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Math Help - Solving Heat Equation

  1. #1
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    Solving Heat Equation

    Find a solution to the heat equation,

    \frac{{\partial u}}{{\partial t}}-k\frac{{\partial^2\ u}}{{\partial x^2}}=0

    (k>0 is a constant)
    which takes the form

    u(x,t)=f(x-kt)

    for some f(s) which is twice differentiable in s.

    (Hint: find out the form that f(s) must take by substituting u(x,t)=f(k-xt) into the equation.)

    I...dono what i'm supposed to do. We went over deriving the heat equation in class and proving a solution is unique, but not so much the actual solving part. How do I go about solving such kinds of problems?
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  2. #2
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    The heat equation is generally solved by separation of variables: Assume a solution of the form: u(x,y)=X(x)T(t). That is, a product of a function of x and a function of t. Now, substitute the expression X(x)T(t) into the PDE and obtain separate ODEs in terms of t and x. Solve those ODEs under suitable "well-posed" conditions (boundary and initial conditions) to arrive at a solution. I'd recommend "Basic Partial Differential Equations" by D. Bleecker and G. CSordas. Here's a well-posed problem:

    \textbf{D.E.}\quad u_t=ku_{xx}\quad 0\leq x\leq L,\; t\geq 0

    \textbf{B.C}\quad u(0,t)=A(t)\quad u(L,t)=B(t)

    \textbf{I.C}\quad u(x,0)=f(x)

    If you get into PDEs, always try to "pose the problem well" (correctly define the boundary and initial conditions). The style of these will change with the PDE.
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  3. #3
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    Quote Originally Posted by Avitar View Post
    Find a solution to the heat equation,

    \frac{{\partial u}}{{\partial t}}-k\frac{{\partial^2\ u}}{{\partial x^2}}=0

    (k>0 is a constant)
    which takes the form

    u(x,t)=f(x-kt)

    for some f(s) which is twice differentiable in s.

    (Hint: find out the form that f(s) must take by substituting u(x,t)=f(k-xt) into the equation.)
    u_t = -kf'(x-kt) and u_x = f'(x-kt) and u_{xx} = f''(x-kt).

    Substitute that,
    k f'(x-kt) - kf''(x-kt) = 0 \implies f'(x-kt) = f''(x-kt).
    Since x-kt can be made any number is means,
    f ' ( z) = f'' (z) for any number z.
    Thus, f ' (z ) = ae^z \implies f(z) = ae^z + b.
    Thus, f(x-kt) = ae^{x-kt} + b for some constants a,b.
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  4. #4
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    Alright, I see what you did there. That helps enormously, thanks. Although, I have to ask what happened to the negative sign from finding the partial with respect to u during the substitution? Is there something I don't know about that lets us disregard it, or should it still be there?
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