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Homework Help: Solving the Helmholtz Equation for a Point Source

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    By integrating (2-55), over a small volume containing the origin, substituting ψ = Ce-jβr/r, and letting r approach zero, show that C = 1/4π, thus proving (2-58).

    2. Relevant equations

    (2-55): ∇2ψ + β2ψ = -δ(x)δ(y)δ(z)
    (2-58): ψ = e-jβr/(4πr)

    3. The attempt at a solution

    I am SO lost on this.

    I integrated the RHS first. I would assume I have to do this in spherical coordinates. I know from the internet that to convert the delta function between coordinate systems you need to incorporate the jacobian, but because it's at the origin (i.e., x,y,z = 0) I just assumed that:

    ∫∫∫(-δ(x)δ(y)δ(z))r2sin(θ)dr dθ d∅
    = -∫∫∫r2sin(θ)dr dθ d∅
    = -(4π/3)r03

    Where I've integrated from r = 0 to r0, θ from 0 to pi, and ∅ from 0 to 2pi. Is this even correct?

    For the next part, the LHS, I'm totally lost. I'm trying to follow the instructions chronologically, thus

    ∫∫∫(∇2 + β2)ψ r2sin(θ)dr dθ d∅

    Where I'm using the same integration limits as before. I'm really, really unsure how to even do this, but I assumed that psi is regarded as scalar (or otherwise does not have dependency on r, theta or phi). Thus, starting with the r component:

    ∫∫∫(1/r2)(∂/∂r)[r2(∂/∂r)] (ψ r2sin(θ)dr dθ d∅)
    = ∫∫∫(1/r2)(∂/∂r)[r2 (ψ 2r sin(θ)dr dθ d∅)]
    = ∫∫∫(1/r2)(∂/∂r)[2r3 (ψ sin(θ)dr dθ d∅)]
    = ∫∫∫(1/r2)[6r2 (ψ sin(θ)dr dθ d∅)]
    = ∫∫∫[6 (ψ sin(θ)dr dθ d∅)]
    = 4ψ∏r0

    For the next part (θ component):
    ∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[sin(θ)(∂/∂θ)] (ψ r2sin(θ)dr dθ d∅)
    =∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[sin(θ)cos(θ)] (ψ r2dr dθ d∅)
    =∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[(1/2)sin(2θ)] (ψ r2dr dθ d∅)
    =∫∫∫(1/[r2 sin(θ)])[cos(2θ)] (ψ r2dr dθ d∅)
    =∫∫∫(cos(2θ)/ sin(θ)]) (ψ dr dθ d∅)

    But I can't integrate this because (cos(2θ)/ sin(θ)]) doesn't converge. I'm really lost on where to go from here and don't fully understand what I'm being asked to do.

    ANY help is appreciated. Thank you. Please note that I'm an electrical eng so our notation may be different, and I may have some difficulty with more of the hardcore physics notation.
  2. jcsd
  3. Feb 7, 2013 #2
    No takers? I understand this is a bit complicated and I haven't found anything on the internet that I've found helpful (or potentially have understood to be). Again, any help is appreciated.
  4. Feb 7, 2013 #3


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    The key is to write the laplacian operator in the first term as the divergence of the gradient. Apply Gauss's law for integrating a divergence. There is no angular dependence in the function, so the gradient will be simple in spherical coordinates. Choose a small sphere for the shape of your small volume.
  5. Feb 7, 2013 #4
    Thanks TSny, I really appreciate the help.

    That makes more sense. In "There is no angular dependence in the function, so the gradient will be simple in spherical coordinates," does this refer to a lack of angular dependence in the differential volume dV, or the function in question, ψ?

    Sorry, I'm trying to be as clear with what I mean as possible. Essentially what I'm asking is, do I apply the operators (del, grad, div, and differential operators) to ψ, to dV (= r2 sin[θ] dr dθ d∅), or both?

    Finally, were my assumptions/solution for the RHS (the delta dirac functions) correct?
  6. Feb 7, 2013 #5


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    The laplacian is applied only to the function ψ(r) and then you integrate over the small spherical volume. Note ∇2ψ = ∇##\cdot##(∇ψ). Use Gauss's divergence theorem to write the integral of this terms as

    ∫∇##\cdot##(∇ψ) dV = ∫∇ψ##\cdot##d[itex]\vec{a}[/itex]

    where the right hand side is an integral over the surface of the small spherical region of integration.
  7. Feb 7, 2013 #6
    Okay, I think I have a better idea of how to proceed. Now I just need to find the time to get back to it. Thanks again!
  8. Feb 7, 2013 #7


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    No. When you integrate the delta functions over a small spherical volume centered at the origin, the result will be 1 according to the basic properties of the delta function.
  9. Feb 7, 2013 #8
    I understand this logically, as because as the sphere condenses towards r=0, only the delta functions would remain at the origin.

    However, I normalized by the jacobian as such:

    -δ(x)δ(y)δ(z) = δ(r)/(r2sin[θ])

    And then integrated:

    ∫∫∫-[δ(r')/(r'2sin[θ])]r'2sin[θ] dr' dθ d∅

    -∫∫∫δ(r') dr' dθ d∅

    Integrating from θ = 0 to π, ∅ = 0 to 2π, and r' = 0 to r, gets me, well, I assume:

    = -2π2r

    Which would seem to, puzzlingly, approach zero as r->0. So I'm not sure if I'm doing this part right. Notably, also, from the first part, I had a much easier time, but I'm finding that both terms of the LHS:

    2ψ = -4πC(jβr + 1)e-jβr
    β2ψ = 4πC(jβr + 1)e-jβr

    Which, considering ∇2ψ + β2ψ = 0 for all r ≠ 0.

    But even as r→0,
    2ψ = -4πC
    β2ψ = 4πC

    Which suggests also the LHS is zero as well.

    So I'm thinking I did the integration of the delta functions wrong, and that I have a sign error somewhere in my integration of either the laplacian or the second (beta) term. Any thoughts? I can supply my integration more rigorously but I'd rather spare both of us if it's not necessary.

    Again, thanks for all your help.
  10. Feb 7, 2013 #9


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    You can find expressions for the delta function in spherical coordinates https://webspace.utexas.edu/vjg297/www/files/Misc/Delta_function_in_orthogonal_coords.pdf [Broken] .

    Note that for the delta function where the origin is the point of singularity (as you have in your exercise) the expression is gioven as δ(r)/(4πr2). It appears you really have to be careful with writing the delta function in curvilinear coordinates. However, for your problem it seems to me to be easier to stick with cartesian coordinates for the integration over the delta functions. You are going to integrate over a small sphere. Imagine introducing a small cube centered at the origin that fits entirely inside the sphere. Then the integral over the sphere can be broken up into integration over the cube plus integration over the exterior of the cube that's still inside the sphere. But the delta functions are clearly zero over the region ouside the cube. So, you just need to integrate over the cube which can easily be seen to yield 1 if you stick with Cartesian coordinates.

    You should be able to convence yourself that the integral of the β2ψ term will go to zero in the limit of letting the size of the spherical integration region shrink to zero.

    The laplacian term can be evaluated using Gauss' theorem as suggested before. You will just need to integrate the gradient of ψ over the surface of the sphere and then let the radius of the sphere go to zero.
    Last edited by a moderator: May 6, 2017
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