Deriving the heat equation in cylindrical coordinates

[nettle404]

Homework Statement

Consider heat flow in a long circular cylinder where the temperature depends only on [itex]t[/itex] and on the distance [itex]r[/itex] to the axis of the cylinder. Here [itex]r=\sqrt{x^2+y^2}[/itex] is the cylindrical coordinate. From the three-dimensional heat equation derive the equation [itex]U_t=k(U_{rr}+2U_r/r)[/itex].

Homework Equations

The standard heat equation is

[tex]c\rho\frac{\partial}{\partial t}U(x,y,z,t)=\kappa\nabla^2U(x,y,z,t)[/tex]

The Attempt at a Solution

Attempted to work backwards from [itex]U_t=k(U_{rr}+U_r/r)[/itex] with the chain rule, but that did not produce anything of value. I can also probably solve the problem by deriving the heat equation starting in cylindrical coordinates, but the question asks to specifically "transplant" cylindrical coordinates onto the Cartesian coordinate heat equation by some tricky variable substitution algebra I can't imagine performing. That is: Where should I start?

 

[Geofleur]

You could start with, e.g., ## \frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial U}{\partial z}\frac{\partial z}{\partial x} ## (the last term vanishes, why?). Once you have ## \frac{\partial U}{\partial x} ## in terms of ## r, \theta,## and ##z##, you can get ## \frac{\partial^2 U}{\partial x^2} = \frac{\partial}{\partial r}\left( \frac{\partial U}{\partial x} \right) \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\left(\frac{\partial U}{\partial x} \right)\frac{\partial \theta}{\partial x} ##. You can assemble ##\nabla^2## going on this way. There are faster methods, but this is the one that requires the least sophisticated mathematics (as far as I know).

 

[nettle404]

You could start with, e.g., ## \frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial U}{\partial z}\frac{\partial z}{\partial x} ## (the last term vanishes, why?). Once you have ## \frac{\partial U}{\partial x} ## in terms of ## r, \theta,## and ##z##, you can get ## \frac{\partial^2 U}{\partial x^2} = \frac{\partial}{\partial r}\left( \frac{\partial U}{\partial x} \right) \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\left(\frac{\partial U}{\partial x} \right)\frac{\partial \theta}{\partial x} ##. You can assemble ##\nabla^2## going on this way. There are faster methods, but this is the one that requires the least sophisticated mathematics (as far as I know).

The last term vanishes because $z$ is unaffected by $x$? I'm not entirely convinced that's necessarily the case, but it's a bit late where I am and I need some sleep. I'll return to the problem tomorrow and try to solve it: I think your hint has more or less cleared the issue for me.

Out of curiosity, what are these faster methods you're speaking of?

 

[Geofleur]

Indeed, ## z ## and ## x ## are independent of one another. The other methods:

(1) The expressions for ##\nabla ##, ## \nabla^2 ##, ## \nabla \cdot ##, and ## \nabla \times ## can all be derived in curvilinear coordinates by using the divergence and Stokes' theorems. See, e.g., Fujita and Godoy, Mathematical Physics, pg. 73.

(2) Using tensor calculus and the covariant derivative, one can derive general expressions for these operators in terms of the metric tensor ##g##. ##\nabla^2##, for example, turns out to be:

## \nabla^2\Psi = \frac{1}{|g|^{1/2}}\frac{\partial}{\partial x^i}\left[ |g|^{1/2}g^{ik}\frac{\partial \Psi}{\partial x^k} \right] ##.

Above, when an index appears once raised and once lowered, it is to be summed over (known as the Einstein summation convention).