
MHB Master
#1
December 14th, 2016,
16:12
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
$L_{n1} M=0, M(0,x,g)=g(x), M_t(0,x,g)=0$.
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?

December 14th, 2016 16:12
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MHB Master
#2
December 14th, 2016,
16:41
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 $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.

MHB Master
#3
December 16th, 2016,
16:53
Thread Author
Originally Posted by
Euge
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 $(\cos \phi, \sin \phi)$ ?
Originally Posted by
Euge
The last equation follows from the divergence theorem, since $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.
I see...

MHB Master
#4
December 16th, 2016,
19:22
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 $\Delta M= \frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1 \xi_1} \cos^2{\phi}+g_{\xi_2 \xi_2} \sin^2{\phi}) d{\phi}$ ?

MHB Master
#5
December 18th, 2016,
17:21
Originally Posted by
evinda
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 $\nu(\xi) = \frac{t(\cos \phi, \sin \phi)}{t} = (\cos \phi, \sin \phi)$.
Originally Posted by
evinda
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.
Originally Posted by
evinda
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.
Originally Posted by
evinda
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?

MHB Master
#6
December 19th, 2016,
07:01
Thread Author
Originally Posted by
Euge
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...
Originally Posted by
Euge
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...
Originally Posted by
Euge
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?
Originally Posted by
Euge
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?

MHB Master
#7
December 19th, 2016,
22:50
Originally Posted by
evinda
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
$$\lim_{\Delta t\to 0} \int_0^{2\pi} \int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi)\, r\, dr\, d\phi$$
Originally Posted by
evinda
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.

MHB Master
#8
December 20th, 2016,
10:03
Thread Author
Originally Posted by
Euge
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?
Originally Posted by
Euge
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?

MHB Master
#9
December 20th, 2016,
10:45
Originally Posted by
evinda
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
$$\int_{t < \lvert x  \xi \rvert < t + \Delta t} \Delta_\xi g\, d\xi = \int_0^{2\pi}\int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi) \,r\, dr\, d\phi$$
Originally Posted by
evinda
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.