# Fluid Mechanics for Low Reynolds Number

I've tried to answer this question but I seem to get a really weird (and suggestively incorrect) answer.

The question is: "Fluid is injected radially and slowly into a circle, radius R, with the velocity on the surface of the circle being $u_{\rho} = K \cos{\phi}$, in polar coordinates. Determine the streamfunction for the flow inside the circle.

I've used the biharmonic equation $$\nabla^2(\nabla^2 \psi)= 0$$ With the flow satisfying $$u_\rho = \frac{1}{\rho}\frac{\partial \psi}{\partial\phi} , \quad u_\phi = -\frac{\partial \psi}{\partial \rho}$$

and boundary condition $$u_\rho =\frac{1}{R}\frac{\partial \psi}{\partial\phi}= K \cos \phi, \quad \rho = R$$

to attain the general form of $\psi$ as $$\psi(\rho,\phi) = \sin{\phi} \cdot(A\rho +\frac{B}{\rho} + C\rho \ln\rho +D\rho^3)$$

But now my problem is using Boundary conditions - when i set $$\psi(R,\phi) = KR\sin{\phi}$$

This gives $$K = A$$ $$B=C=D=0$$ $$\Longrightarrow \psi(\rho,\phi) = K \rho \sin \phi$$ This looks really wrong - have I misinterpreted the question? any assistance is greatly appreciated

• I didn't check your solutions to the differential equations, but shouldn't you be using the derivative of $\psi$ for the boundary condition? Commented Dec 3, 2014 at 21:40
• I used that $\frac{1}{R}\frac{\partial \psi}{\partial \phi}=K\cos \phi$ on the boundary ie $\frac{\partial \psi}{\partial \phi}=KR \cos \phi$ Commented Dec 3, 2014 at 21:57

The problem is you neglected the boundary condition for $u_{\phi}$.

The general solution (which will reduce to your solution) is

$$\psi(\rho,\phi)=\left(A\rho + \frac{B}{\rho}+ C \rho \ln \rho + D\rho^3\right)\sin \phi\\+\left(a\rho + \frac{b}{\rho}+ c \rho \ln \rho + d\rho^3\right)\cos \phi \\ + \sum_{n=2}^{\infty}\left(A_n\rho^n + B_n\rho^{-n}+ C_n \rho^{n+2} + D_n\rho^{-n+2}\right)\sin (n\phi) \\ + \sum_{n=2}^{\infty}\left(a_n\rho^n + b_n\rho^{-n}+ c_n \rho^{n+2} + d_n\rho^{-n+2}\right)\cos (n\phi).$$

The boundary conditions must specify both $u_{\rho}$ and $u_{\phi}$ on the surface. Since the flow is radial, the conditions are:

$$u_\rho =\frac{1}{R}\frac{\partial \psi}{\partial\phi}= K \cos \phi, \quad \rho = R \tag{1}$$ $$u_\phi =-\frac{\partial \psi}{\partial\rho}= 0, \quad\quad \quad \quad \rho = R \tag{2}$$

Given the form of boundary condition (1) we immediately find many of the coefficients are equal to $0$:

$$a = b = c = d = 0,\\A_n=B_n=C_n = D_n = 0,\\a_n=b_n=c_n = d_n = 0.$$

$$\psi(\rho,\phi)=\left(A\rho + \frac{B}{\rho}+ C \rho \ln \rho + D\rho^3\right)\sin \phi.$$

The velocity must be finite at $\rho=0$, which implies $B=C=0$ and

$$\psi(\rho,\phi)=\left(A\rho + D\rho^3\right)\sin \phi.$$

Now applying both boundary conditions (1) and (2) we get:

$$A +DR^2 =K,\\A + 3DR^2 = 0.$$

Solving for the coefficients we obtain

$$A = \frac{3}{2}K, \,\,\, D = -\frac{1}{2R^2}K,$$

and the streamfunction is

$$\psi(\rho,\phi) = \left(\frac{3}{2}K\rho - \frac{K}{2R^2}\rho^3\right) \sin \phi.$$