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Separation of heat equation

  1. Mar 31, 2012 #1

    So in my book they separate out a solution the heat equation into a function of x and a function of t and then separate the t and x dependencies, giving

    [tex] -\frac{T''}{c^2T} = - \frac {X''}{X} = \lambda [/tex]

    The book says that the proof is because

    [tex] \frac {\partial \lambda}{\partial x} = 0[/tex]


    [tex] \frac {\partial \lambda}{\partial t} = 0[/tex]

    but if X = X(x) and T = T(t), how do you get zero out when you take d/dx and d/dt? Why don't you use the chain rule on -X''/X to get something like

    [tex] \frac{\partial}{\partial x} (\frac{-X''}{X}) = (-\frac{X'''}{X} + \frac{X''(X')}{X^2}) [/tex]

  2. jcsd
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