Green's function approach using Lebesgue integration

[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?

 

[jostpuur]

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.

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.

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:

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

<br /> \partial_x \int dx&#039;\; \theta(x-x&#039;) f(x&#039;) = \int dx&#039;\; \partial_x \theta(x-x&#039;) f(x&#039;),<br />

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

<br /> \Big|\frac{\theta(x+\Delta x - x&#039;) - \theta(x - x&#039;)}{\Delta x}\Big| = \frac{1}{|\Delta x|} \chi_{[x,x+\Delta x]}(x&#039;) \leq h(x&#039;)<br />

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

\nabla^2 \phi(\vec{x})=\frac{-1}{4\pi}\int\nabla^2\left(\frac{1}{\left|\vec{x}-\vec{x}&#039;\right|}\right)f(\vec{x}&#039;)d^3\vec{x}&#039;

But now it looks like this Lebesgue integral is NOT well-defined! How do we deal with \nabla^2(1/|x-x&#039;|) at the singular point?

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.

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?

One way is do a variable change like this

<br /> \nabla^2_x \int \frac{f(x&#039;)}{|x-x&#039;|} d^3x&#039; = \nabla^2_x \int \frac{f(x-u)}{|u|} d^3u<br />

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.

Is there a way to use the Dirac measure, or the Dirac Delta as a linear functional?

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.

 

[bdforbes]

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

 

[bdforbes]

One way is do a variable change like this

<br /> \nabla^2_x \int \frac{f(x&#039;)}{|x-x&#039;|} d^3x&#039; = \nabla^2_x \int \frac{f(x-u)}{|u|} d^3u<br />

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.

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&#039;|f(x&#039;)dx&#039; 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<br /> <br /> =\left[\frac{1}{2}|u|\frac{\partial}{\partial x}f(x-u)\right]^{x-a}_{x-b}<br /> <br /> =\frac{1}{2}|x-a|f&#039;(a)-\frac{1}{2}|x-b|f&#039;(b)

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

 

[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&#039; too) will have to be sufficiently zero somewhere for the integration by parts to work.

 

[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.

 

[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

<br /> \frac{d^2}{dx^2} \int\limits_{x-b}^{x-a} \cdots du<br />

The boundaries will give some extra terms if |a|,|b|&lt;\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.

 

[jostpuur]

I'll now set a=-\infty and b=\infty, and assume that f has sufficient properties so that f(\pm\infty) and f&#039;(\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&#039;.

<br /> \frac{d^2}{dx^2} \int\limits_{-\infty}^{\infty} \frac{1}{2}|x-x&#039;| f(x&#039;)dx&#039; = \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<br />

Now substitute

<br /> 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).<br />

<br /> \cdots = \infty f&#039;(\infty) - \infty f&#039;(-\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)<br />

 

[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
<br /> \frac{1}{2}|x-a|f&#039;(a)-\frac{1}{2}|x-b|f&#039;(b)<br />

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.

 

[jostpuur]

One question, why do we turn d^2/dx^2 into d^2/du^2?

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.

 

[bdforbes]

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

 

[jostpuur]

<br /> \frac{d}{dx} f(x-u) = f&#039;(x-u)<br />

<br /> \frac{d}{du} f(x-u) = -f&#039;(x-u)<br />

Then take second derivatives the same way.

 

[bdforbes]

Thanks, that makes sense.

 

[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.

 

[jostpuur]

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.

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

One way is to do a variable change like this...

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

I learned the change of variable trick u=x-x&#039; from the https://www.amazon.com/dp/0821807722/?tag=pfamazon01-20 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/dp/047130932X/?tag=pfamazon01-20 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

<br /> \nabla_x^2 \int d^3x&#039;\; \frac{\rho(x&#039;)}{\sqrt{|x-x&#039;|^2}} = \lim_{a\to 0} \int d^3x&#039;\; \nabla^2_x \frac{\rho(x&#039;)}{\sqrt{|x-x&#039;|^2 + a^2}} = \lim_{a\to 0} \int d^3x&#039;\; \nabla_x\cdot\Big(-\frac{(x-x&#039;)\rho(x&#039;)}{(|x-x&#039;|^2 + a^2)^{3/2}}\Big)<br />
<br /> = -\lim_{a\to 0} \int d^3x&#039;\; \frac{3a^2 \rho(x&#039;)}{(|x-x&#039;|^2 + a^2)^{5/2}} = -4\pi \rho(x)<br />

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&#039; 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&#039;. 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&#039; does lead to a paradox.

