# Green's function approach using Lebesgue integration

1. Jul 6, 2009

### bdforbes

I can't figure out how to use the Green's function approach rigorously, i.e., taking into account the fact that the Dirac Delta function is not a function on the reals.

Suppose we have Laplace's Equation:

$$\nabla^2 \phi(\vec{x})=f(\vec{x})$$

The solution, for "well-behaved" $f(\vec{x})$ is

$$\phi(\vec{x})=\frac{-1}{4\pi}\int \frac{f(\vec{x}')}{\left|\vec{x}-\vec{x}'\right|}d^3\vec{x}'$$

It is my understanding that this integral is well-defined as both a Riemann-Darboux and Lebesgue integral. If we treat it as a Lebesgue integral, I believe the limiting operations can be exchanged, i.e., we can apply the Laplacian to the integrand:

$$\nabla^2 \phi(\vec{x})=\frac{-1}{4\pi}\int\nabla^2\left(\frac{1}{\left|\vec{x}-\vec{x}'\right|}\right)f(\vec{x}')d^3\vec{x}'$$

But now it looks like this Lebesgue integral is NOT well-defined! How do we deal with $\nabla^2(1/|x-x'|)$ at the singular point?
If we naively apply the divergence theorem, we can arrive at the desired result, but that is not good enough for me.

How can we do this rigorously? Is there a way to use the Dirac measure, or the Dirac Delta as a linear functional?

2. Jul 6, 2009

### jostpuur

According to my experience, we are usually interested in such functions $f$, that it doesn't matter what definition of the integral we are using. They all agree on the relevant functions $f$.

That's a mistake. If you actually tried to find some dominating function so that you could use the Lebesgue's dominated convergence, you would notice that you cannot find suitable dominating function. Take a look at the simpler example, $\partial_x \theta(x-x') = \delta(x-x')$. If you try to do this

$$\partial_x \int dx'\; \theta(x-x') f(x') = \int dx'\; \partial_x \theta(x-x') f(x'),$$

you would find yourself trying to find such integrable function $h$ that

$$\Big|\frac{\theta(x+\Delta x - x') - \theta(x - x')}{\Delta x}\Big| = \frac{1}{|\Delta x|} \chi_{[x,x+\Delta x]}(x') \leq h(x')$$

for all $\Delta x$, and that's not possible.

Actually the integrals are well defined, because in Lebesgue integration you can always ignore values of functions at single points (or other sets of zero measure). The integral in the right side is simply zero. On the other hand the integral on the left side is usually not zero. So both sides of the equation are well defined, but the equation it self is not correct.

One way is do a variable change like this

$$\nabla^2_x \int \frac{f(x')}{|x-x'|} d^3x' = \nabla^2_x \int \frac{f(x-u)}{|u|} d^3u$$

and then use niceness properties of $f$ for commutation of derivation and integration. After commutation some calculus trickery with divergence theorem and integration by parts will be needed.

According to my experience, Dirac measure and delta distribution are useful for defining some properties by definition, or for stating results once they have been proven, but not useful for actually proving anything.

3. Jul 6, 2009

### bdforbes

Thanks for those insights, that's very helpful. I'll work through the variable change method and see what I get.

4. Jul 6, 2009

### bdforbes

I've been trying to do this for the simpler one dimensional case, but I don't get the desired result. Here is my working:

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

$$\phi(x)=\int_a^b \frac{1}{2}|x-x'|f(x')dx'$$ with a<x<b

Let u=x-x'

$$\frac{\partial^2}{\partial x^2}\phi(x)=-\int_{x-a}^{x-b}\frac{1}{2}|u|\frac{\partial^2}{\partial x^2}f(x-u)du =\left[\frac{1}{2}|u|\frac{\partial}{\partial x}f(x-u)\right]^{x-a}_{x-b} =\frac{1}{2}|x-a|f'(a)-\frac{1}{2}|x-b|f'(b)$$

The final result almost looks right, but it would only work for a and b very close to x, wouldn't it?

5. Jul 6, 2009

### jostpuur

Those are calculation mistakes, made in too quick calculation. Notice that you need pay attention how to switch operators $$\frac{\partial}{\partial x}$$ and $$\frac{\partial}{\partial u}$$. (It could be you would notice this soon anyway...)

Not any $a,b,f$ are going to be fine. $f$ ($f'$ too) will have to be sufficiently zero somewhere for the integration by parts to work.

