How to prove that $u(r)=k \frac{1}{r}$ is the only solution for the integral equation $\int_{V'}\rho'\ u(r)\ dV' = constant$? Consider a hollow spherical charge with density $\rho'$ continuously varying only with respect to distance from the center $O$.

$V'=$ yellow volume
$k \in \mathbb {R}$
$\forall$ point $P$ inside the hollow sphere:
\begin{align}
\vec{E}_P &=\displaystyle\int_{V'}\rho'\ \vec{f}(r)\ dV'\\
&=\int_{V'}\rho'\ [-\nabla u(r)]\ dV'\\ 
&=-\nabla \int_{V'}\rho'\ u(r)\ dV' = 0\\
\implies \int_{V'}\rho'\ u(r)\ dV' &= constant\\
\implies u(r) &=k \dfrac{1}{r}\\
\end{align}
For the proof of last equation, see below:


*

*Logically deducing inverse square law (physics)

*https://www.docdroid.net/Wg59yXo/v-pdf

Question:
At first we do not know how many solutions are there. So let the solutions be  $u_1(r), u_2(r), u_3(r),......$
We started from $\int_{V'}\rho'\ u(r)\ dV' = constant$ and reached $u_1(r)= k \dfrac{1}{r}$
Now if we start from $\int_{V'}\rho'\ u(r)\ dV' = constant$ and by a different approach (i.e. through a series of different reasoning), we reach another solution $u_2(r)$
Again we start from $\int_{V'}\rho'\ u(r)\ dV' = constant$ and by a third approach, we reach third solution $u_3(r)$
Why other such solutions are not possible?

Please read the grey highlighted sentence in the following Wikipedia article: Shell theorem which says:
"If we further constrain the force by requiring that the second part of the theorem also holds, namely that there is no force inside a hollow ball, we exclude the possibility of the additional term, and the inverse square law is indeed the unique force law satisfying the theorem."




How shall we prove that $u(r)=k \dfrac{1}{r}$ is the only solution for the integral equation $\displaystyle\int_{V'}\rho'\ u(r)\ dV' = constant$?
Shall we require high level mathematics to do so?

$$\text{ }$$
I am a mathematics graduate. So please try to answer at my level.
 A: The Wikipedia is right. According to the experiment with the hollow sphere, the inverse square law is indeed the unique force law satisfying the theorem.
Your problem seems to lie in the mathematical logic. I have worked out the PDF you posted.
I will use $C_1$, $C_2$, $C_3$... as constants
As you described in your question, you seem to know everything up to $\psi=C_1$. We need to find the solution for $\dfrac{f(r)}{r}$
Now let me begin:
Let the solutions be $\dfrac{f_1(r)}{r}$, $\dfrac{f_2(r)}{r}$, $\dfrac{f_3(r)}{r}$......
For some time, let us concentrate only on $\dfrac{f_1(r)}{r}$
Since $\dfrac{f_1(r)}{r}$ is a solution,
$\psi = \int_V\ \rho\ \dfrac{f_1(r)}{r} dV = C_1$
This must be true.
Now continue your calculations in the PDF you posted. And finally, 
$\dfrac{f_1(r)}{r} = C_2 + \dfrac{C_3}{r}$
Following the implications in the PDF, this also must be true.
Now let us concentrate on $\dfrac{f_2(r)}{r}$
Following the exact same reasoning, we see $\dfrac{f_2(r)}{r} = C_4 + \dfrac{C_5}{r}$ must be true.
This same result apply $\forall\ \dfrac{f_n(r)}{r}\ (n\in \mathbb{N})$
Now continuing the last calculations in your PDF (i.e. taking gradient), we can see that inverse square law is the only force law satisfying the if condition of inverse of shell theorem.
