how to show: Is $u$ harmonic, so are following identities true
$begingroup$
Let $V subset mathbb{R}^n , 2 leq n $
be a set, where you can apply Gauß's Theorem.
To show: Is $ u in C^{(2)}( bar{V} ) $ harmonic on $V$, then:
$$ int_{ partial V} frac{ partial u }{ partial nu } do =0 $$
and
$$ int_V | nabla u(x)|^2 dx = int_{ partial V} u frac{ partial u }{partial nu} do $$
where $ nu $ is the unit normal on $ partial V $
__
Probably it's a bit remodelling, but I still can not get around.
Okay, so the Gauß's Theorem says
$$ int_{ Omega} nabla v dx = int_{ partial Omega} v n ds $$
and If $u in C^{(2)} $ is harmonic then :
$$ Delta u = frac{ partial^2 u}{ partial x^2} + frac{ partial^2 u}{ partial y^2}=0 $$
real-analysis integration gaussian-integral
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
Let $V subset mathbb{R}^n , 2 leq n $
be a set, where you can apply Gauß's Theorem.
To show: Is $ u in C^{(2)}( bar{V} ) $ harmonic on $V$, then:
$$ int_{ partial V} frac{ partial u }{ partial nu } do =0 $$
and
$$ int_V | nabla u(x)|^2 dx = int_{ partial V} u frac{ partial u }{partial nu} do $$
where $ nu $ is the unit normal on $ partial V $
__
Probably it's a bit remodelling, but I still can not get around.
Okay, so the Gauß's Theorem says
$$ int_{ Omega} nabla v dx = int_{ partial Omega} v n ds $$
and If $u in C^{(2)} $ is harmonic then :
$$ Delta u = frac{ partial^2 u}{ partial x^2} + frac{ partial^2 u}{ partial y^2}=0 $$
real-analysis integration gaussian-integral
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
Let $V subset mathbb{R}^n , 2 leq n $
be a set, where you can apply Gauß's Theorem.
To show: Is $ u in C^{(2)}( bar{V} ) $ harmonic on $V$, then:
$$ int_{ partial V} frac{ partial u }{ partial nu } do =0 $$
and
$$ int_V | nabla u(x)|^2 dx = int_{ partial V} u frac{ partial u }{partial nu} do $$
where $ nu $ is the unit normal on $ partial V $
__
Probably it's a bit remodelling, but I still can not get around.
Okay, so the Gauß's Theorem says
$$ int_{ Omega} nabla v dx = int_{ partial Omega} v n ds $$
and If $u in C^{(2)} $ is harmonic then :
$$ Delta u = frac{ partial^2 u}{ partial x^2} + frac{ partial^2 u}{ partial y^2}=0 $$
real-analysis integration gaussian-integral
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
$endgroup$
Let $V subset mathbb{R}^n , 2 leq n $
be a set, where you can apply Gauß's Theorem.
To show: Is $ u in C^{(2)}( bar{V} ) $ harmonic on $V$, then:
$$ int_{ partial V} frac{ partial u }{ partial nu } do =0 $$
and
$$ int_V | nabla u(x)|^2 dx = int_{ partial V} u frac{ partial u }{partial nu} do $$
where $ nu $ is the unit normal on $ partial V $
__
Probably it's a bit remodelling, but I still can not get around.
