Green's function approach using Lebesgue integration


by bdforbes
Tags: function, green, integration, lebesgue
bdforbes
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#37
Jul14-09, 05:11 PM
P: 152
Quote Quote by jostpuur View Post
At this point it is nicer to do a variable change [itex]x'=x+ar[/itex], so that the next integral is

[tex]
\int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr
[/tex]

Then there will be no need for Taylor series of [itex]f[/itex].
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
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#38
Jul14-09, 05:13 PM
P: 152
Quote Quote by jostpuur View Post
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
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#39
Jul14-09, 05:15 PM
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P: 532
Jostpuur/bdforbes: Your function [itex]\phi_a[/itex] is a convolution [itex]\phi_a=\phi\star\delta_a[/itex], where [itex]\delta_a(r) = a(a^2+r^2)^{-\frac{3}{2}}/2[/itex] is an approximation to the Dirac delta. The integral of [itex]\delta_a(x)[/itex] is one, and its weight becomes concentrated towards 0 as a->0. So, [itex]\phi_a\to\phi * \delta=\phi[/itex] as a->0. As Jostpuur mentiones, this is not enough to guarantee convergence of the second derivatives. However, it will converge as long as [itex]\phi[/itex] it twice continuously differentiable.
[tex]
\nabla^2\phi_a=(\nabla^2\phi)*\delta_a\to(\nabla^2\phi)*\delta=\nabla^2 \phi.
[/itex]
So it is the same issue as I was addressing in my posts.
Also, convergence always holds in distributions, giving
- the PDE is satisfied in distribution.
- if [itex]\phi[/itex] is twice continuously differentiable, then the pde is satisfied using the standard 'pointwise' derivatives.
We have really arrived at the same point but using a slightly different method.
Not surprising really, because the method of distributions involves integrating vs an arbitrary smooth function whereas you effectively convolved with specific smooth functions.
jostpuur
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#40
Jul14-09, 05:17 PM
P: 1,983
Quote Quote by bdforbes View Post
Quote Quote by jostpuur View Post
At this point it is nicer to do a variable change [itex]x'=x+ar[/itex], so that the next integral is

[tex]
\int\limits_{-\infty}^{\infty} \frac{1}{2}\frac{1}{(r^2 + 1)^{3/2}} f(x+ar) dr
[/tex]

Then there will be no need for Taylor series of [itex]f[/itex].
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 [itex]a\to 0[/itex].

[tex]
\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
[/tex]
bdforbes
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#41
Jul14-09, 05:19 PM
P: 152
Quote Quote by gel View Post
So the question still remains as to whether a continuous f means that [itex]\phi[/itex] is twice continuously differentiable. In fact, this is false.
Do you mean to imply that [itex]\nabla^2\phi=f[/itex] 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
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#42
Jul14-09, 05:22 PM
P: 152
Quote Quote by jostpuur View Post
The next step is to take the limit [itex]a\to 0[/itex].

[tex]
\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
[/tex]
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
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#43
Jul14-09, 05:24 PM
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P: 532
Quote Quote by bdforbes View Post
Do you mean to imply that [itex]\nabla^2\phi=f[/itex] 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
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#44
Jul14-09, 05:27 PM
P: 152
Quote Quote by gel View Post
Jostpuur/bdforbes: Your function [itex]\phi_a[/itex] is a convolution [itex]\phi_a=\phi\star\delta_a[/itex], where [itex]\delta_a(r) = a(a^2+r^2)^{-\frac{3}{2}}/2[/itex] is an approximation to the Dirac delta. The integral of [itex]\delta_a(x)[/itex] is one, and its weight becomes concentrated towards 0 as a->0. So, [itex]\phi_a\to\phi * \delta=\phi[/itex] as a->0. As Jostpuur mentiones, this is not enough to guarantee convergence of the second derivatives. However, it will converge as long as [itex]\phi[/itex] it twice continuously differentiable.
[tex]
\nabla^2\phi_a=(\nabla^2\phi)*\delta_a\to(\nabla^2\phi)*\delta=\nabla^2 \phi.
[/itex]
So it is the same issue as I was addressing in my posts.
Also, convergence always holds in distributions, giving
- the PDE is satisfied in distribution.
- if [itex]\phi[/itex] is twice continuously differentiable, then the pde is satisfied using the standard 'pointwise' derivatives.
We have really arrived at the same point but using a slightly different method.
Not surprising really, because the method of distributions involves integrating vs an arbitrary smooth function whereas you effectively convolved with specific smooth functions.
Great! It's a good sign when we arrive at the result through different paths.

Is the condition on [itex]\phi[/itex] equivalent to requiring f be continuously differentiable? I can't figure out the conclusion to your large post earlier.
gel
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#45
Jul14-09, 05:33 PM
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P: 532
Quote Quote by bdforbes View Post
Great! It's a good sign when we arrive at the result through different paths.

