How to derive boundary conditions for variational problems?

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I have a variational problem of the form



$$
E(u) = int_Omega F(x,y,u,u_x,u_y)dxdy
$$



which leads me



$$
fracdEdu = int_Omega left( fracpartial Fpartial u - fracddxfracpartial Fpartial u_x - fracddyfracpartial Fpartial u_y right)h dxdy + int_Gamma left(F_u_xdy - F_u_ydx right)h
$$



Assuming the increment function $h$ is not fixed,it can be arbitrary, how do I get from this term



$$
int_Gamma left(F_u_xdy - F_u_ydx right)h = 0
$$



the boundary conditions of my problem when I try to apply gradient descent? Namely how do I fill the system



$$
left{
beginarrayl
fracpartial upartial t = - fracpartial Fpartial u + fracddxfracpartial Fpartial u_x + fracddyfracpartial Fpartial u_y \
textboundary conditions?
endarray
right.
$$



I should end up having the Von Neumann boundary conditions, but I'm slightly confused how to pass from the line integral to the boundary conditions.



My guess is that we should pose $F_u_x = 0, F_u_y = 0$ on the boundary, I can't prove it though.



Thank you




pde calculus-of-variations boundary-value-problem




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  • $h=h(x)$? if so, not clear in your question
    – phdmba7of12
    Aug 3 at 16:28










  • $h=0$ on the boudary.
    – Rafa Budría
    Aug 3 at 16:36






  • 1




    @phdmba7of12, yes it's h=h(x)
    – user8469759
    Aug 3 at 17:04






  • 1




    @Rafa, it's not 0.
    – user8469759
    Aug 4 at 13:19














up vote
1
down vote

favorite
1












I have a variational problem of the form



$$
E(u) = int_Omega F(x,y,u,u_x,u_y)dxdy
$$



which leads me



$$
fracdEdu = int_Omega left( fracpartial Fpartial u - fracddxfracpartial Fpartial u_x - fracddyfracpartial Fpartial u_y right)h dxdy + int_Gamma left(F_u_xdy - F_u_ydx right)h
$$



Assuming the increment function $h$ is not fixed,it can be arbitrary, how do I get from this term



$$
int_Gamma left(F_u_xdy - F_u_ydx right)h = 0
$$



the boundary conditions of my problem when I try to apply gradient descent? Namely how do I fill the system



$$
left{
beginarrayl
fracpartial upartial t = - fracpartial Fpartial u + fracddxfracpartial Fpartial u_x + fracddyfracpartial Fpartial u_y \
textboundary conditions?
endarray
right.
$$



I should end up having the Von Neumann boundary conditions, but I'm slightly confused how to pass from the line integral to the boundary conditions.



My guess is that we should pose $F_u_x = 0, F_u_y = 0$ on the boundary, I can't prove it though.



Thank you




pde calculus-of-variations boundary-value-problem




share|cite|improve this question





















  • $h=h(x)$? if so, not clear in your question
    – phdmba7of12
    Aug 3 at 16:28










  • $h=0$ on the boudary.
    – Rafa Budría
    Aug 3 at 16:36






  • 1




    @phdmba7of12, yes it's h=h(x)
    – user8469759
    Aug 3 at 17:04






  • 1




    @Rafa, it's not 0.
    – user8469759
    Aug 4 at 13:19












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I have a variational problem of the form



$$
E(u) = int_Omega F(x,y,u,u_x,u_y)dxdy
$$



which leads me



$$
fracdEdu = int_Omega left( fracpartial Fpartial u - fracddxfracpartial Fpartial u_x - fracddyfracpartial Fpartial u_y right)h dxdy + int_Gamma left(F_u_xdy - F_u_ydx right)h
$$



Assuming the increment function $h$ is not fixed,it can be arbitrary, how do I get from this term



$$
int_Gamma left(F_u_xdy - F_u_ydx right)h = 0
$$



the boundary conditions of my problem when I try to apply gradient descent? Namely how do I fill the system



$$
left{
beginarrayl
fracpartial upartial t = - fracpartial Fpartial u + fracddxfracpartial Fpartial u_x + fracddyfracpartial Fpartial u_y \
textboundary conditions?
endarray
right.
$$



I should end up having the Von Neumann boundary conditions, but I'm slightly confused how to pass from the line integral to the boundary conditions.



My guess is that we should pose $F_u_x = 0, F_u_y = 0$ on the boundary, I can't prove it though.



Thank you




pde calculus-of-variations boundary-value-problem




share|cite|improve this question













I have a variational problem of the form



$$
E(u) = int_Omega F(x,y,u,u_x,u_y)dxdy
$$



which leads me



$$
fracdEdu = int_Omega left( fracpartial Fpartial u - fracddxfracpartial Fpartial u_x - fracddyfracpartial Fpartial u_y right)h dxdy + int_Gamma left(F_u_xdy - F_u_ydx right)h
$$



Assuming the increment function $h$ is not fixed,it can be arbitrary, how do I get from this term



$$
int_Gamma left(F_u_xdy - F_u_ydx right)h = 0
$$



the boundary conditions of my problem when I try to apply gradient descent? Namely how do I fill the system



$$
left{
beginarrayl
fracpartial upartial t = - fracpartial Fpartial u + fracddxfracpartial Fpartial u_x + fracddyfracpartial Fpartial u_y \
textboundary conditions?
endarray
right.
$$



I should end up having the Von Neumann boundary conditions, but I'm slightly confused how to pass from the line integral to the boundary conditions.



My guess is that we should pose $F_u_x = 0, F_u_y = 0$ on the boundary, I can't prove it though.



Thank you





pde calculus-of-variations boundary-value-problem





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share|cite|improve this question




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edited Aug 2 at 16:46
























asked Aug 2 at 14:40









user8469759

1,4271513




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  • $h=h(x)$? if so, not clear in your question
    – phdmba7of12
    Aug 3 at 16:28










  • $h=0$ on the boudary.
    – Rafa Budría
    Aug 3 at 16:36






  • 1




    @phdmba7of12, yes it's h=h(x)
    – user8469759
    Aug 3 at 17:04






  • 1




    @Rafa, it's not 0.
    – user8469759
    Aug 4 at 13:19
















  • $h=h(x)$? if so, not clear in your question
    – phdmba7of12
    Aug 3 at 16:28










  • $h=0$ on the boudary.
    – Rafa Budría
    Aug 3 at 16:36






  • 1




    @phdmba7of12, yes it's h=h(x)
    – user8469759
    Aug 3 at 17:04






  • 1




    @Rafa, it's not 0.
    – user8469759
    Aug 4 at 13:19















$h=h(x)$? if so, not clear in your question
– phdmba7of12
Aug 3 at 16:28




$h=h(x)$? if so, not clear in your question
– phdmba7of12
Aug 3 at 16:28












$h=0$ on the boudary.
– Rafa Budría
Aug 3 at 16:36




$h=0$ on the boudary.
– Rafa Budría
Aug 3 at 16:36




1




1




@phdmba7of12, yes it's h=h(x)
– user8469759
Aug 3 at 17:04




@phdmba7of12, yes it's h=h(x)
– user8469759
Aug 3 at 17:04




1




1




@Rafa, it's not 0.
– user8469759
Aug 4 at 13:19




@Rafa, it's not 0.
– user8469759
Aug 4 at 13:19















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