Green's function approach using Lebesgue integration

[gel]

That must be true, since the "second derivative" of phi IS f. What would be the implication of this situation? Could it be that the closed integral form for phi I quoted in the OP would not work then?

First, I don't think it is obvious without looking at an example. If f wasn't continuous then clearly phi couldn't be twice continuously differentiable.
If f is continuous, but not continuously differentiable, then it is not clear to me except by looking at an example. My example above shows that in fact phi might not be twice differentiable. In this case \nabla^2\phi=f only makes sense in a distributional sense, or restricted to the regions where f is differentiable.

 

[gel]

Could this occur in a physical situation? We often get point sources and surface discontinuities in real applications. How should we interpret the fact that the differential equation only holds in the sense of distributions?

A point source is tricky. Then you have the density f(x)=\delta(x) being a distribution.
If f is continuously differentiable everywhere except for on some surface, where it is continuous, then you can say that \nabla^2\phi=f away from this surface, and that \phi is once continuously differentiable (across the surface). (Maybe even twice diff - see my Edit above). If f is discontinuous at the surface, then phi will be continuous but not continuously differentiable here (Edit: maybe it is continuously differentiable. Not sure).

Anyway, way past my bedtime, so that's enough from me for today.

 

[bdforbes]

Here is my working for Jackson's approach to the one-dimensional Laplace equation.

\frac{\partial^2}{\partial x^2}\phi(x)=f(x)

\phi(x)=\int_{-\infty}^\infty \frac{1}{2}|x-x'|f(x')dx'

\phi_a(x)=\int_{-\infty}^\infty \frac{1}{2}\sqrt{(x-x')^2+a^2}f(x')dx'

<br /> \begin{align*}<br /> \frac{\partial^2}{\partial x^2}\phi_a(x)&amp;=\int_{-\infty}^\infty \frac{1}{2}\frac{a^2}{\left[(x-x&#039;)^2+a^2\right]^{3/2}}f(x&#039;)dx&#039;\\<br /> &amp;=\int_{-\infty}^\infty \frac{1}{2}\frac{a^2}{(r^2+a^2)^{3/2}}f(x+r)dr\\<br /> &amp;=\int_{|r|&gt;R} + \int_{-R}^{R} \frac{1}{2}\frac{a^2}{(r^2+a^2)^{3/2}}f(x+r)dr\\<br /> &amp;=\int_{-R}^{R} \frac{1}{2}\frac{a^2}{(r^2+a^2)^{3/2}}\left[\Sigma \frac{f^{(n)}(x)}{n!}r^n\right]dr\\<br /> &amp;=\left[\frac{r}{2\sqrt{r^2+a^2}}f(x)+\cdots\right]_{-R}^R\\<br /> &amp;=\frac{R}{\sqrt{R^2+a^2}}f(x)+\cdots\\<br /> &amp;\rightarrow f(x) \mbox{ as }a\rightarrow 0<br /> \end{align*}<br />

The contribution from |r|>R goes like a^2, so it vanishes as a->0.
The other terms in the series are O(a^2), so they vanish.
I am slightly uncomfortable with assuming the function has a Taylor series expansion. Is it general enough?

Here is the key part I guess:

\phi(x)=\lim_{a\rightarrow 0}\phi_a(x)

<br /> \frac{\partial^2}{\partial x^2}\phi(x)=\frac{\partial^2}{\partial x^2}\lim_{a\rightarrow 0}\phi_a(x)=\lim_{a\rightarrow 0}\frac{\partial^2}{\partial x^2}\phi_a(x)=f(x)

I think this exchange of limiting operations requires that \phi_a(x)\rightarrow \phi(x) uniformly. Provided f(x) either has compact support or falls off nicely for large x, and the increasing powers of a dominate any divergent behaviour in the derivatives of f(x), shouldn't we expect uniform convergence?

Does this method of solution imply that the key issue about proving the integral form for phi is related to our ability to exchange these limiting operations? We used the trick of substituting a nicer function phi_a in order to alter our task from taking derivatives of an integral, to exchanging derivatives and limits of sequences of functions.

 

[jostpuur]

<br /> \frac{\partial^2}{\partial x^2}\phi_a(x)=\int_{-\infty}^\infty \frac{1}{2}\frac{a^2}{\left[(x-x&#039;)^2+a^2\right]^{3/2}}f(x&#039;)dx&#039;<br />

At this point it is nicer to do a variable change x&#039;=x+ar, so that the next integral is

<br /> \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr<br />

Then there will be no need for Taylor series of f.