The last steps uses a delta function identity

<br /> \frac{3a^2}{(|x-x&#039;|^2 + a^2)^{5/2}} \underset{a\to 0}{\to} 4\pi \delta^3(x-x&#039;)<br />

It can be showed with a variable change x&#039;=x+au.

<br /> \int d^3x&#039;\; \frac{3a^2\rho(x&#039;)}{(|x-x&#039;|^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}}<br />
<br /> \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)<br />

 

[gel]

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

You can. First, you need the identity

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

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.

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

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

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

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

You want to prove

<br /> \nabla^2_{x^\prime}\,\phi(x^\prime)=f(x^\prime)<br />

which, in distributional form, is equivalent to

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

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

 

[bdforbes]

I learned the change of variable trick u=x-x&#039; from the https://www.amazon.com/dp/0821807722/?tag=pfamazon01-20 some years ago.

I will definitely take a look at that!

I was reading the https://www.amazon.com/dp/047130932X/?tag=pfamazon01-20 this spring

I have been reading it too by coincidence.

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&#039; 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&#039;. 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&#039; does lead to a paradox.

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

 

[bdforbes]

You can. First, you need the identity

<br /> \nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)=\delta(x-x^\prime).<br /> \eqno{(1)}<br />

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.

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.

<br /> \int u(x)\nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)\,dx=\int u(x)\delta(x-x^\prime)\,dx.<br />

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

<br /> \int \left(\nabla_x^2u(x)\right)\frac{1}{|x-x^\prime|}\,dx=u(x^\prime).\eqno{(2)}<br />

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

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.

You want to prove

<br /> \nabla^2_{x^\prime}\,\phi(x^\prime)=f(x^\prime)<br />

which, in distributional form, is equivalent to

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

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

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.

 

[jostpuur]

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

Do you mean that we can prove statement (2) is true, or that it is equivalent to (1)?

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).

 

[bdforbes]

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).

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

 

[gel]

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).

Yes, (2) is just the definition of (1). Can break it into two steps. First, we can prove
<br /> \nabla\left(\frac{1}{|x|}\right)=-\frac{\hat x}{|x|^2}. \eqno{(3a)}<br />
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
<br /> \int \left(\nabla u(x)\right)\frac{1}{|x|}\,dx=\int u(x)\frac{\hat x}{|x|^2}\,dx\eqno{(3b)}<br />
for all smooth u with compact support.
Choose r>0 and let B_r be a ball of radius r, with B^c_r being its complement.
<br /> \begin{align*}<br /> \int \left(\nabla u(x)\right)\frac{1}{|x|}\,dx &amp;=\lim_{r\rightarrow 0}\int_{B_r^c}\left(\nabla u(x)\right)\frac{1}{|x|}\,dx\\<br /> &amp;= \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\\<br /> &amp;= -\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<br /> \end{align*}<br />

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 S_r 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
<br /> \nabla \cdot\left(\frac{\hat x}{|x|^2}\right)=4\pi\delta(x).\eqno{(4a)}<br />
Again, this is differentiation in the sense of distributions and is equivalent to
<br /> \int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx=-4\pi u(0).\eqno{(4b)}<br />
Can prove this in a similar way as above.
<br /> \begin{align*}<br /> \int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx &amp;=\lim_{r\to 0}\int_{B_r^c} \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx\\<br /> &amp;=\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<br /> \end{align*}<br />
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).
<br /> \begin{align*}<br /> \int \nabla(u(x))\cdot\left(\frac{\hat x}{|x|^2}\right)\,dx<br /> &amp;=-\lim_{r\to 0}\int_{S_r}\hat x\cdot\left(u(x)\frac{\hat x}{|x|^2}\right)\,d\sigma(x)\\<br /> &amp;=-\lim_{r\to 0}\int_{S_r}u(x)\frac{1}{|x|^2}\,d\sigma(x)<br /> \end{align*}<br />
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

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

 

[gel]

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.

