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Integration Proof

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that
    [tex]\frac{\partial}{\partial t} \int_{B(x,t)} \nabla^2 p(x') dx'=\int_{S(x,t)} \nabla^2 p(x') d\sigma_t[/tex]

    [Hint: Introduce spherical coordinates.]

    2. Relevant equations



    3. The attempt at a solution

    I thought the divergence thm would be necessary to get from the ball to the surface of the ball and so I will need to construct a unit normal to the surface
    [tex]x'=x+t\alpha[/tex]
    where x is the vector to the centre of the ball, t is the radius and [itex]\alpha[/itex] is the unit vector in the direction of the radius, so x' is the vector to the surface of the ball
    that makes [itex]\alpha[/itex] the unit normal, so using the divergence thm,

    [tex]\frac{\partial}{\partial t}\int_{B(x,t)}\nabla^2p(x')dx' =\frac{\partial}{\partial t}\int_{S(x,t)} \nabla p(x+t \alpha ). \alpha d\sigma_t[/tex]

    this is where I get (more?) lost...

    converting to spherical polars...
    [tex]d \sigma_t = r^2 sin\phi d\theta d\phi[/tex]

    so
    [tex] = \frac{\partial}{\partial t}\int_{S(x,t)} \nabla p(x+t \alpha ). \alpha r^2 sin\phi d\theta d\phi[/tex]

    and really.... I'm stuck....
    what is r in terms of t ? I must be going the wrong way here.
     
    Last edited: Jun 8, 2010
  2. jcsd
  3. Jun 8, 2010 #2

    lanedance

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    can you explain your notation in the OP... i'm guessing:
    p(x') is a scalar function, is there anything else special about p?
    B(x,t) is a 3D ball
    S(x,t) is the closed surface of the ball
    dx' = dV a volume element

    what is [itex] d \sigma_t[/itex]?
     
  4. Jun 8, 2010 #3
    Well, in the text before the question, p is part of the initial conditions of the wave equation but by the way the question was worded I thought that this was just a general identity.
    In the text,

    [tex]\nabla^2u(x,t)-u_{tt}(x,t)=0[/tex]
    where [itex]u(x,0)=0[/itex] and [itex]u_t(x,0)=p(x)[/itex]

    [itex] d \sigma_t[/itex] is the surface element of the ball with radius t.

    Your other guesses are spot on...

    Also, rather annoyingly, the text says "it is an easy exercise to compute the derivative on the left side of the equation - see problems" so maybe this is a much simpler thing than I am thinking....
     
  5. Jun 8, 2010 #4

    lanedance

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    ok, got you now... & i think i can explain why its so though.... imagine changing the radius of the ball by dt, the addition to the volume integral will be the thin spherical shell multiplied by p(x) giving the integral on the right hand side.

    Now we need to try & put into identities, so first start by assuming a ball at the origin to make thing easier, we can generalise later if need be

    note you cannot just move the t derivative under the integral while the bound of the integral is also dependent on t.

    now a couple of points on your attempts so far, not 100% where it will lead, buts lets have a try, the divergence theorem should be applied as follows:

    [tex]\int_{B(x,t)}\nabla^2p(x')dx' =\int_{B(x,t)}\nabla \bullet (\nabla p(x')dx')
    =\int_{S(x,t)}\nabla p(x') \bullet \textbf{da} [/tex]

    where [itex] \textbf{da} [/itex] is a unit area element pointing outward normal from the ball

    also this should be an area element not an volume element?
     
    Last edited: Jun 8, 2010
  6. Jun 8, 2010 #5

    lanedance

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    in this wave equation isn't t time, so which is it???
     
    Last edited: Jun 8, 2010
  7. Jun 8, 2010 #6
    Yes, t is time... so our ball is getting bigger!

    I have seriously messed up the spherical polars bit as I was sort of half thinking we had a ball centered at O and so r would be constant (at a single moment of time).

