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Need help in solving an equation involving volume, single and double layer potentials
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Let be $V subset mathbb{R}^n $, $ 3leq n$ an open set, where you can apply Gauß's Theorem.
To show is, that for all $ U in C^{(1)} ( bar{V} ) cap C^{(2)} (V) $ with bounded 2nd derivatives the following equation for $y in V $:
$$ (n-2) omega_{n-1} U(y) = int_{ partial V} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x)- int_V frac{ Delta U(x)}{ |x-y|^{n-2}} dx $$
where
$ nu_x $ is the outer normal unit vector on $xin partial V$ and
$w_{n-1} := frac{n pi^{n/2}}{ Gamma( frac{n}{2} +1) }$
Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$.
Therefore, instead of integrating over $V$ , first integrate over $V_{epsilon}:= V$ $ K_{epsilon}(y) $ and use the limes $epsilon rightarrow 0+ $
Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.
Any help is therefore very appreciated !!
real-analysis multivariable-calculus multiple-integral gaussian-integral potential-theory
asked Dec 19 '18 at 19:31
constant94constant94
6310
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add a comment |
$begingroup$
Let be $V subset mathbb{R}^n $, $ 3leq n$ an open set, where you can apply Gauß's Theorem.
To show is, that for all $ U in C^{(1)} ( bar{V} ) cap C^{(2)} (V) $ with bounded 2nd derivatives the following equation for $y in V $:
$$ (n-2) omega_{n-1} U(y) = int_{ partial V} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x)- int_V frac{ Delta U(x)}{ |x-y|^{n-2}} dx $$
where
$ nu_x $ is the outer normal unit vector on $xin partial V$ and
$w_{n-1} := frac{n pi^{n/2}}{ Gamma( frac{n}{2} +1) }$
Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$.
Therefore, instead of integrating over $V$ , first integrate over $V_{epsilon}:= V$ $ K_{epsilon}(y) $ and use the limes $epsilon rightarrow 0+ $
Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.
Any help is therefore very appreciated !!
real-analysis multivariable-calculus multiple-integral gaussian-integral potential-theory
asked Dec 19 '18 at 19:31
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
Let be $V subset mathbb{R}^n $, $ 3leq n$ an open set, where you can apply Gauß's Theorem.
To show is, that for all $ U in C^{(1)} ( bar{V} ) cap C^{(2)} (V) $ with bounded 2nd derivatives the following equation for $y in V $:
$$ (n-2) omega_{n-1} U(y) = int_{ partial V} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x)- int_V frac{ Delta U(x)}{ |x-y|^{n-2}} dx $$
where
$ nu_x $ is the outer normal unit vector on $xin partial V$ and
$w_{n-1} := frac{n pi^{n/2}}{ Gamma( frac{n}{2} +1) }$
Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$.
Therefore, instead of integrating over $V$ , first integrate over $V_{epsilon}:= V$ $ K_{epsilon}(y) $ and use the limes $epsilon rightarrow 0+ $
Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.
Any help is therefore very appreciated !!
real-analysis multivariable-calculus multiple-integral gaussian-integral potential-theory
asked Dec 19 '18 at 19:31
constant94constant94
6310
$endgroup$
Let be $V subset mathbb{R}^n $, $ 3leq n$ an open set, where you can apply Gauß's Theorem.
To show is, that for all $ U in C^{(1)} ( bar{V} ) cap C^{(2)} (V) $ with bounded 2nd derivatives the following equation for $y in V $:
$$ (n-2) omega_{n-1} U(y) = int_{ partial V} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x)- int_V frac{ Delta U(x)}{ |x-y|^{n-2}} dx $$
where
$ nu_x $ is the outer normal unit vector on $xin partial V$ and
$w_{n-1} := frac{n pi^{n/2}}{ Gamma( frac{n}{2} +1) }$
Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$.
Therefore, instead of integrating over $V$ , first integrate over $V_{epsilon}:= V$ $ K_{epsilon}(y) $ and use the limes $epsilon rightarrow 0+ $
Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.
