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joshmccraney

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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.