most recent 30 from http://mathoverflow.net 2013-05-25T10:46:34Z http://mathoverflow.net/feeds/user/19243 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89179/is-the-feynman-kac-formula-valid-for-a-time-dependent-potential alexsuse 2012-02-22T11:50:42Z 2012-02-22T11:50:42Z

<p>So I'm looking at a diffusion process with killing with a state- and time-dependent killing rate. This is described in Oksendal's Stochastic differential equations pages 143-145 "The Feynman-Kac Formula. Killing". Basically, you have a generator $$ L f = -\sum_i \frac{\partial }{\partial x_i} A_i(x,t) f(x) + \frac{1}{2} \sum_{i,j} \frac{\partial^2}{\partial x_i \partial x_j} B_{i,j}(x,t) f(x) - c(x,t)f(x), $$ and this corresponds to a process with drift $A$, diffusion $B$ and a killing rate given by $c(x,t)$.</p> <p>However, no one mentions if the killing rate can be time-dependent. The demonstration uses the stochastic process $$ Z_t = exp(-\int_0^t c(X_s)) ds, $$ with $dZ_t$ given by $$ dZ_t = -Z_t c(X_t) dt. $$ However, if $c$ is a function of time too, there would be an additional term in the differential, right? Wikipedia's Feynman-Kac formula page (http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula) states the problem with a time-dependent potential $V(x,t)$ but then goes on to drop this dependence throughout the page. Can i go on to use the Feynman-Kac formula if the potential is time-dependent? Do I have to include some additional terms somewhere? My hunch is yes, but I'm not sure how to derive the correct formulation!</p> <p>Thanks to everyone! Man, stochastic calculus can sure put knots in your brain!</p>

http://mathoverflow.net/questions/87796/reparametrizing-characteristic-curves-for-pdes alexsuse 2012-02-07T14:06:38Z 2012-02-08T10:10:11Z

<p>I'm looking for solutions for a PDE that looks like this $$ \nabla u(\vec x) \cdot f(\vec x) = k. $$ For some clarification, $u$ is a log-probability. And this arises from a Fokker-Plank-like equation, and I'd like to find the marginals $u(x_1)$ for example. The method of characteristics tells me that if i take characteristic curves parametrized by $$ \frac{d \vec x}{dt} = f(\vec x), $$ the variation of $u(\vec x)$ along these curves will be $$ \frac{du}{dt}= k. $$ But in my particular case, $f$ determines a dynamic system $\dot{\vec x} = f$ which converges to a fixed point only in infinite time. Since I'd like to be able to have $u(x)$ instead of $u(t;x_0)$ I've come to the idea of parametrizing the characteristic curves with constant velocity, that is, take $$ \frac{d \vec x}{ds} = c \frac{f(\vec x)}{\|f(\vec x)\|}, $$ which will result in a new parametrization s which goes from 0 to the length of the characteristic curve from the boundary condition to the equilibrium point. This will lead to an equation for $u(s)$ $$ \frac{du}{ds}= \frac{ck}{\|f(\vec x)\|}. $$ Clearly this only corresponds to the original equation for a correct choice of the constant $c$, yet it seems to me that the PDE is satisfied for any value of $c$. I'm a little confused. Am I doing something terribly stupid along the way, or am I just missing the place where I should constrain the constant $c$.</p>

http://mathoverflow.net/questions/80877/matrix-derivative-with-respect-to-the-pseudo-inverse alexsuse 2011-11-14T09:31:04Z 2011-11-14T13:16:22Z

<p>Hi everyone, I'm trying to find a expression for the matrix derivative with respect to the pseudo-inverse of a matrix. So, i have some function $f(A)$ of a matrix $A$, which is singular. If it weren't I could use that $$ \frac{df(A)}{dA^{-1}} = -A^{-1}\frac{df(A)}{dA}A^{-1}, $$ but I can't right? So does anyone know where I could find a pseudo-inverse version of this? So basically I want an expression for $\frac{df(A)}{dA^+}$, and yes I reckon it won't be as cleas and simple as the one above. Also, does anyone know where I could find pseudo-inverse generalizations of all those classic matrix inversion lemmas?</p> <p>Thanks in advance for any comments!</p>

http://mathoverflow.net/questions/87796/reparametrizing-characteristic-curves-for-pdes/87828#87828 alexsuse 2012-02-08T10:14:15Z 2012-02-08T10:14:15Z

Thanks for pointing out the typo! But why doesn't the second set of equations solve the pde? If i have $u(s)$ given by the solution to the last equation, and just take $u(t) = u(t(s))$ where $t(s)$ is the change of variables, then $u(t(s))$ will solve the first equation as far as I could think it through.

http://mathoverflow.net/questions/80877/matrix-derivative-with-respect-to-the-pseudo-inverse/80891#80891 alexsuse 2012-02-07T13:55:20Z 2012-02-07T13:55:20Z

Cool, thanks for the links. This seems to answer the second question pretty well!