Okay, so the Gauß's Theorem says
$$ int_{ Omega} nabla v dx = int_{ partial Omega} v n ds $$
and If $u in C^{(2)} $ is harmonic then :
$$ Delta u = frac{ partial^2 u}{ partial x^2} + frac{ partial^2 u}{ partial y^2}=0 $$
real-analysis integration gaussian-integral
real-analysis integration gaussian-integral
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
edited Dec 19 '18 at 21:35
Robert Lewis
47.2k23067
47.2k23067
asked Dec 19 '18 at 21:33
constant94constant94
6310
asked Dec 19 '18 at 21:33
constant94constant94
6310
asked Dec 19 '18 at 21:33
constant94constant94
6310
6310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that
$Delta u = nabla^2 u = nabla cdot nabla u; tag 1$
if $u$ is harmonic on $bar V$, then
$Delta u = nabla^2 u(x) = 0, ; x in bar V; tag 2$
we may apply the divergence theorem of Gauss to the integral
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do; tag 3$
on $partial V$ we have
$dfrac{partial u}{partial nu} = nabla u cdot mathbf n, tag 4$
where $mathbf n$ is the outward-pointing unit normal vector field on $partial V$; then using the divergence theorem we find
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do = displaystyle int_{partial V} nabla u cdot mathbf n ; do = int_V nabla cdot nabla u ; dV = int_V nabla^2 u ; dV = 0, tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$nabla cdot (u nabla u) = nabla u cdot nabla u + u nabla cdot nabla u = vert nabla u vert^2 + unabla^2u = vert nabla u vert^2 tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$displaystyle int_V vert nabla u vert^2 ; dV = int_V nabla cdot (u nabla u) ; dV; tag 7$
another application of the divergence theorem similar to the above yields
$displaystyle int_V nabla cdot (u nabla u) ; dV = int_{partial V} u nabla u cdot mathbf n ; do = int_{partial V} udfrac{partial u}{partial nu} ; do; tag 8$
combining (7) and (8) yields
$displaystyle int_V vert nabla u vert^2 ; dV = displaystyle int u dfrac{partial u}{partial nu} ; do, tag 9$
the requisite result. $OEDelta$.
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that
$Delta u = nabla^2 u = nabla cdot nabla u; tag 1$
if $u$ is harmonic on $bar V$, then
$Delta u = nabla^2 u(x) = 0, ; x in bar V; tag 2$
we may apply the divergence theorem of Gauss to the integral
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do; tag 3$
on $partial V$ we have
$dfrac{partial u}{partial nu} = nabla u cdot mathbf n, tag 4$
where $mathbf n$ is the outward-pointing unit normal vector field on $partial V$; then using the divergence theorem we find
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do = displaystyle int_{partial V} nabla u cdot mathbf n ; do = int_V nabla cdot nabla u ; dV = int_V nabla^2 u ; dV = 0, tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$nabla cdot (u nabla u) = nabla u cdot nabla u + u nabla cdot nabla u = vert nabla u vert^2 + unabla^2u = vert nabla u vert^2 tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$displaystyle int_V vert nabla u vert^2 ; dV = int_V nabla cdot (u nabla u) ; dV; tag 7$
another application of the divergence theorem similar to the above yields
$displaystyle int_V nabla cdot (u nabla u) ; dV = int_{partial V} u nabla u cdot mathbf n ; do = int_{partial V} udfrac{partial u}{partial nu} ; do; tag 8$
combining (7) and (8) yields
$displaystyle int_V vert nabla u vert^2 ; dV = displaystyle int u dfrac{partial u}{partial nu} ; do, tag 9$
the requisite result. $OEDelta$.
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
$endgroup$
add a comment |
$begingroup$
Recall that
$Delta u = nabla^2 u = nabla cdot nabla u; tag 1$
if $u$ is harmonic on $bar V$, then
$Delta u = nabla^2 u(x) = 0, ; x in bar V; tag 2$
we may apply the divergence theorem of Gauss to the integral
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do; tag 3$
on $partial V$ we have
$dfrac{partial u}{partial nu} = nabla u cdot mathbf n, tag 4$
where $mathbf n$ is the outward-pointing unit normal vector field on $partial V$; then using the divergence theorem we find
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do = displaystyle int_{partial V} nabla u cdot mathbf n ; do = int_V nabla cdot nabla u ; dV = int_V nabla^2 u ; dV = 0, tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$nabla cdot (u nabla u) = nabla u cdot nabla u + u nabla cdot nabla u = vert nabla u vert^2 + unabla^2u = vert nabla u vert^2 tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$displaystyle int_V vert nabla u vert^2 ; dV = int_V nabla cdot (u nabla u) ; dV; tag 7$
another application of the divergence theorem similar to the above yields
$displaystyle int_V nabla cdot (u nabla u) ; dV = int_{partial V} u nabla u cdot mathbf n ; do = int_{partial V} udfrac{partial u}{partial nu} ; do; tag 8$
combining (7) and (8) yields
$displaystyle int_V vert nabla u vert^2 ; dV = displaystyle int u dfrac{partial u}{partial nu} ; do, tag 9$
the requisite result. $OEDelta$.