6. Jul 6, 2009

### bdforbes

I'm not sure what you mean. If we view u as depending on x, why are we still allowed to commute the integration and differentiation? I thought the point was switch the x dependence to the function f because we can assume it is a nice function.

7. Jul 8, 2009

### jostpuur

It could be a good idea to set $a=-\infty$ and $b=\infty$ in the beginning. It will save from some trouble when the order of derivation and integration are supposed to be changed in

$$\frac{d^2}{dx^2} \int\limits_{x-b}^{x-a} \cdots du$$

The boundaries will give some extra terms if $|a|,|b|<\infty$. Actually I've never tried to carry out calculations like the in full rigor. It is not awfully difficult to get right answers, though.

Last edited: Jul 8, 2009
8. Jul 8, 2009

### jostpuur

I'll now set $a=-\infty$ and $b=\infty$, and assume that $f$ has sufficient properties so that $f(\pm\infty)$ and $f'(\pm\infty)$ will bring all else to zero in integration by parts, and also so that $\frac{d^2}{dx^2}$ and $\int du$ can be commuted after the change of variable $u=x-x'$.

$$\frac{d^2}{dx^2} \int\limits_{-\infty}^{\infty} \frac{1}{2}|x-x'| f(x')dx' = \int\limits_{-\infty}^{\infty} \frac{1}{2}|u| \frac{d^2}{dx^2} f(x-u) du = -\frac{1}{2}\int\limits_{-\infty}^0 u \frac{d^2}{du^2} f(x-u) du + \frac{1}{2}\int\limits_0^{\infty} u \frac{d^2}{du^2} f(x-u)du = \cdots$$

Now substitute

$$u \frac{d^2}{du^2} f(x-u) = \frac{d}{du}\Big( u \frac{d}{du} f(x-u)\Big) - \frac{d}{du} f(x-u).$$

$$\cdots = \infty f'(\infty) - \infty f'(-\infty) + \underbrace{\frac{1}{2}\int\limits_{-\infty}^0 \frac{d}{du}f(x-u) du}_{=\frac{1}{2}f(x) - \frac{1}{2}f(-\infty)} - \underbrace{\frac{1}{2}\int\limits_0^{\infty} \frac{d}{du} f(x-u) du}_{=-\frac{1}{2}f(\infty) + \frac{1}{2}f(x)} = f(x)$$

Last edited: Jul 8, 2009
9. Jul 8, 2009

### bdforbes

Thanks, it does look like this can be done rigorously. One question, why do we turn $d^2/dx^2$ into $d^2/du^2$?

I believe my result
$$\frac{1}{2}|x-a|f'(a)-\frac{1}{2}|x-b|f'(b)$$

is also good, if we only consider a very small interval around the singularity. The other contributions would go to zero anyway since if we avoid the singularity, we can take the double derivative of |x-x'| which is zero. My answer approaches f(x) in the limit a->x<-b.

10. Jul 8, 2009

### jostpuur

For the purpose of using the fundamental theorem of calculus, and integration by parts. We are integrating over $u$, so we want derivatives with respect to $u$ inside the integral too.

11. Jul 8, 2009

### bdforbes

But why is it valid to do it? It's not immediately obvious to me.

12. Jul 8, 2009

### jostpuur

$$\frac{d}{dx} f(x-u) = f'(x-u)$$

$$\frac{d}{du} f(x-u) = -f'(x-u)$$

Then take second derivatives the same way.

13. Jul 9, 2009

### bdforbes

Thanks, that makes sense.

14. Jul 10, 2009

### bdforbes

What if I now wanted to use a generalised function approach? Ie, write the Green's function as a limit of ordinary functions, which would enable me to commute the differentiation and integration. If I chose the sequence cleverly, perhaps the second derivative would give a sequence equivalent to the Dirac delta, and I would thus arrive at the result. It feels like the usual physicist approach to Green's functions, whereby one flails around wildly until a reasonable result if found, is basically shorthand for using generalised functions. The method you have given above appears to be qualitatively different.

15. Jul 11, 2009

### jostpuur

Good thing that you asked. For some reason I didn't bother trying to mention the other way I already knew about. Notice that I was careful to say

in my original response. I didn't claim it would be "the" way.

