Why isn't c_1 eliminated in the stream function boundary condition?

[member 428835]

Hi PF!

So my book has boiled the problem down to $$\psi(r,\theta) = c_1 \ln \frac{r}{a}+\sum_{n=1}^\infty A_n\left(r^n-\frac{a^{2n}}{r^n}\right)\sin n\theta$$ subject to $\psi \approx Ur\sin\theta$ as $r$ get big. The book then writes
$$\psi(r,\theta) = c_1 \ln \frac{r}{a}+U\left(r-\frac{a^2}{r}\right) \sin \theta$$ I understand $A_n=0\, \forall\, n \geq 2$ and $A_1=U$ but why isn't $c_1$ eliminated too? To me it seems the natural log does not allow the "boundary condition" to be satisfied.

 

[OrangeDog]

Whats the context of the problem?

 

[member 428835]

Fluid flow in 2D passing a cylinder of radius $a$. The flow satisfies continuity, so $\nabla \cdot \vec{v} = 0$. Define $\psi \equiv \nabla \times \vec{v}$ so that continuity is satisfied. Then $v_x = \psi_y$ and $v_y=-\psi_x$. The resulting PDE is then $\nabla^2 \psi = 0$. Solving this in polar coordinates yields the above solution I posted, the infinite series. In the far wake, velocity in the $x$ direction is constant $U$, which suggests $v_x = \psi_y = U \implies \psi = Uy$ (I guess I don't understand why there is no constant too, which is to say why isn't this true: $\psi = Uy + C_0(x)$).

 

[member 428835]

Any ideas? My thought process is since $r$ grows much larger than $\ln r$ we may leave the natural logarithm? Can anyone confirm this?