Any help is therefore very appreciated !!
real-analysis multivariable-calculus multiple-integral gaussian-integral potential-theory
real-analysis multivariable-calculus multiple-integral gaussian-integral potential-theory
asked Dec 19 '18 at 19:31
constant94constant94
6310
asked Dec 19 '18 at 19:31
constant94constant94
6310
edited Jan 9 at 11:46
constant94
asked Dec 19 '18 at 19:31
constant94constant94
6310
asked Dec 19 '18 at 19:31
constant94constant94
6310
asked Dec 19 '18 at 19:31
constant94constant94
6310
6310
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2 Answers
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I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity
$$
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x = intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_xlabel{1}tag{1}
$$
However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,epsilon)triangleq Vsetminus B(y,epsilon)$, where $B(y,epsilon)$ is the $n$-dimensional ball of radius $epsilon>0$ centered at $yinBbb R^n$. It is sufficient to recall that, for $ngeq 3$,
$$
Delta frac{1}{|y-x|^{n-2}}=-(n-2)omega_{n-1} delta(x-y)=-(n-2)omega_{n-1} delta(y-x)label{2}tag{2}
$$
where $delta$ is the usual Dirac distribution supported at $yinBbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting
$$
W(x)=frac{1}{|x-y|^{n-2}}quadtext{for an arbitrary }yinBbb R^n
$$
in formula eqref{1}, at the left side we get
$$
begin{split}
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x &= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x \
&= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x\
&= -(n-2)omega_{n-1}intlimits_V U(x)delta(y-x)mathrm{d}x - intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x\
&= -(n-2)omega_{n-1}U(y)- intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x
end{split}
$$
while at the second side we simply have
$$
intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_x =intlimits_{ partial V} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
From these two relations, you immediately get the sought for formula.
How to prove directly, without using distribution theory, the following equality, key relation of the whole proof?
$$
intlimits_V U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=-(n-2)omega_{n-1}U(y)quadforall yin Vlabel{3}tag{3}
$$
I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression
$$
lim_{epsilonto 0} intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x,
$$
since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e.
$$
Delta frac{1}{|x-y|^{n-2}}=0text{ on }V(y,epsilon)implies intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=0quadforallepsilon>0.label{a}tag{*}
$$
However we can use eqref{a} and eqref{1} by defining, for each $B(y,epsilon)Subset V$,
$$
begin{split}
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x & triangleq lim_{epsilonto 0} intlimits_{B(y,epsilon)} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x \
triangleq lim_{epsilonto 0},&
left[ ,intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x + intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)right]
end{split}
$$
and, since
$$
intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}xunderset{epsilon to 0}{longrightarrow} 0
$$
because $Uin C^2(V)iff Delta Uin C^0(V)$ and $|x-y|^{2-n} in L^1_mathrm{loc}(Bbb R^n)$ for $n ge 2$, we finally get
$$
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x triangleq lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $xmapsto (r,theta_1,ldots,theta_{n-1})$: for the first one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} mathrm{d}sigma(x)&=lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U frac{ partial }{partial r} frac{1}{r^{n-2}}mathrm{d}sigma\
&=-lim_{epsilonto 0} frac{n-2}{epsilon^{n-1}} intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma\
&=-(n-2) omega_{n-1} lim_{epsilonto 0} left[frac{1}{omega_{n-1}epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma right]\
&= -(n-2) omega_{n-1} U(y)
end{split}
$$
since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $partial B(y,epsilon)$. For the second one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}mathrm{d}sigma(x) &= lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{r^{n-2}} frac{ partial U }{partial r} mathrm{d}sigma \
&= lim_{epsilonto 0} frac{1}{epsilon^{n-2}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigma \
&= omega_{n-1} lim_{epsilonto 0} epsilon cdot left[frac{1}{omega_{n-1} epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigmaright] \
&=0
end{split}
$$
since the integral within the square brackets is the spherical mean of the $C^1$ function $nabla U(x)$ over the sphere $partial B(y,epsilon)$. This last step proves equation eqref{3} and thus the formula in the OP question.