 

[gel]

Been thinking about this problem a bit more, and I think I can give a fairly complete answer, using the method of distributions. Recall that we have \phi(x)=(-1/4\pi)\int f(y)/|x-y|\,dy and want to show that it solves the pde \nabla^2\phi=f. I assume that f has compact support for simplicity (we have to assume some boundary conditions, otherwise the integral won't be defined).

In my previous posts I showed that the PDE is satisfied in the sense of distributions. Jostpuur/bdforbes showed that the PDE is satisfied if we can assume that some limit commutes with the differentiation. In fact, limits always commute with differentiation if they are done in the sense of distributions, so we have the same result.

The question remains whether \phi is twice continuously differentiable, so that the PDE will be satisfied in the standard sense and not just 'in distribution'. If f is continuously differentiable, then this will be the case (see my previous post). What about if f is continuous, but not necessarily differentiable?
One method is to smooth f by convolving it with a smooth function which is close to a dirac delta. Inside the integral, this is the same as convolving 1/|x-y| with a smooth function and reduces to the same method as jostpuur/bdforbes, where the smooth dirac delta approximation is a(r^2+a^2)^{-3/2}/2 (in 1d, at least), and you then need to know whether the limit a->0 commutes with differentiation.

So the question still remains as to whether a continuous f means that \phi is twice continuously differentiable. In fact, this is false.

We know the following cases.
i) For a ball of uniform density, it is a standard result that the gravitational field (=-\nabla\phi) varies linearly with distance to the centre inside the sphere, then follows the inverse square rule outside. So its derivative jumps (and is undefined) at the boundary. This is not too bad, as the derivative is at least bounded and we already know that it must be discontinuous wherever f is.
ii) In a previous post I tried to construct a counterexample example with a non-differentiable spherically symmetric mass density, but \phi was still twice continuously differentiable.

Now consider a solid object with uniform density f inside and f=0 outside. If the surface is smooth, then the behaviour is like in (i) above with discontinuous but bounded second derivatives for \phi.
However, if the boundary has any ridges or spikes, then at these points the second derivative of \phi will diverge to infinity. Can prove this in a separate post (this one is getting a bit long).

Now suppose that you have a solid body and again that the density f is 0 outside, but inside we have the density decreasing suddenly (but continuously) to 0 as you approach the boundary. Decreasing as one over the log will do the trick,

<br /> f(x) = 1/\max(1,\log(-r))\textrm{\ \ (r=distance to edge of body)}<br />

Then it is still the case that the derivative of the gravitational field (second derivative of \phi) will diverge to infinity along any ridges or spikes on the surface. Have to finish here, but can show this in a separate post.

 

[jostpuur]

I think this exchange of limiting operations requires that \phi_a(x)\rightarrow \phi(x) uniformly.

In general uniform convergence does not mean that limit and derivative operator could be commutated. I'm not sure what's the right argument for commutation here right now.

 

[bdforbes]

At this point it is nicer to do a variable change x&#039;=x+ar, so that the next integral is

<br /> \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr<br />

Then there will be no need for Taylor series of f.

But the integrand has a factor f(x+ar), how do we deal with that? That was the whole point of the Taylor series, to give us integrals we can solve, and an asymptotic series in a.

 

[bdforbes]

In general uniform convergence does not mean that limit and derivative operator could be commutated. I'm not sure what's the right argument for commutation here right now.

Would it be safe to assume that unless the source distribution is pathological, the commutation is possible?

 

[gel]

Jostpuur/bdforbes: Your function \phi_a is a convolution \phi_a=\phi\star\delta_a, where \delta_a(r) = a(a^2+r^2)^{-\frac{3}{2}}/2 is an approximation to the Dirac delta. The integral of \delta_a(x) is one, and its weight becomes concentrated towards 0 as a->0. So, \phi_a\to\phi * \delta=\phi as a->0. As Jostpuur mentiones, this is not enough to guarantee convergence of the second derivatives. However, it will converge as long as \phi it twice continuously differentiable.
<br /> \nabla^2\phi_a=(\nabla^2\phi)*\delta_a\to(\nabla^2\phi)*\delta=\nabla^2\phi.<br /> [/itex]<br /> So it is the same issue as I was addressing in my posts.<br /> Also, convergence always holds in distributions, giving<br /> - the PDE is satisfied in distribution.<br /> - if \phi is twice continuously differentiable, then the pde is satisfied using the standard &#039;pointwise&#039; derivatives.<br /> We have really arrived at the same point but using a slightly different method.<br /> Not surprising really, because the method of distributions involves integrating vs an arbitrary smooth function whereas you effectively convolved with specific smooth functions.