More explicitly
<br /> \begin{align*}<br /> \int \left(\nabla_{x^\prime}^2\,u(x^\prime)\right)\phi(x^\prime)\,dx^\prime<br /> &amp;=-\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\\<br /> &amp;=-\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)}\\<br /> &amp;=-\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\\<br /> &amp;=-\frac{1}{4\pi}\int f(x) (-4\pi u(x))\,dx = \int u(x)f(x)\,dx<br /> \end{align*}<br />

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.

 

[jostpuur]

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

<br /> \nabla^2_x \int \frac{f(x&#039;)}{|x-x&#039;|} d^3x&#039; = \nabla^2_x \int \frac{f(x-u)}{|u|} d^3u<br />

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

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

Suppose we have Laplace's Equation:

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

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

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

\phi(\vec{x})=\frac{-1}{4\pi}\int \frac{f(\vec{x}&#039;)}{\left|\vec{x}-\vec{x}&#039;\right|}d^3\vec{x}&#039;

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

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

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.

 

[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
<br /> \begin{align*}<br /> \nabla_{x}^2\phi(x)&amp;=-\frac{1}{4\pi}\int f(x^\prime)\nabla_x^2\left(\frac{1}{|x-x^\prime|}\right)\,dx^\prime\\<br /> &amp;= -\frac{1}{4\pi}\int f(x^\prime)(-4\pi)\delta(x-x^\prime)\,dx^\prime\\<br /> &amp;= f(x).<br /> \end{align*}<br />
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.

 

[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

 

[gel]

I'm not about to buy the book mentioned in Friedel's post, but from Wikipedia, Green's 2nd identity is

\int_U \left( \psi \nabla^2 \varphi - \varphi \nabla^2 \psi\right)\, dV = \oint_{\partial U} \left( \psi {\partial \varphi \over \partial n} - \varphi {\partial \psi \over \partial n}\right)\, dS.

which you can use with U=B_r^c to get something like I had using divergence theorem above, if you put my 2 steps together

 

[gel]

Thinking about this a bit more. My approach showed that \nabla^2\phi = f in the sense of distributions. To be rigorous, you would also have to show that \phi is twice differentiable. Maybe this isn't true if f is only assumed to be continuous (and compact support, for simplicity).
If you assume that f is continuously differentiable, then you can show

<br /> \nabla_i\nabla_j\phi(x) = \frac{-1}{4\pi}\int \left(\nabla_i f(x+y)\right)\frac{\hat y}{|y|^2}\,dy.<br />

This can be proven in the same way as my workings out above (i.e. prove 'in the sense of distributions' by integrating vs a smooth function). The right hand side is continuous if f is continuously differentiable, so \phi is twice continuously differentiable.

If f isn't continuously differentiable, then maybe \phi isn't twice continuously differentiable?

 

[gel]

Consider the example f(x) = 1_{\{|x|&lt;1\}}\sqrt{(1-|x|^3)}. Here f is continuous, but not differentiable. As it is spherically symmetric, we can solve by concentrating the mass at the origin (http://en.wikipedia.org/wiki/Shell_theorem" )
<br /> \nabla \phi(x) = \frac{M_{|x|}\hat x}{4\pi|x|^2}.<br />

Here, M_r is the total mass inside radius r. For r <= 1,
<br /> M_r = 4\pi\int_0^r\sqrt{1-s^3}s^2\,ds=\frac{8\pi}{9}-\frac{8\pi}{9}(1-r^3)^{\frac{3}{2}}<br />
and, for r>=1,
<br /> M_r=M_1=\frac{8\pi}{9}<br />
This is continuous, but not differentiable at |x|=1, so strictly speaking \nabla^2\phi=f fails except in the sense of distributions.

Edit: M_r is continuously differentiable at r=1. interesting.

 

[bdforbes]

If f isn't continuously differentiable, then maybe \phi isn't twice continuously differentiable?

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?

PS: I still have to work through your posts in detail, but it all looks good.

EDIT: You answered me while I was drafting my response.

 

[bdforbes]

This is continuous, but not differentiable at |x|=1, so strictly speaking \nabla^2\phi=f fails except in the sense of distributions.

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?

 

[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&#039;|f(x&#039;)dx&#039;

\phi_a(x)=\int_{-\infty}^\infty \frac{1}{2}\sqrt{(x-x&#039;)^2+a^2}f(x&#039;)dx&#039;

<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.