Solving the Helmholtz Equation for a Point Source

[bladesong]

Homework Statement

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).

Homework Equations

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

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. I have posted this on the intro physics board but nothing doing, so I'm not sure if maybe this is more advanced. If it isn't, my apologies.

 

[fzero]

Homework Statement

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).

Homework Equations

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

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?

No, it's not correct. You've dropped the delta function in the second line ($\delta(\vec{x})\neq 1$ everywhere). There's no particular reason to integrate over spherical variables here on the RHS, but if you do, then you will actually need to use the Jacobian.

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).

$\psi$ is a scalar function, but it definitely depends on $r$. You are told in the problem to substitute the expression for it in 2-58 into 2-55. Why don't you actually try doing that?

 

[clamtrox]

∫∫∫(-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?

Do the right hand side in Cartesian coordinates (you don't need to worry about limits as long as origin is inside the volume).

∫∫∫(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

That's not correct...
The radial part of the Laplacian is $$\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \psi}{\partial r}\right)$$ The derivatives hit $\psi$.

The key here is to notice that ψ diverges in the origin, so you can't just plug it into the formulas and differentiate. The most intuitively clear way to proceed is to use the divergence theorem on the first term. You should have something like $$\int dV \nabla^2 \psi = \int dS \nabla \psi \cdot \hat{n}.$$ Take the surface to be a sphere, assume that ψ does not depend on the angles and you can evaluate the integral easily.