User mahdiyar - MathOverflow

Solve |\nabla u|^2=1

2011-12-12T19:45:19Z

I need all solutions of $(\partial_x u)^2+(\partial_y u)^2=1$ for the function $u(x,y)$. Of course I know simple solutions like $u=ax \pm \sqrt{1-a^2}y + c$, or $u=\sqrt{x^2+y^2}+c$; but what's the general solution?

More generally, I'd like to know how to tackle a PDE of the form $|\nabla u|^2=f^2(u)$ where $f$ is some given function. Again a simple solution is to assume that $u$ is a function of $r=\sqrt{x^2+y^2}$ and integrate the equation. But what's the most general solution?

Can a self-adjoint operator have a continuous set of eigenvalues?

2010-09-25T01:41:11Z

This should be a trivial question for mathematicians but not for typical physicists.

I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.

Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?

I appreciate any help.

Markov random field with continuous index set

2011-03-12T22:50:11Z

There's Markov random field (MRF) which, by my Wikipedia-based knowledge, is an extension of Markov chain. I'd like to think of it as going from 1D to higher dimensional spaces. Inherent in its definition though, the space a MRF lives on (i.e., the index set of the stochastic process) is a discrete graph. So it's actually a lattice and not the continuum of a Euclidean space (or some manifold for that matter). I'm wondering if there exists a MRF of the latter form, in other words, an extension of the Markov process (as opposed to the Markov chain) to higher dimensions.

I know that a similar extension exists for the Poisson process, namely the spatial Poisson process. But for MRF, I'm not even sure how I'd define the Markov property.

Any references or remarks are appreciated.

Answer by Mahdiyar for Jokes in the sense of Littlewood: examples?

2010-10-03T07:28:51Z

I recall that the following simple "proof" of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is attributed to Euler:

Begin with the fact that for a polynomial $a_0 + a_1 x + \cdots + a_N x^N$ the sum of the inverses of the roots is given by $\sum_{n=1}^N \frac{1}{x_n} = -\frac{a_1}{a_0}$. (If you only remember the formula for the sum of the roots just make a change of variable $y=1/x$). Now consider the "polynomial" $$\frac{\sin\sqrt{x}}{\sqrt{x}} = 1 - \frac{x}{3!} + \cdots$$ whose roots are $x_n = (n\pi)^2$ for $n\in N$. By applying the aforementioned fact the desired result is immediate.

Comment by Mahdiyar

2011-12-13T10:18:24Z

Yes, I've seen in it classical mechanics, but I didn't recognize it! Nevertheless, even if I did I wouldn't know the general solution Robert talked about. In fact, I was trying to find a canonical transformation to make $2H=A_1(q)p_1^2+A_2(q)p_2^2+2V(q)$ look like $P_1^2+P_2^2+2U(Q)$, that I came up with the original equation.

Comment by Mahdiyar

2011-12-13T10:10:41Z

Thanks Robert. I guess you meant $F'=f$ not $=1/f$. I would've picked this as my answer if it were not a comment.

Comment by Mahdiyar

2011-03-14T20:32:55Z

In fact, I'm interested in applying the Markov property to a process taking place on a Lorentzian manifold (spacetime). An event taking place at a point can then only depend on events in its past light cone.

Comment by Mahdiyar

2011-03-14T20:29:54Z

I need to learn some preliminaries even to understand the Introduction, but it definitely looks like an interesting paper. I was particularly amused by the partial order structured of the index set mentioned in the introduction. I don't know how important a role this plays in their Markov property. But I like this ordering to be somehow reflected in the property (the domain Markov property mentioned by John and Yuri doesn't seem to care about it).

Comment by Mahdiyar

2010-10-04T03:48:27Z

Oh yes; you're right. I did skim through all responses before submitting mine but somehow it didn't capture my attention since it didn't have a big $\pi^2/6$ right at the center! Thanks for the comment and sorry for duplicate answer.

Comment by Mahdiyar

2010-09-26T07:03:18Z

Such a simple answer! Thanks. This is exactly what I expected for the "simpler version" of my question. Actually Piero D'Ancona also provided the same answer to my question but this one was more to the point and simpler to understand (for someone like me!).

Comment by Mahdiyar

2010-09-26T06:40:01Z

Thanks for your reply. Maybe I haven't been very clear. But I tried to explicitly say that I do know that the spectrum and the set of eigenvalues are two different things and that the former can be a continuum. By the way, as pointed out by Yemon Choi "discrete" is vague; I meant countable. Anyway I think you made a point about part of the answer, that compactness is a sufficient condition for eigenvalues to be countable (can you give a reference?).

Comment by Mahdiyar

2010-09-26T06:16:55Z

Thanks for both of your good comments. Yes I do assume separable Hilbert spaces. The example is also pretty illuminating.