Mean value theorem for harmonic functions

**arbolis** · March 26th 2010, 08:08 AM

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.

**xxp9** · March 26th 2010, 05:27 PM

Note that $\frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\phi$ , thus it equals $<grad(u),\phi>$ . So the integral evaluates
$ \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV =\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.

**arbolis** · March 27th 2010, 06:20 AM

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 $<grad(u),\phi>$ . So the integral evaluates
$ \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV =\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?

**xxp9** · March 27th 2010, 06:51 AM

<,> 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?

**arbolis** · March 27th 2010, 11:39 AM

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
$ \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV $

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.

**xxp9** · March 27th 2010, 03:59 PM

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.

**arbolis** · March 27th 2010, 05:59 PM

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...

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