How to approach the following boundary value problem on the unit ball: $\Delta u(x,y)=| x|$, $u(x,y)=0$ for $x^{2}+y^{2}=1$

Question

I think the title is pretty self-explanatory but I'm elaborating just in case: let $u\in C^{2}(D(0,1),\mathbb{R})$ (where $D(0,1)$ is the (open) unit disk centered at $0$) such that the laplacian of $u$ equals $\mid x\mid$ on $D(0,1)$ and $u=0$ on $\partial D(0,1)$. I know the solution can be computed using Green's function but I'm not entirely sure how to do it in this example. I would very much appreciate some suggestions.

Answer 1

Let me sketch the solution. First, notice that $v(x,y):=\frac{|x|^3}{6}$ satisfies the equation $\Delta v(x,y)=|x|$, but not the boundary condition $v(x,y)=0$ at $r:=\sqrt{x^2+y^2}=1$. Instead, it satisfies (in polar coordinates) the boundary condition $v(r=1,\phi)=\frac{1}{6}|\cos\phi\,|^3$. Let's then consider the function $w:=u-v$. From the definition of $v$, it follows that $w$ satisfies Laplace's equation $$ \Delta w=\Delta u - \Delta v = 0 \tag{1} $$ with boundary condition $$ w(r=1,\phi)=u(r=1,\phi)-v(r=1,\phi)=-\frac{1}{6}|\cos\phi\,|^3. \tag{2} $$ The general solution to $(1)$ in polar coordinates is given by$^{(*)}$ $$ w(r,\phi)=a_0+\sum_{n=1}^{\infty}r^n(a_n\cos n\phi + b_n\sin n\phi), \tag{3} $$ which has the form of a Fourier series. To determine its coefficients, we use the boundary condition $(2)$, which yields \begin{align} a_0&=\frac{1}{2\pi}\int_0^{2\pi}\left(-\frac{1}{6}|\cos\phi\,|^3\right) dx, \tag{4a} \\ a_n&=\frac{1}{\pi}\int_0^{2\pi}\left(-\frac{1}{6}|\cos\phi\,|^3\right)\cos n\phi\, dx, \tag{4b} \\ b_n&=\frac{1}{\pi}\int_0^{2\pi}\left(-\frac{1}{6}|\cos\phi\,|^3\right)\sin n\phi\, dx. \tag{4c} \end{align} Once these integrals are calculated, the solution to the original PDE is given by $u(r,\phi)=v(r,\phi)+w(r,\phi)$, where $v(r,\phi)=\frac{r^3}{6}|\cos\phi\,|^3$ and $w(r,\phi)$ is given by $(3)$ and $(4)$.

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$^{(*)}$ See, for instance, https://en.wikipedia.org/wiki/Laplace%27s_equation#In_two_dimensions.