Notes
In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).
The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
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Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_epsilon$.
If we also use that $Delta W(x)=0$ for $xneq y$ then the equation becomes
$$
(n-2) omega_{n-1} U(y) = int_{ partial K_epsilon} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x).
$$
Note that the terms involving integrals over $V$ or $partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $epsilonto0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
left|int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x) dsigma(x)right|
leq
C int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} dsigma(x)
= C epsilon^{2-n} omega_{n-1} epsilon^{n-1}to 0,
$$
where we used that the surface area of a ball with radius $epsilon>0$ is given by $omega_{n-1}epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}
= -(n-2) frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
int_{ partial K_epsilon} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} dsigma(x)
=
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
left|
(n-2)omega_{n-1} U(y)
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x)
right|
=left|
(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(y)-U(x) dsigma(x)
right|
leq sup_{xin partial K_epsilon} |U(y)-U(x)| (n-2)epsilon^{-(n-1)}omega_{n-1} epsilon^{n-1}
to0
$$
where the $sup$-term converges to $0$ because $U$ is continuous.
answered Jan 9 at 13:06
supinfsupinf
6,2161028
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firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
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– constant94
Jan 9 at 15:49
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@constant94 i added some more details.
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– supinf
Jan 10 at 7:42
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2 Answers
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$begingroup$
I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity
$$
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x = intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_xlabel{1}tag{1}
$$
However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,epsilon)triangleq Vsetminus B(y,epsilon)$, where $B(y,epsilon)$ is the $n$-dimensional ball of radius $epsilon>0$ centered at $yinBbb R^n$. It is sufficient to recall that, for $ngeq 3$,
$$
Delta frac{1}{|y-x|^{n-2}}=-(n-2)omega_{n-1} delta(x-y)=-(n-2)omega_{n-1} delta(y-x)label{2}tag{2}
$$
where $delta$ is the usual Dirac distribution supported at $yinBbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting
$$
W(x)=frac{1}{|x-y|^{n-2}}quadtext{for an arbitrary }yinBbb R^n
$$
in formula eqref{1}, at the left side we get
$$
begin{split}
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x &= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x \
&= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x\
&= -(n-2)omega_{n-1}intlimits_V U(x)delta(y-x)mathrm{d}x - intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x\
&= -(n-2)omega_{n-1}U(y)- intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x
end{split}
$$
while at the second side we simply have
$$
intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_x =intlimits_{ partial V} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
From these two relations, you immediately get the sought for formula.
How to prove directly, without using distribution theory, the following equality, key relation of the whole proof?
$$
intlimits_V U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=-(n-2)omega_{n-1}U(y)quadforall yin Vlabel{3}tag{3}
$$
I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression
$$
lim_{epsilonto 0} intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x,
$$
since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e.
$$
Delta frac{1}{|x-y|^{n-2}}=0text{ on }V(y,epsilon)implies intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=0quadforallepsilon>0.label{a}tag{*}
$$
However we can use eqref{a} and eqref{1} by defining, for each $B(y,epsilon)Subset V$,
$$
begin{split}
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x & triangleq lim_{epsilonto 0} intlimits_{B(y,epsilon)} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x \
triangleq lim_{epsilonto 0},&
left[ ,intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x + intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)right]
end{split}
$$
and, since
$$
intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}xunderset{epsilon to 0}{longrightarrow} 0
$$
because $Uin C^2(V)iff Delta Uin C^0(V)$ and $|x-y|^{2-n} in L^1_mathrm{loc}(Bbb R^n)$ for $n ge 2$, we finally get
$$
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x triangleq lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $xmapsto (r,theta_1,ldots,theta_{n-1})$: for the first one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} mathrm{d}sigma(x)&=lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U frac{ partial }{partial r} frac{1}{r^{n-2}}mathrm{d}sigma\
&=-lim_{epsilonto 0} frac{n-2}{epsilon^{n-1}} intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma\
&=-(n-2) omega_{n-1} lim_{epsilonto 0} left[frac{1}{omega_{n-1}epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma right]\
&= -(n-2) omega_{n-1} U(y)
end{split}
$$
since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $partial B(y,epsilon)$. For the second one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}mathrm{d}sigma(x) &= lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{r^{n-2}} frac{ partial U }{partial r} mathrm{d}sigma \
&= lim_{epsilonto 0} frac{1}{epsilon^{n-2}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigma \
&= omega_{n-1} lim_{epsilonto 0} epsilon cdot left[frac{1}{omega_{n-1} epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigmaright] \
&=0
end{split}
$$
since the integral within the square brackets is the spherical mean of the $C^1$ function $nabla U(x)$ over the sphere $partial B(y,epsilon)$. This last step proves equation eqref{3} and thus the formula in the OP question.
