Solution of the transient heat equation for an infinite domain with a circular hole and angular symmetry

Question

I mean to solve the heat equation in 2D $$ \frac{\partial T}{\partial t} = \alpha \frac 1r \frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right) \tag{1} $$

in an infinite domain with a circular hole $D = \{ (x,y) : x^2 + y^2 \ge b^2 \}$. The boundary conditions (BC) are fixed $$T(b,t) = T_1 \tag{2a},$$ $$T(r,t) = T_0 \quad \text{as} \quad r \rightarrow \infty \tag{2b},$$ for all $t$. The initial condition (IC) is $$T(r,0) = T_0 \tag{3}$$ for all $b \le r$.

This is what I did (I am posting all details in case it helps).
First I changed variables $U = (T − T_0) / (T_1 − T_0)$. So $$ \frac 1r \frac{\partial}{\partial r}\left(r \frac{\partial U}{\partial r}\right) = \frac 1\alpha \frac{\partial U}{\partial t} \tag{4}, $$ $$U(b,t) = 1 \tag{5a},$$ $$U(r,t) = 0 \quad \text{as} \quad r \rightarrow \infty \tag{5b},$$ $$U(r,0) = 0 \tag{6}.$$

Then I used the Laplace transform $\bar{U} = {\cal L}[U]$, obtaining (the IC gets embedded in the Laplace-transformed ODE) $$ \frac{d^2 \bar{U}}{d r^2} + \frac 1r \frac{d \bar{U}}{d r} = \frac 1\alpha \left( s \bar{U} - U(r,0) \right) \tag{7}, $$ $$\bar{U}(b,s) = 1/s \tag{8a},$$ $$\bar{U}(r,s) = 0 \quad \text{as} \quad r \rightarrow \infty \tag{8b}$$

Using the IC (6) I get $$ r^2 \frac{d^2 \bar{U}}{d r^2} + r \frac{d \bar{U}}{d r} - \frac {r^2 s}{\alpha} \bar{U} = 0 \tag{9}. $$

The solution is $$ \bar{U}(r,s) = A \, I_0 \left(\sqrt{s/\alpha} \, r \right) + B \, K_0 \left(\sqrt{s/\alpha} \, r \right) \tag{10}. $$

Using (8b), $A=0$. Using (8a), $$ B = \frac{1}{s K_0 \left(\sqrt{s/\alpha} \, b \right)} \tag{11}. $$

Then $$ \bar{U}(r,s) = \frac{1}{s} \frac{K_0 \left(\sqrt{s/\alpha} \, r \right)}{K_0 \left(\sqrt{s/\alpha} \, b \right)} \tag{12}. $$

This is where I am stuck. The last step is an inverse transform. The two hurdles I find are:

1. How to get rid of $\sqrt{s}$.
2. How to deal with $K_0$ in the denominator. For $K_0$ in the numerator, I guess I could use ${\cal L}^{-1}[K_0(\sqrt{s} r)] = e^{-r^2/4t} / 2t$ [1], or a series expansion, once I handle the above.

I found two related cases, with solutions, which contain similar difficulties as 1 and 2 above. None of them explains how to proceed:

1. https://people.bath.ac.uk/masrs/ma20010/chap5.pdf , Eqns. (5.50), (5.51) (using the present symbols): $$ \bar{U}(r,s) = \frac{A}{s} \frac{\sinh \left(\sqrt{s/\alpha} \, r \right)}{\sinh \left(\sqrt{s/\alpha} \, b \right)} \tag{5.50}. $$ $$ U(r,t) = A \left( \frac rb + \frac 2\pi \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin \left( n\pi \frac rb \right) e^{-n^2 \pi^2 \alpha t / b^2} \right) \tag{5.51}. $$

2. https://www.researchgate.net/file.PostFileLoader.html?id=56882e7a614325f8078b457c&assetKey=AS%3A313540840230912%401451765369569 , Sec. 7.1.2 $$ \bar{U}(r,s) = \frac{A}{s} - \frac{Ab}{rs} \frac{\sinh \left(\sqrt{s \alpha} \, r \right)}{\sinh \left(\sqrt{s \alpha} \, b \right)} . $$ $$ U(r,t) = A - \frac{Ab}{r} \sum_{n=0}^{\infty} \left[ \text{erfc} \left( \frac{b (2n+1) - r}{2 \sqrt{\alpha t}} \right) - \text{erfc} \left( \frac{b (2n+1) + r}{2 \sqrt{\alpha t}} \right) \right] . $$

Can someone comment: (1) how to invert Eqn. (12), (2) is there is an alternative general solution method?

Related:

1. Heat equation in cylindrical coordinates with Neumann boundary condition
2. Laplace transform of heat conduction PDE in cylindrical coordinates.
3. Inverse Laplace Transform of the modified Bessel function [1]
4. Inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$
5. https://planetmath.org/usinglaplacetransformtosolveheatequation

Answer 1

One solution of the present problem via the Laplace transform method as suggested in the OP is shown in Inverse Laplace transform of $ \bar{U}(r,s) = \frac{1}{s} \frac{K_0 \left(\sqrt{s/\alpha} \, r \right)}{K_0 \left(\sqrt{s/\alpha} \, b \right)}$

There might be other methods for inverting the Laplace transform along the lines of the examples quoted (series expansion), or also methods other than Laplace transform.