Hello!!!

We set $L_k \equiv u_{tt}+\frac{k}{t}-\Delta u=0, k \in \mathbb{N}_0$.

I am looking at the proof of the following proposition:

We suppose that $g(x)$ is twice differentiable in $\mathbb{R}^n$. Then the mean value of $g(x)$ at the sphere with radius $t$ and center at $x$, which we symbolize with $M(t,x,g)$, is a solution of the problem

$L_{n-1} M=0, M(0,x,g)=g(x), M_t(0,x,g)=0$.

I am given the proof for $n=2$, but I have some questions.

For $n=2$ we have $M(t,x,g)=\frac{1}{2 \pi t} \int_{|x-\xi|=t} g(\xi) ds=\frac{1}{2 \pi} \int_0^{2 \pi} g(x_1+t \cos{\phi}, x_2+ t \sin{\phi}) d{\phi}$.

Let $\xi_1=x_1+ t \cos{\phi}$ and $\xi_2=x_2+ t \sin{\phi}$.

We have:

$$\frac{\partial{M}}{\partial{t}}=\frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1} \cos{\phi}+g_{\xi_2} \sin{\phi}) d{\phi}=\frac{1}{2 \pi} \int_0^{2 \pi} \nabla_{\xi}g \cdot (\cos{\phi}, \sin{\phi}) d{\phi}=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \nabla_{\xi}g \cdot \mathcal{v} ds=\frac{1}{2 \pi} \int_{|x-\xi| \leq t} \Delta_{\xi}gd{\xi}$$

Could you explain to me how we get the last two relations?