Notes
In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).
The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
$endgroup$
add a comment |
$begingroup$
I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity
$$
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x = intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_xlabel{1}tag{1}
$$
However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,epsilon)triangleq Vsetminus B(y,epsilon)$, where $B(y,epsilon)$ is the $n$-dimensional ball of radius $epsilon>0$ centered at $yinBbb R^n$. It is sufficient to recall that, for $ngeq 3$,
$$
Delta frac{1}{|y-x|^{n-2}}=-(n-2)omega_{n-1} delta(x-y)=-(n-2)omega_{n-1} delta(y-x)label{2}tag{2}
$$
where $delta$ is the usual Dirac distribution supported at $yinBbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting
$$
W(x)=frac{1}{|x-y|^{n-2}}quadtext{for an arbitrary }yinBbb R^n
$$
in formula eqref{1}, at the left side we get
$$
begin{split}
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x &= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x \
&= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x\
&= -(n-2)omega_{n-1}intlimits_V U(x)delta(y-x)mathrm{d}x - intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x\
&= -(n-2)omega_{n-1}U(y)- intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x
end{split}
$$
while at the second side we simply have
$$
intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_x =intlimits_{ partial V} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
From these two relations, you immediately get the sought for formula.
How to prove directly, without using distribution theory, the following equality, key relation of the whole proof?
$$
intlimits_V U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=-(n-2)omega_{n-1}U(y)quadforall yin Vlabel{3}tag{3}
$$
I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression
$$
lim_{epsilonto 0} intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x,
$$
since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e.
$$
Delta frac{1}{|x-y|^{n-2}}=0text{ on }V(y,epsilon)implies intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=0quadforallepsilon>0.label{a}tag{*}
$$
However we can use eqref{a} and eqref{1} by defining, for each $B(y,epsilon)Subset V$,
$$
begin{split}
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x & triangleq lim_{epsilonto 0} intlimits_{B(y,epsilon)} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x \
triangleq lim_{epsilonto 0},&
left[ ,intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x + intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)right]
end{split}
$$
and, since
$$
intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}xunderset{epsilon to 0}{longrightarrow} 0
$$
because $Uin C^2(V)iff Delta Uin C^0(V)$ and $|x-y|^{2-n} in L^1_mathrm{loc}(Bbb R^n)$ for $n ge 2$, we finally get
$$
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x triangleq lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $xmapsto (r,theta_1,ldots,theta_{n-1})$: for the first one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} mathrm{d}sigma(x)&=lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U frac{ partial }{partial r} frac{1}{r^{n-2}}mathrm{d}sigma\
&=-lim_{epsilonto 0} frac{n-2}{epsilon^{n-1}} intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma\
&=-(n-2) omega_{n-1} lim_{epsilonto 0} left[frac{1}{omega_{n-1}epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma right]\
&= -(n-2) omega_{n-1} U(y)
end{split}
$$
since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $partial B(y,epsilon)$. For the second one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}mathrm{d}sigma(x) &= lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{r^{n-2}} frac{ partial U }{partial r} mathrm{d}sigma \
&= lim_{epsilonto 0} frac{1}{epsilon^{n-2}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigma \
&= omega_{n-1} lim_{epsilonto 0} epsilon cdot left[frac{1}{omega_{n-1} epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigmaright] \
&=0
end{split}
$$
since the integral within the square brackets is the spherical mean of the $C^1$ function $nabla U(x)$ over the sphere $partial B(y,epsilon)$. This last step proves equation eqref{3} and thus the formula in the OP question.
