Intuition behind Pohozaev identity
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I'm wondering if the Pohozaev identity:
$$nint_Omegaint_0^{u(x)}f(t)operatorname{d}toperatorname{d}x-frac{n-2}{2}int_Omega u(x)f(u(x))operatorname{d}x=frac{1}{2}int_{partialOmega}left|frac{partial u}{partialnu}(x)right|^2xcdotnu(x)operatorname{d}S(x)$$
where $Omega$ is a non-empty open bounded subset of $mathbb{R}^n$ with smooth boundary, $nu:partialOmegatomathbb{R}^n$ is the outer normal, $S$ is the surface measure, $fin C^1(mathbb{R})$ and $uin C^2(Omega)cap C^1_0(barOmega)$ is such that $-Delta u=f(u)$,
is just a "black magic formula" or if there some kind of intuition behind it... I looked into its proof, but actually it seems just a lot of dirty tricks about rewriting terms in order to use divergence theorem several times, leaving me a bit confused about the meaning of this formula... Can anyone give some kind of (geometric?) insight about this formula and/or can show an intuitive way to prove it?
calculus-of-variations variational-analysis
asked Nov 28 '18 at 17:07
BobBob
1,5381625
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add a comment |
$begingroup$
I'm wondering if the Pohozaev identity:
$$nint_Omegaint_0^{u(x)}f(t)operatorname{d}toperatorname{d}x-frac{n-2}{2}int_Omega u(x)f(u(x))operatorname{d}x=frac{1}{2}int_{partialOmega}left|frac{partial u}{partialnu}(x)right|^2xcdotnu(x)operatorname{d}S(x)$$
where $Omega$ is a non-empty open bounded subset of $mathbb{R}^n$ with smooth boundary, $nu:partialOmegatomathbb{R}^n$ is the outer normal, $S$ is the surface measure, $fin C^1(mathbb{R})$ and $uin C^2(Omega)cap C^1_0(barOmega)$ is such that $-Delta u=f(u)$,
is just a "black magic formula" or if there some kind of intuition behind it... I looked into its proof, but actually it seems just a lot of dirty tricks about rewriting terms in order to use divergence theorem several times, leaving me a bit confused about the meaning of this formula... Can anyone give some kind of (geometric?) insight about this formula and/or can show an intuitive way to prove it?
calculus-of-variations variational-analysis
asked Nov 28 '18 at 17:07
BobBob
1,5381625
$endgroup$
add a comment |
$begingroup$
I'm wondering if the Pohozaev identity:
$$nint_Omegaint_0^{u(x)}f(t)operatorname{d}toperatorname{d}x-frac{n-2}{2}int_Omega u(x)f(u(x))operatorname{d}x=frac{1}{2}int_{partialOmega}left|frac{partial u}{partialnu}(x)right|^2xcdotnu(x)operatorname{d}S(x)$$
where $Omega$ is a non-empty open bounded subset of $mathbb{R}^n$ with smooth boundary, $nu:partialOmegatomathbb{R}^n$ is the outer normal, $S$ is the surface measure, $fin C^1(mathbb{R})$ and $uin C^2(Omega)cap C^1_0(barOmega)$ is such that $-Delta u=f(u)$,
is just a "black magic formula" or if there some kind of intuition behind it... I looked into its proof, but actually it seems just a lot of dirty tricks about rewriting terms in order to use divergence theorem several times, leaving me a bit confused about the meaning of this formula... Can anyone give some kind of (geometric?) insight about this formula and/or can show an intuitive way to prove it?
calculus-of-variations variational-analysis
asked Nov 28 '18 at 17:07
BobBob
1,5381625
$endgroup$
I'm wondering if the Pohozaev identity:
$$nint_Omegaint_0^{u(x)}f(t)operatorname{d}toperatorname{d}x-frac{n-2}{2}int_Omega u(x)f(u(x))operatorname{d}x=frac{1}{2}int_{partialOmega}left|frac{partial u}{partialnu}(x)right|^2xcdotnu(x)operatorname{d}S(x)$$
where $Omega$ is a non-empty open bounded subset of $mathbb{R}^n$ with smooth boundary, $nu:partialOmegatomathbb{R}^n$ is the outer normal, $S$ is the surface measure, $fin C^1(mathbb{R})$ and $uin C^2(Omega)cap C^1_0(barOmega)$ is such that $-Delta u=f(u)$,
is just a "black magic formula" or if there some kind of intuition behind it... I looked into its proof, but actually it seems just a lot of dirty tricks about rewriting terms in order to use divergence theorem several times, leaving me a bit confused about the meaning of this formula... Can anyone give some kind of (geometric?) insight about this formula and/or can show an intuitive way to prove it?
calculus-of-variations variational-analysis
calculus-of-variations variational-analysis
asked Nov 28 '18 at 17:07
BobBob
1,5381625
asked Nov 28 '18 at 17:07
BobBob
1,5381625
edited Nov 28 '18 at 17:43
Bob
asked Nov 28 '18 at 17:07
BobBob
1,5381625
asked Nov 28 '18 at 17:07
BobBob
1,5381625
asked Nov 28 '18 at 17:07
BobBob
1,5381625
1,5381625
add a comment |
add a comment |
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