How to proof the equality $\int_{S_{a}^{n-1}}\nabla u.d\sigma= c *vol (S^{n-1})$.
Hi all,
I am reading a text in portugues about PDE, is about Laplace operator and Coloumb Potential my specifics questions (i have 2 question). So, we want to find the general solution for the equation $\Delta u= 0$ when $u$ present a spherical symmetric $u(x)=u(r)$ (radial solution) where $r^2=\sum_{i=1}^n x_i^2$. For this function we have that
$$\Delta u (x)= u''(r)+\frac{n-1}{r}u'(r)=0$$ with solution $u(r)= -\frac{c}{(n-2)r^{n-2}}+c'$ whrn $n>2$, $u(r)=c\ln r +c'$ when $n=2$ or $u(r)=cr+c'$ when $n=1$, where $c,c'$ are constants. Until here is ok! My problem star right here The book says, "($\star$) $\frac{\partial u}{\partial x_i}(x)= \frac{c}{r^{n-1}}\frac{x_i}{r}$"
Then book say, "Using that $\int_{aK}f(x)d\sigma(x)=a^{n-1}\int_K f(ax)d\sigma(x)$ for any function $f$ continuous over the sphere with ratio $a$ and any compact subset $K$ in unitary sphere, and $\star$ we can show that $\int_{S_{a}^{n-1}}\nabla u.d\sigma= c *vol (S^{n-1})$ "
My attempt for the second question is
\begin{equation} \begin{split} \int_{S_{a}^{n-1}}\nabla u.d\sigma&= \int_{aS^{n-1}}\nabla u.d\sigma\\ (\text{Using the hint of the book})&= a^{n-1} \int_{S^{n-1}}\nabla u (ax).d\sigma(x)\\ (\text{using variable change y=ax})&= a^{-1}\int_{S^{n-1}}\nabla u (y).d\sigma(y)\\ &=\frac{1}{a}\int_{S^{n-1}} \nabla u.(1,0,\ldots,0)d\sigma(y)\\ (\text{Using Gauss´s theorem})&=\frac{1}{a}\int_{B^{n}}div \nabla u dy\\ &= \frac{1}{a}\int_{B^{n}} \Delta u(y)dy\\ (\text{What is wrong?})&= 0 \end{split} \end{equation} And what about the first question? Please i will appreciate any hint or help or solution, thank you
Best
The first and second questions are not well explicit for me. But as far as I understand, the first question is to prove $(\star)$ which is just taking the $x_i$ derivative to the radial expression of $u$, since $\frac{\partial r}{\partial x_i}=\frac{x_i}r$.
For the second one, your computation went wrong when you replace $\nabla u$ by $\nabla u\cdot(1,0,\ldots,0)$, or I miss something.
Rather, you should use the fact that $u$ is a radial function, thus $\nabla u$ is along $\vec{r}$, namely your formula ($\star$) means $\nabla u=\frac{c}{r^{n-1}}\frac{\vec{r}} r$. Since the equality to show gives a scalar value to $$\int_{S_a^{n-1}}\nabla u \cdot d\sigma$$ it seems that this $d\sigma$ stands for $\vec{n}ds$ where $n$ is the outward normal and $ds$ the surface measure. Using the fact that for a sphere of radius $a$, $\vec{n}=\frac{\vec{r}}{a}$, we see that on this sphere (where $r=a$) we have $\nabla u \cdot d\sigma=\frac{c}{a^{n-1}}ds$. Thus $$\int_{S_a^{n-1}}\nabla u \cdot d\sigma=\frac{c}{a^{n-1}}vol(S_a^{n-1}).$$ Now applying your scaling formula for $f=1$ gives $vol(S_a^{n-1})=a^{n-1}vol(S^{n-1})$ and you are done.