Notes
In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).
The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
$endgroup$
add a comment |
$begingroup$
I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity
$$
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x = intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_xlabel{1}tag{1}
$$
However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,epsilon)triangleq Vsetminus B(y,epsilon)$, where $B(y,epsilon)$ is the $n$-dimensional ball of radius $epsilon>0$ centered at $yinBbb R^n$. It is sufficient to recall that, for $ngeq 3$,
$$
Delta frac{1}{|y-x|^{n-2}}=-(n-2)omega_{n-1} delta(x-y)=-(n-2)omega_{n-1} delta(y-x)label{2}tag{2}
$$
where $delta$ is the usual Dirac distribution supported at $yinBbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting
$$
W(x)=frac{1}{|x-y|^{n-2}}quadtext{for an arbitrary }yinBbb R^n
$$
in formula eqref{1}, at the left side we get
$$
begin{split}
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x &= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x \
&= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x\
&= -(n-2)omega_{n-1}intlimits_V U(x)delta(y-x)mathrm{d}x - intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x\
&= -(n-2)omega_{n-1}U(y)- intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x
end{split}
$$
while at the second side we simply have
$$
intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_x =intlimits_{ partial V} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
From these two relations, you immediately get the sought for formula.
How to prove directly, without using distribution theory, the following equality, key relation of the whole proof?
$$
intlimits_V U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=-(n-2)omega_{n-1}U(y)quadforall yin Vlabel{3}tag{3}
$$
I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression
$$
lim_{epsilonto 0} intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x,
$$
since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e.
$$
Delta frac{1}{|x-y|^{n-2}}=0text{ on }V(y,epsilon)implies intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=0quadforallepsilon>0.label{a}tag{*}
$$
However we can use eqref{a} and eqref{1} by defining, for each $B(y,epsilon)Subset V$,
$$
begin{split}
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x & triangleq lim_{epsilonto 0} intlimits_{B(y,epsilon)} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x \
triangleq lim_{epsilonto 0},&
left[ ,intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x + intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)right]
end{split}
$$
and, since
$$
intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}xunderset{epsilon to 0}{longrightarrow} 0
$$
because $Uin C^2(V)iff Delta Uin C^0(V)$ and $|x-y|^{2-n} in L^1_mathrm{loc}(Bbb R^n)$ for $n ge 2$, we finally get
$$
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x triangleq lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $xmapsto (r,theta_1,ldots,theta_{n-1})$: for the first one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} mathrm{d}sigma(x)&=lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U frac{ partial }{partial r} frac{1}{r^{n-2}}mathrm{d}sigma\
&=-lim_{epsilonto 0} frac{n-2}{epsilon^{n-1}} intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma\
&=-(n-2) omega_{n-1} lim_{epsilonto 0} left[frac{1}{omega_{n-1}epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma right]\
&= -(n-2) omega_{n-1} U(y)
end{split}
$$
since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $partial B(y,epsilon)$. For the second one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}mathrm{d}sigma(x) &= lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{r^{n-2}} frac{ partial U }{partial r} mathrm{d}sigma \
&= lim_{epsilonto 0} frac{1}{epsilon^{n-2}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigma \
&= omega_{n-1} lim_{epsilonto 0} epsilon cdot left[frac{1}{omega_{n-1} epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigmaright] \
&=0
end{split}
$$
since the integral within the square brackets is the spherical mean of the $C^1$ function $nabla U(x)$ over the sphere $partial B(y,epsilon)$. This last step proves equation eqref{3} and thus the formula in the OP question.
Notes
In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).