I learned the change of variable trick $u=x-x'$ from the https://www.amazon.com/Partial-Differential-Equations-Graduate-Mathematics/dp/0821807722 some years ago. Evans wasn't particularly speaking about Green's functions, but instead only stated that solutions to some PDEs could be written as some integral expressions. When looking at the proofs, I recognized a solution also to the Green's function problem that had been bothering me already earlier.

I was reading the https://www.amazon.com/Classical-Electrodynamics-Third-David-Jackson/dp/047130932X this spring, and to my positive surprise I noticed that Jackson too comments this same problem. On page 35 (of 3th edition) he shows a following calculation

$$\nabla_x^2 \int d^3x'\; \frac{\rho(x')}{\sqrt{|x-x'|^2}} = \lim_{a\to 0} \int d^3x'\; \nabla^2_x \frac{\rho(x')}{\sqrt{|x-x'|^2 + a^2}} = \lim_{a\to 0} \int d^3x'\; \nabla_x\cdot\Big(-\frac{(x-x')\rho(x')}{(|x-x'|^2 + a^2)^{3/2}}\Big)$$
$$= -\lim_{a\to 0} \int d^3x'\; \frac{3a^2 \rho(x')}{(|x-x'|^2 + a^2)^{5/2}} = -4\pi \rho(x)$$

If you are interested in rigor, this is not necessarily easier than the trick I got from Evans' book. You would first need to justify the commutation of $\int d^3x'$ and $\lim_{a\to 0}$, then the commutation of $\nabla^2_x$ and $\lim_{a\to 0}$, and then the commutation of $\nabla^2_x$ and $\int d^3x'$. So there's lot to do. I have never tried to figure out what arguments one should use to justify these (obviously Jackson doesn't speak about justifying them either, because his is a physics book), but I believe that they can be justified, because these steps do not lead to paradoxes. Like for example direct commutation of $\nabla^2_x$ and $\int d^3x'$ does lead to a paradox.

The last steps uses a delta function identity

$$\frac{3a^2}{(|x-x'|^2 + a^2)^{5/2}} \underset{a\to 0}{\to} 4\pi \delta^3(x-x')$$

It can be showed with a variable change $x'=x+au$.

$$\int d^3x'\; \frac{3a^2\rho(x')}{(|x-x'|^2 + a^2)^{5/2}} = \int d^3u\; a^3 \frac{3a^2\rho(x+au)}{(|au|^2 + a^2)^{5/2}} = 3 \int d^3u\; \frac{\rho(x+au)}{(|u|^2 + 1)^{5/2}}$$
$$\underset{a\to 0}{\to} 3\int d^3u\; \frac{\rho(x)}{(|u|^2 + 1)^{5/2}} = 12\pi \rho(x) \underbrace{\int\limits_0^{\infty} \frac{r^2 dr}{(r^2 + 1)^{5/2}}}_{\cdots = 1/3} = 4\pi \rho(x)$$

Last edited by a moderator: May 4, 2017
16. Jul 11, 2009

### gel

You can. First, you need the identity

$$\nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)=-4\pi\delta(x-x^\prime). \eqno{(1)}$$

Both sides of this equality are to be regarded as distributions. That is, they are linear functionals mapping the smooth real valued functions of compact support $u\colon\mathbb{R}^3\to\mathbb{R}$ to the real numbers, with the mapping being written as the integral against u.

$$\int u(x)\nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)\,dx=4\pi\int u(x)\delta(x-x^\prime)\,dx.$$

The right hand side is simply $u(x^\prime)$ by definition of the Dirac delta. The derivative 'in the sense of distributions' on the left hand side is defined via integration by parts

$$\int \left(\nabla_x^2u(x)\right)\frac{1}{|x-x^\prime|}\,dx=4\pi u(x^\prime).\eqno{(2)}$$

Equation (2) is then an equivalent statement of (1). Can prove it by using the divergence theorem.

You want to prove

$$\nabla^2_{x^\prime}\,\phi(x^\prime)=f(x^\prime)$$

which, in distributional form, is equivalent to

$$\int \left(\nabla^2_{x^\prime}\,u(x^\prime)\right)\phi(x^\prime)\,dx^\prime=\int u(x^\prime)f(x^\prime)\,dx^\prime.$$

To prove this, substitute in the expression for $\phi$, commute the order of integration, and apply (1) or equivalently (2).