 

[jostpuur]

At this point it is nicer to do a variable change x&#039;=x+ar, so that the next integral is

<br /> \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr<br />

Then there will be no need for Taylor series of f.

But the integrand has a factor f(x+ar), how do we deal with that? That was the whole point of the Taylor series, to give us integrals we can solve, and an asymptotic series in a.

The next step is to take the limit a\to 0.

<br /> \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr \underset{a\to 0}{\to} f(x) \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} dr<br />

 

[bdforbes]

So the question still remains as to whether a continuous f means that \phi is twice continuously differentiable. In fact, this is false.

Do you mean to imply that \nabla^2\phi=f won't hold for some continuous f, which perhaps are not differentiable? This might correspond to jostpuur's method where a change of integration variables results in the Laplacian acting of f instead of the Green's function, thus requiring differentiability of f.

 

[bdforbes]

The next step is to take the limit a\to 0.

<br /> \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr \underset{a\to 0}{\to} f(x) \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} dr<br />

Aren't you implicitly assuming the convergence of f(x+ar) to f(x) is "nice" in some sense, in order to commute limit and integration?

 

[gel]

Do you mean to imply that \nabla^2\phi=f won't hold for some continuous f, which perhaps are not differentiable? This might correspond to jostpuur's method where a change of integration variables results in the Laplacian acting of f instead of the Green's function, thus requiring differentiability of f.

Yes. Jostpuur and my methods do correspond with each other, as I mentioned. I used distributions, by integrating vs an arbitrary smooth function. His method involved convolving vs specific smooth functions for which the integral (convolution vs the Green's function) can be done explicitly.

And, your other point. Dominated convergence allows you to commute integration and limits. All that is required is that the integrands are bounded by (the same) integrable function.

 

[bdforbes]

Jostpuur/bdforbes: Your function \phi_a is a convolution \phi_a=\phi\star\delta_a, where \delta_a(r) = a(a^2+r^2)^{-\frac{3}{2}}/2 is an approximation to the Dirac delta. The integral of \delta_a(x) is one, and its weight becomes concentrated towards 0 as a->0. So, \phi_a\to\phi * \delta=\phi as a->0. As Jostpuur mentiones, this is not enough to guarantee convergence of the second derivatives. However, it will converge as long as \phi it twice continuously differentiable.
<br /> \nabla^2\phi_a=(\nabla^2\phi)*\delta_a\to(\nabla^2\phi)*\delta=\nabla^2\phi.<br /> [/itex]<br /> So it is the same issue as I was addressing in my posts.<br /> Also, convergence always holds in distributions, giving<br /> - the PDE is satisfied in distribution.<br /> - if \phi is twice continuously differentiable, then the pde is satisfied using the standard &#039;pointwise&#039; derivatives.<br /> We have really arrived at the same point but using a slightly different method.<br /> Not surprising really, because the method of distributions involves integrating vs an arbitrary smooth function whereas you effectively convolved with specific smooth functions.

<br /> <br /> Great! It&#039;s a good sign when we arrive at the result through different paths.<br /> <br /> Is the condition on \phi equivalent to requiring f be continuously differentiable? I can&#039;t figure out the conclusion to your large post earlier.

 

[gel]

Great! It's a good sign when we arrive at the result through different paths.

Is the condition on \phi equivalent to requiring f be continuously differentiable? I can't figure out the conclusion to your large post earlier.

Almost.
f continuously differentiable => \phi is twice continuously differentiability => PDE holds everywhere.
The inverse implications don't quite hold. For many continuous f, \phi will still be twice continuously differentiable. However, there are counterexamples to this, which was my point above (long post) about bodies with ridges on the surface. The existence of such counterexamples was the conclusion to that post (although I didn't prove it there). So, you can't just drop the requirement for f to be continuously differentiable.
Similarly, there may be twice differentiable \phi, but not continuously so, for which the pde still holds.

 

[jostpuur]

Aren't you implicitly assuming the convergence of f(x+ar) to f(x) is "nice" in some sense, in order to commute limit and integration?

