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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.

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  • $\begingroup$ This question has been answered and should be closed. $\endgroup$ – cpraveen Aug 22 '18 at 11:03
  • $\begingroup$ @PraveenChandrashekar, that's not how this site works. $\endgroup$ – nicoguaro Aug 23 '18 at 2:06
  • $\begingroup$ This is from Aug 2017. If the questioner has dissappeared, whats the point of bumping these questions again and again. It is just noise. And I see this happening with a lot of questions. $\endgroup$ – cpraveen Aug 23 '18 at 3:48
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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: $$ \frac{\partial\Psi}{\partial y}=v_0\qquad \frac{\partial\Psi}{\partial x}=0 $$

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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.

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  • $\begingroup$ I had some doubt that psi_y =u0 does not mean psi = constant, it can be psi=u0y $\endgroup$ – Eda Suresh Reddy Aug 23 '18 at 12:16
  • $\begingroup$ On top boundary $\psi = const$ implies $\psi$ does not depend on $x$. $y$ is already fixed on the top boundary.. The other condition says $\psi_y(x,y_{top}) = u_0$ but you cannot integrate this as you seem to do, because that condition holds only at $y=y_{top}$. The thing is you have two bc on the top, a dirichlet and neumann. I am suggesting to use dirichlet bc. The solution you get must satisfy the other bc, maybe only in a weak sense. $\endgroup$ – cpraveen Aug 24 '18 at 3:27

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