The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
$endgroup$
I changed a bit the notation from the original post, and added sections in order to give a better answer.
Let's start by recalling, as supinf does in his answer, that the above formula is a consequence of Green's second identity
$$
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x = intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_xlabel{1}tag{1}
$$
However, it is not needed to consider a deleted neighborhood of $y$, i.e. $V(y,epsilon)triangleq Vsetminus B(y,epsilon)$, where $B(y,epsilon)$ is the $n$-dimensional ball of radius $epsilon>0$ centered at $yinBbb R^n$. It is sufficient to recall that, for $ngeq 3$,
$$
Delta frac{1}{|y-x|^{n-2}}=-(n-2)omega_{n-1} delta(x-y)=-(n-2)omega_{n-1} delta(y-x)label{2}tag{2}
$$
where $delta$ is the usual Dirac distribution supported at $yinBbb R^n$, and the laplacian is calculated respect to the $x$ variable. Then, putting
$$
W(x)=frac{1}{|x-y|^{n-2}}quadtext{for an arbitrary }yinBbb R^n
$$
in formula eqref{1}, at the left side we get
$$
begin{split}
intlimits_V big[U(x)Delta W(x) - W(x)Delta U(x)big]mathrm{d}x &= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x \
&= intlimits_V bigg[U(x)Delta frac{1}{|x-y|^{n-2}} - frac{Delta U(x)}{|x-y|^{n-2}}bigg]mathrm{d}x\
&= -(n-2)omega_{n-1}intlimits_V U(x)delta(y-x)mathrm{d}x - intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x\
&= -(n-2)omega_{n-1}U(y)- intlimits_Vfrac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x
end{split}
$$
while at the second side we simply have
$$
intlimits_{partial V}bigg[U(x)frac{ partial W(x)}{ partial nu_x}-W(x)frac{ partial U(x)}{ partial nu_x}bigg] mathrm{d}sigma_x =intlimits_{ partial V} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
From these two relations, you immediately get the sought for formula.
How to prove directly, without using distribution theory, the following equality, key relation of the whole proof?
$$
intlimits_V U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=-(n-2)omega_{n-1}U(y)quadforall yin Vlabel{3}tag{3}
$$
I added this section since it seems that the main problem posed the OP is how to prove this fact (and thus the above statement) by using a classical "hard analysis" argument.
The first thing to note is that the integral at the left side of eqref{3} should be intended in a generalized sense: it cannot be evaluated by using the limit expression
$$
lim_{epsilonto 0} intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x,
$$
since $|x-y|^{2-n}$ is harmonic on every deleted neighborhood of $y$, i.e.
$$
Delta frac{1}{|x-y|^{n-2}}=0text{ on }V(y,epsilon)implies intlimits_{V_epsilon} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x=0quadforallepsilon>0.label{a}tag{*}
$$
However we can use eqref{a} and eqref{1} by defining, for each $B(y,epsilon)Subset V$,
$$
begin{split}
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x & triangleq lim_{epsilonto 0} intlimits_{B(y,epsilon)} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x \
triangleq lim_{epsilonto 0},&
left[ ,intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}x + intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)right]
end{split}
$$
and, since
$$
intlimits_{B(y,epsilon)}frac{Delta U(x)}{|x-y|^{n-2}}mathrm{d}xunderset{epsilon to 0}{longrightarrow} 0
$$
because $Uin C^2(V)iff Delta Uin C^0(V)$ and $|x-y|^{2-n} in L^1_mathrm{loc}(Bbb R^n)$ for $n ge 2$, we finally get
$$
intlimits_{V} U(x)Delta frac{1}{|x-y|^{n-2}}mathrm{d}x triangleq lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} left[U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} - frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}right] mathrm{d}sigma(x)
$$
Now let's evaluate each term of the right member of this "definition", by passing to hyperspherical coordinates $xmapsto (r,theta_1,ldots,theta_{n-1})$: for the first one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} mathrm{d}sigma(x)&=lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} U frac{ partial }{partial r} frac{1}{r^{n-2}}mathrm{d}sigma\
