Mixing
$int_V left( gnabla^2f-f nabla^2g right) dV=int_S left( gnabla f-f nabla g right) cdot u_n , dS$ not depend on time
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I' m wondering why the following relationship, known as Green's identity, doesn't depends on time.
Let
$f(x,y,z,t)$ and $g=frace^ikrr$
so
$$int_V left( gnabla^2f-f nabla^2g right) dV=int_S left( gnabla f-f nabla g right) cdot u_n , dS$$
integration laplacian greens-theorem gradient-flows
asked Jul 21 at 12:49
Stefano Barone
379110
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up vote
0
down vote
favorite
I' m wondering why the following relationship, known as Green's identity, doesn't depends on time.
Let
$f(x,y,z,t)$ and $g=frace^ikrr$
so
$$int_V left( gnabla^2f-f nabla^2g right) dV=int_S left( gnabla f-f nabla g right) cdot u_n , dS$$
integration laplacian greens-theorem gradient-flows
asked Jul 21 at 12:49
Stefano Barone
379110
1
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I' m wondering why the following relationship, known as Green's identity, doesn't depends on time.
Let
$f(x,y,z,t)$ and $g=frace^ikrr$
so
$$int_V left( gnabla^2f-f nabla^2g right) dV=int_S left( gnabla f-f nabla g right) cdot u_n , dS$$
integration laplacian greens-theorem gradient-flows
asked Jul 21 at 12:49
Stefano Barone
379110
I' m wondering why the following relationship, known as Green's identity, doesn't depends on time.
Let
$f(x,y,z,t)$ and $g=frace^ikrr$
so
$$int_V left( gnabla^2f-f nabla^2g right) dV=int_S left( gnabla f-f nabla g right) cdot u_n , dS$$
integration laplacian greens-theorem gradient-flows
asked Jul 21 at 12:49
Stefano Barone
379110
asked Jul 21 at 12:49
Stefano Barone
379110
asked Jul 21 at 12:49
Stefano Barone
379110
asked Jul 21 at 12:49
Stefano Barone
379110
379110
1
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03
add a comment |Â
1
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03
1
1
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03
add a comment |Â
1 Answer
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The identity is easily shown using the divergence theorem:
$$
int_V left( g , nabla^2 f - f , nabla^2 g right) dV
= int_V nabla cdot left( g , nabla f - f , nabla g right) dV
= oint_S left( g , nabla f - f , nabla g right) cdot u_n , dS
$$
Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $int cdot , dV$ being only over space, not over time.
answered Jul 21 at 13:10
md2perpe
5,87511022
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The identity is easily shown using the divergence theorem:
$$
int_V left( g , nabla^2 f - f , nabla^2 g right) dV
= int_V nabla cdot left( g , nabla f - f , nabla g right) dV
= oint_S left( g , nabla f - f , nabla g right) cdot u_n , dS
$$
Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $int cdot , dV$ being only over space, not over time.
answered Jul 21 at 13:10
md2perpe
5,87511022
add a comment |Â
up vote
1
down vote
The identity is easily shown using the divergence theorem:
$$
int_V left( g , nabla^2 f - f , nabla^2 g right) dV
= int_V nabla cdot left( g , nabla f - f , nabla g right) dV
= oint_S left( g , nabla f - f , nabla g right) cdot u_n , dS
$$
Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $int cdot , dV$ being only over space, not over time.
answered Jul 21 at 13:10
md2perpe
5,87511022
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The identity is easily shown using the divergence theorem:
$$
int_V left( g , nabla^2 f - f , nabla^2 g right) dV
= int_V nabla cdot left( g , nabla f - f , nabla g right) dV
= oint_S left( g , nabla f - f , nabla g right) cdot u_n , dS
$$
Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $int cdot , dV$ being only over space, not over time.
answered Jul 21 at 13:10
md2perpe
5,87511022
The identity is easily shown using the divergence theorem:
$$
int_V left( g , nabla^2 f - f , nabla^2 g right) dV
= int_V nabla cdot left( g , nabla f - f , nabla g right) dV
= oint_S left( g , nabla f - f , nabla g right) cdot u_n , dS
$$
Both $f$ and $g$ may depend on $t$, but this dependency does not enter the identity. There will be no derivatives w.r.t. $t$. That is related to the integral $int cdot , dV$ being only over space, not over time.
answered Jul 21 at 13:10
md2perpe
5,87511022
answered Jul 21 at 13:10
md2perpe
5,87511022
answered Jul 21 at 13:10
md2perpe
5,87511022
answered Jul 21 at 13:10
md2perpe
5,87511022
5,87511022
add a comment |Â
add a comment |Â
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1
Assuming $dV$ is the volume form in $xyz$-space, both the quantity on the right-hand side of your equality and the quantity on the left-hand side of your equality depend on $t$. Thus, your equality should be interpreted as equality of functions of $t$ (i.e. for each fixed $t$ the left-hand side equals the right-hand side).
– BindersFull
Jul 21 at 13:03