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How do you solve the ff double integral?

  1. Sep 1, 2006 #1
    given [tex]\phi[/tex] to be a function of x and t, how do you solve

    [tex]2\int_{0}^{\infty}\int_{x}^{\infty}\frac{\partial^{2}\partial\phi}{\partial t^{2}} dt dx - 2\int_{0}^{\infty}\int_{0}^{t}\frac{\partial^{2}\partial\phi}{\partial x^{2}} dx dt[/tex]

    any hints would be great.
    Last edited: Sep 1, 2006
  2. jcsd
  3. Sep 1, 2006 #2
    phi is a function of x and t, you mean?
  4. Sep 1, 2006 #3
    yeah, sorry bout that, I'll edit my post above.
  5. Sep 1, 2006 #4


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    Homework Helper

    Assuming x and t are independent, you have, eg:

    [tex]\int_a^b \frac{\partial f(x,t)}{\partial t} dt = f(x,b)-f(x,a)[/tex]

    Does that help?
  6. Sep 1, 2006 #5
    I would have:
    [tex]2\int_{0}^{\infty}\frac{\partial\phi(\infty, x)}{\partial t} - \frac{\partial\phi(x,x)}{\partial t} dx - 2\int_{0}^{\infty}\frac{\partial\phi(\infty, t)}{\partial x} - \frac{\partial\phi(t,t)}{\partial x} dt [/tex]

    this is correct right?

    but then I don't know how to process from there.
  7. Sep 1, 2006 #6
    it is given in the original problem that [tex]\phi[/tex] is twice continously differentiable with compact support.

    so does that mean
    [tex]\frac{\partial \phi (\infty, x)}{\partial t} = \frac{\partial \phi (\infty, t)}{\partial x} = 0[/tex]?

    or that only
    [tex]\phi(\infty, \infty) = 0[/tex]?

    how about for any x, or t,
    [tex]\phi(\infty, t) = \phi(x, \infty) = 0[/tex]?
  8. Sep 1, 2006 #7


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    I'll assume you know what "compact" and "support" mean. Then it suffices to know that the image of a compact set under a continuous map is another compact set, and that the projection maps onto each coordinate are continuous. So if the function has compact support in the (x,t) plane, then the projection of this set onto either the x or t axes is also compact. I hope you can proceed from here.
  9. Sep 2, 2006 #8
    just to make sure my understanding of compact support is correct...
    is it okay then to say that

    [tex]\frac{\partial \phi (\infty, x)}{\partial t} = \frac{\partial \phi (\infty, t)}{\partial x} = 0[/tex]

    [tex]\phi(\infty, \infty) = 0[/tex]

    [tex]\phi(\infty, t) = \phi(x, \infty) = 0[/tex]

    I apologize if this is a dumb question...but admittedly, my grasp of real analysis concepts is not that firm.

    thanks for the help again.
  10. Sep 2, 2006 #9


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    "Compact support" simply means that [itex]\phi[/itex] is only non-zero on a compact set- a closed and bounded set. Yes, [itex]\phi[/itex] is 0 at and on a "neighborhood" of [itex]\infty[/itex] (so its derivative is 0 at [itex]\infty[/itex] also).
  11. Sep 2, 2006 #10
    hi, HallsofIvy, thanks for the clarification.

    so now I have:
    [tex]-2\int_{0}^{\infty} \frac{\partial\phi(x,x)}{\partial t} dx + 2\int_{0}^{\infty} \frac{\partial\phi(t,t)}{\partial x} dt [/tex]

    Is this the final answer? or is it possible to simplify this further?

    thanks again for the help.
  12. Sep 2, 2006 #11

    for some reason, I have the feeling the answer is going to be zero where the two integrals of the partial over infinity would be be zero because of compact support and the two integrals of the partial over 0 would cancel each other out.

    Would this be correct? If so, my prolem is how to write this out mathematically. Thanks again.
  13. Sep 2, 2006 #12


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    Well, [itex]\partial \phi(x,x)/ \partial t=0[/tex], since there is no explicit t dependence. Maybe I'm being picky, but what you should have there is something like:

    [tex]\frac{\partial \phi(t,x)}{\partial t} |_{t=x}[/tex]

    That way you see the derivative is w.r.t. the first "slot". At this point I don't think you can go any further without knowing something about [itex]\phi(t,x)[/itex]. Note that:

    [tex]\int_{0}^{\infty} \frac{\partial\phi(t,x)}{\partial t} |_{t=x} dx \neq \phi(\infty,\infty)-\phi(0,0) [/tex]

    As I think is clearer when you write the derivative this way.
  14. Sep 2, 2006 #13
    I get what you are saying StatusX.

    So I'm left with:
    [tex]-2\int_{0}^{\infty} \frac{\partial\phi(t,x)}{\partial t} |_{t=x} dx + 2\int_{0}^{\infty} \frac{\partial\phi(t,x)}{\partial x} |_{x=t} dt[/tex]

    for some reason, though this answer looks technically correct...I don't think this is the answer the question is looking for, as the answer is not simplified enough, if you get what I mean. I may have to clarify with my professor on this.

    BTW, the topic we are studying is distributions (generalized function), and this question was asked so that we can compare its answer to a similar problem using distributions as a way to solve.

    Here's the original question, maybe I did something wrong...

    Consider the function Y defined by

    Y(x,t) = 1, if |x| <= t, Y(x,t) = 0 if |x|> t or t<= 0.

    For any [tex]\phi \in \mathfrak{D} (\mathbb{R}^{2})[/tex]where [tex]\mathfrak{D}[/tex] means functions which are continuously differentiable with compact support)

    solve for

    [tex]\int_{\mathbb{R}}\int_{\mathbb{R}}Y(x,t) (\frac{\partial^{2}\phi}{\partial t^{2}} - \frac{\partial^{2}\phi}{\partial x^{2}}) (x,t) dx dt[/tex]
  15. Sep 7, 2006 #14
    based on the question on the post before this one, I belive that the answer that the question is looking for is this:

    [tex]\int_{\mathbb{R}}\int_{\mathbb{R}} (\frac{\partial^{2}Y}{\partial t^{2}} - \frac{\partial^{2}Y}{\partial x^{2}}) (x,t) \phi(x,t)dx dt[/tex]

    so any ideas how I can manipulate:
    [tex]\int_{\mathbb{R}}\int_{\mathbb{R}}Y(x,t) (\frac{\partial^{2}\phi}{\partial t^{2}} - \frac{\partial^{2}\phi}{\partial x^{2}}) (x,t) dx dt[/tex]

    [tex]\int_{\mathbb{R}}\int_{\mathbb{R}} (\frac{\partial^{2}Y}{\partial t^{2}} - \frac{\partial^{2}Y}{\partial x^{2}}) (x,t) \phi(x,t)dx dt[/tex]

    any ideas?

    Last edited: Sep 7, 2006
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