Last edited: Jul 12, 2009
17. Jul 11, 2009

### bdforbes

I will definately take a look at that!

I have been reading it too by coincidence.

Perhaps we need to appeal to absolute continuity or uniform convergence?

Last edited by a moderator: May 4, 2017
18. Jul 11, 2009

### bdforbes

I can believe this, it is the result obtained formally by ignoring the singularity and applying the divergence theorem, which must be equivalent to treating the objects as distributions.

Do you mean that we can prove statement (2) is true, or that it is equivalent to (1)? It seems to me that the latter has already been proven above.
Also, could you demonstrate how to use the divergence theorem rigorously here? Every time I see the divergence theorem used in this context, the author ignores the second bounding surface which arises from the integral being improper. This second bounding surface always cancels out the contribution from the first. Their mistake was really to commute differentiation and integration with a singular integrand.

This seems very reasonable. Are there any conditions required for commuting the order of integration? Also, did the Laplacian find itself acting on u(x') via integration by parts, as above? I assume compact support is essential for this to work.

19. Jul 12, 2009

### jostpuur

I hope he meant that the equation (2) can be proven using divergence theorem (and possibly some other calculus stuff). You should not attempt to prove that equations (1) and (2) are equivalent. gel's equation (2) is the definition of the equation (1). The equation (1) should be considered to be notation for the equation (2).

20. Jul 12, 2009

### bdforbes

That makes sense. I guess using the divergence theorem here would be similar to the method you showed earlier, right jostpuur?

21. Jul 12, 2009

### gel

Yes, (2) is just the definition of (1). Can break it into two steps. First, we can prove
$$\nabla\left(\frac{1}{|x|}\right)=-\frac{\hat x}{|x|^2}. \eqno{(3a)}$$
This is easily calculated for x != 0, but we need to show that it holds everywhere in the sense of distributions. In the sense of distributions, (3a) is an equivalent statement to
$$\int \left(\nabla u(x)\right)\frac{1}{|x|}\,dx=\int u(x)\frac{\hat x}{|x|^2}\,dx\eqno{(3b)}$$
for all smooth u with compact support.
Choose r>0 and let Br be a ball of radius r, with Bcr being its complement.
\begin{align*} \int \left(\nabla u(x)\right)\frac{1}{|x|}\,dx &=\lim_{r\rightarrow 0}\int_{B_r^c}\left(\nabla u(x)\right)\frac{1}{|x|}\,dx\\ &= \lim_{r\rightarrow 0}\int_{B_r^c}\nabla\left(u(x)/|x|\right)\,dx - \lim_{r\to 0}\int_{B_r^c}u(x)\nabla\left(\frac{1}{|x|}\right)\,dx\\ &= -\lim_{r\to 0}\int_{S_r} \hat x u(x)/|x|\,d\sigma(x) + \lim_{r\to 0}\int_{B_r^c} u(x)\frac{\hat x}{|x|^2}\,dx \end{align*}

Here, the divergence theorem has been applied to the first integral on the RHS to write it as the surface integral $d\sigma(x)$ over the sphere Sr of radius r.
Note that there is no 'second surface' because u is chosen to have compact support. The surface integral is bounded by $\Vert u\Vert 4\pi r$ which vanishes as r->0, so we get (3b).

The second part is to prove
$$\nabla \cdot\left(\frac{\hat x}{|x|^2}\right)=4\pi\delta(x).\eqno{(4a)}$$
Again, this is differentiation in the sense of distributions and is equivalent to
$$\int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx=-4\pi u(0).\eqno{(4b)}$$
Can prove this in a similar way as above.
\begin{align*} \int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx &=\lim_{r\to 0}\int_{B_r^c} \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx\\ &=\lim_{r\to 0}\int_{B_r^c} \nabla\cdot \left(u(x)\frac{\hat x}{|x|^2}\right)\,dx-\lim_{r\to 0}\int_{B_r^c}u(x)\nabla\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx \end{align*}
The second integral on the RHS is 0, since $\nabla\cdot(\hat x/|x|^2)=0$ for x!=0. Use the divergence theorem for the first integral (again, no second surface as u has compact support).
\begin{align*} \int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx &=-\lim_{r\to 0}\int_{S_r}\hat x\cdot\left(u(x)\frac{\hat x}{|x|^2}\right)\,d\sigma(x)\\ &=-\lim_{r\to 0}\int_{S_r}u(x)\frac{1}{|x|^2}\,d\sigma(x) \end{align*}
As |x|->0 then u(x)->u(0), so this integral converges to $-u(0)\int_{S_r}(1/r^2)\,d\sigma(x)=-4\pi u(0)$, giving (4b).