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
$endgroup$
add a comment |
$begingroup$
Recall that
$Delta u = nabla^2 u = nabla cdot nabla u; tag 1$
if $u$ is harmonic on $bar V$, then
$Delta u = nabla^2 u(x) = 0, ; x in bar V; tag 2$
we may apply the divergence theorem of Gauss to the integral
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do; tag 3$
on $partial V$ we have
$dfrac{partial u}{partial nu} = nabla u cdot mathbf n, tag 4$
where $mathbf n$ is the outward-pointing unit normal vector field on $partial V$; then using the divergence theorem we find
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do = displaystyle int_{partial V} nabla u cdot mathbf n ; do = int_V nabla cdot nabla u ; dV = int_V nabla^2 u ; dV = 0, tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$nabla cdot (u nabla u) = nabla u cdot nabla u + u nabla cdot nabla u = vert nabla u vert^2 + unabla^2u = vert nabla u vert^2 tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$displaystyle int_V vert nabla u vert^2 ; dV = int_V nabla cdot (u nabla u) ; dV; tag 7$
another application of the divergence theorem similar to the above yields
$displaystyle int_V nabla cdot (u nabla u) ; dV = int_{partial V} u nabla u cdot mathbf n ; do = int_{partial V} udfrac{partial u}{partial nu} ; do; tag 8$
combining (7) and (8) yields
$displaystyle int_V vert nabla u vert^2 ; dV = displaystyle int u dfrac{partial u}{partial nu} ; do, tag 9$
the requisite result. $OEDelta$.
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
$endgroup$
Recall that
$Delta u = nabla^2 u = nabla cdot nabla u; tag 1$
if $u$ is harmonic on $bar V$, then
$Delta u = nabla^2 u(x) = 0, ; x in bar V; tag 2$
we may apply the divergence theorem of Gauss to the integral
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do; tag 3$
on $partial V$ we have
$dfrac{partial u}{partial nu} = nabla u cdot mathbf n, tag 4$
where $mathbf n$ is the outward-pointing unit normal vector field on $partial V$; then using the divergence theorem we find
$displaystyle int_{partial V} dfrac{partial u}{partial nu} ; do = displaystyle int_{partial V} nabla u cdot mathbf n ; do = int_V nabla cdot nabla u ; dV = int_V nabla^2 u ; dV = 0, tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$nabla cdot (u nabla u) = nabla u cdot nabla u + u nabla cdot nabla u = vert nabla u vert^2 + unabla^2u = vert nabla u vert^2 tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$displaystyle int_V vert nabla u vert^2 ; dV = int_V nabla cdot (u nabla u) ; dV; tag 7$
another application of the divergence theorem similar to the above yields
$displaystyle int_V nabla cdot (u nabla u) ; dV = int_{partial V} u nabla u cdot mathbf n ; do = int_{partial V} udfrac{partial u}{partial nu} ; do; tag 8$
combining (7) and (8) yields
$displaystyle int_V vert nabla u vert^2 ; dV = displaystyle int u dfrac{partial u}{partial nu} ; do, tag 9$
the requisite result. $OEDelta$.
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
edited Dec 19 '18 at 23:05
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
answered Dec 19 '18 at 22:52
Robert LewisRobert Lewis
47.2k23067
47.2k23067
add a comment |
add a comment |
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