Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral (Simplifying it)

  1. Apr 18, 2006 #1
    I don't understand this simplification given in this problem:

    Q: At [itex] t=0 [/itex] a hot gas is on one side and a cold gas is on the other: [itex] u_0 = 1 [/itex] for [itex] x>0 [/itex], [itex] -1 [/itex] for [itex] x>0 [/itex]. Write down the solution:
    [tex] u(t,x)=\int_{\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy[/tex]

    to [itex] u_{tt}=u_{xx} [/itex] and with [itex] z = y-x [/itex] simplify to:
    [tex] u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz [/tex]



    So my attempt thus far (with very little progress):

    [tex] u(t,x)=\int_{\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy[/tex]
    [tex] u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz [/tex]

    [tex] z = y-x \Rightarrow -z = -y+x \Rightarrow y = z+x[/tex], [tex] dy=dz+dx [/tex]

    [tex] u(t,x)=\frac{1}{2\sqrt{\pi t}} \int_{\infty}^{\infty} e^{-(-z)^2/4t}u_0(z+x)(dz+dx)[/tex]

    [tex] u(t,x)=\frac{1}{2\sqrt{\pi t}} \int_{\infty}^{\infty} e^{-z^2/4t}u_0(z+x)(dz+dx)[/tex]

    Things get sketchy here (assuming things are even right above):
    [tex]u_0(z+x) = \left\{ \begin{array}{c} 1 \,\,\,\, ,0 > z+x \\ -1 \,\, ,0<z+x \end{array}[/tex]

    [tex]u(t,x) = \frac{1}{\sqrt{\pi t}}\left( \int_{-\infty}^{0}e^{-z^2/4t}(-1)(dz+dx) + \int_{0}^{\infty}e^{-z^2/4t}(dz+dx) \right)[/tex]

    and that's it folks...

    I don't understand how the bounds of the integration made the jump from [itex] (\infty,-\infty) [/itex] to [tex] (-x,x) [/itex]. I definitely need some help :)

    Thanks in advance.
     
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. x must be treated as a constant. So dy=dz
    So in the first case, you have 0>z+x so z<-x. So you must integrate from -infinity to -x. In the second case, z>-x so you integrate from -x to plus infinity. a quick glance at the integrals gives me the impression that after rewriting the integrals a little and using the parity of the integrands, you will get a difference of two integrals that will cancel outside of the interval [-x,x]. But I did not work it out explicitly.
     
  4. Apr 18, 2006 #3
    Why is this so? Sorry, I'm not seeing it. I understand why differentiating a constant is 0. But why is it held as a constant?

    I'll work through it. I'm slapping myself in the face for this part:

    Thanks man. I appreciate it.
     
    Last edited: Apr 18, 2006
  5. Apr 18, 2006 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Because you are evaluating u(t,x). So you must think of the integral as providing the value of u(t,x) for a certain fixed value of x and t.

    Patrick
     
  6. Apr 18, 2006 #5
    Ahh beautiful. Thank you!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook