# question on Poisson equation with solution not in $H_0^1$

I'm considering the following quesion about Poisson equation:

$$-\Delta u=f$$ in a ball radius $$1$$ in $$3$$ dimension, if $$f\in L^{2}$$, then the theory of elliptic PDE says that the above equation exists unique solution $$u\in H_{0}^{1}$$. Then I suppose that $$f=r-1$$, where $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$, this is a radial function, so we can suppose that the solution $$u$$ must be a radial function, i.e. $$u\left(x\right)=u\left(r\right)$$, then I considered $$u''\left(r\right)+\frac{n-1}{r}u'\left(r\right)=1-r.$$ The solution of this ODE with boundary condition $$u\left(1\right)=0$$ and $$u'\left(1\right)=0$$ added is $$u\left(r\right)=-\frac{\left(r+1\right)\left(r-1\right)^{3}}{12r}.$$ It is true that $$u\left(r\right)$$ is a solution of the Poisson equation, but by calculating $$\nabla u=u'\left(r\right)\cdot\frac{x}{r}=\frac{-1+\left(4-3r\right)r^{3}}{12r^{2}}\cdot\frac{x}{r}$$ and $$\left|\nabla u\right|=\frac{1-\left(4-3r\right)r^{3}}{12r^{2}}$$, but $$\|\nabla u\|_{L^{2}}=\infty$$, this contradicts $$u\in H_{0}^{1}$$. what am I get wrong in somewhere?

• The equation can describe an electrostatic potential $u$ with a charge density $f$. Your boundary condition says that the electric field is zero at $r=1$. That implies that the net charge is $0$. Your solution is telling you that there must be a point charge of opposite sign at the origin to cancel the field of the charge distribution. – Keith McClary Nov 16 at 20:13

The problem is coming from your enforcement of the second boundary condition $$u'(1)=0$$, which is not something that must be true. The general solution of the ODE with the BC $$u(1) =0$$ is $$u(r) = C\left(1 - \frac{1}{r}\right) - \frac{1}{12} + \frac{r^2}{6} -\frac{r^3}{12}.$$ for some constant $$C$$. If you enforce the condition $$u'(1)=0$$ then you get a nontrivial $$C$$, which is causing the problems with the gradient. If instead you pick $$C$$ to enforce the gradient condition then you'll get $$C=0$$, which kills the singular term at the origin. Roughly speaking, what's going on here is that the second boundary condition is at $$r=0$$ and enforces the condition that your solution is not too singular.