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Show integral is independent

  1. Jul 26, 2010 #1
    du/dt = d2u/dx2

    u(x,t) = (t^a) * (g(e)) where e = x/sqrt(t) and a is a constant

    Show that

    integral from -inf to inf [ u(x,t) ] dx = integral from -inf to inf [ (t^a) * g(e) ] dx

    is independent of t only if a=-0.5

    My attempt was to diff both sides by t (sorry not x) giving

    integral from -inf to inf [d2u/dx2 ] dx = integral from -inf to inf [at^(a-1) g(e) + t^a dg(g)/dt ] dx

    Not sure if this is correct and cant see where to go from here...any help most appreciated. Thanks
     
    Last edited: Jul 26, 2010
  2. jcsd
  3. Jul 26, 2010 #2

    lanedance

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    so just to write it out in tex your starting point is:
    [tex] f(t) = \int^{\infty}_{-\infty} u(x,t) dx = \int^{\infty}_{-\infty} (t^a) g(\frac{x}{\sqrt{t}})dx [/tex]

    i don't think its ok here to differentiate both sides w.r.t. x, you always need to be careful bringing it differntiation inside an integral, and here in effect its just a dummy integration variable...
     
    Last edited: Jul 26, 2010
  4. Jul 26, 2010 #3

    lanedance

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    but you can differentitae w.r.t. t no worries
    [tex] f'(t) = \frac{\partial }{\partial t} \int^{\infty}_{-\infty} u(x,t) dx = \frac{\partial }{\partial t}\int^{\infty}_{-\infty} (t^a) g(\frac{x}{\sqrt{t}})dx [/tex]

    [tex] f'(t) = \int^{\infty}_{-\infty} u_t(x,t) dx = \int^{\infty}_{-\infty} u_{xx}(x,t) dx = \int^{\infty}_{-\infty} \frac{\partial }{\partial t}(t^a) g(\frac{x}{\sqrt{t}})dx [/tex]

    [tex] f'(t)
    = \int^{\infty}_{-\infty} \left( a t^{a-1} g(\frac{x}{\sqrt{t}})
    - \frac{xt ^{a-3/2}}{2}g'(\frac{x}{\sqrt{t}}) \right)dx
    [/tex]

    assuming my algebra is correct, and which appears close to what you presented

    if you can show f'(t) = zero, then its clear the integral is independent of t
     
    Last edited: Jul 26, 2010
  5. Jul 26, 2010 #4

    lanedance

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    then you can look at the x derivatives
    [tex] u_x(x,t) = \frac{\partial }{\partial x}t^a g(\frac{x}{\sqrt{t}}) [/tex]

    [tex] u_x(x,t) = t^a g'(\frac{x}{\sqrt{t}}) \frac{1}{\sqrt{t}}= t^{a-1/2} g'(\frac{x}{\sqrt{t}}) [/tex]

    and carry on from there, you may have to consider a=1/2 and otherwise separatley

    PS - i haven't fully worked it but this is what i would try
     
    Last edited: Jul 26, 2010
  6. Jul 26, 2010 #5

    hunt_mat

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    So you have an integral:
    [tex]
    \int_{\infty}^{\infty}t^{a}g(x/\sqrt{t})dx
    [/tex]
    use [tex]v=x/\sqrt{t}[/tex] as a substitution to obtain (after a little algerbra)
    [tex]
    \int_{\infty}^{\infty}t^{a+1/2}g(v)dv
    [/tex]
    This will ne infependent of t if and only if a=-1/2. I will leave you to fill in the details.

    Mat
     
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