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:
$\displaystyle h\left[\Delta U\sin(\omega t) - T|_{r=R}\right] = -k\frac{\partial T}{\partial r}\Bigg|_{r=R}$
$T|_{t=0} = 0$
$T|_{r=\infty}=0$
where $\Delta U, h, \omega, \alpha, k,$ and$R$ are constants
Any help transforming this would be appreciated!
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
$$ B(s) = \left[hK_0\left(\sqrt{\frac{s}{\alpha}}R\right)+k\sqrt{\frac{s}{\alpha}}K_1\left(\sqrt{\frac{s}{\alpha}}R\right)\right]^{-1}h\Delta U\frac{\omega}{s^2+\omega^2} $$