Mixing
derivative of surface integral over sphere
$begingroup$
let $u$ be harmonic in the domain $U subset mathbb{R}^n$ and $B_R(0) subset U$ and $u(0)=0, uneq 0$. Let $0<r<R$. Define $a(r):= frac{1}{r^{n-1}} int_{partial B_r(0)} u^2dS, b(r):= frac{1}{r^{n-2}} int_{B_r(0)} |nabla u|^2dx$. The monotony-formula for $u$ harmonic is $b'(r)= frac{2}{r^{n-2}} int_{partial B_r(0)} u_r^2dS$.
Show that $a'= frac{2}{r^{n-1}} int_{partial B_r(0)} u u_rdS=frac{2}{r}b$.
I was able to show the second equality, but with the first one I have a little trouble. I used the transformation formula twice with the diffeomorphism $Phi : B_1(0) rightarrow B_r(0), y mapsto ry=:x$ and $Phi^{-1}$ with $|det(DPhi(y))|=r$ to get:
$a'= (frac{r}{r^{n-1}} int_{partial B_1(0)} u(ry)^2dS(y))' \
= frac{-(n-2)r^{n-3}}{(r^{n-2})^2} int_{partial B_1(0)} u(ry)^2dS + frac{1}{r^{n-2}}int_{partial B_1(0)} 2u(ry) u_r(ry)dS mathrm{;(with ; product ; rule)}\
= frac{-(n-2)}{r^{n-1}} int_{partial B_r(0)} u^2 frac{1}{r}dS +
frac{1}{r^{n-2}}int_{partial B_1(0)} 2u u_r frac{1}{r}dS$.
The last term is what I want to get but the first term should vanish. However I can't see why the first term should be zero, so I'm thinking that maybe my calculations are wrong but I don't know where. Maybe someone could tell me what is wrong in my conclusions.
integration analysis derivatives integral-transforms
asked Jan 20 at 17:48
mathstumathstu
315
$endgroup$
add a comment |
$begingroup$
let $u$ be harmonic in the domain $U subset mathbb{R}^n$ and $B_R(0) subset U$ and $u(0)=0, uneq 0$. Let $0<r<R$. Define $a(r):= frac{1}{r^{n-1}} int_{partial B_r(0)} u^2dS, b(r):= frac{1}{r^{n-2}} int_{B_r(0)} |nabla u|^2dx$. The monotony-formula for $u$ harmonic is $b'(r)= frac{2}{r^{n-2}} int_{partial B_r(0)} u_r^2dS$.
Show that $a'= frac{2}{r^{n-1}} int_{partial B_r(0)} u u_rdS=frac{2}{r}b$.
I was able to show the second equality, but with the first one I have a little trouble. I used the transformation formula twice with the diffeomorphism $Phi : B_1(0) rightarrow B_r(0), y mapsto ry=:x$ and $Phi^{-1}$ with $|det(DPhi(y))|=r$ to get:
$a'= (frac{r}{r^{n-1}} int_{partial B_1(0)} u(ry)^2dS(y))' \
= frac{-(n-2)r^{n-3}}{(r^{n-2})^2} int_{partial B_1(0)} u(ry)^2dS + frac{1}{r^{n-2}}int_{partial B_1(0)} 2u(ry) u_r(ry)dS mathrm{;(with ; product ; rule)}\
= frac{-(n-2)}{r^{n-1}} int_{partial B_r(0)} u^2 frac{1}{r}dS +
frac{1}{r^{n-2}}int_{partial B_1(0)} 2u u_r frac{1}{r}dS$.
