Partial DE - heated cylinder

  • Thread starter skrat
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  • #1
skrat
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Homework Statement


A very long homogeneous cylinder is cut in half along its axis. One half s than equally heated, while the other half is equally cooled. How does the temperature change when the two parts are joined back together, if the cylinder is well isolated?

Homework Equations




The Attempt at a Solution



Hmmm,
I know I can work with $$T(r,\varphi ,t)=\sum _{m,n}J_{m,n}(\xi _{m,n}\frac r R)[B_m cos(m\varphi )+C_msin(m\varphi )]e^{-i\omega t}$$where ##\xi _{m,n}## is m-th zero of n-th Bessel function, but what I do not understand at this point is why all ##B_m## are zero?

Because if I am not mistaken, one boundary condition is $$j=-\lambda \frac{\partial }{\partial r}T(r=R,\varphi t)=0$$ and the other should be something at ##\varphi =0## or ##\varphi =\pi ##. I guess? I assume this second bondary condition will also tell me why the ##cos## is gone in the first equation.
 

Answers and Replies

  • #2
Orodruin
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After you rejoin the cylinder, there really is no boundary in the ##\phi## coordinate. If the ##B_m## or ##C_m## are zero depends on how you put your coordinate system. The type of argument to be used to see this is based on symmetry (or actually performing the integrals to solve for the coefficients).
 

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