&=-lim_{epsilonto 0} frac{n-2}{epsilon^{n-1}} intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma\
&=-(n-2) omega_{n-1} lim_{epsilonto 0} left[frac{1}{omega_{n-1}epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} U mathrm{d}sigma right]\
&= -(n-2) omega_{n-1} U(y)
end{split}
$$
since the integral within square brackets is nothing less than the spherical mean of the $C^2$ function $U(x)$ over the sphere $partial B(y,epsilon)$. For the second one we have
$$
begin{split}
lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{|x-y|^{n-2}} frac{ partial U(x)}{ partial nu_x}mathrm{d}sigma(x) &= lim_{epsilonto 0} intlimits_{ partial B(y,epsilon)} frac{1}{r^{n-2}} frac{ partial U }{partial r} mathrm{d}sigma \
&= lim_{epsilonto 0} frac{1}{epsilon^{n-2}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigma \
&= omega_{n-1} lim_{epsilonto 0} epsilon cdot left[frac{1}{omega_{n-1} epsilon^{n-1}}intlimits_{ partial B(y,epsilon)} frac{ partial U}{partial r} mathrm{d}sigmaright] \
&=0
end{split}
$$
since the integral within the square brackets is the spherical mean of the $C^1$ function $nabla U(x)$ over the sphere $partial B(y,epsilon)$. This last step proves equation eqref{3} and thus the formula in the OP question.
Notes
In many mathematical physics texts, the OP's formula is called Green's formula (see for example [1], chapter IV, §2.1, p. 318 and [2], §4.9.2, pp. 69).
The "soft analysis" (distribution theory) approach described in the first part of the answer is similar to the one proposed by Vladimirov ([2], §4.9.2, pp. 69-70), though adapted to the question notation, while the "hard analysis" proof of eqref{3} is inspired to the one given by Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318) who, however, deal only with the case $n=3$ and work on the whole formula eqref{1} in order to avoid the difficulties intrinsically inherent to the definition of the first member of eqref{3}. Each of these approaches has its own merits, the firs one being conceptually simpler but requiring considerable more background, while the second one requiring some more analytical machinery, however interesting per se.
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
edited Jan 13 at 17:21
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
answered Jan 10 at 22:38
Daniele TampieriDaniele Tampieri
2,1291621
2,1291621
add a comment |
add a comment |
$begingroup$
Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_epsilon$.
If we also use that $Delta W(x)=0$ for $xneq y$ then the equation becomes
$$
(n-2) omega_{n-1} U(y) = int_{ partial K_epsilon} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x).
$$
Note that the terms involving integrals over $V$ or $partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $epsilonto0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
left|int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x) dsigma(x)right|
leq
C int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} dsigma(x)
= C epsilon^{2-n} omega_{n-1} epsilon^{n-1}to 0,
$$
where we used that the surface area of a ball with radius $epsilon>0$ is given by $omega_{n-1}epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}
= -(n-2) frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
int_{ partial K_epsilon} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} dsigma(x)
=
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
left|
(n-2)omega_{n-1} U(y)
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x)
right|
=left|
(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(y)-U(x) dsigma(x)
right|
leq sup_{xin partial K_epsilon} |U(y)-U(x)| (n-2)epsilon^{-(n-1)}omega_{n-1} epsilon^{n-1}
to0
$$
where the $sup$-term converges to $0$ because $U$ is continuous.
answered Jan 9 at 13:06
supinfsupinf
6,2161028
$endgroup$
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
add a comment |
$begingroup$
Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_epsilon$.
