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Show integral is independent of time

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Given that [itex] u(x,t) [/itex] satisfies [itex] u_t-6uu_x+u_{xxx}=0 [/itex] (*) and [itex] u, u_x, u_{xx} \to 0 [/itex] as [itex] |x| \to \infty [/itex] show that

    [itex] I_1 = \int_{-\infty}^{\infty}u^2dx [/itex]

    is independent of time ([itex] \frac{\partial I_1}{\partial t}=0 [/itex]).

    2. Relevant equations

    -

    3. The attempt at a solution

    I guess it would suffice to show that [itex] I_1 [/itex] is equal to some function of (only) [itex] x [/itex]. I can't think of a way to do this which involves the equation (*). My attempt so far has been to do some kind of substitution so that I integrate with respect to [itex] u [/itex] rather than [itex] x [/itex]. Something like

    [itex] I_1 = \int_{-\infty}^{\infty}u^2dx = \int_{u(-\infty,t)}^{u(\infty, t)}u^2 \frac{dx}{du} du = \int_{0}^{0} \frac{u^2}{u_x} dx = 0 [/itex]

    Am I on the right track?
     
  2. jcsd
  3. Dec 10, 2014 #2

    ShayanJ

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    That doesn't seem right to me. Use Leibniz integral rule for differentiating the derivative w.r.t. time and then use the differential equation for substitution.
     
  4. Dec 10, 2014 #3

    Ray Vickson

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    As "Shyan" has already suggested, you can show that ##I_1## is a constant by showing that ##dI_1/dt = 0##.
     
  5. Dec 10, 2014 #4
    Thanks guys.

    I've given your advice a go. So far I have used Leibniz integral rule and [itex] f=u^2 [/itex]:

    [itex] \frac{\partial I_1}{\partial t} = \frac{\partial}{\partial t}(\int_{-\infty}^{\infty} f(x,t)dx) = \int_{-\infty}^{\infty} \frac{\partial f}{\partial t} dx = \int_{-\infty}^{\infty} uu_tdx[/itex].

    Where do I go from here?
     
  6. Dec 10, 2014 #5

    Ray Vickson

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    Are you forgetting that ##u## satisfies a partial differential equation? You have all the information you need; it is just a matter of applying it.
     
  7. Dec 11, 2014 #6
    I thought about subbing in [itex] u_t=6uu_x-u_{xxx} [/itex] to get

    [itex] 6\int_{-\infty}^{\infty}u^2u_xdx-\int_{-\infty}^{\infty}uu_{xxx}dx [/itex].

    I can compute the first integral to get [itex] \frac{1}{3}u^3 [/itex] evaluated at [itex] -\infty [/itex] to [itex] \infty [/itex] (so that integral is equal to 0) but I can't do anything with the second integral. I had this problem last night but decided to sleep on it - unfortunately that didn't help!
     
  8. Dec 11, 2014 #7

    ShayanJ

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    Integrate by parts!
     
  9. Dec 11, 2014 #8
    Right, I got it thanks!

    There's another part to the question, it's pretty much the same as part one but now for

    [itex] I_2=\int_{-\infty)^{\infty}(u^3+\frac{1}{2}u_x^2)dx [/itex].

    I got it down to

    [itex] \frac{\partial I_2}{\partial t}=3\int_{-\infty)^{\infty}u^2u_tdx + \int_{-\infty)^{\infty}u_xu_{xt} [/itex].

    Now subbing in [itex] u_t=6uu_x-u_{xxx} [/itex] and [itex] u_{tx}=6u_x^2+6u_{xx}-u_{xxxx} [/itex] gives a few difficult integrals to solve. I can do three of them but the other two I can't get past are

    [itex] \int_{-\infty)^{\infty}u^2u_{xxx}dx [/itex] and [itex] \int_{-\infty)^{\infty}u_x^3dx [/itex].

    Have I made a mistake to get to this point? If not, are these integrals possible?
     