The last term is what I want to get but the first term should vanish. However I can't see why the first term should be zero, so I'm thinking that maybe my calculations are wrong but I don't know where. Maybe someone could tell me what is wrong in my conclusions.
integration analysis derivatives integral-transforms
asked Jan 20 at 17:48
mathstumathstu
315
$endgroup$
add a comment |
$begingroup$
let $u$ be harmonic in the domain $U subset mathbb{R}^n$ and $B_R(0) subset U$ and $u(0)=0, uneq 0$. Let $0<r<R$. Define $a(r):= frac{1}{r^{n-1}} int_{partial B_r(0)} u^2dS, b(r):= frac{1}{r^{n-2}} int_{B_r(0)} |nabla u|^2dx$. The monotony-formula for $u$ harmonic is $b'(r)= frac{2}{r^{n-2}} int_{partial B_r(0)} u_r^2dS$.
Show that $a'= frac{2}{r^{n-1}} int_{partial B_r(0)} u u_rdS=frac{2}{r}b$.
I was able to show the second equality, but with the first one I have a little trouble. I used the transformation formula twice with the diffeomorphism $Phi : B_1(0) rightarrow B_r(0), y mapsto ry=:x$ and $Phi^{-1}$ with $|det(DPhi(y))|=r$ to get:
$a'= (frac{r}{r^{n-1}} int_{partial B_1(0)} u(ry)^2dS(y))' \
= frac{-(n-2)r^{n-3}}{(r^{n-2})^2} int_{partial B_1(0)} u(ry)^2dS + frac{1}{r^{n-2}}int_{partial B_1(0)} 2u(ry) u_r(ry)dS mathrm{;(with ; product ; rule)}\
= frac{-(n-2)}{r^{n-1}} int_{partial B_r(0)} u^2 frac{1}{r}dS +
frac{1}{r^{n-2}}int_{partial B_1(0)} 2u u_r frac{1}{r}dS$.
The last term is what I want to get but the first term should vanish. However I can't see why the first term should be zero, so I'm thinking that maybe my calculations are wrong but I don't know where. Maybe someone could tell me what is wrong in my conclusions.
integration analysis derivatives integral-transforms
asked Jan 20 at 17:48
mathstumathstu
315
$endgroup$
let $u$ be harmonic in the domain $U subset mathbb{R}^n$ and $B_R(0) subset U$ and $u(0)=0, uneq 0$. Let $0<r<R$. Define $a(r):= frac{1}{r^{n-1}} int_{partial B_r(0)} u^2dS, b(r):= frac{1}{r^{n-2}} int_{B_r(0)} |nabla u|^2dx$. The monotony-formula for $u$ harmonic is $b'(r)= frac{2}{r^{n-2}} int_{partial B_r(0)} u_r^2dS$.
Show that $a'= frac{2}{r^{n-1}} int_{partial B_r(0)} u u_rdS=frac{2}{r}b$.
I was able to show the second equality, but with the first one I have a little trouble. I used the transformation formula twice with the diffeomorphism $Phi : B_1(0) rightarrow B_r(0), y mapsto ry=:x$ and $Phi^{-1}$ with $|det(DPhi(y))|=r$ to get:
$a'= (frac{r}{r^{n-1}} int_{partial B_1(0)} u(ry)^2dS(y))' \
= frac{-(n-2)r^{n-3}}{(r^{n-2})^2} int_{partial B_1(0)} u(ry)^2dS + frac{1}{r^{n-2}}int_{partial B_1(0)} 2u(ry) u_r(ry)dS mathrm{;(with ; product ; rule)}\
= frac{-(n-2)}{r^{n-1}} int_{partial B_r(0)} u^2 frac{1}{r}dS +
frac{1}{r^{n-2}}int_{partial B_1(0)} 2u u_r frac{1}{r}dS$.
The last term is what I want to get but the first term should vanish. However I can't see why the first term should be zero, so I'm thinking that maybe my calculations are wrong but I don't know where. Maybe someone could tell me what is wrong in my conclusions.
integration analysis derivatives integral-transforms
integration analysis derivatives integral-transforms
asked Jan 20 at 17:48
mathstumathstu
315
asked Jan 20 at 17:48
mathstumathstu
315
asked Jan 20 at 17:48
mathstumathstu
315
asked Jan 20 at 17:48
mathstumathstu
315
asked Jan 20 at 17:48
mathstumathstu
315
315
add a comment |
add a comment |
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