Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Verifying the Helmholtz Green’s function.

Posted by peeterjoot on December 9, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


In class this week, looking at an instance of the Helmholtz equation

\begin{aligned}\left( \boldsymbol{\nabla}^2 + \mathbf{k}^2\right) \psi_\mathbf{k}(\mathbf{r}) = s(\mathbf{r}).\end{aligned} \hspace{\stretch{1}}(1.1)

We were told that the Green’s function

\begin{aligned}\left( \boldsymbol{\nabla}^2 + \mathbf{k}^2\right) G^0(\mathbf{r}, \mathbf{r}') = \delta(\mathbf{r}- \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(1.2)

that can be used to solve for a particular solution this differential equation via convolution

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) = \int G^0(\mathbf{r}, \mathbf{r}') s(\mathbf{r}') d^3 \mathbf{r}',\end{aligned} \hspace{\stretch{1}}(1.3)

had the value

\begin{aligned}G^0(\mathbf{r}, \mathbf{r}') = - \frac{1}{{4 \pi}} \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} }{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(1.4)

Let’s try to verify this.


Application of the Helmholtz differential operator \boldsymbol{\nabla}^2 + \mathbf{k}^2 on the presumed solution gives

\begin{aligned}(\boldsymbol{\nabla}^2 + \mathbf{k}^2) \psi_\mathbf{k}(\mathbf{r}) = - \frac{1}{{4 \pi}} \int (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} }{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}s(\mathbf{r}') d^3 \mathbf{r}'.\end{aligned} \hspace{\stretch{1}}(2.5)

When \mathbf{r} \ne \mathbf{r}'.

To proceed we’ll need to evaluate

\begin{aligned}\boldsymbol{\nabla}^2 \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} }{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.6)

Writing \mu = {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} we start with the computation of

\begin{aligned}\frac{\partial {}}{\partial {x}} \frac{e^{i k \mu} }{\mu}&=\frac{\partial {\mu}}{\partial {x}} \left( \frac{i k}{\mu} - \frac{1}{{\mu^2}} \right) e^{i k \mu} \\ &=\frac{\partial {\mu}}{\partial {x}} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}\end{aligned}

We see that we’ll have

\begin{aligned}\boldsymbol{\nabla} \frac{e^{i k \mu} }{\mu} = \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \boldsymbol{\nabla} \mu.\end{aligned} \hspace{\stretch{1}}(2.7)

Taking second derivatives with respect to x we find

\begin{aligned}\frac{\partial^2 {{}}}{\partial {{x}}^2} \frac{e^{i k \mu} }{\mu}&=\frac{\partial^2 {{\mu}}}{\partial {{x}}^2} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}+\frac{\partial {\mu}}{\partial {x}} \frac{\partial {\mu}}{\partial {x}} \frac{1}{{\mu^2}} \frac{e^{i k \mu}}{\mu}+\left( \frac{\partial {\mu}}{\partial {x}} \right)^2 \left( i k - \frac{1}{{\mu}} \right)^2 \frac{e^{i k \mu}}{\mu} \\ &=\frac{\partial^2 {{\mu}}}{\partial {{x}}^2} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}+\left( \frac{\partial {\mu}}{\partial {x}} \right)^2 \left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu}.\end{aligned}

Our Laplacian is then

\begin{aligned}\boldsymbol{\nabla}^2\frac{e^{i k \mu} }{\mu} =\left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \boldsymbol{\nabla}^2 \mu+\left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu} (\boldsymbol{\nabla} \mu)^2.\end{aligned} \hspace{\stretch{1}}(2.8)

Now lets calculate the derivatives of \mu. Working on x again, we have

\begin{aligned}\frac{\partial {}}{\partial {x}} \mu&=\frac{\partial {}}{\partial {x}} \sqrt{ (x - x')^2 +(y - y')^2 +(z - z')^2 } \\ &=\frac{1}{{2}} 2 (x - x')\frac{1}{{\sqrt{ (x - x')^2 +(y - y')^2 +(z - z')^2 }}} \\ &=\frac{x - x'}{\mu}.\end{aligned}

So we have

\begin{aligned}\boldsymbol{\nabla} \mu &= \frac{\mathbf{r} - \mathbf{r}'}{\mu} \\ (\boldsymbol{\nabla} \mu)^2 &= 1 \end{aligned} \hspace{\stretch{1}}(2.9)

