Consider the heat equation in spherical coordinates (considering only variation in the radial direction):
$$\frac{\partial\phi}{\partial t} = \frac{\alpha}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial\phi}{\partial r}\right).$$
Using the following variable substitution,
$$\theta = r\phi$$
it is possible to convert this equation into a "Cartesian-like" form:
$$\frac{\partial\theta}{\partial t} = \alpha\frac{\partial^2\theta}{\partial r^2}.$$
My question is:
Is it possible to begin with the heat equation in cylindrical coordinates (again only considering variation in the radial direction),
$$\frac{\partial\phi}{\partial t} = \frac{\alpha}{r} \frac{\partial}{\partial r}\left(r \frac{\partial\phi}{\partial r}\right)$$
and, using a similar variable substitution, achieve this same "Cartesian-like" end form? If it is possible, what transformation achieves this?
$\endgroup$1 Answer
$\begingroup$Use the substitution $z = r\phi$. We have $\frac{\partial \phi}{\partial r} = \frac{-z}{r^2}$ and $\frac{\partial^2 \phi}{\partial r^2} = \frac{z}{r^3}$. Then:$$\frac{\partial \phi}{\partial t} = \frac{\alpha}{r} \frac{\partial}{\partial r} \left[r \cdot \left(\frac{-z}{r^2}\right)\right] = \frac{\alpha}{r} \cdot \frac{z}{r^2} = \alpha \cdot \frac{z}{r^3} = \alpha \frac{\partial^2 \phi}{\partial r^2}.$$
$\endgroup$ 5
