I am trying to solve Lid driven square cavity flow problem of Stokes equation using finite element method. I have boundary conditions for velocity as zeros on every boundary but u=1 on top boundary. can you please help me with the boundary conditions of stream function and how to obtain ? especially on top boundary.
I have gone through many literature but why most of them considered as zero on every boundary.
The velocity and stream function are related by $$ u = \psi_y, \qquad v = - \psi_x $$ This can be related to the vorticity $$ -\Delta \psi = \omega = v_y - u_x $$ On left and right, we have $(u,v)=(0,0)$ $$ \psi = const, \quad \psi_x = 0 $$ On bottom $(u,v) = (0,0)$ $$ \psi_y = 0, \qquad \psi = const $$ On top $(u,v)=(u_0,0)$ $$ \psi_y = u_0, \qquad \psi = const $$ You cannot enforce two boundary conditions. Let us use Dirichlet bc $$ \psi = const $$ on all boundaries. Since $\psi$ must be continuous (assuming $\omega$ in $L^2$ atleast), the constant must be same on all sides which you can set to zero.
For the streamline formulation the divergence equation for the velocity $\vec{v}=(u,v)^{T}$ $$\mathrm{div}\,\vec{v}=0$$ holds identically if $$ \vec{v} =(u,v)^{T}= \left( \frac{\partial \Psi}{\partial y},-\frac{\partial \Psi}{\partial x}\right)^{T}\tag{*}$$
For boundaries in which the velocity is $\vec{v}=\vec{0}$, it is easy to see from $(*)$ that $\Psi=constant$ and $\vec{n}\cdot\vec{\mathrm{grad}}\,\Psi=0$. Without any loss of generality you can set this constant to $0$.
On the other hand, on the top of the cavity in which $\vec{v}=(u_0,0)^{T}$ the streamline function $\Psi$ must fulfill: $$ \frac{\partial\Psi}{\partial y}=v_0\qquad \frac{\partial\Psi}{\partial x}=0 $$
