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Heat equation

  1. Mar 30, 2010 #1
    Consider a heat equation for the temperature u of a rod of length 1:
    ut = uxx, 0 < x < 1, t > 0 with boundary conditions ux(0,t) = 0 & u(1,t) = 0. I derived Xn(x) = cos((n+1/2)[tex]\pi[/tex]x) using separation of variables.
    How do I show that [tex]\int_{0}^1[/tex] Xn(x)Xm(x) dx = 1/2 if m = n and 0 if m [tex]\neq[/tex] n.
    I used the product to sum formula: cos(A)cos(B) = cos(A+B)/2 + cos(A-B)/2 to get 1/2cos((n+m+1)[tex]\pi[/tex]x) 1/2cos((n-m)[tex]\pi[/tex]x) but I am stuck after that. Someone help, am I even on the right track.
     
  2. jcsd
  3. Apr 1, 2010 #2

    LCKurtz

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    Yes, you are on the right track. When m = n use the formula

    [tex]\cos^2(\theta) = \frac {1 + \cos(2\theta)}{2}[/tex]

    which is easy to integrate. I'm not sure why you are stuck on the others.

    [tex]\frac 1 2 \cos((n+m+1)\pi x)[/tex]

    is just as easy to integrate is [itex]\cos(kx)[/itex].

    n and m are integers, you know.
     
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