Natural Boundary Condition?

[Srihari Sritharan]

Hi,

Just for my clarification, I understand that essential boundary
condition means that you prescribe u=some value on the boundary, i.e.
Dirichlet condition. What does "natural boundary condition" mean?

Does it mean we prescribe the derivative to be something, or does it
mean we prescribe nothing, and let the solution be whatever it turns
out to be there? If it is the latter, how does that make sense in
terms of formulation/solution integrity?

Just a little confused,
Sri

[Pavel Solin]

Hi Sri,

On Sat, Jul 2, 2011 at 9:17 PM, Srihari Sritharan <sriha...@gmail.com> wrote:

Hi,

Just for my clarification, I understand that essential boundary
condition means that you prescribe u=some value on the boundary, i.e.
Dirichlet condition. What does "natural boundary condition" mean?

Natural means that the solution is unknown on the part of the bdy 

and yet is has to fulfill some constraint. For example, the derivative 

has to be something. Or the derivative plus the solution value, etc. 

This is incorporated into the weak formulation (after you do the 

Green's theorem, there are some surface integrals - those are adjusted

using the relation for the derivative etc.). 

Pavel

 

Does it mean we prescribe the derivative to be something, or does it
mean we prescribe nothing, and let the solution be whatever it turns
out to be there? If it is the latter, how does that make sense in
terms of formulation/solution integrity?

Just a little confused,
Sri

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[Srihari Sritharan]

Hi,

Oh, so that means that I cannot simply leave it as natural boundary
condition and be on my way, I need to still attend to the boundary
somewhere else, like adding a surface integral to the weak term?

Sri

On Jul 3, 5:00 am, Pavel Solin <so...@unr.edu> wrote:
> Hi Sri,
>

[Pavel Solin]

On Sun, Jul 3, 2011 at 12:45 PM, Srihari Sritharan <sriha...@gmail.com> wrote:

Hi,

Oh, so that means that I cannot simply leave it as natural boundary
condition and be on my way, I need to still attend to the boundary
somewhere else, like adding a surface integral to the weak term?

It depends on the equation that you have. If you have 

second derivatives, then most likely you need to apply the 

Green's theorem and this will automatically generate 

a surface integral. Then you use the natural BC for it. 

Example: you have du/dn = 5.7, If the boundary integral is 

\int_{\Gamma} du/dn v dx dy dz then it will become 

\int_{\Gamma} 5.7 v dx dy dz and it goes to the right-hand 

side. 

Pavel

[Srihari Sritharan]

Hi,

I'm solving Schrodinger's equation, which has 2nd derivatives, so I'll
look into utilizing green's theorem. Thanks!

Sri

On Jul 3, 1:00 pm, Pavel Solin <so...@unr.edu> wrote: