User psyduck - MathOverflow

Dirac Delta function with a complex argument

2013-01-05T01:11:13Z

According to:

Dirac, P. A. M. (1927). "The physical interpretation of the quantum dynamics." Proceedings of the Royal Society of London. Series A, Containing Papers of a Mathematical and Physical Character 113(765), pp.621–641.

For any $y \in \mathbb{C}$, if $f$ is analytic, then $\int_{-\infty}^\infty f(x) \delta(y-x)dx = f(y)$.

And, according to Wolfram Mathematica, the Fourier transform: $$ \int_{-\infty}^\infty \frac{e^{i \lambda x}}{\sqrt{2\pi}} e^{cx} dx = \sqrt{2\pi} \delta(\lambda - i c). $$ which essentially means that for any $\lambda \in \mathbb{C}$ we have $$ \int_{-\infty}^\infty \frac{e^{i \lambda x}}{2\pi}dx = \delta(\lambda). $$ This all seems consistent since $$ e^{cx} = \int_{-\infty}^\infty \frac{e^{-i \lambda x}}{\sqrt{2\pi}} \sqrt{2\pi} \delta(\lambda - i c) d\lambda = e^{-i (i c)x} = e^{cx} . $$ But, I am wondering if anybody knows where I can find some sort of justification for these formal manipulations. I'd like to have a better understanding of what is going on here. And neither Mathematica nor Dirac provide any sort of justification for the above results.

conditions for a phillips class perturbation

2012-07-27T02:32:43Z

If $A$ be the generator of a one parameter semigroup $T_t$ on a Banach space $B$, then we say that an operator $Z$ is a "class P perturbation of $A$" if the following hold: (i) the operator $Z$ is closed, (ii) $\bigcup_{t>0} T_t(B) \subseteq {\rm dom}(Z)$, (iii) $\int_0^1 || Z T_t|| < \infty$. Then according to "Linear operators and their spectra" Davies (2007) Theorem 11.4.5, the operator $A+Z$ is the generator of a semigroup $S_t$.

Now, I have a very practical situation. For me the operator $A = D^2 - D$, where $D$ denotes differentiation. Note that $A$ is the generator of a semigroup $T_t$ on $L^2(R)$. The operator $Z = \eta * ( D^2 - D )$, where $\eta$ is some function. I would like to find conditions on $\eta$ for which the perturbation $B$ is "class P". I would prefer to work in the Hilbert space $L^2(R)$, but if results can be proved for a smaller Banach space e.g. $C_0^\infty(R)$ then I would be happy with that. Also, I do not really care for the most general conditions on $\eta$. I just would the simplest condition to understand (other than $\eta = $ constant).

Finally, please forgive me in advance. I have not had a course in functional analysis. So, I am not as familiar with showing operator bounds as I should be -- especially with differential operators. But, by seeing how to approach the above (hopefully simple) case, I hope to be able to work with more complex operators. Thanks in advance for any advice you can offer.

showing $ \int f_{n+1} dx / \int f_n dx\to 0 $

2012-07-14T02:41:59Z

I have formally derived a solution to a PDE as a power series $$ u = \sum_{n=0}^\infty \epsilon^n u_n . $$ I would like to show that the radius of convergence for is $\mathbb{R}$. I assume that the easiest way to do that is to show $$ \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}| = 0 . $$ The difficulty in showing this is that the $u_n$ are of the form $$ u_n = \int_\mathbb{R} f_n(x) dx $$ So far, my best strategy is to introduce $$ g_{n+1} := \frac{f_{n+1}}{f_n} $$ so that $$ |\frac{u_{n+1}}{u_n}|^2 = \frac{| \int g_{n+1}f_n dx |^2}{|\int f_n dx|^2} \leq \frac{\int | g_{n+1} |^2 dx \cdot \int | f_n |^2 dx}{|\int f_n dx|^2} $$ and $$ \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}|^2 \leq \limsup_{n \to \infty} \int | g_{n+1} |^2 dx \cdot \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} $$ $$ \leq \int \limsup_{n \to \infty} | g_{n+1} |^2 dx \cdot \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} $$ Thus I need to show $$ \limsup_{n \to \infty} | g_{n+1} |^2 = 0 \qquad (A) $$ and $$ \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} < \infty \qquad (B) $$ This seems more complicated to me than I think it should be. Intuitively, it seems to me like I should just need to show (A). Anyway, if anybody has a simpler strategy, I would greatly appreciate it.