Yes! My calculation is the same thing as changing the order of integral and limit. I didn't mention what I'm assuming of f, but now when you are asking it, I'll mention that for example

<br /> \|f\|_{\textrm{sup}} := \sup_{x\in\mathbb{R}} |f(x)| &lt; \infty<br />

will be sufficient, assuming that f is also continuous at x. If we define

<br /> h(r) := \frac{1}{2}\frac{\|f\|_{\textrm{sup}}}{(r^2 + 1)^{3/2}}<br />

then

<br /> \int\limits_{-\infty}^{\infty} h(r)dr &lt; \infty<br />

and

<br /> \Big|\frac{1}{2}\frac{f(x+ar)}{(r^2+1)^{3/2}}\Big| \leq h(r)<br />

for all a. So

<br /> \lim_{a\to 0} \int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{f(x+ar)}{(r^2+1)^{3/2}} dr<br /> = \int\limits_{-\infty}^{\infty} \lim_{a\to 0}\frac{1}{2}\frac{f(x+ar)}{(r^2+1)^{3/2}} dr<br />

so is 100% justified. I'm not aware how similar rigor could be achieved in the Taylor series way.

The boundedness assumption \|f\|_{\textrm{sup}}&lt;\infty is only an example. I thought it is often satisfied, but it does not seem necessary. If f is not bounded, but satisfies some other nice properties, it can be that suitable dominating functions h(r) can still be found.

 

[bdforbes]

Dominated convergence allows you to commute integration and limits. All that is required is that the integrands are bounded by (the same) integrable function.

Do you mean these integrands?

\frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar)

and

f(x)\frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}}

I guess finding an integrable function to bound these by would involve the requirement that f be continuously differentiable, so that f(x+ar) approaches f(x) nicely as a->0.

 

[gel]

I guess finding an integrable function to bound these by would involve the requirement that f be continuously differentiable, so that f(x+ar) approaches f(x) nicely as a->0.

Yes, kind of. Jostpuur just did that. It just requires f to be bounded (not continuous) to show that the integrands are dominated. The continuous requirement for f is just so that the integrands converge. Differentiability is required to commute differentiation with the limit on the other side of the equality.

 

[bdforbes]

Looks like we have this nailed down except for pathological sources. Thanks for your help jostpuur and gel. I will write this all up and bounce it off anyone in my department who will listen.

 

[gel]

Cool. I'll also sketch the "pathological" cases (wouldn't quite as far to call it that) in a mo.

 

[gel]

For the counterexample, where f is continuous but the second derivatives of \phi blow up. This will occur, e.g., on the edges of a uniform density cube and also if the density in the cube drops off inversely proportional to 1/log(distance to surface of cube), which is continuous. More generally you can replace 'cube' with any solid body whose surface has ridges.

The details are rather involved, but I'll try my best. I don't know a simple and quick method. First we can derive a general expression for the second derivatives. The following derivative is easily calculated for x != 0. I'm using \delta_{ij} for the (non-Dirac) delta function equal to 1 if i=j and 0 otherwise.
<br /> \nabla_i\nabla_j (1/|x|) = -\nabla_i(\hat x_j/|x|^2) = (3\hat x_i\hat x_j-\delta_{ij})/|x|^3.<br />
Then, the derivatives in the sense of distributions (including point at 0) can be calculated using the divergence theorem as
<br /> \nabla_i\nabla_j (1/|x|) = -\frac{4}{3}\pi\delta_{ij}\delta(x)+\lim_{r\to 0}1_{\{|x|&gt;r\}}(3\hat x_i\hat x_j-\delta_{ij})/|x|^3.<br />
This is an equality of distributions, meaning that you integrate vs a smooth function. The limit r->0 on the rhs is similarly a limit in distribution so that the limit is taken after the integration. You can't just set r to 0, because that term wouldn't be locally integrable.

Plugging this into the definition of \phi gives the following
<br /> \nabla_i\nabla_j\phi(x)=\frac{1}{3}\delta_{ij}f(x)+\lim_{r\to 0}\frac{1}{4\pi}\int_{|y|&gt;r} f(x+y)\frac{\delta_{ij}-3\hat y_i\hat y_j}{|y|^3}\,dy.<br />
Letting d\sigma(y) be the area integral on S - the sphere of radius 1 centered at 0, we can change variables.
<br /> \nabla_i\nabla_j\phi(x)=\frac{1}{3}\delta_{ij}f(x)+\lim_{r\to 0}\frac{1}{4\pi}\int_{r}^\infty \int_S f(x+ty)(\delta_{ij}-3 y_i y_j)\,d\sigma(y)\,\frac{dt}{t}.<br />

The integral dt/t will blow up as the lower limit r->0. However, the integral of \delta_{ij}-3y_iy_j over the sphere S is zero, so if you Taylor expand f about x, the zeroth order term will drop out of the integral above and it remains finite as r->0. Actually, the integral of \delta_{ij}-3y_iy_j is also zero over a hemisphere, so the same applies if x is on a smooth boundary surface between two regions where f is smooth. E.g. the derivatives of the gravitational field don't blow up on the boundary of a uniform density ball.

