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Reducing surface integral to line integral?

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    hi im trying to adjust this general integral to my problem,

    my problem consists of a semi-infinite rod, i.e. x in [0,∞)

    the primed variables are the integration variables



    2. Relevant equations

    http://img339.imageshack.us/img339/5038/42247711.jpg [Broken]

    3. The attempt at a solution

    so since its a 1D problem it means that dS' = dx' right?

    but the integral is a surface integral so its evaluated the the surface right? which is at x'=0

    im a bit confused here, whether it should be (int dx' u(x',t') dG/dx'(x,t,x',t'), {x'=0...x})

    or (int dx' u(0,t') dG/dx'(x,t,x',t')|x'=0, {x'=0...x}) (evaluated at the surface then integrated)

    but if this is the case then youll just get x times u(0,t') dG/dx' at x'=0 which doesnt seem right..

    would there even be an integral involved? would it just be

    y(x,t) = int_{t,t_0} dt u(0,t') dG/dx'(x,t|x',t')|x'=0 ? < this seems like the correct way to do it to me.. im just not sure the x' integral will go away..

    which leads me to another point of confusion,

    i know that dG/dn' = - something, but is it = -dG/dx' or = -dG/dx ?

    im pretty sure its -dG/dx'
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 21, 2013 #2

    mfb

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    Where is S defined?
    It might be necessary to modify it before you can reduce it to a 1-dimensional integral.
     
  4. Apr 21, 2013 #3

    jambaugh

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    It would help if you were to state "your problem" as well as the source for your integral formula to judge whether and how it ought to be applied.

    With that said I here will speculate on some points.

    The dG/dn' would likely be the directional derivative of G in the direction normal to the surface. That would make it equivalent to the gradient of G dotted with the the unit surface normal.
    [tex] \int dt' \iint y(r',t') \nabla' G(r,t;r',t') \bullet d\mathbf{S}[/tex]
    You will likely have apply Gauss's Theorem (Divergence Theorem) to rewrite the surface integral as a volume integral. Then the integral over the rod will be nominally a volume integral with the cross sectional area component trivial. This though sounds like "undoing" whatever was used to derive the formula you stated. Again without context I'm speculating wildly.

    But if I'm close then this formula is derived from some prior principle formula more applicable to your problem. It looks like a Green's function integral for some differential equation satisfied by y(r,t) but with additional application specific identities applied.

    Even a subject context would help, is this a QM problem? An Electrodynamics problem? Gravitation? A pure math problem?
     
  5. Apr 21, 2013 #4
    thanks for your replies!

    ok sorry, i thought i could leave it general,

    the formula i have given is the boundary value term of the general solution to the heat equation,

    there a 3 terms, one associated with the driving function, one associated to initial conditions and one associated with boundary conditions.

    in my question there is no driving function of initial conditions so only the boundary condition term is required.

    the problem is one dimensional i.e. the differential equation is, k d^2u/dx^2 = du/dt, partials of course.

    the boundary condition given is y(0,t)= A + Bsin(t)

    the normal derivative to the greens function is negative because the rods 'surface' is at x=0 and opposing the direction of the rod, so in the -x direction.

    its dG/dn' over the primed variables thats why i assumed it was -dG/dx'

    i take it the surface S is supposed to be analogous to the point x=0 of the rod,

    this is why im getting confused,

    if you have int f(r) dS' and the surface is defined to be a point in 1D, adjusting this to the 1D for, it wouldnt be an integral anymore? theres nothing to integrate? its just f(0) right?

    thats what i am thinking is the correct thing to do but all the primed variables and stuff have made me pretty confused
     
  6. Apr 21, 2013 #5

    jambaugh

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    OK, now we're getting somewhere. Yes your assumptions are correct.

    The Divergence theorem reduces in 1 (spatial) dimension to the FTC. Remember the "outward normal" for an interval [a,b] on the x axis then becomes [itex]-\hat{i}[/itex] at x=a, and [itex] \hat{i}[/itex] at x=b. And the gradient becomes simply [itex]\hat{i}\partial_x[/itex]. Presumably your x=b is at infinity and the functions are zero in that limit so there's no contribution. But I'll be more generic here but understand "evaluation at infinites" will be limits at infinity.

    The "surface integral" for the zero dimensional "surface" of the 1-D "volume" becomes simple evaluation at that end point. So...

    [tex] \int dt' \iint y(r',t') \nabla' G(r,t;r',t') \bullet d\mathbf{S}[/tex]
    becomes...
    [tex] \int dt' \left[ y(x',t') \partial/\partial x' G(x,t;x',t') \right]_{x=a}^{x=b}= \int dt' \left[y(b,t') \partial/\partial x' G(x,t;b,t') -y(a,t') \partial/\partial x' G(x,t;a,t')\right][/tex]

    Remember the primed variables are the variables of integration, you are summing over one set of positions and times (r',t') via the integrals to find the values at another specific but unspecified position and time (r,t).
     
  7. Apr 21, 2013 #6
    thanks alot!
     
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