Taking second derivatives with respect to x we find

\begin{aligned}\frac{\partial^2 {{}}}{\partial {{x}}^2} \mu&= \frac{\partial {}}{\partial {x}}\frac{x - x'}{\mu} \\ &= \frac{1}{\mu} - (x - x') \frac{\partial {\mu}}{\partial {x}} \frac{1}{{\mu^2}}\\ &=\frac{1}{\mu} - (x - x') \frac{x - x'}{\mu} \frac{1}{{\mu^2}}\\ &=\frac{1}{\mu} - (x - x')^2 \frac{1}{{\mu^3}}.\end{aligned}

So we find

\begin{aligned}\boldsymbol{\nabla}^2 \mu = \frac{3}{\mu} - \frac{1}{{\mu}},\end{aligned} \hspace{\stretch{1}}(2.11)


\begin{aligned}\boldsymbol{\nabla}^2 \mu = \frac{2}{\mu}.\end{aligned} \hspace{\stretch{1}}(2.12)

Inserting this and (\boldsymbol{\nabla} \mu)^2 into 2.8 we find

\begin{aligned}\begin{aligned}\boldsymbol{\nabla}^2\frac{e^{i k \mu} }{\mu} &=\left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \frac{2}{\mu}+\left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu}&=-k^2 \frac{e^{i k \mu}}{\mu}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.13)

This shows us that provided \mathbf{r} \ne \mathbf{r}' we have

\begin{aligned}(\boldsymbol{\nabla}^2 + \mathbf{k}^2) G^0(\mathbf{r}, \mathbf{r}') = 0.\end{aligned} \hspace{\stretch{1}}(2.14)

In the neighborhood of {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon.

Having shown that we end up with zero everywhere that \mathbf{r} \ne \mathbf{r}' we are left to consider a neighborhood of the volume surrounding the point \mathbf{r} in our integral. Following the Coulomb treatment in section 2.2 of [1] we use a spherical volume element centered around \mathbf{r} of radius \epsilon, and then convert a divergence to a surface area to evaluate the integral away from the problematic point

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}'=-\frac{1}{{4\pi}} \int_{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}'\end{aligned} \hspace{\stretch{1}}(2.15)

We make the change of variables \mathbf{r}' = \mathbf{r} + \mathbf{a}. We add an explicit \mathbf{r} suffix to our Laplacian at the same time to remind us that it is taking derivatives with respect to the coordinates of \mathbf{r} = (x, y, z), and not the coordinates of our integration variable \mathbf{a} = (a_x, a_y, a_z). Assuming sufficient continuity and “well behavedness” of s(\mathbf{r}') we’ll be able to pull it out of the integral, giving

\begin{aligned}-\frac{1}{{4\pi}} \int_{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}'&= -\frac{1}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} s(\mathbf{r} + \mathbf{a}) d^3 \mathbf{a} \\ &= -\frac{s(\mathbf{r})}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} \end{aligned}

Recalling the dependencies on the derivatives of {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} in our previous gradient evaluations, we note that we have

\begin{aligned}\boldsymbol{\nabla}_\mathbf{r} {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} &= -\boldsymbol{\nabla}_\mathbf{a} {\left\lvert{\mathbf{a}}\right\rvert} \\ (\boldsymbol{\nabla}_\mathbf{r} {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert})^2 &= (\boldsymbol{\nabla}_\mathbf{a} {\left\lvert{\mathbf{a}}\right\rvert})^2 \\ \boldsymbol{\nabla}_\mathbf{r}^2 {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} &= \boldsymbol{\nabla}_\mathbf{a}^2 {\left\lvert{\mathbf{a}}\right\rvert},\end{aligned} \hspace{\stretch{1}}(2.16)

so with \mathbf{a} = \mathbf{r} - \mathbf{r}', we can rewrite our Laplacian as

\begin{aligned}\boldsymbol{\nabla}_\mathbf{r}^2 \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} = \boldsymbol{\nabla}_\mathbf{a}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} = \boldsymbol{\nabla}_\mathbf{a} \cdot \left(\boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right)\end{aligned} \hspace{\stretch{1}}(2.19)