* Additional Information not in my original post *

The integrals written above in terms of $x$ in the notation of my paper are $$ \int_\mathbb{R} f_n(\lambda) d\lambda $$ The specific form of $f_n(\lambda)$ is as follows $$ f_n(\lambda) = \left( \sum_{k=0}^n \frac{e^{t \phi_{\lambda-ik\beta}}} {\prod_{j\neq k}^n (\phi_{\lambda-ik\beta}-\phi_{\lambda-ij\beta})} \right) \left( \prod_{k=0}^{n-1} \chi_{\lambda-ik\beta}\right) (\psi_\lambda, h) \psi_{\lambda} $$ where $$ \phi_\lambda =\frac{1}{2} a_0^2 ( -\lambda^2 - i \lambda ) - \int_\mathbb{R} \nu_0(dz) ( e^{z} - 1 - z ) i \lambda + \int_\mathbb{R} \nu_0(dz) ( e^{ i \lambda z} - 1 - i \lambda z ) - c_0 $$ and $$ \chi_\lambda =\frac{1}{2} a_1^2 ( -\lambda^2 - i \lambda ) - \int_\mathbb{R} \nu_1(dz) ( e^{z} - 1 - z ) i \lambda + \int_\mathbb{R} \nu_1(dz) ( e^{ i \lambda z} - 1 - i \lambda z ) - c_1 $$ the $a_i$ and $c_i$ are real positive constants. The $\nu_i$ are Levy measures. $$ \psi_\lambda = e^{i \lambda y} \qquad (\psi_\lambda,h) = \int_\mathbb{R} \psi_\lambda(y) h(y) dy $$ The domain of $f(\lambda)$, $\psi_\lambda$ and $\chi_\lambda$ are $\mathbb{C}$. For simplicity, assume that $h$ is $L^2(\mathbb{R},dy)$

When can a perturbation be treated as a regular perturbation?

2012-07-07T19:38:57Z

I am working with cauchy problem of the form $$ ( - \partial_t + A^\delta) u^\delta = 0 , \qquad u^\delta(0,x) = h(x), $$ where the domain of $u^\delta$ is $[0,\infty) \times \mathbb{R}$. The operator $A^\delta$ can be written as follows: $$ A^\delta = A_0 + \delta e^{b x} A_1 , $$ where $b<0$ and $\delta \geq 0$. I have found a formal solution to the above problem by assuming that $u^\delta$ has a power series expansion in $\delta$. That is $$ u^\delta = \sum_{n=0}^\infty \delta^n u_n .$$ Plugging the expansion into the Cauchy problem yields $$ ( - \partial_t + A_0) u_0 = 0 , \qquad u_0(0,x) = h(x) , $$ and $$ ( - \partial_t + A_1) u_n = -A_1 u_{n-1} , \qquad u_n(0,x) = 0 $$ I can solve these equations explicity for a given $u_n$. But the solution for each $u_n$ is written as an integral (a Fourier Transform). And it is not easy to check if my series solution converges.

So, it would be very nice if I had some sort of theorem that would tell me that my assumption -- that $u^\delta$ can be expanded power series in $\delta$ -- is justified. If I had such a theorem, then I wouldn't need to check convergence of the sum.

So my question is: When can I assume that $u^\delta$ has a power series expansion in $\delta$? Specifically, is this a valid assumption in my problem? And, more generally, are there conditions I can check in the future to know if a perturbation can be treated as regular?

If it helps, both $A_0$ and $A_1$ are infinitesimal generators of Levy processes. i.e. $$ A_i = \mu_i \partial x + \frac{1}{2} \sigma_i^2 \partial_x^2 + \int_{\mathbb{R}} \nu_i(dz) ( e^{z \partial_x} - 1 - z \partial_x )$$ where $\mu_i$ and $\sigma_i$ and constants, $\nu_i(dz)$ is Levy measure and $e^{z \partial_x} f(x) = f(x+z)$ for any $C^\infty$ function.