If we also use that $Delta W(x)=0$ for $xneq y$ then the equation becomes
$$
(n-2) omega_{n-1} U(y) = int_{ partial K_epsilon} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x).
$$
Note that the terms involving integrals over $V$ or $partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $epsilonto0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
left|int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x) dsigma(x)right|
leq
C int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} dsigma(x)
= C epsilon^{2-n} omega_{n-1} epsilon^{n-1}to 0,
$$
where we used that the surface area of a ball with radius $epsilon>0$ is given by $omega_{n-1}epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}
= -(n-2) frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
int_{ partial K_epsilon} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} dsigma(x)
=
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
left|
(n-2)omega_{n-1} U(y)
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x)
right|
=left|
(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(y)-U(x) dsigma(x)
right|
leq sup_{xin partial K_epsilon} |U(y)-U(x)| (n-2)epsilon^{-(n-1)}omega_{n-1} epsilon^{n-1}
to0
$$
where the $sup$-term converges to $0$ because $U$ is continuous.
answered Jan 9 at 13:06
supinfsupinf
6,2161028
$endgroup$
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
add a comment |
$begingroup$
Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_epsilon$.
If we also use that $Delta W(x)=0$ for $xneq y$ then the equation becomes
$$
(n-2) omega_{n-1} U(y) = int_{ partial K_epsilon} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x).
$$
Note that the terms involving integrals over $V$ or $partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $epsilonto0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
left|int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x) dsigma(x)right|
leq
C int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} dsigma(x)
= C epsilon^{2-n} omega_{n-1} epsilon^{n-1}to 0,
$$
where we used that the surface area of a ball with radius $epsilon>0$ is given by $omega_{n-1}epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}
= -(n-2) frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
int_{ partial K_epsilon} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} dsigma(x)
=
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
left|
(n-2)omega_{n-1} U(y)
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x)
right|
=left|
(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(y)-U(x) dsigma(x)
right|
leq sup_{xin partial K_epsilon} |U(y)-U(x)| (n-2)epsilon^{-(n-1)}omega_{n-1} epsilon^{n-1}
to0
$$
where the $sup$-term converges to $0$ because $U$ is continuous.
answered Jan 9 at 13:06
supinfsupinf
6,2161028
$endgroup$
Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_epsilon$.
If we also use that $Delta W(x)=0$ for $xneq y$ then the equation becomes
$$
(n-2) omega_{n-1} U(y) = int_{ partial K_epsilon} left[frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x)-U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}right] dsigma(x).
$$
Note that the terms involving integrals over $V$ or $partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $epsilonto0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
left|int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} frac{ partial U}{ partial nu} (x) dsigma(x)right|
leq
C int_{ partial K_epsilon} frac{1}{|x-y|^{n-2}} dsigma(x)
= C epsilon^{2-n} omega_{n-1} epsilon^{n-1}to 0,
$$
where we used that the surface area of a ball with radius $epsilon>0$ is given by $omega_{n-1}epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}}
= -(n-2) frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
int_{ partial K_epsilon} U(x) frac{ partial }{partial nu_x} frac{1}{|x-y|^{n-2}} dsigma(x)
=
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
left|
(n-2)omega_{n-1} U(y)
-(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(x) dsigma(x)
right|
=left|
(n-2) epsilon^{-(n-1)} int_{ partial K_epsilon} U(y)-U(x) dsigma(x)
right|
leq sup_{xin partial K_epsilon} |U(y)-U(x)| (n-2)epsilon^{-(n-1)}omega_{n-1} epsilon^{n-1}
to0
$$
where the $sup$-term converges to $0$ because $U$ is continuous.
answered Jan 9 at 13:06
supinfsupinf
6,2161028
edited Jan 10 at 7:41
answered Jan 9 at 13:06
supinfsupinf
6,2161028
answered Jan 9 at 13:06
supinfsupinf
6,2161028
answered Jan 9 at 13:06
supinfsupinf
6,2161028
6,2161028
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
add a comment |
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
firstly; thank you so much !. I got it till the part where to show that one of the two right-hand side terms are left. I don't know how to proceed showing that one Term is converging to 0 and for the other part as well..It would be awesome if you could show me some details ! :)
$endgroup$
– constant94
Jan 9 at 15:49
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
$begingroup$
@constant94 i added some more details.
$endgroup$
– supinf
Jan 10 at 7:42
add a comment |
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