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Question. Let $f: \mathbf{R} \to \mathbf{R}$ be an analytic function. Is there a harmonic function $u$ on the circular cylinder $D \times \mathbf{R} \subset \mathbf{R}^3$ so that $u = f$ along the axis $\{ (0,0) \} \times \mathbf{R}$?

  • The problem is obviously ill-posed in the sense of Hadamard because it is very underdetermined. Although it doesn't make sense as a 'Dirichlet problem', I think a (positive or negative) answer is possible.
  • Probabilistic arguments seem tricky because the axis is too small for hitting times of Brownian motion to be defined.
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Assuming the Taylor series of $f$ has an infinite radius of convergence, the sum $$ \sum_{k=0}^\infty \left(x^2+y^2\right)^kf^{(2k)}(z)\cdot \frac{(-1)^k}{4^k k!^2} $$ converges absolutely and locally uniformly on $\mathbb{R}^3$ to a function $u(x,y,z)$ which is harmonic and satisfies $u(0,0,z)=f(z)$

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  • $\begingroup$ That's very neat - thanks! I checked the formal calculation, and it works out for $u$ to be harmonic. To be clear, if you want $u$ to be defined for $\rho < 1$, you want the radius of convergence of the series $\sum_k f^{(k)}(z) \rho^k / k!$ to be at least $1$ for each $z \in \mathbf{R}$, right? Is it obvious that this is also necessary - that there can't be a solution if the radius of convergence were strictly smaller? Also, would you mind explaining how you got the formula? I'll admit I haven't dealt with Taylor series much (or at all) in a while, so I'm not familiar with the expression... $\endgroup$
    – Leo Moos
    Nov 9, 2022 at 19:19
  • $\begingroup$ Oops, forgot to say that I used the shorthand notation where $\rho = (x^2 + y^2)^{1/2}$ in the comment above. $\endgroup$
    – Leo Moos
    Nov 9, 2022 at 19:20
  • $\begingroup$ @LeoMoos I just searched for a solution of the form $u(x,y,z) = \sum_k \left(x^2+y^2\right)^k u_k(z)$ (so that it would be rotationally symmetric along the z axis) with $u_0 = f$, by expanding the equation $\Delta u=0$ to a recurrence relation on $u_k$. Also I'm not sure if a large enough radius of convergence is a necessary condition. $\endgroup$
    – user49822
    Nov 9, 2022 at 20:00
  • $\begingroup$ Thanks again - I appreciate the clarification! $\endgroup$
    – Leo Moos
    Nov 9, 2022 at 20:25

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