# Volume potential (show continuity)

Consider $$\Omega:=B_{R_2}(0)\setminus \overline{B_{R_1}}(0)\subset\mathbb{R}^3$$ with $0<R_1<r_2<\infty$ (hollow ball) and $$U_3(x):=\frac{1}{4\pi}\int_{\Omega}\frac{1}{\lVert x-y\rVert}\, dy\text{ for }x\in\mathbb{R}^3$$ (volume potential). Calculate $U_3(x)$ and check its continuity..

Hello, first of all I calculated a lot and got $$U_3(0)=\frac{1}{2}(R_2^2-R_1^2), ~~~~~U_3(x)=\begin{cases}\frac{1}{2}(R_2^2-R_1^2), & 0<\lVert x\rVert\leq R_1\\\frac{R_2^2}{2}-\frac{\lVert x\rVert^2}{6}-\frac{R_1^3}{3\lVert x\rVert}, & R_1<\lVert x\rVert <R_2\\\frac{R_2^3}{3\lVert x\rVert}-\frac{R_1^3}{3\lVert x\rVert}, & \lVert x\rVert\geq R_2\end{cases}$$ Fortunately, the professor gave the solutions so that I can at least tell you, that my results are right.

Now I still have to show that this function (which can be seen as depending on $\lVert x\rVert$) is continious and that to do so I "only" have to show the continuity in the "problem cases" that are $\lVert x\rVert=R_1$ and $\lVert x\rVert=R_2$.

But I do not know how to do so. I was not able to handle it.

You just want to check that the limits of $U_{3}(x)$ are equal as $\|x\|\uparrow R_{1}$ and $\|x\|\downarrow R_{1}$, and the same thing for $R_{2}$. Nothing else is required because your functions only depend on $r=\|x\|$.
If you don't expected any surface charge layers, then you should also check that the derivatives with respect to $r$ are also continuous across the interfaces at $r=R_{1}$ and $r=R_{2}$. For example, the derivative with respect to $r$ is 0 for $r < R_{1}$, and the derivative with respect to $r$ for $R_{1} < r < R_{2}$ is $-2r/6+R_{1}^{2}/(3r^{2})$, which tends to 0 as $r\downarrow R_{1}$. So, by Gauss' law, there is no surface charge layer at $r=R_{1}$.