I am working on a 2D problem with the potential flow around a circular cylinder. In this symmetrical problem, a cylinder is placed in a flow with uniform velocity $U_\infty$. There are 2 obvious stagnation points: A and B, which are aligned on the x-axis. The stream function is as follows: $$\psi(r,\theta)=U_\infty(1-R^2/r^2)r\sin\theta$$ I assume the Point Of Interest at $r=R$ and $\theta=0$.
I have written out the Taylor expansion as follow: $$Y(r,\theta)=\psi(R,0)+\frac{\partial \psi}{\partial r}(R,0)(r-r_0)+\frac{\partial \psi}{\partial \theta}(R,0)(\theta-\theta_0)+\frac{1}{2!}\frac{\partial^2 \psi}{\partial r^2}(R,0)(r-r_0)^2+\frac{1}{2!}\frac{\partial^2 \psi}{\partial \theta^2}(R,0)(\theta-\theta_0)^2+\frac{1}{2!}\frac{\partial^2 \psi}{\partial r\partial \theta}(R,0)(r-r_0)(\theta-\theta_0)+\frac{1}{2!}\frac{\partial^2 \psi}{\partial \theta\partial r}(R,0)(\theta-\theta_0)(r-r_0).$$
Here, $r_0=R$ and $\theta_0=0$, since the stagnation points are on the wall of the cylinder and aligned along the $x$ axis.
I want to prove, by application of the Taylor expansion series, that the stream function at the stagnation point can be approximated with $\psi=Axy$, where $x=R\theta$ and $y=r-R$ and A is related to the parameters of the overall flow field ie $U_\infty$ and $R$.
In order to find the derivatives, I worked with the following gradient formulas: $$\nabla \cdot \psi= \begin{pmatrix} \frac{1}{r}\frac{\partial}{\partial r}(r\psi) \\ \frac{1}{r}\frac{\partial \psi}{\partial \theta} \\ 0\end{pmatrix}$$ The $0$ is there since the problem is in 2D. Using this, I found the following derivatives and approximations at the POI $r=R, \theta=0$
- $\psi(r,\theta)=U_\infty(1-R^2/r^2)r\sin\theta$ and $\psi(R,0)=0$
- $\psi_r=\frac{2}{r}U_\infty \sin\theta$ and $\psi_r(R,0)=0$
- $\psi_{rr}=\frac{2}{r}U_\infty \sin \theta$ and $\psi_{rr}(R,0)=0$
- $\psi_\theta=U_\infty(1-\frac{R^2}{r^2})\cos\theta$ and $\psi_\theta(R,0)=0$
- $\psi_{\theta\theta}=\frac{1}{r}U_\infty (\frac{R^2}{r}-1)\sin\theta$ and $\psi_{\theta\theta}(R,0)=0$
- $\psi_{r \theta}=\frac{2}{r^2}U_\infty \cos \theta$ and $\psi_{r\theta}(R,0)=\frac{2}{R^2}U_\infty$
- $\psi_{r \theta}=\frac{U_\infty}{r} \cos \theta+U_\infty \frac{R^2}{r^3}\cos\theta$ and $\psi_{\theta r}(R,0)=\frac{U_\infty}{R}+\frac{U_\infty}{R}=2\frac{U_\infty}{R}$
When I insert the derivatives and approximations in the POI, I find the following approximation for the stream function nearby the point $(R,0)$ : $$Y(r,\theta)=\frac{1}{2!}\frac{2}{R^2}U_\infty (r-R)\theta+\frac{1}{2!}\frac{2}{R}U\infty(r-R)\theta$$ which is equivalent to $$Y(r,\theta)=\frac{U_\infty}{R}\left(\frac{1}{R}+1\right)(r-R)\theta.$$
Here is where I get stuck. The approximity around this point $(R,0)$ should be in terms of $\psi=Axy$ where A is some kind of combination of $U_\infty$ and $R$. The $y$ is there but the $x$ isn't. Plus, the solution looks odd. Could you guys help me out where I go wrong? And subsequently, how does this tell me something about the shape and size of the area around the POI?
Many thanks in advance.
The derivation for $\psi_r$ was off, as it does not carry the dimension of velocity. It should have been something like $U_\infty(1+\frac{R^2}{r^2})\sin\theta.$