However, it would blow up at the edges of a uniform density cube. Suppose that x is on such an edge with the x₁ direction pointing inwards and bisecting the angle of the two adjacent faces.

<br /> \left(\frac{\partial}{\partial x_1}\right)^2\phi(x)=\frac{1}{3}f(x)+\lim_{r\to 0}\frac{1}{4\pi}\int_{r}^\infty \int_S f(x+ty)(1-3 y_1^2)\,d\sigma(y)\,\frac{dt}{t}.<br />

For small t, f(x+ty) will then be nonzero on a 90 degree 'wedge' on the surface of the sphere y\in S, and for y₁>= 0. This will cause the integral of f(x+ty)(1-3y₁²) to be strictly negative and nonvanishing as t->0.
So the integral above will be negatively proportional to \int_{r}^\cdot dt/t in the limit as r->0, which blows up at rate log(r).
Finally suppose that the density f inside the cube drops off at rate -1/log(u) towards the surface of the cube where u = distance to edge. Then, f is continuous, and the integral of f(x+ty)(1-3y₁²) will be negative and going to zero at rate 1/log(t) as t-> 0. So, the integral above will be negatively proportional to \int_r^\cdot dt/|t\log t| as r->0, which blows up at rate log(-log(r)). The second derivative above will diverge to negative infinity.

 

[George Jones]

Looks like we have this nailed down except for pathological sources. Thanks for your help jostpuur and gel. I will write this all up and bounce it off anyone in my department who will listen.

A couple of books that give a bit of distributional treatment of Green's functions are Fourier Analysis and Its Applications by Gerald B. Folland and Mathematics for Physics and Physicists by Walter Appel. To understand their notation and conventions, it might first be necessary to scan the treatments of distributions by these books.

 

[bdforbes]

A couple of books that give a bit of distributional treatment of Green's functions are Fourier Analysis and Its Applications by Gerald B. Folland and Mathematics for Physics and Physicists by Walter Appel. To understand their notation and conventions, it might first be necessary to scan the treatments of distributions by these books.

Thanks George, I checked out the Folland text at http://books.google.com/books?id=id...nd+fourier+analysis&ei=w85fSuiJA4zSkASinYHUCg, it looks promising. Unfortunately my library doesn't have either text but I will look elsewhere.

 

[George Jones]

Thanks George, I checked out the Folland text at http://books.google.com/books?id=id...nd+fourier+analysis&ei=w85fSuiJA4zSkASinYHUCg, it looks promising.

Folland doesn't required any prior familiarity with Lebesgue integration as long as the reader is willing to accept a few results (e.g., Lebesgue dominated convergence theorem) without proof.

Unfortunately my library doesn't have either text but I will look elsewhere.

If you want, you probably can get Appel through inter-library loan. Although not as rigourous as Folland, it is a very interesting book,

http://press.princeton.edu/titles/8452.html,

and it is still much more rigourous than many physics books. Its treatment of Green's functions starts with a pedagogical example of the Green's functions for a driven harmonic oscillator (I think; out of town right now so I'm not sure) that is basic, but illuminating.

 

[bdforbes]

I put in a request for an inter-library loan of Appel, and I got "Intro to PDEs" by Folland. His approach to the Laplace operator is rigorous right from the start, it's very good.

 

[bdforbes]

I'm now trying to determine the fundamental solution to the Helmholtz wave equation. My starting point is

(\nabla^2+k^2)G(\vec{r})=-4\pi\delta(\vec{r}).

Treating this as a distributional statement, it is equivalent to

\int G(\vec{r})(\nabla^2+k^2)e^{-2\pi i\vec{\xi}\cdot\vec{r}}d^3\vec{r}=-4\pi

This leads to the desired result after some contour integration.

