – Stephan F. Kroneck Apr 23 '11 at 16:53

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