Given a Levy Exponent find the jump-measure and drift

2011-11-10T13:57:36Z

A Levy subordinator is an finite variation Levy process with non-negative drift and positive jumps. The Levy exponent is given by

$\phi(\lambda) = \gamma \lambda + \int_0^\infty ( 1 - e^{-\lambda s} ) \nu(ds)$

where $\gamma>0$ is the drift of the subordinator and $\nu$ is the jump measure (Levy measure). If the jumps are a compound Poisson process with (net) jump intensity $\alpha$ and jump-size distribution $\mu$ then $\nu = \alpha \mu $ and the levy exponent becomes

$\phi(\lambda) = \gamma \lambda + \alpha (1 - \widehat{\mu}( \lambda ) )$

where

$\widehat{\mu}( \lambda ) = \int_0^\infty e^{- \lambda s} \mu(ds)$

My Questions are as follows:

Given a function $\phi(\lambda)$, how do I know if there is a $\nu$ and a $\gamma$ that generates it?
For a given $\phi$ is the pair that generates it $(\gamma,\nu)$ unique?
Assume the jumps are a compound Poisson process. If you are given $\phi(\lambda)$ can you find $\alpha$ and $\gamma$? Finding $\alpha$ and $\gamma$ would uniquely determine $\widehat{\mu}( \lambda )$ and allow us to reconstruct $\mu(ds)$ from the inverse Laplace transform. Then $\nu(ds) = \alpha \mu(ds)$.
More generally, given $\phi(\lambda)$, can you find $\nu$ and $\gamma$.

The reason for these questions is that I am going to numerically construct $\phi(\lambda)$ from data. Ideally, I would like to then construct $\gamma$ and $\nu$ (or $\alpha$ for a Poisson process) as well. At this point, it isn't clear to me that I actually need $\gamma$ and $\nu$ for my calculations. It may be that $\phi(\lambda)$ is enough (this project is in its nascent stage at the moment). But, even if I don't need $\nu$ and $\gamma$ I am curious to see if I can construct them. And an existance and uniqueness result would definitely strengthen my paper.

....

So, I have a partial answer to the construction of $(\gamma,\nu)$ from $\phi$. Clearly

$\gamma = \lim_{\lambda \to \infty} \phi(\lambda)/\lambda$

Still looking for a construction of $\nu$ at the moment.

Answer by psyduck for Existence of power series expansion for implicitly defined function

2012-06-28T20:51:29Z

I found an acceptable answer to my particular question (not the general one).

The inverse of an invertible analytic function whose derivative is nowhere zero is analytic.
The composition of two analytic functions is analytic.

$F[ g(x) ] = y(x)$

$F[g]$ satisfies 1 in my case. And $y(x)$ is analytic. Therefore

$g(x) = F^{-1}[ y (x)]$ is analytic.

Existence of power series expansion for implicitly defined function

2012-06-28T00:28:53Z

I am trying to solve the following implicit equation for $g(x)$.

$F[ g(x) ] = y(x)$

For simplicity assume that $F$, $g$ and $y$ all map $\mathbb{R} \to \mathbb{R}$. It is known that, for every $x$ there exists a unique (I added "unique") number $g(x)$ such that $F[g(x)] = y(x)$. So the equation is well-posed.

The function $y(x)$ has a power series expansion in $x$ valid for all real $x$. The function $F[ g ]$ has a power series expansion in $g$ valid for all $g$. I would like to argue that $g(x)$ must therefore have a power series expansion in $x$ valid for all real $x$.

This seems like sound logic. But, before I put this in a paper, I would like to be sure that this is correct. As always, thanks in advance for any advice you can offer.

EDIT: I should have phrased things slightly differently. Instead of saying "F is invertible" I should have said, for every $x$ there exists a number $g(x)$ such that $F[g(x)] = y(x)$. This is obviously not the same thing as saying $F$ is invertible. I made this edit above.

