I'm solving the following exercise: find $u\in C^2(B^\circ(0,R))\cap C(B(0,R))$ such that: \begin{equation} \begin{cases} \Delta u=|x|^\beta \ \ \text{on} \ B^\circ(0,R)\\ u(x)=0 \ \ \ \text{on} \ \partial B(0,R) \end{cases} \end{equation} where with $B^\circ(0,R)$ I'm intending the internal of the ball of radius $R$ centered at 0.
MY ATTEMPT: For the whole computation $x$ is a vector on $\mathbb{R}^3$.Using a hint given in the exercise I look for radial solutions of the form $u(x)=\phi(|x|)$ thus I need to compute first the gradient of $u$ that is: \begin{equation} \nabla u(x)=\dfrac{\phi'(|x|)}{|x|}x \end{equation} and the laplacian is: \begin{equation} \Delta u(x)=\bigg(\dfrac{n-1}{|x|}\bigg)\phi'(|x|)+\phi''(|x|) \end{equation} (of course $n=3$). Replacing now the r(x)=|x| we get that $\phi$ must satisfy the following ODE: \begin{equation} \begin{cases} \phi''(r)+\bigg(\dfrac{2}{r}\bigg)\phi'(r)=r^\beta\\ \phi(R)=0 \end{cases} \end{equation} to find a solution I multiply the first equation by the quantity $r^2$ to obtain: \begin{equation} r^2\phi''(r)+\dfrac{2r^2}{r}\phi'(r)=r^{2+\beta}\Rightarrow r^2\phi''(r)+2r\phi'(r)=r^{2+\beta}\Rightarrow \dfrac{d}{dr}(r^2\phi'(r))=r^{2+\beta} \end{equation} Now integrating on both sides we get: \begin{equation} r^2\phi'(r)=\dfrac{r^{3+\beta}}{3+\beta}+C\Rightarrow \phi'(r)=\dfrac{r^{1+\beta}}{3+\beta}+\dfrac{C}{r^2}\Rightarrow \phi(r)=\dfrac{r^{2+\beta}}{(3+\beta)(2+\beta)}-\dfrac{C}{r}+D \end{equation} Thus substituting back we can conclude that: \begin{equation} u(x)=\dfrac{|x|^{2+\beta}}{(3+\beta)(2+\beta)}-\dfrac{C}{|x|}+D \ \ \ x\in B^{\circ}(0,R)-\{0\} \end{equation} Is this computation fine?
You have correctly calculated that $$ u(x) = \frac{\vert x \vert^{\beta+2}}{(\beta+3)(\beta+2)}+\frac C {\vert x \vert} +D$$ for some $C,D\in \mathbb R$.
Hint: Firstly, for what values of $C$ is $u$ differentiable (or even continuous) at the origin? Then from here choose $D$ based on the fact $u(x)=0$ when $\vert x \vert = R$.