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Math Help - Solution to heat equation

  1. #1
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    Solution to heat equation

    u_t = 2u_{xx} \quad

    T'(t) = -2 \lambda T(t) \quad
    and
    X''(x) = - \lambda X(x). \quad

    Does it make any difference if choose constant to be \lambda and not -\lambda as above

    Thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    u_t = 2u_{xx} \quad

    T'(t) = -2 \lambda T(t) \quad
    and
    X''(x) = - \lambda X(x). \quad

    Does it make any difference if choose constant to be \lambda and not -\lambda as above

    Thanks
    It can be shown that the seperation constant must be negative or you will only get the trivial solution.

    Here is the proof for homogenoious Dirichlet or Neumann boundry contdtions

    u_t=au_{xx} after seperating we get

    \frac{T'}{aT}=\frac{X''}{X}=\beta

    So we get

    X''-\beta X=0

    Now multiply this equation by X and integrate the equation.

    XX''-\beta X^2=0 \iff \int_{0}^{L}XX''dx -\int_{0}^{L}\beta X^2dx=0

    Now integrate the first term by parts with u=X \implies du=X' and dv=X'' \implies v=X' to get

    XX'\bigg|_{0}^{L}-\int_{0}^{L}(X')^2dx-\int_{0}^{L}X^2dx =0

    Now if we have the homogenious condtions mentioned above either

    X(0)=X'(L)=0 \text{ or } X'(0)=X'(L)=0 in either case the first term above is equal to zero when evaluated at the end points. So we get

    -\int_{0}^{L} [(X')^2+\beta (X)^2]dx=0

    Since both (X')^2 \text{ and } (X)^2 are always non negative the only when this integral can be zero is if \beta < 0

    So the seperation constant must be negative in the heat equation.

    This can be shown for the mixed and other boundry conditions but it is not as clean. I hope this helps.
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