# 2-D steady incompressible inviscid flow Problem

For 2-D steady, incompressible inviscid flow, subject to a conservative body force $\underline{X} = -\nabla\chi$, Helmholtz's vorticity equation reduces to $$\nabla^2\psi = -\Omega(\psi)\quad \text{where}\quad \underline{\omega} = \underline{\nabla}\times\underline{u} = \Omega(x,y,t)\underline{\hat{k}}$$ is the vorticity vector and $\psi$ is the $2$-D stream function. Consider $2$-D channel flow through a constriction with rigid walls $y = 0$ and $y = f(x)$ such that far upstream $(x\to -\infty)$ we have $f(x)\to h$ and far downstream $(x\to+\infty)$ we have $f(x)\to bh$ $(0<b<1)$.

Assume that far upstream we have $\psi\to\psi_L(y),$ $\underline{u}\to u_L(y)\underline{\hat{\imath}}$ and $\Omega\to\Omega_L(y)$ as $x\to-\infty$. Find $\psi_L(y)$ and $\Omega_L(y)$ for a velocity profile has the no-slip form $$\underline{u}\to u_L(y)\underline{\hat{\imath}}\quad\text{where}\quad u_L(y) = U_0\sin\left(\frac{\pi y}{h}\right)$$ where w.l.o.g take $\psi = 0$ on $y = 0$. Hence, find the function $\Omega = \Omega(\psi)$ and the vorticity equation to be satisfied by $\psi$ for all $x$. By assuming $\psi\to\psi_R(y)$ as $x\to+\infty$, solve for $\psi_R(y)$ and show that $\underline{u} \to u_R(y)\underline{\hat{\imath}}$ where $$u_R(y) = U_0\left[\sin\left(\frac{\pi y}{h}\right) + \frac{1 + \cos(\pi b)}{\sin(\pi b)}\cos\left(\frac{\pi y}{h}\right)\right]$$

So I have that $\underline{u} = \underline{\nabla}\psi\times\underline{\hat{k}}$. Then I found that $$\underline{\nabla}\times(\underline{\nabla}\psi\times\underline{\hat{k}}) = -\nabla^2\psi\underline{\hat{k}} = \underline{\omega} = \Omega(x,y,t)\underline{\hat{k}} \Rightarrow -\nabla^2\psi = \Omega(x,y,t) \Rightarrow \nabla^2\psi = -\Omega(\psi)$$ Now I need to find $\psi_L(y)$ and $\Omega_L(y)$. How would I do this?

And once I have those, I should find that as $(x\to-\infty)$ the streamfunction $\psi$ is a function of the vorticity $\Omega$. Assuming this function holds for all $x$, substitute into Helmoltz vorticity equation. This gives a $2^{\mathrm{nd}}$- order linear equation for the streamfunction.

Now assume the streamfunction $\psi$ tends to $\psi_{R} (y)$ as $x$ tends to $+\infty$. The vorticity equation should reduce to a $2^{\mathrm{nd}}$- order ODE. I then need to solve for $\psi_{R} (y)$ and confirm $u_{R} (y)$.

I have all this process layed out but I need guidence how to follow it through.

To get you started, far upstream, we have $u_L(y) = U_0 \sin \frac{\pi y}{h}$. Since we have only a non-zero $x-$component of velocity that is a function of $y$ at this location, the upstream vorticity is given by

$$\Omega_L(y) = - \frac{\partial u_L}{\partial y} = -\frac{U_0\pi}{h} \cos \frac{\pi y}{h}.\tag1$$

To find the streamfunction we use $u_L = \frac{\partial \psi_L}{\partial y}$ and integrate using $\psi_L(0) = 0$ to obtain

$$\psi_L(y)= U_0\int_0^y \sin \frac{\pi \xi}{h} \, d \xi = \frac{U_0h}{\pi}\left(1 - \cos \frac{\pi y}{h}\right) = \frac{U_0h}{\pi} + \frac{h^2}{\pi^2}\left(- \frac{U_0\pi}{h} \cos \frac{\pi y}{h} \right) \\ = \frac{U_0h}{\pi} + \frac{h^2}{\pi^2}\Omega_L(y).\tag2$$

We can now solve for the vorticity in terms of the streamfunction and obtain

$$\Omega_L(y) = \frac{\pi^2}{h^2} \psi_L(y) - \frac{U_0\pi}{h}.\tag3$$

In 2D inviscid flow, the vorticity is constant along a streamline, so at every point downstream we have

$$\Omega(\psi) = \frac{\pi^2}{h^2} \psi - \frac{U_0\pi}{h}.\tag4$$

The streamfunction-vorticity equation becomes

$$\nabla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = \frac{U_0\pi}{h} - \frac{\pi^2}{h^2} \psi ,\tag5$$

or in the form of an inhomogeneous Helmholtz equation,

$$\nabla^2 \psi + \frac{\pi^2}{h^2} \psi = \frac{U_0\pi}{h}.\tag6$$

The general solution would be a combination of a particular solution and a homogeneous solution which could be solved by separation of variables.

Far downstream, the solution tends to a function $\psi_R(y)$ independent of $x$ which satisfies the ODE

$$\frac{d^2\psi_R}{dy^2} + \frac{\pi^2}{h^2} \psi_R = \frac{U_0\pi}{h},$$

and with general solution,

$$\psi_R(y) = A \cos \frac{\pi y}{h} + B \sin \frac{\pi y}{h} + \frac{U_0 h}{\pi}.$$

Applying the boundary condition $\psi_R(0) = 0$ we get

$$\psi_R(y) = \frac{U_0 h}{\pi}\left(1 - \cos \frac{\pi y }{h} \right) + B \sin \frac{\pi y}{h}, \\ u_R(y) = \frac{d \psi_R}{dy} = U_0 \sin \frac{\pi y}{h} + B \cos \frac{\pi y}{h}.$$

It remains to apply another condition to find the constant $B$ to complete the problem. Conservation of mass requires

$$\int_0^h u_L(y) \, dy = \int_0^{bh} u_R(y) \, dy,$$

yielding $B = U_0(1 + \cos \pi b)/\sin \pi b$.

• Yes I continued to work through the night and managed to find the final answer you have for equation $2$. I then rearranged to get equ. $3$ as well. I didn't know the reason for your equ. $4$ so now I know that. And yes I did find that equation $6$ too. So I feel better now that what I thought was wrong is what you have here. Because I thought I was wrong I haven't tried to solve the ODE but I'll try now and get back to you.
– MRT
Commented Apr 10, 2018 at 14:56
• The difficulty is that this is a PDE with boundary condition applied at $y = f(x)$ with $f$ mostly unspecified. Of help is the condition that the solution is finite as $x \to \infty$.
– RRL
Commented Apr 10, 2018 at 15:00
• @MRT: Found a way to dispense with the issue of $x-$ dependence.
– RRL
Commented Apr 10, 2018 at 19:51