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Hom. heat equation in cylindrical coordinates using Fourier & Laplace transforms

 
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Jul16-10, 05:44 PM   #1
 

Hom. heat equation in cylindrical coordinates using Fourier & Laplace transforms


I'm trying to solve the homogeneous heat equation of a semi-infinite cylinder in cylindrical coordinates for a semi-infinite cable (no theta dependence):
[tex]\frac{\partial U}{\partial t}=D\left(\frac{\partial^{2} U}{\partial r^{2}}+\frac{1}{r}\frac{\partial U}{\partial r}+\frac{\partial^{2} U}{\partial z^{2}}\right) [/tex]
with boundary conditions and initial values:
  • [tex](1) U(r,z,0)=0[/tex]
  • [tex](2) U(r,\infty,t)=0[/tex]
  • [tex](3) U(R,z,t)=0[/tex]
  • [tex](4) kU_{z}(r,0,t)=\Phi_{o} [/tex]
Where r,z,t represent what they usually do, D is the thermal diffusivity, k is the thermal conductivity, and [tex]\Phi_{o}[/tex] is a constant flux going into one end of the cylinder at z=0. R is the radius of the cylinder.

I first tried separation of variables, which was working fine with two separation constants, but I had trouble getting the boundary conditions to fit. A friend suggested I try Green's functions, but I wasn't sure how to go about that since there is no internal heat source. So, my next attempt was going to be with Fourier and Laplace transforms.

I've never had to apply two transforms to a PDE before, so I'm a little unsure how to begin.

Correct me if I'm wrong: I should apply the Laplace transform to the time derivative, and the Fourier transform to the position derivatives. I haven't seen any examples of using a Fourier transform on more than one variable, so I'm stuck in this step. Also, the cylinder is semi-infinite, not infinite, so how should I account for this with the Fourier transform?

Thanks!
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Jul19-10, 05:45 PM   #2
 
Anyone?? Just a quick "no, that will not work..." or the like would be very helpful. I haven't been able to work on the problem much since I posted it, but I'm planning on using a Laplace transform for time and z, and a finite Hankel/Fourier-Bessel transform for the radial component.

I'll work through the math tonight and post what I come up with.
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