- #1
Arkuski
- 40
- 0
Usually, when considering the biharmonic equation (given by [itex]Δ^2u=f[/itex], we look for weak solutions in [itex]H^2_0(U)[/itex], which should obviously have Neumann boundary conditions ([itex]u=0[/itex] and [itex]\bigtriangledown u\cdot\nu =0[/itex] where [itex]\nu[/itex] is normal to [itex]U[/itex]).
Now consider that we are looking for solutions [itex]u\in H^1_0(U)\cap H^2(U)[/itex]. Clearly, the boundary condition [itex]u=0[/itex] should still apply since we are considering [itex]H^1_0[/itex], but I'm having trouble deciding what the other boundary condition should be. If we multiply the PDE by a test function [itex]v\in H^1_0(U)\cap H^2(U)[/itex] and integrate by parts, we find the following:
[itex]\int _U (Δ^2u)v dx=-(Dv)(Δu)_{∂U}|+\int _U (Δv)(Δu)dx[/itex]
If we want the boundary term to vanish, we would require that something like [itex](\bigtriangledown u\cdot\nu )(Δu)=0[/itex] on the boundary, but I'm confused as to why we require these terms to vanish, or if this is even a necessity.
Now consider that we are looking for solutions [itex]u\in H^1_0(U)\cap H^2(U)[/itex]. Clearly, the boundary condition [itex]u=0[/itex] should still apply since we are considering [itex]H^1_0[/itex], but I'm having trouble deciding what the other boundary condition should be. If we multiply the PDE by a test function [itex]v\in H^1_0(U)\cap H^2(U)[/itex] and integrate by parts, we find the following:
[itex]\int _U (Δ^2u)v dx=-(Dv)(Δu)_{∂U}|+\int _U (Δv)(Δu)dx[/itex]
If we want the boundary term to vanish, we would require that something like [itex](\bigtriangledown u\cdot\nu )(Δu)=0[/itex] on the boundary, but I'm confused as to why we require these terms to vanish, or if this is even a necessity.