Then, (3a)+(4a) gives

$$\nabla^2\left(\frac{1}{|x|}\right) = -\nabla\cdot\left(\frac{\hat x}{|x|^2}\right)=-4\pi\delta(x).$$
Finally, its just a translation to replace x by x-x'.

Last edited: Jul 12, 2009
22. Jul 12, 2009

### gel

More explicitly
\begin{align*} \int \left(\nabla_{x^\prime}^2\,u(x^\prime)\right)\phi(x^\prime)\,dx^\prime &=-\frac{1}{4\pi}\int\int \left(\nabla_{x^\prime}^2\,u(x^\prime)\right)f(x)\frac{1}{|x-x^\prime|}\,dx\,dx^\prime\\ &=-\frac{1}{4\pi}\int\int \left(\nabla^2_{x^\prime}\,u(x^\prime)\right)f(x)\frac{1}{|x-x^\prime|}\,dx^\prime\,dx\textrm{\ \ (commute order of integration)}\\ &=-\frac{1}{4\pi}\int f(x)\int \left(\nabla_{x^\prime}^2\,u(x^\prime)\right)\frac{1}{|x-x^\prime|}\,dx^\prime\,dx\\ &=-\frac{1}{4\pi}\int f(x) (-4\pi u(x))\,dx = \int u(x)f(x)\,dx \end{align*}

Commuting the order of integration works, because the integrand is integrable (Fubini's theorem). Its easy to show this if u and f have compact support. If f isn't compact support then you'll have to enforce some other boundary condition to ensure integrability.
The result of this is to prove that $\nabla^2\phi=f$ in the sense of distributions.

Last edited: Jul 12, 2009
23. Jul 12, 2009

### jostpuur

That's one way, but it doesn't look optimal. It is easier to begin like this

$$\cdots = \int \nabla^2_x \frac{f(x-u)}{|u|} d^3u$$

and then carry out the calculation to the end mostly like was shown in the post #21.

Also recall the beginning of the problem. We want to solve a PDE

and here $f$ is usually a function, not a distribution, and $\nabla^2$ is a derivative operator defined by limit. The claim is that if we set

where the integral is a Riemann integral, then this $\phi$ will satisfy the original PDE. If the proof goes through a step that looks like this

it creates an appearance that the nabla-operator in the original PDE problem would have been something else than an ordinary derivative operator defined by limit.

24. Jul 12, 2009

### gel

Yeah, my proof shows that $\nabla^2\phi = f$ in the sense of distributions. The 'in the sense of distributions' can be dropped if you know that phi is indeed twice differentiable. Haven't looked through your proof in enough detail to tell if it proves twice differentiability or assumes it (does it?).
Doing it in terms of distributions is a bit more optimal than it looks as I wrote it out, because I was careful to write it out in 'dual' form (integrated vs u). Really, all that is happening is that you are applying
\begin{align*} \nabla_{x}^2\phi(x)&=-\frac{1}{4\pi}\int f(x^\prime)\nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)\,dx^\prime\\ &= -\frac{1}{4\pi}\int f(x^\prime)(-4\pi)\delta(x-x^\prime)\,dx^\prime\\ &= f(x). \end{align*}
Then, I was integrating vs a smooth function of compact support u(x) in order to make both steps rigorous.

Edit: Also, the identity $\nabla^2(1/|x|)=-4\pi\delta(x)$ only makes sense in terms of distributions. So, if you want to use that, then you have introduced distributions from the start.

Last edited: Jul 12, 2009
25. Jul 12, 2009

### Friedel

I did not follow the whole discussion, but it seems to me that you approach the problem from the wrong direction. I recommend my book 'Introduction to Boundary Elements', Springer-Verlag, 1989 p. 108 the section with the title: The influence function for the deflection u(x) (that is the potential).

There you see how by considering Green's second identity on a punctured domain, you spare the source point x (where the Dirac delta is located), the influence function for the potential is easily derived.

The same simple approach works in 3-D. No need to discuss Lebesgue or Riemann etc.

Friedel Hartmann