Closed Luv-crypto closed 1 month ago
I can apply two of the boundary conditions for x in the form of applied transform: $\psi=\phi (x+1) (x-1)+ cos(\pi y) (x-1)$. So at x=1, $\psi=0$ and at x=-1, $\psi=-2cos(\pi y)$. However, I do not know how to apply the third boundary conditions to a hard constraint. You may have to apply soft constraint for the third boundary condition and apply high loss weight for the boundary condition.
Hi @Luv-crypto, My solution was wrong. It did not satisfy all the boundary conditions and I am deleting it.