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Math Help - Heat Conduction, Finding Cn

  1. #1
    idrinkjuice
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    Heat Conduction, Finding Cn

    Hello, I'm having difficulty finding the Cn for a heat conduction problem, and any help would be appreciated.

    The problem is #8, Section 10.5 in Boyce and DiPrima if you have it.

    Uxx = 4Ut, 0 < x < 2, t>0
    U(0,t) = 0 = U(2,t), t>0
    U(x,0) = 2sin(pi x / 2) - sin(pi x) + 4sin(2pi x)

    Thanks.
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  2. #2
    Senior Member
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    General solution

    To solve \dfrac{\partial^{2}U}{\partial x^2} = 4 \dfrac{\partial U}{\partial t}, you use separation of variables U(x,t)=X(x)T(t) to get
    \dfrac{4}{T}\dfrac{dT}{dt}=\dfrac{1}{X}\dfrac{d^{2  }X}{dx^2}=-\dfrac{1}{\lambda^2} with lambda a constant (of the same units as x) and the negative sign chosen to make the solution finite for all t>0. Solving these separately, one gets an exponential for T(t) and a sine wave function for X(x). Using the x boundary conditions, one determines the discrete values \lambda_n allowed for \lambda to get the U_{n}(x,t), and

    U(x,t)=\sum_{n=1}^\infty c_{n}U_{n}(x,t)

    Then we compare this at t=0 with the boundary condition U(x,0)=2\sin (\tfrac{\pi}{2}x) - \sin (\pi x) + 4\sin (2\pi x). The c_n should then become quite apparent.

    --Kevin C.
    Last edited by TwistedOne151; December 4th 2007 at 04:32 PM. Reason: Formatting
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  3. #3
    Global Moderator

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    No need to solve this problem from first principles. There is a general formula for the coefficients that is derived from seperation of variables. But when I solve a PDE I do not do it from first principles, I identity what equation it it ands its BVP and apply the known formula. It saves a lot of time.
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