An elliptic PDE problem of Laplace function Let $u$ be a harmonic function and we define
$$ q(r)=\int_{\partial B(0,r)}u^2(x)\,dx $$
I want to prove that $q$ is monotone and convex.
I tried the usual trick, by changing of variable, we have
$$ q(r)=\int_{\partial B(0,r)}u^2(x)\,dx =r^{N-1}\int_{\partial B(0,1)}u^2(ry)\,dy $$
Hence I could compute $q'(r)$ and I have 
$$ q'(r)=(N-1)r^{N-2}\int_{\partial B(0,1)}u^2(ry)\,dy+r^{N-1}\int_{\partial B(0,1)}2u(ry)\nabla u(ry)\cdot y\,dy $$
and hence we have
$$q'(r)=(N-1)\frac{1}{r} \int_{\partial B(0,r)}u^2(x)\,dx+\int_{\partial B(0,r)}\partial_{\gamma}(u^2(x))\,dx $$
where $\gamma$ is the out normal vector of $B(0,r)$.
I was trying to prove $q'(r)$ is either positive or negative but I can't do it... Any hint would be very welcome!
The next question is about to prove that $q(r)$ is log-convex, i.e., I want to show $\log q(r)$ is convex function of $\log r$
I compute 
$$ \frac{d^2}{(d\log r)^2} \log q(r) = r^2\frac{q''(r)}{q(r)}+r\frac{q'(r)}{q(r)}-\left(r\frac{q'(r)}{q(r)}\right)^2 $$ 
I was hoping to prove that 
$$ r\frac{q'(r)}{q(r)}\geq\left(r\frac{q'(r)}{q(r)}\right)^2 $$ 
i.e., 
$$r q'(r)\geq q(r)$$
However, unfortunately, I proved the opposite direction...
 A: I'll assume $N \ge 2$, and out of preference I'll write
$$q'(r) = \frac{1}{r} \left\{(N - 1)q(r) + 2r \int_{\partial B(0,r)} u\nabla u \cdot \mathbf{n}\, dx\right\}$$
By the divergence theorem, since $u$ is harmonic,
$$\int_{\partial B(0,r)} u\nabla u \cdot \mathbf{n}\, dx = \int_{B(0,r)} \text{div}(u\nabla u)\, dV = \int_{B(0,r)} (u\Delta u + |\nabla u|^2)\, dV = \int_{B(0,r)} |\nabla u|^2\, dV$$
Therefore
$$q'(r) = \frac{1}{r}\left\{(N - 1)q(r) + 2r \int_{B(0,r)} |\nabla u|^2\, dV\right\} \ge 0$$
This shows that $q$ is monotone increasing. By the produt rule
$$q''(r) = -\frac{1}{r^2}\left\{(N - 1)q(r) + 2r \int_{B(0,r)} |\nabla u|^2\, dV\right\} + \frac{1}{r}\left\{(N - 1)q'(r) + 2\int_{B(0,r)} |\nabla u|^2\, dV + 2r \frac{d}{dr} \int_{B(0,r)} |\nabla u|^2\, dV\right\}$$
Note
$$\frac{d}{dr} \int_{B(0,r)} |\nabla u|^2 \, dV = \frac{d}{dr} \int_0^r \int_{\partial B(0,1)} |\nabla u(\rho y)|^2\, \rho^{N-1}\, dy\, d\rho = r^{N-1}\int_{\partial B(0,1)} |\nabla u(ry)|^2\, dy$$
Hence
$$q''(r) = \frac{(N-1)}{r^2}\{rq'(r) - q(r)\} + 2r^{N-1} \int_{\partial B(0,1)} |\nabla u(ry)|^2\, dy$$
As 
$$rq'(r) = (N - 1)q(r) + 2r\int_{B(0,r)} |\nabla u|^2\, dV \ge (N - 1)q(r)$$
it follows that
$$q''(r) \ge \frac{(N - 1)(N - 2)}{r^2}q(r) + 2r^{N-1} \int_{\partial B(0,1)} |\nabla u(ry)|^2\, dy \ge 0$$
Therefore, $q$ is convex.
