Question: How do I transform the boundary conditions of the speed formulation $(u, \ v)$ into the vortex-stream formulation $(\psi, \ w)$?
Cavity flow problem:
A incompressible Newtonian fluid is in a square $\Omega = \left[0, \ 1\right] \times \left[0, \ 1\right]$. Only the upper wall is moving at speed $U_{up}(x)$.
- \eqref{1} is continuity equation
- \eqref{2} is the navier's equation
$$\nabla \cdot \vec{u} = 0\tag{1}\label{1}$$
$$\dfrac{\partial \vec{u}}{\partial t} + \left(\vec{u} \cdot \nabla\right) \vec{v} = -\nabla p + \mu \Delta \vec{u}\tag{2}\label{2}$$
The boundary conditions are
$$\begin{cases}u(0, y) = 0 \\ u(1, \ y) = 0 \\ u(x, \ 0) = 0 \\ u(x, 1) = U_{up}(x)\end{cases} \ \ \ \ \ \ \ \ v = 0 \ \ \text{on} \ \partial \Omega\label{3} \tag{3}$$
Vortex-Streamline formulation
Another formulation is made by transforming the pair of variables $(u, \ v)$ into $\left(\psi, \ w\right)$.
$$u = \dfrac{\partial \psi}{\partial y} \ \ \ \ \ \ \ \ \ \ v = -\dfrac{\partial \psi}{ \partial x} \label{4} \tag{4}$$
$$w = \nabla \times \vec{u} \tag{5}\label{5}$$
Then, the new equations to solve are:
$$w + \dfrac{\partial^2 \psi}{\partial x^2} + \dfrac{\partial^2 \psi}{\partial y^2} = 0\label{6}\tag{6}$$
$$\dfrac{\partial w}{\partial t} + \dfrac{\partial \psi}{\partial y} \cdot \dfrac{\partial w}{\partial x} – \dfrac{\partial \psi}{\partial x} \cdot \dfrac{\partial w}{\partial y} = \mu \left(\dfrac{\partial^2 w}{\partial x^2} + \dfrac{\partial^2 w}{\partial y^2}\right)\label{7}\tag{7}$$
My problem now is finding the boundary conditions for $\psi$ and $w$.
So far I found, from \eqref{4} for the lower, left and right wall:
$$u=v=0 \Rightarrow \dfrac{\partial \psi}{\partial y} = \dfrac{-\partial \psi}{\partial x} = 0 \Rightarrow \psi = \text{const}$$
$$\begin{cases}\psi(0, \ y) = 0 \\ \psi(x, \ 0) = 0 \\ \psi(1, \ y) = 0\end{cases}\label{8}\tag{8}$$
For the upper wall
$$\begin{cases}u(x, \ 1) = U_{up}(x) \\ v(x, \ 1) = 0\end{cases} \Rightarrow \begin{cases}\left[\dfrac{\partial \psi}{\partial x}\right]_{y=1} = 0 \\ \left[\dfrac{\partial \psi}{\partial y}\right]_{y=1} = U_{up}(x)\end{cases}\label{9}\tag{9}$$
Problem 1: Now I don't know which of these two (from \eqref{9}) I should use for upper boundary
Problem 2: What for $w$? I tryied to use \eqref{5}, but I get the same equation \eqref{6}, but applied on the boundary.
Best Answer
Stream Function Boundary Condition
The boundary condition for the stream function on the upper boundary, where $y = 1$, similarly must be
$$\psi(x,1) = 0$$
Recall that for two-dimensional incompressible flow, the stream function $\psi$ is defined by a line integral along any path $C$ joining some arbitrary reference point $(0,0)$ where $\psi(0,0) := 0$ to $(x,y)$, that is
$$\psi(x,y) = \int_C \mathbb{u} \cdot \mathbb{n} dl = \int_C u \, dy - v \, dx.$$
Using the incompressiblity condition $\nabla \cdot \mathbb{u} = 0$ and Green's theorem it follows that the line integral is independent of path and $\psi$ is a well-defined function.
Furthermore, since $ \int_C \mathbb{u} \cdot \mathbb{n} dl $ is the volumetric flux of fluid through the curve $C$, the difference between the stream function at two points $(x_1,y_1)$ and $(x_2,y_2)$ is the flux through any curve joining those two points. Hence, any curve to which the velocity field is tangential (a streamline) must be a level curve of the stream function where it takes a constant value. (The flux through the curve between any two points on it is zero and so is difference in the value of the stream function at those points).
This implies that on the upper wall, we have $\psi(x,1) = C$. We can show that $C = 0$ by considering the fact that the net flux through the segment from $(x,0)$ on the lower wall to $(x,1)$ on the upper wall is $0$. (Fluid cannot accumulate on one side of this segment by conservation of mass). Thus, $\psi(x,1) = \psi(x,0) = 0$ for $x \in [0,1]$.
Vorticity Boundary Condition
There are no natural boundary conditions for vorticity on no-slip surfaces (where vorticity is generated by viscous stress). The conditions that apply are induced by the connection to the stream function through the Poisson equation
$$\omega = -\nabla^2\psi$$
Since the derivatives of the stream function along a wall must vanish we get
$$\omega(0,y) = - \left.\frac{\partial^2 \psi}{\partial x^2}\right|_{x = 0}, \quad \omega(1,y) = - \left.\frac{\partial^2 \psi}{\partial x^2}\right|_{x = 1},$$
and $$\omega(x,0) = - \left.\frac{\partial^2 \psi}{\partial y^2}\right|_{y = 0}, \quad \omega(x,1) = - \left.\frac{\partial^2 \psi}{\partial y^2}\right|_{y = 1}$$
This facilitates solution of equations (6) and (7) by an iterative procedure. Typically, when a finite-difference method is used to obtain a numerical solution, the vorticity boundary conditions are applied approximately using a finite-difference approximation of the second-order partial derivatives of $\psi$.