    As you can probably tell, I'm very confused.
    Your divergence theorem bit is the same as mine right? I just split your .da into [itex].\alpha d\sigma_t[/itex]
     
  8. Jun 8, 2010 #7

    lanedance

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    if t is the radius of the ball consider converting the volume integral directly to spherical coordinates as such...
    [tex]
    \frac{\partial}{\partial t} \int_{B(x,t)} \nabla^2 p(x') dx'
    = \frac{\partial}{\partial t}
    \int_0^t dr \int_0^{\pi} d \theta \int_0^{2 \pi} d \phi .r^2 .sin\theta. \nabla^2 p(x')
    [/tex]
     
    Last edited: Jun 8, 2010
  9. Jun 8, 2010 #8

    lanedance

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    ok if i understand correct r is a linear function of t, say [itex] r = \alpha t [/itex],

    try this then...
    [tex]\frac{\partial}{\partial t} \int_{B(x,t)} \nabla^2 p(x') dx'= \frac{\partial}{\partial t} \int_0^{\alpha t} dr \int_0^{\pi} d \theta \int_0^{2 \pi} d \phi .r^2 .sin\theta .\nabla^2 p(x') [/tex]

    you should be able to do this derivative with a simple change of variable
     
    Last edited: Jun 8, 2010
  10. Jun 8, 2010 #9

    lanedance

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    not quite, you have to retain the dot product, as you don't know the direction of grad(p), that is unless p is spherically symmetric...

    anyhow, looks like there is a better way to do it as per the last post ;)
     
  11. Jun 8, 2010 #10

    lanedance

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    updated above with the correct volume contirbution for the integrand r^2 sin(theta)
     
  12. Jun 8, 2010 #11

    lanedance

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    PS - sorry for the multi-multi-posts, but its shows if you can avoid confusion laying out the problem, it may make life easier ;)
     
  13. Jun 8, 2010 #12
    is
    [tex]
    \frac{\partial}{\partial t} \int_0^{\alpha t} dr \int_0^{\pi} d \theta \int_0^{2 \pi} d \phi .r^2 .sin\phi .\nabla^2 p(x')
    [/tex]

    the same as
    [tex]
    \frac{\partial}{\partial t} \int_0^{\alpha t} \int_0^{\pi}\int_0^{2 \pi} r^2 sin\phi \nabla^2 p(x') d \phi d \theta dr
    [/tex]
    if so, don't you have to change p(x') to polars too so that you can integrate it? isnt x' dependent on r, theta and phi?
     
  14. Jun 8, 2010 #13
    Ahh... Having had a bit of time to mull it over, p is not dependant on t (or r), so integrating wrt r first
    [tex]

    \frac{\partial}{\partial t} \int_0^ t \int_0^{\pi}\int_0^{2 \pi} r^2 sin\phi \nabla^2 p(x') d \phi d \theta dr

    [/tex]

    [tex]

    =\frac{\partial}{\partial t} \int_0^{\pi}\int_0^{2 \pi} \frac{1}{3}t^3 sin\phi \nabla^2 p(x') d \phi d \theta

    [/tex]

    [tex]

    = \int_0^{\pi}\int_0^{2 \pi} t^2 sin\phi \nabla^2 p(x') d \phi d \theta

    [/tex]

    and then back to cartesian coords gives the result... wow, thanks lanedance...that's really helped, but this is for r between 0 and t though, ie ball centred at O. for a ball centred at x, though, I suppose it could range between [itex] \left| \textbf{x} \right| -t [/itex] and [itex] \left| \textbf{x} \right| +t [/itex] and for that it doesn't behave at all.
     
  15. Jun 8, 2010 #14

    lanedance

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    it should behave eaxctly the same, just use a variable change to shift to the centre of the ball before changing to spherical coords
     
  16. Jun 9, 2010 #15
    ah, yes so a simple substitution X=x'-x will shift the ball to the origin. when x'=x X=0.
    dx'=dX so there is no difference.

    Thanks for your time Lanedance
     
  17. Jun 9, 2010 #16

    lanedance

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    no worries ;)
     
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