  10. Dec 11, 2014 #9
    Right, I got it thanks!

    There's another part to the question, it's pretty much the same as part one but now for

    [itex] I_2 = \int_{-\infty}^{\infty}(u^3+\frac{1}{2}u_x^2)dx [/itex].

    I got it down to

    [itex] \frac{\partial I_2}{\partial t}=3\int_{-\infty}^{\infty}u^2u_tdx + \int_{-\infty}^{\infty}u_xu_{xt} [/itex].

    Now subbing in [itex] u_t=6uu_x-u_{xxx} [/itex] and [itex] u_{tx}=6u_x^2+6u_{xx}-u_{xxxx} [/itex] gives a few difficult integrals to solve. I can do three of them but the other two I can't get past are

    [itex] \int_{-\infty}^{\infty}u^2u_{xxx}dx [/itex] and [itex] \int_{-\infty}^{\infty}u_x^3dx [/itex].

    Have I made a mistake to get to this point? If not, are these integrals possible?
     
  11. Dec 11, 2014 #10

    ShayanJ

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    That seems right. But you can also use [itex] u_{xt}=u_{tx} [/itex] and integrate by parts to eliminate the x derivative. Then substitute using the differential equation. But I'm stuck at the end too! My problem is I can't deal with [itex] \int_{-\infty}^{\infty} u u_x u_{xx} dx [/itex]. If you prove this is zero, then its finished.

    EDIT: I found it. There are two such integrals which cancel each other and give zero. This was nice!
     
  12. Dec 11, 2014 #11
    I'm really sorry, I don't follow what you're saying. How is [itex] u_{xt} = u_{tx} [/itex] useful and what should be integrated by parts?
     
  13. Dec 11, 2014 #12

    ShayanJ

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    [itex]\int_{-\infty}^{\infty} u_x u_{tx}dx=-\int_{-\infty}^{\infty} u_{xx} u_t dx [/itex].
     
  14. Dec 11, 2014 #13

    PeroK

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    I just wanted to point out that ##I_1## is actually a function of ##t## only. So, you shouldn't be using the partial derivative wrt ##t##.

    You need to be careful to check what sort of function you have - especially when you're using condensed notation. It's often better to put the variable(s) in so that can see what you're doing:

    [itex] I_1(t) = \int_{-\infty}^{\infty}u(x, t)^2dx [/itex]
     
  15. Dec 11, 2014 #14
    So far I have
    [itex] \frac{\partial I_2}{\partial t} = \frac{\partial}{\partial t}\int_{-\infty}^{\infty}(u^3+\frac{1}{2}u_x^2)dx = 3\int_{-\infty}^{\infty}u^2u_tdx + \int_{-\infty}^{\infty}u_xu_{xt} = 3\int_{-\infty}^{\infty}u^2u_tdx - \int_{-\infty}^{\infty}u_{xx}u_tdx [/itex]

    [itex] \int_{-\infty}^{\infty}u^2u_tdx = 18\int_{-\infty}^{\infty}u^3u_xdx-3\int_{-\infty}^{\infty}u^2u_{xxx}dx [/itex].

    [itex] \int_{-\infty}^{\infty}u_{xx}u_tdx= -6 \int_{-\infty}^{\infty}u_{xx}u_xdx + \int_{-\infty}^{\infty}u_{xx}u_{xxx}dx [/itex].

    I can compute all of these integrals (to be 0) but not [itex] \int_{-\infty}^{\infty}u^2u_{xxx}dx [/itex]. If I can compute that integral then I'm done.

    I didn't come across [itex] \int_{-\infty}^{\infty}uu_xu_{xx}dx [/itex] at all.
     
  16. Dec 11, 2014 #15

    ShayanJ

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    As PeroK noted, you should write [itex] \frac{dI_2}{dt} [/itex] because a definite integral on x of a function of x and t, is no more a function of x.
    Anyway, you did it wrong! [itex] u_{xx}u_x [/itex] should be [itex] u u_{xx} u_x [/itex]. And the term you can't evaluate, can be integrated by parts to give a similar term.
     
  17. Dec 11, 2014 #16
    I've got it! Thanks so much for all of the help, I certainly needed it!
     
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