But if I want to be a little more rigorous and not rely on distributional techniques, I run into trouble. Integrating both sides over punctured \textbf{R}^3:

<br /> \begin{align*}<br /> 0&amp;=\lim_{\epsilon\to 0}\int_{B^c_\epsilon}e^{-2\pi i\vec{\xi}\cdot\vec{r}}(\nabla^2+k^2)G(\vec{r})d^3\vec{r}\\<br /> &amp;=\lim_{\epsilon\to 0}\int_{B^c_\epsilon}\left[\nabla^2\left(G(\vec{r})e^{-2\pi i\vec{\xi}\cdot\vec{r}}\right)-G(\vec{r})\nabla^2e^{-2\pi i\vec{\xi}\cdot\vec{r}}-2\nabla G(\vec{r})\cdot\nabla e^{-2\pi i\vec{\xi}\cdot\vec{r}}\right]d^3\vec{r}+k^2\tilde{G}(\vec{\xi})\\<br /> &amp;=\left[4\pi^2\xi^2+k^2\right]\tilde{G}(\vec{\xi})+\lim_{\epsilon\to 0}\left\{<br /> \int_{\partial B^c_\epsilon}\nabla\left(G(\vec{r})e^{-2\pi i\vec{\xi}\cdot\vec{r}}\right)\cdot d\vec{A}+4\pi i\vec{\xi}\cdot\int_{B^c_\epsilon}e^{-2\pi i\vec{\xi}\cdot\vec{r}}\nabla G(\vec{r})d^3\vec{r}\right\}\\<br /> &amp;=\left[k^2-4\pi^2\xi^2\right]\tilde{G}(\vec{\xi})+\lim_{\epsilon\to 0}\left\{\int_{\partial B^c_\epsilon}\nabla\left(G(\vec{r})e^{-2\pi i\vec{\xi}\cdot\vec{r}}\right)\cdot d\vec{A}+4\pi i\vec{\xi}\cdot\int_{\partial B^c_\epsilon}G(\vec{r})e^{-2\pi i\vec{\xi}\cdot\vec{r}}d\vec{A}\right\}<br /> \end{align*}<br />

The problem is that I want the remaining integrals to evaluate to a constant, but it appears that it will depend on

\lim_{\epsilon\to 0}G(\epsilon)

and

\lim_{\epsilon\to 0}\left.\frac{\partial G}{\partial r}\eval\right|_{r=\epsilon}

I can solve the integrals by assuming spherical symmetry of G(r), but I just end up with functions of \xi, \epsilon and G, and it is unclear if there is even convergence in the \epsilon\to 0 limit.

How can I proceed from this point?

EDIT:

I just had a thought. We are dealing with a second order PDE, so there are two degrees of freedom to the solution. Could we choose G(0) and G'(0) in such a way as to get reduce the remaining integrals to a normalization constant?

 

[bdforbes]

If we set that whole mess to some constant A independent of \xi, then proceed with the contour integration, we will arrive at the correct result. We can then substitute the Green's function back in and confirm that the limit does exist.

Is it a problem to assume there is no \xi dependence?

 

[bdforbes]

Upon further expansion we get:

<br /> \begin{align*}<br /> 0&amp;=\left[k^2-4\pi^2\xi^2\right]\tilde{G}(\vec{\xi})+\lim_{\epsilon\to 0}\left\{\int_{\partial B^c_\epsilon}e^{-2\pi i\vec{\xi}\cdot\vec{r}}\nabla G(\vec{r})\cdot d\vec{A}+2\pi i\vec{\xi}\cdot\int_{\partial B^c_\epsilon}G(\vec{r})e^{-2\pi i\vec{\xi}\cdot\vec{r}}d\vec{A}\right\}<br /> \end{align*}<br />

The first integral might not be a problem in terms of \xi dependence since the magnitude of \vec{r} goes to zero. But the second integral has \xi out the front. I think the second integral shouldn't be there. It was nearly canceled out, but for a factor of two in one term. I can't work out where a stray factor of two could be, but hopefully it is somewhere.

 

[bdforbes]

When I substitute G(r)=e^{ikr}/r into the first integral, I get

\frac{2}{\xi\epsilon^3}\left(ik\epsilon e^{ik\epsilon}-e^{ik\epsilon}\right)sin(2\pi\xi\epsilon)

This diverges as \epsilon\to 0, so I must have made an error early on. Should I not have ignored the surface at infinity when I applied the divergence theorem?

 

[bdforbes]

I've noticed another big problem with the Fourier transform approach. The Fourier transform of

G(r)=\frac{e^{ikr}}{r}

does not actually exist, it diverges. I assume there is a sense in which we are able to work with the object \tilde{G}(\vec{\xi}) and still achieve reasonable results, perhaps by appealing to Cauchy principle values etc.

I am very frustrated by the fact that hand wavy techniques which ignore so many mathematical issues, are able to produce the desired results so easily. Is there any simple, intuitive explanation as to why this happens?