# Solving Heat Transfer PDE using Laplace Transform

I have the following PDE in the time domain:

$$\frac{r}{\alpha}\frac{\partial T}{\partial t} =\frac{\partial T}{\partial r}+r\frac{\partial ^2T}{\partial r^2}$$

Where temperature ($$T$$) is a function of time ($$t$$) and radial distance ($$r$$)

My goal is to transform this PDE into an ODE in the $$s$$-domain using the Laplace transform, but I am having trouble transforming each term, specifically $$r\frac{\partial^2 T}{\partial r^2}$$

Boundary conditions if needed:

1. $$\displaystyle h\left[\Delta U\sin(\omega t) - T|_{r=R}\right] = -k\frac{\partial T}{\partial r}\Bigg|_{r=R}$$

2. $$T|_{t=0} = 0$$

3. $$T|_{r=\infty}=0$$

where $$\Delta U, h, \omega, \alpha, k,$$ and$$R$$ are constants

Any help transforming this would be appreciated!

• Those terms will be unaffected, since all of the operations are in $r$. The only thing that changes is the time derivative. – Dylan Mar 12 at 4:04

Denote $$\tilde T (r,s)$$ to be the Laplace transform of $$T(r,t)$$ we get

$$\frac{rs}{\alpha}\tilde T = \frac{\partial \tilde T}{\partial r} + r\frac{\partial^2\tilde T}{\partial r^2}$$

Rewrite as

$$r^2\frac{\partial^2 \tilde T}{\partial r^2} + r\frac{\partial \tilde T}{\partial r} - \frac{s}{\alpha}r^2\tilde T = 0$$

This has solutions in the form of modified Bessel functions.

$$\tilde T(r,s) = A(s)I_0\left(\sqrt{\frac{s}{\alpha}}r\right) + B(s)K_0\left(\sqrt{\frac{s}{\alpha}}r\right)$$

Since the domain is $$R, we don't want to solution to blow up at $$\infty$$, so $$A(s)=0$$.

You should be able to transform the boundary condition to get

$$h\tilde T(R,s) - k \frac{\partial \tilde T}{\partial r}(R,s) = h\Delta U \frac{\omega}{s^2+\omega^2}$$

which gives

$$B(s) = \left[hK_0\left(\sqrt{\frac{s}{\alpha}}R\right)+k\sqrt{\frac{s}{\alpha}}K_1\left(\sqrt{\frac{s}{\alpha}}R\right)\right]^{-1}h\Delta U\frac{\omega}{s^2+\omega^2}$$