Is the condition on [itex]\phi[/itex] equivalent to requiring f be continuously differentiable? I can't figure out the conclusion to your large post earlier.
Almost.
f continuously differentiable => [itex]\phi[/itex] is twice continuously differentiability => PDE holds everywhere.
The inverse implications don't quite hold. For many continuous f, [itex]\phi[/itex] 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 [itex]\phi[/itex], but not continuously so, for which the pde still holds.
jostpuur
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#46
Jul14-09, 05:35 PM
P: 1,983
Quote Quote by bdforbes View Post
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 [itex]f[/itex], but now when you are asking it, I'll mention that for example

[tex]
\|f\|_{\textrm{sup}} := \sup_{x\in\mathbb{R}} |f(x)| < \infty
[/tex]

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

[tex]
h(r) := \frac{1}{2}\frac{\|f\|_{\textrm{sup}}}{(r^2 + 1)^{3/2}}
[/tex]

then

[tex]
\int\limits_{-\infty}^{\infty} h(r)dr < \infty
[/tex]

and

[tex]
\Big|\frac{1}{2}\frac{f(x+ar)}{(r^2+1)^{3/2}}\Big| \leq h(r)
[/tex]

for all [itex]a[/itex]. So

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

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

The boundedness assumption [itex]\|f\|_{\textrm{sup}}<\infty[/itex] is only an example. I thought it is often satisfied, but it does not seem necessary. If [itex]f[/itex] is not bounded, but satisfies some other nice properties, it can be that suitable dominating functions [itex]h(r)[/itex] can still be found.
bdforbes
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#47
Jul14-09, 05:37 PM
P: 152
Quote Quote by gel View Post
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?

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

and

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

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
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#48
Jul14-09, 05:38 PM
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P: 532
Quote Quote by bdforbes View Post
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
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#49
Jul14-09, 05:50 PM
P: 152
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
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#50
Jul14-09, 06:01 PM
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P: 532
Cool. I'll also sketch the "pathological" cases (wouldn't quite as far to call it that) in a mo.
gel
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#51
Jul14-09, 06:51 PM
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For the counterexample, where f is continuous but the second derivatives of [itex]\phi[/itex] 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 [itex]\delta_{ij}[/itex] for the (non-Dirac) delta function equal to 1 if i=j and 0 otherwise.
[tex]
\nabla_i\nabla_j (1/|x|) = -\nabla_i(\hat x_j/|x|^2) = (3\hat x_i\hat x_j-\delta_{ij})/|x|^3.
[/tex]
Then, the derivatives in the sense of distributions (including point at 0) can be calculated using the divergence theorem as
[tex]
\nabla_i\nabla_j (1/|x|) = -\frac{4}{3}\pi\delta_{ij}\delta(x)+\lim_{r\to 0}1_{\{|x|>r\}}(3\hat x_i\hat x_j-\delta_{ij})/|x|^3.
[/tex]
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 [itex]\phi[/itex] gives the following
[tex]
\nabla_i\nabla_j\phi(x)=\frac{1}{3}\delta_{ij}f(x)+\lim_{r\to 0}\frac{1}{4\pi}\int_{|y|>r} f(x+y)\frac{\delta_{ij}-3\hat y_i\hat y_j}{|y|^3}\,dy.
[/tex]
Letting [itex]d\sigma(y)[/itex] be the area integral on S - the sphere of radius 1 centered at 0, we can change variables.
[tex]
\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}.
[/tex]

The integral dt/t will blow up as the lower limit r->0. However, the integral of [itex]\delta_{ij}-3y_iy_j[/itex] 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 [itex]\delta_{ij}-3y_iy_j[/itex] 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.

[tex]
\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}.
[/tex]

For small t, f(x+ty) will then be nonzero on a 90 degree 'wedge' on the surface of the sphere [itex]y\in S[/itex], 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 [itex]\int_{r}^\cdot dt/t[/itex] 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 [itex]\int_r^\cdot dt/|t\log t|[/itex] as r->0, which blows up at rate log(-log(r)). The second derivative above will diverge to negative infinity.
George Jones
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#52
Jul16-09, 08:22 AM
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Quote Quote by bdforbes View Post
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
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#53
Jul16-09, 08:12 PM
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Quote Quote by George Jones View Post
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=idA...A4zSkASinYHUCg, it looks promising. Unfortunately my library doesn't have either text but I will look elsewhere.
George Jones
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#54
Jul17-09, 10:58 AM
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Quote Quote by bdforbes View Post
Thanks George, I checked out the Folland text at http://books.google.com/books?id=idA...A4zSkASinYHUCg, 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.
Quote Quote by bdforbes View Post
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.


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