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Thread: Mean value theorem for harmonic functions

  1. #1
    MHF Contributor arbolis's Avatar
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    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 $\displaystyle u: \mathbb{R}^3 \to \mathbb{R}$. We define the spherical mean of $\displaystyle f$ as $\displaystyle M_t(u)(x)=\frac{1}{4 \pi} \int _{|\psi |=1} u(x+t \psi ) dS_{\psi}$.
    Demonstrate that if $\displaystyle u$ is a harmonic function (i.e. $\displaystyle \triangle u =0$) then u satisfies the equation $\displaystyle u(x)=M_t (u)(x)$ for all $\displaystyle t$.
    Hint: Use $\displaystyle \partial _t M_t (u)$.

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

    So I guess I should start with $\displaystyle \partial _t M _t (u)(x)= \frac{1}{4 \pi} \int _{|\psi|=1} \partial _t u (x+t \psi) dS _ \psi$.
    Now I'm not really sure how to proceed. I guess there's a trick with $\displaystyle \partial _t u(x+t \psi)$. 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.
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    Note that $\displaystyle \frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\displaystyle \phi$, thus it equals $\displaystyle <grad(u),\phi>$. So the integral evaluates
    $\displaystyle
    \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV
    =\int_{|\phi|=1} \Delta{u} dV = 0
    $
    So $\displaystyle M_t(u)(x)$ is constant for t. And obviously when t is 0 the equation holds.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by xxp9 View Post
    Note that $\displaystyle \frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\displaystyle \phi$, thus it equals $\displaystyle <grad(u),\phi>$. So the integral evaluates
    $\displaystyle
    \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV
    =\int_{|\phi|=1} \Delta{u} dV = 0
    $
    So $\displaystyle M_t(u)(x)$ 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 $\displaystyle <,>$ notation here.
    By the way I'm also stuck to understand which unit vector is $\displaystyle \phi$. Is it the unit vector of the cylindrical or spherical coordinate system? Or is it associated with the solution of the wave equation?
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    <,> is the inner product of the two vectors. $\displaystyle \phi$ is the one that appears in the integral domain $\displaystyle \int_{|\phi|=1}$
    Quote Originally Posted by arbolis View Post
    Thank you very much for the reply. I think I could understand everything if someone tells me what is the $\displaystyle <,>$ notation here.
    By the way I'm also stuck to understand which unit vector is $\displaystyle \phi$. Is it the unit vector of the cylindrical or spherical coordinate system? Or is it associated with the solution of the wave equation?
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  5. #5
    MHF Contributor arbolis's Avatar
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    Ok perfect.
    It starts to make sense to me.



    Quote Originally Posted by xxp9 View Post
    Note that $\displaystyle \frac {\partial{u}}{\partial{t}}$ is the directional derivative of u in the direction of the unit vector $\displaystyle \phi$
    Do you know any book where I could visualize similar things with figures?
    Quote Originally Posted by xxp9
    So the integral evaluates
    $\displaystyle
    \int_{|\phi|=1} <grad(u),\phi> dS = \int_{|\phi|=1} div(grad(u))dV

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

    Quote Originally Posted by xxp9
    $\displaystyle =\int_{|\phi|=1} \Delta{u} dV = 0$ So $\displaystyle M_t(u)(x)$ is constant for t. And obviously when t is 0 the equation holds.
    Ok for this part.
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    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 $\displaystyle \frac{df(p+tv)}{dt} = \frac{\partial{f}}{\partial{x}}a+\frac{\partial{f} }{\partial{y}}b+\frac{\partial{f}}{\partial{z}}c=< grad(f),v>$
    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

    Quote Originally Posted by arbolis View Post
    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.
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  7. #7
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by xxp9 View Post
    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 $\displaystyle \frac{df(p+tv)}{dt} = \frac{\partial{f}}{\partial{x}}a+\frac{\partial{f} }{\partial{y}}b+\frac{\partial{f}}{\partial{z}}c=< grad(f),v>$
    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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