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!

  • $\begingroup$ Those terms will be unaffected, since all of the operations are in $r$. The only thing that changes is the time derivative. $\endgroup$
    – Dylan
    Mar 12, 2019 at 4:04

1 Answer 1


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 $R<r<\infty$, 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

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


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