----------------------- EDIT -----------------------------------------

I greatly appreciate all of the comments that people have left. Certainly, my understanding of what can go wrong with my initial assumption has been greatly clarified. In an effort to see if my solution to a specific problem is valid, I will write it below. As always, many thanks for your comments. And, thank you for being patient who does not have a rigorous mathematical background.

I would like to find $\sigma^\epsilon$ that solves $v(\sigma^\epsilon) = u^\epsilon$ where $$u^\epsilon = \int d \lambda e^{\phi_0 (\lambda) + \epsilon \phi_1 (\lambda)} h(\lambda,k) \qquad h(\lambda,k) = \frac{-e^{k-i k \lambda }}{\sqrt{2 \pi } \left(i \lambda +\lambda ^2\right)}$$ and $$v(\sigma^\epsilon) = \int d\lambda e^{ \phi(\lambda;\sigma^\epsilon)} h(\lambda,k) , \qquad \phi(\lambda;\sigma^\epsilon) = \frac{1}{2}(\sigma^\epsilon)^2(-\lambda^2 - i \lambda)$$

All functions of $\lambda$ are analytic on the set {Im$(\lambda)<1$}. Integration is over a line parallel to the real axis in the stip of analyticity.

The way that I "solved" this was by expanding both sides in powers of $\epsilon$ ASSUMING that $\sigma^\epsilon$ has a power series expansion $$\sigma^\epsilon = \sigma_0 + \epsilon \sigma_1 + \cdots$$. I then matched terms of like powers of $\epsilon$ to find the cofficients {$\sigma_n$}.

My "solution" is indistinguishable for values of $k$ near $0$ (for basically any size $\epsilon$. But, as $k$ moves away from $x$ the convergence is quite bad. I don't know if this is a problem with my series solution or my numerical intergration scheme. So, if I somehow knew that expanding $\sigma^\epsilon$ in powers of $\epsilon$ were valid, then I would know that it the numerical integration scheme that is causing problems, and not my series solution.

combinatorics problem: $$ \sum_n a_n (\sum_k b_k \epsilon^k )^n $$

2012-06-24T15:53:42Z

I would like to rewrite the following series $$ \sum_{n=0}^\infty \frac{1}{n!}(\Delta^\epsilon)^n a_n , \qquad \Delta^\epsilon=\sum_{k=1}^\infty \epsilon^k b_k $$ As a series in $\epsilon$ $$ \sum_{n=0}^\infty c_n \epsilon^n $$ (i.e. I need to find the c_n).

Obviously, I can do this term-by-term. But the general case seems quite difficult. Any help would be greatly appreciated.

Perturbative solution to an Eigenvalue Problem with a continuous spectrum

2011-07-29T19:19:31Z

I am trying to find an approximate solution to an eigenvalue equation using techniques from perturbation theory. Roughly speaking, the problem is as follows

$L^s \phi_q^s = \lambda_q^s \phi_q^s$

The operator $L^s$ can be written

$L^s = L^0 + s L^1$

where $s>0$ is small and $L^0$ is self-adjoint acting on some Hilbert space. I am trying to find solutions of the form

$\phi_q^s = \phi_q^0 + s \phi_q^1 + \ldots$

$\lambda_q^s = \lambda_q^0 + s \lambda_q^1 + \ldots$

Inserting the above expansions into the eigenvalue equation, and collecting terms of like orders in $s$ yields

$( L^0 - \lambda_q^0 ) \phi_q^0 = 0$ (1)

$( L^0 - \lambda_q^0 ) \phi_q^1 = (\lambda_q^1 - L^1 ) \phi_q^0 = 0$ (2)

When the Hilbert space is $L^2((a,b))$ ($-\infty < a < b < \infty$) the spectrum of $L^0$ is simple and purely discrete. Solving (1) yields a complete set of orthonormal basis functions with the orthogonality relation $<\phi_n^0,\phi^0_k>=\delta_{n,k}$. By the Fredholm alternative, in order for a solution $\phi_n^1$ of (2) to exist, the RHS of (2) must satisfy

