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Please note this is a homework question, and I just want some discussion on the choice of the separation parameter.

Suppose we have a cylinder which satisfies the following steady-state heat equation:

$$u_{rr}+\frac{1}{r}u_r+u_{zz}=0$$,

with the following boundary conditions:

\begin{align} u(r,0)&=0\\ u(r,20)&=20\\ [u_r+u]|_{r=4}&=0 \end{align}

Separation gives two very standard ODEs - the radial part gives a parametric Bessel equation of order $0$, and the other gives either a hyperbolic equation or a trigonometric equation. In other examples available to me it is stated that making $Z(z)$ be hyperbolic "makes sense in the context," with no explanation of what exactly is meant by that. I'm pretty sure the choice here depends entirely on how we expect $Z(z)$ to behave. If we expect the $Z$ part to be hyperbolic I have absolutely no idea why we expect it to be hyperbolic.

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If one insists that the $Z(z)$ function is oscillatory, then $R(r)$ obeys the modified Bessel equation of order 0 instead which does have solutions $I_0(\lambda x), K_0(\lambda x)$. The second solution diverges at the origin so we can't use it to build a set of solutions. The problem with the first solution is that, unfortunately, it doesn't exhibit oscillatory behavior and in fact it is always positive. As a result the 3rd boundary condition cannot be satisfied or has a finite number of functions that can satisfy it, which in turn stiffens the boundary condition in the z-direction, in the sense that most boundary problems you come up with have no solution.

All in all, there is a give and a take in elliptical equations in which direction is oscillatory and which is not. We always choose critically which one will be given the oscillatory behavior given how constraining the boundary conditions are. In this case, the radial direction needs the oscillatory behavior of a Bessel function so that it can satisfy the radial boundary condition. In other cases one has to think about both possibilities to make a decision, since there is no general rule of thumb.

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  • $\begingroup$ Would you say that for my purposes currently that I should never get a Bessel function with an imaginary argument? I'm unfamiliar with Bessel functions at best, and they've only been introduced as a solution to the ODE we need for our separation of variables. $\endgroup$
    – Algebraic
    Nov 18, 2020 at 9:56
  • $\begingroup$ Sure. It is indeed unlikely, especially since the field fluctuating (temperature) needs to be positive definite. $\endgroup$ Nov 18, 2020 at 16:05

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