I am trying to prove that the potential of a spherically symmetric stellar system is non-decreasing (source of problem 2016 Paper 4 Question 7, Astrophysics Tripos).
What I did is the following.
We want to show: $$\frac{d \Phi}{dr} \geqslant 0 \quad \forall r$$ We know: $$\nabla^{2} \Phi=4 \pi G \rho$$ Expand: $$\frac{1}{r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r} \Phi\right)=4 \pi G\rho$$ Multiply up by $r^2$, integrate up, divide by $r^2$, get: $$\frac{d}{d r} \Phi=\frac{4 \pi G}{r^{2}}\left(\int_{0}^{r} \rho\left(r^{\prime}\right) r^{\prime 2} d r^{\prime}+C\right)$$ Where $C$ is free to choose. And if I choose $C$ to be a large negative number, then $\frac{d\Phi}{dr}<0$, the opposite of what I was trying to prove.
What am I doing wrong?
Let's call $F : r\mapsto r^2 \dfrac{\mathrm{d}\Phi}{\mathrm{d}r} $ you have $F'(r) = 4\pi G \rho(r) r^2$ thus by integrating :
$$ \displaystyle \int_0^rF'(\hat{r})\mathrm{d}\hat{r}=4\pi G\int_0^r \rho(\hat{r}) \hat{r}^2 \mathrm{d}\hat{r}$$
But $$ \displaystyle \int_0^rF'(\hat{r})\mathrm{d}\hat{r} = F(r)-F(0) = r^2 \dfrac{\mathrm{d}\Phi}{\mathrm{d}r} -0 $$
Therefore :
$$ r^2 \dfrac{\mathrm{d}\Phi}{\mathrm{d}r} = 4\pi G\int_0^r \rho(\hat{r}) \hat{r}^2 \mathrm{d}\hat{r}$$