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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 October 9, 2019, 8:38 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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