$< \phi_n^0, (\lambda_n^1 - L^1 ) \phi_n^0 > = 0$

Thus $\lambda_n^1$ is given by

$\lambda_n^1 = < \phi_n^0, L^1 \phi_n^0 >$

And $\phi_n^1$ is given by applying the resolvent operator to the RHS of (2), which yields

$\phi_n^1 = \sum_{k \neq n} \frac{< \phi_k^0, L^1 \phi_n^0 >}{\lambda_n - \lambda_k} \phi_k^0$

When the Hilbert space is $L^2(-\infty,\infty)$ the spectrum of $L^0$ is purely absolutely continuous. Solving (1) still yeilds a complete set of orthonormal basis functions with the orthogonality relation $<\phi_p^0,\phi^0_q>=\delta(p-q)$. However, solving (2) is now more complicated. When the operator $L^1$ is such that

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + g^1(p,q)$ (3)

Trying a solution analgous to the discrete case works works

$\lambda_q^1 = f^1(q)$

$\psi_q^1 = \int \frac{g^1(p,q)}{\lambda_q - \lambda_p} \phi_p^0 dp$ (the integral converges). (4)

However, I am looking at various $L^1$. And, for certain $L^1$ I do not have (3). Rather, I have

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + h(q) \delta'(p-q) + g^1(p,q)$

(yes, that's a derivative of a delta function). Frankly, at this point, I am totally stuck. I've tried a solution of the form (4) with

$g^1(p,q) \to h(q) \delta'(p-q) + g^1(p,q)$.

But that solution blows up. My sense is that I should be looking for some sort of condition on $\lambda_q^1$ which would guarantee that a solution $\phi_q^1$ of (2) exists. But, I know of no such condition.

If it helps, you can think of $L^0$ as $-d^2/dx^2$ so that the eigenfunctions are $\phi_q^0(x)=e^{iqx}/\sqrt{2\pi}$. And $L_1 \phi_q^0 = x \phi_k^0(x)$. If you're wondering, the derivative of the delta function comes about as follows

$< \phi_q^0, x \phi_p^0 > = (1/2 \pi) \int x e^{i(p-q)x} dx = (1/2 \pi i) (d/dp) \int e^{i(p-q)x} dx = (1/2 \pi i)\delta'(p-q)$.

Any guidance on solving this problem would be greatly greatly appreciated.

showing convergence of a function recursion relation

2012-05-26T22:31:12Z

I have obtained (formally) a perturbative solution $$ H(y) = \sum_{n=0}^\infty \delta^n H_n(y) $$ to the following integro-differential equation ($\delta$ is a small constant, $\nu$ is a L\'evy measure and $F$ is a function of $y$) $$ 0 = \delta F ( d_y H - H + 2 ) + \int \nu(dz) \left( \frac{e^{z d_y } - 1}{d_y} - ( e^z - 1 ) \right) H + \int \nu(dz) z^2 $$ Namely $$ H_1 = A_0^{-1} F , \qquad H_n = -A_0^{-1} F A_1 H_{n-1} = ( - A_0^{-1} F A_1 )^{n-1} H_1 . $$ where the operators $A_0$ and $A_1$ are given by $$ A_0 = \int \nu(dz) \left( \frac{e^{z d_y } - 1}{d_y} - ( e^z - 1 ) \right), \qquad A_1 = d_y-1 . $$ The inverse operator (resolvent) $A_0^{-1}$ can be defined through $$ A_0^{-1} F = \int d \lambda \frac{1}{\phi_\lambda} (\psi_\lambda,F ) \psi_\lambda \qquad (F,G) = \int \overline{F}(y) G(y) dy $$ where $$ A_0 \psi_\lambda = \phi_\lambda \psi_\lambda, \qquad \phi_\lambda =\int \nu(dz) \left( \frac{e^{ i \lambda z} - 1}{i\lambda} - ( e^z - 1 ) \right) , \qquad \psi_\lambda =\frac{1}{\sqrt{2\pi}}e^{i\lambda y} $$ Explicitly $$ H_n = (-1)^{n-1} \underbrace{ \int \cdots \int }_{n} d \lambda_n \frac{\psi_{\lambda_n}}{\phi_{\lambda_n}} (\psi_{\lambda_1},F) \prod_{k=1}^{n-1} d \lambda_k \frac{\chi_{\lambda_k}}{\phi_{\lambda_k}} ( \psi_{\lambda_{k+1}},F \psi_{\lambda_k} ) \qquad $$ where $$ A_1 \psi_\lambda = \chi_\lambda \psi_\lambda, \qquad \chi_\lambda = i \lambda - 1 $$ As stated above, my solution is purely formal. In order to show convergence of the infinite sum I need to show that either (i) $$ \lim_{n \to \infty} \frac{|H_{n+1}|}{|H_n|} = 0 $$ or (ii) that the $H_n(y)$ has alternating signs, $\delta^n H_n$ is strictly decreasing, and $\lim_{n\to\infty}H_n=0$. Obviously, showing either of these will depend on $F$ and possibly on $\nu$. But at the moment, I really do not even know how to begin showing either thing. Any advice or references would be appreciated.

