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Leibniz rule for differentiating an integral w.r.t a parameter

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    I have the function[tex]u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(\xi)d\xi[/tex]where g is continuously differentiable and c is a constant. I need to verify that this is a solution to the wave equation.

    2. Relevant equations

    My prof gave me the formula[tex]\frac{d}{dt}\int^{b(t)}_{a(t)}F(y,t)dy=\int^{b(t)}_{a(t)}F_t(y,t)dy+F(b(t),t)b'(t)-F(a(t),t)a'(t)[/tex]

    3. The attempt at a solution

    I really don't see how I can use this rule when trying to take these derivatives. The [itex]\xi[/itex] is really messing me up. Do I need to use this rule right off the bat on my first derivative (I am trying to take the x derivatives first). It seems like this question is just as easy as using a formula, but I cannot seem to get an answer that works out...
     
  2. jcsd
  3. Mar 25, 2014 #2

    LCKurtz

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    That ##xi## is a dummy variable. Would it lessen your confusion if you called it ##y##?$$
    u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(y)dy$$Now the integrand looks just like your Leibnitz formula except it's simpler, having no ##t## variable.
     
  4. Mar 25, 2014 #3

    Ray Vickson

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    Yes, but it does have both x and t in the integration limits. Maybe that is causing the OP's headaches.
     
  5. Mar 25, 2014 #4

    LCKurtz

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    Could be. He may need an Excedrin before he's done.
     
  6. Mar 25, 2014 #5
    so the y still didn't do much for me. If I have g(y only), then how all of a sudden I have g(b(t),t)?
     
  7. Mar 25, 2014 #6

    pasmith

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    More useful is the special case
    [tex]
    \frac{\partial}{\partial x} \int_{a(x,t)}^{b(x,t)} f(y)\,dy = f(b(x,t))\frac{\partial b}{\partial x} - f(a(x,t))\frac{\partial a}{\partial x}[/tex]
     
  8. Mar 25, 2014 #7
    ok, so first of all, how can you say the integral in the general formula is zero? because F is a function of y only?

    second, working this all out, I get something that looks like the wave equation kinda, but I don't see how I exactly have[itex]u_{tt}=c^2u_{xx}[/itex]

    Is it okay if I just upload an image instead of typing out my work?
     
  9. Mar 25, 2014 #8
    http://media.newschoolers.com/uploads/images/17/00/70/22/72/702272.jpeg [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. Mar 25, 2014 #9
    see on the last two lines, I don't think I have an equality here? also the denominator in the last line is a typo. I know it should be c^2/2c
     
  11. Mar 25, 2014 #10

    LCKurtz

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    Yes, that's why the integrand with the ##F_t## is zero.

    That's why I don't like images. It makes us retype things. In the step where you have$$
    \frac 1 {2c}(g_t(x+ct)c + g_t(x-ct)c$$you should have$$
    \frac 1 {2c}g'(x+ct)c - g'(x-ct)(-c)$$The second g should have a negative sign in all your calculations because it came from the lower limit and the last -c is from the chain rule.
     
    Last edited by a moderator: May 6, 2017
  12. Mar 25, 2014 #11
    sorry. I can type stuff out from now on. I'm confused by your primed notation now. what does that mean? even accounting for the negative sign I missed, I do not see how these can be equal, because one I am differentiating wrt x and the other one I am differentiating wrt t.

    [tex]u_{tt}=\frac{c^2}{2c} \left( g_t(x+ct)-g_t(x-ct) \right)[/tex][tex]c^2u_{xx}=\frac{c^2}{2c}\left( g_x(x+ct)+g_x(x-ct) \right)[/tex]I don't see [itex]c^2u_{xx}=u_{tt}[/itex] here...
     
  13. Mar 25, 2014 #12
    wait. am I messing up the chain rule here?
     
  14. Mar 25, 2014 #13
    ok. I see where I went wrong with the chain rule now, I had to write the chain rule in leibniz notation (with ANOTHER dummy variable) and now I think I have the answer. Thanks a lot for your help.It seems so easy now. Although I guess it's always easy once you have figured it out haha.
     
  15. Mar 25, 2014 #14

    LCKurtz

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    Once you get the chain rule correct in your ##u_t## and ##u_{tt}## derivatives you will have a + sign where I have the red and your two right sides will be identical.

    About the prime, ##g## is a function of one variable. Think of it as ##g(u)## where ##u=x\pm ct## (either way). If you want to differentiate ##g## with respect to ##t## you would use the chain rule$$
    g_t = g'(u)u_t = g'(x\pm ct)(\pm c)$$
    [Edit]:Although your answers will agree once you fix the chain rule thing, both signs between the two g terms should be negative once you take into account my comment about lower limits (the last sentence in post #10).
     
    Last edited: Mar 25, 2014
  16. Mar 25, 2014 #15

    LCKurtz

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    Please note the editing in post #14. Your final result should have$$
    \frac{c^2}{2c} \left( g_t(x+ct)-g_t(x-ct) \right)$$on the right side of both.
     
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