User mahdiyar - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:21:01Z http://mathoverflow.net/feeds/user/9504 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83273/solve-nabla-u21 Solve |\nabla u|^2=1 Mahdiyar 2011-12-12T19:45:19Z 2011-12-13T06:16:24Z <p>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?</p> <p>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?</p> http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues Can a self-adjoint operator have a continuous set of eigenvalues? Mahdiyar 2010-09-25T01:41:11Z 2011-10-06T22:59:21Z <p>This should be a trivial question for mathematicians but not for typical physicists.</p> <p>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.</p> <p>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?</p> <p>I appreciate any help.</p> http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set Markov random field with continuous index set Mahdiyar 2011-03-12T22:50:11Z 2011-03-13T23:40:43Z <p>Hi</p> <p>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.</p> <p>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.</p> <p>Any references or remarks are appreciated.</p> http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/40914#40914 Answer by Mahdiyar for Jokes in the sense of Littlewood: examples? Mahdiyar 2010-10-03T07:28:51Z 2010-10-03T07:28:51Z <p>I recall that the following simple "proof" of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is attributed to Euler:</p> <p>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.</p> http://mathoverflow.net/questions/83273/solve-nabla-u21/83275#83275 Comment by Mahdiyar Mahdiyar 2011-12-13T10:18:24Z 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. http://mathoverflow.net/questions/83273/solve-nabla-u21 Comment by Mahdiyar Mahdiyar 2011-12-13T10:10:41Z 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. http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set/58373#58373 Comment by Mahdiyar Mahdiyar 2011-03-14T20:32:55Z 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. http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set/58373#58373 Comment by Mahdiyar Mahdiyar 2011-03-14T20:29:54Z 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). http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/40914#40914 Comment by Mahdiyar Mahdiyar 2010-10-04T03:48:27Z 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. http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues/39927#39927 Comment by Mahdiyar Mahdiyar 2010-09-26T07:03:18Z 2010-09-26T07:03:18Z Such a simple answer! Thanks. This is exactly what I expected for the &quot;simpler version&quot; 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!). http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues/39924#39924 Comment by Mahdiyar Mahdiyar 2010-09-26T06:40:01Z 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 &quot;discrete&quot; 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?). http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues Comment by Mahdiyar Mahdiyar 2010-09-26T06:16:55Z 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.