This gives us

\begin{aligned}-\frac{s(\mathbf{r})}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{a}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} &=-\frac{s(\mathbf{r})}{4\pi} \int_{dV} \boldsymbol{\nabla}_\mathbf{a} \cdot \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) d^3 \mathbf{a} -\frac{s(\mathbf{r})}{4\pi} \int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a}  \\ &=-\frac{s(\mathbf{r})}{4\pi} \int_{dA} \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}} d^2 \mathbf{a} -\frac{s(\mathbf{r})}{4\pi} \int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} \end{aligned}

To complete these evaluations, we can now employ a spherical coordinate change of variables. Let’s do the \mathbf{k}^2 volume integral first. We have

\begin{aligned}\int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} &=\int_{a = 0}^\epsilon \int_{\theta = 0}^\pi \int_{\phi=0}^{2\pi}\mathbf{k}^2 \frac{e^{i k a}}{a} a^2 da \sin\theta d\theta d\phi \\ &=4\pi k^2\int_{a = 0}^\epsilon a e^{i k a} da  \\ &=4\pi \int_{u = 0}^{k\epsilon}u e^{i u} du  \\ &=4\pi {\left.(-i u + 1) e^{i u} \right\vert}_0^{k \epsilon} \\ &=4 \pi \left( (-i k \epsilon + 1)e^{i k \epsilon} - 1 \right)\end{aligned}

To evaluate the surface integral we note that we’ll require only the radial portion of the gradient, so have

\begin{aligned}\left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}}&=\left( \hat{\mathbf{a}} \frac{\partial {}}{\partial {a}} \frac{e^{i k a}}{a} \right) \cdot \hat{\mathbf{a}} \\ &=\frac{\partial {}}{\partial {a}} \frac{e^{i k a}}{a} \\ &=\left( i k \frac{1}{{a}} - \frac{1}{{a^2}} \right)e^{i k a} \\ &=\left( i k a - 1 \right)\frac{e^{i k a}}{a^2}\end{aligned}

Our area element is a^2 \sin\theta d\theta d\phi, so we are left with

\begin{aligned}\begin{aligned}\int_{dA} \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}} d^2 \mathbf{a} &={\left.{{\int_{\theta = 0}^\pi \int_{\phi=0}^{2\pi}\left( i k a - 1 \right)\frac{e^{i k a}}{a^2}a^2 \sin\theta d\theta d\phi }}\right\vert}_{{a = \epsilon}}\\ &=4 \pi\left( i k \epsilon - 1 \right) e^{i k \epsilon}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.20)

Putting everything back together we have

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}'&=-s(\mathbf{r})\left((-i k \epsilon + 1)e^{i k \epsilon} - 1 +\left( i k \epsilon - 1 \right) e^{i k \epsilon}\right) \\ &=-s(\mathbf{r})\left((-i k \epsilon + 1 + i k \epsilon - 1 )e^{i k \epsilon} - 1 \right) \end{aligned}

But this is just

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}' = s(\mathbf{r}).\end{aligned} \hspace{\stretch{1}}(2.21)

This completes the desired verification of the Green’s function for the Helmholtz operator. Observe the perfect cancellation here, so the limit of \epsilon \rightarrow 0 can be independent of how large k is made. You have to complete the integrals for both the Laplacian and the \mathbf{k}^2 portions of the integrals and add them, before taking any limits, or else you’ll get into trouble (as I did in my first attempt).


[1] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.


3 Responses to “Verifying the Helmholtz Green’s function.”

  1. loiosu said

    here can be applied in more cases

  2. loiosu said

    The case of perfect contact is probably what most applications require when two materials are joined. This case also has the advantage of being simpler than the general case where heat can be obstructed at the boundary and the presentation is not encumbered by tedious algebraic manipulations and considered by prof dr mircea orasanu. The common approach to the overall problem is to form a solution by adding regular solutions from the diffusion equations to this source solution and use the parameters to satisfy boundary and initial conditions

  3. prof dr mircea orasanu said

    Green’s Function for the Helmholtz Equation
    If we fourier transform the wave equation, or alternatively attempt to find solutions with a specified harmonic behavior in time , we convert it into the following spatial form: (for example, from the wave equation above, where , , and by assumption). This is called the inhomogeneous Helmholtz equation (IHE).The Green’s function therefore has to solve the PDE: Once again, the Green’s function satisfies the homogeneous Helmholtz equation (HHE). Furthermore, clearly the Poisson equation is the limit of the Helmholtz equation. It is straightforward to show that there are several functions that are good candidates for .

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: