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Homework Help: Delta Dirac - Question on my final

  1. May 8, 2006 #1
    Delta Dirac - Question on my final :(

    :mad: Man I just had this final, and I didn't study the delta function enough. I totally bombed this question. I think I did ok on the rest of it... but I messed up bad on this question. I want to know the answer though.

    Solve the heat equation:
    [tex] u_t = u_{xx} [/tex] with the initial condition [tex] u_0(x)=\delta(x-a) [/tex] using the fundemental solution:

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

    So all I could remember about the delta function is the identity:

    [tex] \int_{\infty}^{\infty} f(x)\delta(x-a) \,\,dx = f(a) [/tex]

    What confused me was the dummy variable? How do I account for that? I need some hints how to solve this problem, because I'm lost....

    grrr... I'm so mad I missed this.
  2. jcsd
  3. May 8, 2006 #2


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    It seems to me that [itex] u_0(y) [/itex] is simply [itex] \delta(y-a)[/itex] so, using this, the integral is trivial (replace y by a) and you get u(x,t).
  4. May 8, 2006 #3
    yeah what I had was:

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

    But from here is where I am lost. Are you saying:

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

    I don't understand.

    Wait... is it because you hold x,t constant, thus y becomes a and you just get:
    [tex] \frac{1}{2 \sqrt{\pi t}} e^{-(x-a)^2/4t} [/tex]
    Last edited: May 8, 2006
  5. May 9, 2006 #4


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    Yes...you are integrating over y and the delta function sets y=a so the result is that last expression you wrote. It was that simple!
  6. May 9, 2006 #5
    lol... simple, at least it is for you
  7. May 9, 2006 #6
    Well maybe I'll get some credit for it.

    For my final answer I put.
    [tex] \frac{1}{2 \sqrt{\pi t}} e^{-(x)^2/4t} [/tex]

    with a question mark next to it with an explanation that was something like: I know this isn't right, I didn't study the delta function enough. I understand that as you integrate along y the function will spike at a specific point and return that value of the function.

    But he will know that I have no idea :(
    All I can hope for is some mercy points I guess.

    Anyways, the question reminds me of my calc II professor. We were learning integration methods and he had a question like:

    [tex] \int_x^x e^{x^2} \,\, dx [/tex]

    It looks easy. Then you try a u substitution and get nowhere. Then you think of other methods you know I get nowhere.

    Then you notice the limits of integration are the same, and the problem is ridiculously easy. Kind of the same thing here :)

    Anyways, thanks for clearing that up man. I appreciate it.
  8. May 13, 2006 #7


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    Too bad you missed it :frown:
    A dirac delta function inside an integration is a godsend...makes the integral trivial (unless the argument of the delta function is itself a function of the variable but even then it is usually easy).
    That was nasty :biggrin: Although it never made sense to me to use the same variable in the limits as what is being integrated over (I owuld have used x' or anything else than x), but I know what you mean.

    A slightly less trivial and more informative example would be to put the limits between x' and x'+dx', so [itex] \int_{x'}^{x'+dx'} dx \, e^{x^2} [/itex]. This is as trivial to do but requires a certain understanding of the integration process.

    And of course, there is always the odd/even trick (as in "integrate [itex] \int_{-5}^5 dx \, x^3\, e^{x^2} [/itex]...unless the prof has insisted on checking the parity first, a lot of people could spend some time doing it the long way whereas the answer can be seen at a glance)
    You are welcome. At least now, if you ever encounter a Dirac delta function again, you'll know what to do :biggrin:
  9. May 13, 2006 #8
    Yup. I'll :smile: instead of :uhh:
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