Poisson's equation and Green's functions

[jmb]

Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?

A solution u(\mathbf{x}) to Poisson's equation satisfies:

\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}

we can find such a solution in a given domain by evaluating:

u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}

or equivalently (if we are only interested in the electric field):

\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}

We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density \rho(\mathbf{x}).

However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of -\frac{\rho}{\epsilon_0}. What am I missing??

To evaluate the Laplacian/divergence of the above I make use of the fact that \mathbf{x'} is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying \nabla^2 to the integral on the RHS and also means the components of \mathbf{x'} all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.

 

[marcusl]

Not so. You should take the Laplacian with respect to x (after all you are calculating \nabla^2{u(x)}). Since
\nabla^2 \frac{1}{|x-x'|}=\delta(x-x')
you get the right result from the integration.
This expression is derived in many places:
Jackson, Classical Electrodynamics, 2nd ed., Eq. (1.31)
Arfken, Math Methods for Physicists, 2nd ed., sect. 1.15
to give two examples.

 

[jmb]

Thanks marcusl.

Yes I was taking the derivative wrt x (hence the derivatives of all components of x' being zero). But I was erroneously evaluating the Laplacian of \frac{1}{|x-x'|} as zero.

I've since found references to the result you quote, but they all start by searching for the Green's function G that satisfies \nabla^2 G = \delta(x-x') and then deducing it to be the above... out of interest do you know of any way to show it starting from the other end --- i.e. directly differentiating \frac{1}{|x-x'|} and showing its Laplacian to be the delta function? Or do the references you quote do it that way?

I'll look them up when I get the chance. Thanks again.

 

[marcusl]

Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')

 

[samalkhaiat]

Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')

In my 2nd edition, Jackson says on p.40 :" since \nabla^{2}(1/r) = 0 for r \neq 0 and its volume integral is -4\pi, we can write the formal equation, \nabla^{2}(1/r) = -4\pi \delta^{3}(x)..."

In exam, if you reproduce "Jackson's proof", I give you 2 marks out of 10! You would gain the remaining 8 marks, if you prove the underlined sentence in Jackson's statement. I can tell you, it is a tricky one, and Jackson doesn't do it!

regards

sam

 

[jmb]

OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

From the divergence theorem we have,

I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}

Since the RHS is a surface integral we don't need to worry about evaluating \nabla (1/r) at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi

(since \frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}) as required.

 

[marcusl]

Jackson has similar expressions on p. 39 (he calls the integral C instead of I).

 

[samalkhaiat]

OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

From the divergence theorem we have,

I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}

Since the RHS is a surface integral we don't need to worry about evaluating \nabla (1/r) at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi

(since \frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}) as required.

For this, I give you 3 marks out of the 8! Your next step is to show that the result -4\pi holds true for ANY surfase S bounding V (not just spherical). Then and only then you can earn the remaining marks! Have a go at it and let me know.


regards

sam

 

[jmb]

For this, I give you 3 marks out of the 8! Your next step is to show that the result -4\pi holds true for ANY surfase S bounding V (not just spherical). Then and only then you can earn the remaining marks! Have a go at it and let me know.


regards

sam

Doesn't it just follow on from the properties of \nabla^2(1/r)?

We already know that \nabla^2(1/r) is zero everywhere except r=0, thus any volume V (enclosed by some surface S) which encloses r=0 will produce the same value for the integral, regardless of its shape, since the integrand gives no contribution to the integral at any other location. Equally any volume not enclosing r=0 will give zero when integrated. Maybe I'm missing a subtlety?

 

[samalkhaiat]

We already know that \nabla^2(1/r) is zero everywhere except r=0, thus any volume V (enclosed by some surface S) which encloses r=0 will produce the same value for the integral, regardless of its shape, since the integrand gives no contribution to the integral at any other location. Equally any volume not enclosing r=0 will give zero when integrated. Maybe I'm missing a subtlety?

By stating this, you gain the remaining marks. The trick is to surround the point x = 0 by small enough sphere and do the integration in the bounded region x \neq 0:

Let \Sigma be a surface bounding a region V containing the orign, and let \sigma be a small spherical surface enclosing the point x = 0. Therefore, we have x \neq 0 everywhere in the region U bounded by \Sigma + \sigma, i.e.,

\int_{U} \nabla^{2}|\frac{1}{x}| \ d^{3}x = 0 = \int_{\Sigma + \sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S}

From this it follows that the integral over the whole region V is

\int_{V} \nabla^{2}|\frac{1}{x}| \ d^{3}x = \int_{\Sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S} = - \int_{\sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S}

Or,

\int_{V} \nabla^{2}|\frac{1}{x}| \ d^{3}x = \int_{\sigma} \frac{1}{|x|^{2}} \left( \hat{x}. \hat{n}_{\sigma} \right) dS

where \hat{x} is the unit vector along x and \hat{n}_{\sigma} is the unit OUTWARD normal at dS of \sigma;

\hat{x} . \hat{n}_{\sigma} = -1

Thus

\int_{V} \nabla^{2}|1/x| \ d^{3}x = - \frac{1}{R^{2}} \int_{\sigma} dS = -4\pi


regards

sam

 

[jmb]

Cool. Many thanks again to both marcusl and samalkhaiat for your help!