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Let $R$ be the annulus $1/4\leq x^2+y^2\leq 4$ and $u$ satisfy the boundary conditions $u\left(r=2,\theta\right) = \cos3\theta$ and $u\left(r=1/2,\theta\right) = 1$. Find a suitable analytic function in $R$ and deduce the solution for $u$. Note: I can't use separation of variables.

As I understand, determining a complex function $f\left(z\right)$ such that $u\equiv\mathfrak{R}\left[f\left(z\right)\right]$ satisfies Laplace's equation should suffice for this problem. My way of going about it is that $f$ should have two "parts", each corresponding to a specific boundary condition. In particular, for the B.C. at $r=2$, we must have something involving powers of $z$ (so that a $\theta$ dependence is introduced). Meanwhile, for the B.C. at $r=1/2$, we should have something like $C\log z+ D$ so that a $\theta$ dependence is eliminated. So, the overall function should come out (in a general form) to be

\begin{equation} f\left(z\right) = Az^n + \dfrac{B}{z^n}+ C\log z+D. \end{equation}

In fact, if we take the real part of $f$ and compare to the boundary conditions, we can obtain

\begin{equation} u\left(r,\theta\right) \equiv \mathfrak{R}\left[f\right] = \left(\dfrac{512}{4095}r^3 - \dfrac{8}{4095r^3}\right)\cos 3\theta - \dfrac{\ln r}{2\ln 2} +\dfrac{1}{2}. \end{equation}

which is actually harmonic (i.e. satisfies $\Delta u=0$) across the annulus given (and everywhere else except at $z=0$ where the logarithm is undefined anyways). Note, however, that $f$ is NOT analytic on $R$ because $\log z$ is discontinuous across the negative real line (or however else we construct our branch cut). How does one resolve this?

gettingmathy
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  • $\log$ being discontinuous for negative arguments doesn't matter here because we are only giving positive values of $r$. – K.defaoite Oct 02 '22 at 06:05
  • I think you are fine here because the vanishing of the Laplacian is a local condition, so the fact that $\ln r = \Re[\log z]$ no matter the choice of logarithm branch means that it has to be harmonic everywhere the logarithm could be defined (outside $z = 0$) – Andrew Oct 02 '22 at 07:44
  • @Andrew So, I have no question on the harmonicity of $\ln r$; I agree that it has to be harmonic everywhere except $z=0$. My only concern is that the initial $f\left(z\right)$ for which $u$ is the real part is not analytic in the annulus whereas the question seems to attach that requirement to $f$. – gettingmathy Oct 03 '22 at 02:12
  • It's not hard to show that a radially symmetric harmonic function on an annulus not including the origin in $\mathbb{R}^2$ must be of the form $D + C\ln r$ like you used. And since the solution to the Laplace equation with Dirichlet boundary conditions is unique, there is no other possible fully analytic solution $f$ you could have derived. So my conclusion would be that either you misunderstood what the problem was asking for or that it was asked incorrectly. See this problem for more details: https://math.stackexchange.com/questions/435296/harmonic-function-on-an-annulus – Andrew Oct 03 '22 at 02:48
  • @Andrew OK, yeah, that makes sense. The above is the exact statement of the problem and I had initially suspected that it was phrased incorrectly. Thanks for the confirmation! – gettingmathy Oct 03 '22 at 07:03

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