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    #1
    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?

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    #2
    The normal $\nu$ to the circle is the unit radial vector from the origin, so in coordinates it's $(\cos \phi, \sin \phi)$. Also, on the circle, the arc length parameter $s$ satisfies $s = t\phi$, and so $ds = t\, d\phi$, or $d\phi = \frac{1}{t}\, ds$. This is why, altogether, you get

    $$\frac{1}{2\pi}\int_0^{2\pi} \nabla_\xi g\cdot(\cos \phi, \sin \phi)\, d\phi = \frac{1}{2\pi t}\int_{\lvert x - \xi\rvert = t} \nabla_\xi g \cdot \nu \, ds.$$

    The last equation follows from the divergence theorem, since $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.

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    #3 Thread Author
    Quote Originally Posted by Euge View Post
    The normal $\nu$ to the circle is the unit radial vector from the origin, so in coordinates it's $(\cos \phi, \sin \phi)$.
    Could you explain further to me why the coordinates of $\nu$ are $(\cos \phi, \sin \phi)$ ?


    Quote Originally Posted by Euge View Post

    The last equation follows from the divergence theorem, since $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.
    I see...

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    #4 Thread Author
    Also, then it says the following:


    $$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

    First of all, why don't we have :

    $$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|\leq t+ \Delta t} \Delta_{\xi}g d{\xi}$$

    ?

    Secondly, how do we get the following equality?

    $$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

    And... does it hold that $\Delta M= \frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1 \xi_1} \cos^2{\phi}+g_{\xi_2 \xi_2} \sin^2{\phi}) d{\phi}$ ?

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    #5
    Quote Originally Posted by evinda View Post
    Could you explain further to me why the coordinates of $\nu$ are $(\cos \phi, \sin \phi)$ ?
    I made a slight error in my description of $\varphi$. What I meant to say is that for every point $\xi$ on your circle, $\nu(\xi)$ is the unit radial vector from the center of the circle (which is $x$) to $\xi$. Thus $\nu(\xi) = \frac{\xi - x}{\|\xi - x\|}$. Now $\xi = x + t(\cos \phi, \sin \phi)$ for some $\phi$, and so $\nu(\xi) = \frac{t(\cos \phi, \sin \phi)}{t} = (\cos \phi, \sin \phi)$.



    Quote Originally Posted by evinda View Post
    Also, then it says the following:


    $$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

    First of all, why don't we have :

    $$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|\leq t+ \Delta t} \Delta_{\xi}g d{\xi}$$

    ?
    Your equation is correct, but so is the author's. Note the domain of integration in your integral differs from the domain of integration of the authors by a null set.



    Quote Originally Posted by evinda View Post
    Secondly, how do we get the following equality?

    $$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$
    Switch to polar coordinates centered at $x$. At some point you'll compute the limits

    $$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \nabla^2 g(x_1 + r\cos \phi, x_2 +r\sin \phi)\, r\, dr,$$

    which evaluate to $\nabla^2 g(x_1 + r\cos \phi, x_2 + r\sin \phi)\,t$ by the fundamental theorem of calculus.


    Quote Originally Posted by evinda View Post
    And... does it hold that $\Delta M= \frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1 \xi_1} \cos^2{\phi}+g_{\xi_2 \xi_2} \sin^2{\phi}) d{\phi}$ ?
    How did you derive this formula?

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    #6 Thread Author
    Quote Originally Posted by Euge View Post
    I made a slight error in my description of $\varphi$. What I meant to say is that for every point $\xi$ on your circle, $\nu(\xi)$ is the unit radial vector from the center of the circle (which is $x$) to $\xi$. Thus $\nu(\xi) = \frac{\xi - x}{\|\xi - x\|}$. Now $\xi = x + t(\cos \phi, \sin \phi)$ for some $\phi$, and so $\nu(\xi) = \frac{t(\cos \phi, \sin \phi)}{t} = (\cos \phi, \sin \phi)$.
    A ok...


    Quote Originally Posted by Euge View Post
    Your equation is correct, but so is the author's. Note the domain of integration in your integral differs from the domain of integration of the authors by a null set.
    I see...



    Quote Originally Posted by Euge View Post

    Switch to polar coordinates centered at $x$. At some point you'll compute the limits

    $$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \nabla^2 g(x_1 + r\cos \phi, x_2 +r\sin \phi)\, r\, dr,$$

    which evaluate to $\nabla^2 g(x_1 + r\cos \phi, x_2 + r\sin \phi)\,t$ by the fundamental theorem of calculus.
    Switching to polar coordinates we get the limit $\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{t<|r|<t+\Delta t} \Delta_{\xi} g r dr$, which is equal to that what you wrote , right?
    Then from the fundamental theorem of calculus, we have that:

    $$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \Delta_{\xi} g r dr=\lim_{\Delta t \to 0} \frac{1}{\Delta t} \left[\Delta_{\xi}g(x_1+(t+\Delta t) \cos{\phi}, x_2+ (t+\Delta t)\sin{\phi})(t+\Delta t)-\Delta_{\xi}g(x_1+t \cos{\phi}, x_2+t \sin{\phi})t\right]$$

    Right? How do we continue?



    Quote Originally Posted by Euge View Post

    How did you derive this formula?
    I don't remember how I got this.
    Do we have:

    $$\Delta M=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \Delta_{\xi}g d{\xi}$$

    although $\xi$ appears at the limit of integration?

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    #7
    Quote Originally Posted by evinda View Post
    Switching to polar coordinates we get the limit $\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{t<|r|<t+\Delta t} \Delta_{\xi} g r dr$, which is equal to that what you wrote , right?

    No. It should be

    $$\lim_{\Delta t\to 0} \int_0^{2\pi} \int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi)\, r\, dr\, d\phi$$



    Quote Originally Posted by evinda View Post
    I don't remember how I got this.
    Do we have:

    $$\Delta M=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \Delta_{\xi}g d{\xi}$$

    although $\xi$ appears at the limit of integration?


    The $d\xi$ should be $ds$. Otherwise, the formula is correct.

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    #8 Thread Author
    Quote Originally Posted by Euge View Post
    No. It should be

    $$\lim_{\Delta t\to 0} \int_0^{2\pi} \int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi)\, r\, dr\, d\phi$$
    Could you explain further to me how you got this?



    Quote Originally Posted by Euge View Post

    The $d\xi$ should be $ds$. Otherwise, the formula is correct.
    Having $ds$ instead of $d\xi$ , in respect to which variable do we integrate?

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    #9
    Quote Originally Posted by evinda View Post
    Could you explain further to me how you got this?

    The region $t < \lvert x - \xi \lvert < t + \Delta t$ is an annulus centered at $x =(x_1,x_2)$ with inner radius $t$ and outer radius $t + \Delta t$. So the radial variable $r$ ranges from $t$ to $t + \Delta t$ and the polar variable $\phi$ ranges from $0$ to $2\pi$. This leads to the double integral representation

    $$\int_{t < \lvert x - \xi \rvert < t + \Delta t} \Delta_\xi g\, d\xi = \int_0^{2\pi}\int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi) \,r\, dr\, d\phi$$


    Quote Originally Posted by evinda View Post
    Having $ds$ instead of $d\xi$ , in respect to which variable do we integrate?

    The integration is done with respect to arclength; the integral over the circle $\lvert x - \xi\rvert = t$ is a line integral. If you use $d\xi$ instead, the integral would be zero because the circle is a null set in the plane.

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