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# Heat Equation on Circle

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Solve the heat problem on the circle

u_t = ku_{xx}
u(x,0) = phi(x) where phi(x) is the 2l periodic extension of phi
using the separation of variables.

I am able to go as far as
u = XT
-X''/X = lambda where lambda = beta^2

usually the solution for X'' + beta^2 * X = 0 is Ccos(beta * L) + D sin(beta * L) I believe.

using cos = (e^i + e^-i)/2 and sin...

X(x) = C[(e^iBx + e^iBx)/2 + D(e^ibx - e^-iBx)/2i]
T(x) = Ae^B^2kt

B = beta and C, D, and A are coeff

I do not know how to find the coefficients.
There is no directly stated boundary conditions.
I think I'm supposed to assume something.
How do I solve for the coefficients?

© BrainMass Inc. brainmass.com September 20, 2018, 11:59 pm ad1c9bdddf - https://brainmass.com/math/fourier-analysis/heat-equation-circle-157787

#### Solution Summary

The heat equation on circle is investigated. The solution is detailed and well presented.

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