# Thread: Mean value theorem for harmonic functions

1. ## Mean value theorem for harmonic functions

I'm stuck on this problem I've been assigned in an electromagnetism class (it's a mathematical problem though).
Let $u: \mathbb{R}^3 \to \mathbb{R}$. We define the spherical mean of $f$ as $M_t(u)(x)=\frac{1}{4 \pi} \int _{|\psi |=1} u(x+t \psi ) dS_{\psi}$.
Demonstrate that if $u$ is a harmonic function (i.e. $\triangle u =0$) then u satisfies the equation $u(x)=M_t (u)(x)$ for all $t$.
Hint: Use $\partial _t M_t (u)$.

According to my class notes, $\partial _t M_t ( g (\vec x))=\frac{1}{4\pi} \int _{S^2} \partial _t \tilde \phi (\vec x +t \hat n) d \Omega$, where $\tilde \phi (\vec x) =\partial _t \phi (t,\vec x)|_{t=0}$.

So I guess I should start with $\partial _t M _t (u)(x)= \frac{1}{4 \pi} \int _{|\psi|=1} \partial _t u (x+t \psi) dS _ \psi$.
Now I'm not really sure how to proceed. I guess there's a trick with $\partial _t u(x+t \psi)$. I've been stuck here for days. I've seen the gradient appearing on the sheets of some people in my class, but I've no idea how I could introduce it.

Thanks for any help.

2. Note that $\frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\phi$, thus it equals $$. So the integral evaluates
$
=\int_{|\phi|=1} \Delta{u} dV = 0
$

So $M_t(u)(x)$ is constant for t. And obviously when t is 0 the equation holds.

3. Originally Posted by xxp9
Note that $\frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\phi$, thus it equals $$. So the integral evaluates
$
=\int_{|\phi|=1} \Delta{u} dV = 0
$

So $M_t(u)(x)$ is constant for t. And obviously when t is 0 the equation holds.
Thank you very much for the reply. I think I could understand everything if someone tells me what is the $<,>$ notation here.
By the way I'm also stuck to understand which unit vector is $\phi$. Is it the unit vector of the cylindrical or spherical coordinate system? Or is it associated with the solution of the wave equation?

4. <,> is the inner product of the two vectors. $\phi$ is the one that appears in the integral domain $\int_{|\phi|=1}$
Originally Posted by arbolis
Thank you very much for the reply. I think I could understand everything if someone tells me what is the $<,>$ notation here.
By the way I'm also stuck to understand which unit vector is $\phi$. Is it the unit vector of the cylindrical or spherical coordinate system? Or is it associated with the solution of the wave equation?

5. Ok perfect.
It starts to make sense to me.

Originally Posted by xxp9
Note that $\frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\phi$
Do you know any book where I could visualize similar things with figures?
Originally Posted by xxp9
So the integral evaluates
$

$
What did you do exactly here? Used Gauss's theorem?

Originally Posted by xxp9
$=\int_{|\phi|=1} \Delta{u} dV = 0$ So $M_t(u)(x)$ is constant for t. And obviously when t is 0 the equation holds.
Ok for this part.

6. Any calculus book will cover directional derivatives, probably with figures?
According to the chain rule, directional derivative along a vector v=(a, b, c) at point p will be $\frac{df(p+tv)}{dt} = \frac{\partial{f}}{\partial{x}}a+\frac{\partial{f} }{\partial{y}}b+\frac{\partial{f}}{\partial{z}}c=< grad(f),v>$
Yes that's Gauss's theorem or Green's or Stokes, anyway it turns a surface integral to a volume one.
see http://en.wikipedia.org/wiki/Divergence_theorem

Originally Posted by arbolis
Ok perfect.
It starts to make sense to me.

Do you know any book where I could visualize similar things with figures?

What did you do exactly here? Used Gauss's theorem?

Ok for this part.

7. Originally Posted by xxp9
Any calculus book will cover directional derivatives, probably with figures?
According to the chain rule, directional derivative along a vector v=(a, b, c) at point p will be $\frac{df(p+tv)}{dt} = \frac{\partial{f}}{\partial{x}}a+\frac{\partial{f} }{\partial{y}}b+\frac{\partial{f}}{\partial{z}}c=< grad(f),v>$
Yes that's Gauss's theorem or Green's or Stokes, anyway it turns a surface integral to a volume one.
see Divergence theorem - Wikipedia, the free encyclopedia
Thank you very much for everything. I was so stuck...