# What is the boundary condition for Ginzburg Landau equation?

I am trying to do some numerical calculation with Ginzburg-Landau (GL) equation for a superconductor. However, I am confused about the boundary condition of the GL equation.

If we introduce the GL free energy of the superconductor $$F=F_0+\alpha|\varphi|^2+\frac{\beta}{2}|\varphi|^4+\frac{1}{2m}|\vec{p}\varphi|^2+\frac{\vec{B}^2}{2\mu_0}$$ where $\vec{p}=-i\hbar\vec{\nabla}-2e\vec{A}$ and $\vec{A}$ is the vector potential satisying $\vec{\nabla}\times\vec{A}=\vec{B}$.

To minimize this free energy, we obtain the GL equation $$\frac{\delta}{\delta\varphi^*}F=0\implies\alpha\varphi+\beta|\varphi|^2\varphi+\frac{\vec{p}^2}{2m}\varphi=0$$ where the boundary condition should be $$\vec{n}\cdot\vec{p}\varphi=0$$ which ensures $J_s\cdot\vec{n}=0$ that no superconducting current flows out the superconductor.

However, this boundary condition looks somehow strange, I know it should be correct but I am confused.

Why is a simply boundary condition $\varphi=0$ not allowed?

The superconducting electron should be confined within superconductor which implies $\varphi=0$ out of the superconductor and thus $\varphi|_n=0$ should be a simply boundary condition.

Are they equivalent? When are they equivalent?

Ginzburg reports this here, an easier to find paper. The condition ${\bf n}\cdot{\bf J}_s=0$ is obtained immediately by the variation principle that has $\delta\varphi$ arbitrary. This grants that surface contributions coming from the variation of the functional cannot contribute.
In the same paper, Ginzburg states that this condition automatically follows from the variation principle if no auxiliary conditions restrict on the boundary: that is the case considered in the original paper with Landau where this set of equations was firstly formulated. As properly explained by these authors, to put $\varphi=0$ on this system would imply that the problem of the superconducting plate cannot be properly treated unless the thickness has specific values but this is unphysical.