Analytic, perturbative or numerical solution to Integro-differential Equation (OIDE) ?

2012-05-15T16:29:06Z

I am trying to find an analytic, or at least a numerical solution to the following ordinary integro-differential equation (OIDE) $$ 0 =\frac{\sigma^2(y)}{2} ( G''(y) - G'(y) + 2 ) + \eta(y) \int [ G(y+z) - 1 - ( e^z - 1 ) G'(y) ] \nu(dz) + \eta(y)\int z^2 \nu(dz) $$ where the integral is over the support of the measure $\nu$ (which is a Levy measure), $G:\mathbb{R} \to \mathbb{R}$, and $\sigma>0$ and $\eta>0$. When $\eta = A \sigma^2 /2$ then the answer can easily be seen to be $$ G = Q_A * y , \qquad Q_A = \frac{2 + A \int z^2 \nu(dz)}{1 + A \int [e^z-1-z] \nu(dz)} $$ Taking the limit as $A \to 0$ gives the solution when $\eta=0$ (obviously) and taking the limit as $A \to \infty$ gives the answer for $\sigma=0$.

I am interested in finding an analytic, perturbative, or numerical solution to the above OIDE in the case where $eta$ is not some multiple of $\sigma^2$ and neither are zero. Using the fact that $G(z + y) = G(y) + z G'(y) + \cdots$, I can turn the above OIDE into an infinite dimensional ODE $$ 0 = \frac{\sigma^2(y)}{2 \eta(y)} ( G'' - G' + 2 ) - G' \int ( e^z - 1 - z ) \nu(dz) + \sum_{n=2}^\infty \frac{1}{n!} G^{(n)} \int z^n \nu(dz) + \int z^2 \nu(dz) . $$ I do not know how to solve this either. By the way, you can assume the above integrals with respect to $\nu$ are known so that $$ 0 =\frac{\sigma^2(y)}{2 \eta(y)} ( G'' - G' + 2 ) - G' I_0 + \sum_{n=2}^\infty \frac{1}{n!} G^{(n)} I_n + I_2 . $$ $$ I_0 =\int ( e^z - 1 - z ) \nu(dz) , $$ $$ I_n =\int z^n \nu(dz) , \qquad n \geq 1 . $$ This looks less intimidating than the OIDE since the G's are outside of the integrals. But now I have infinitely many derivatives of G to deal with. I had thought to trying to find a power series expansion $$ G(y) = \sum_n Q_n y^n $$ But, this would involves some doube-infinite sums. And I don not know how I would find the coefficients $Q_n$. But, perhaps this would be easy to deal with numerically.

Finally, I had considered trying to find a perturbative solution for small $\sigma$. That is, assume $\delta << 1$ and set $\sigma^2 \to \delta \sigma^2$. Then look for solutions of the form $$ G = G_0 + \delta G_1 + \delta^2 G_2 \cdots $$ I could also look for a small $\eta$ solution. I haven't tried that yet.

Finally, I had considered just specifying a $G$ and a $\nu$ and $\eta$ and solving for $\sigma^2$. But, it is incredibly difficult to find a combination that produces a positive $\sigma^2$.

If anybody can point me to solve any of the above equations either analytically, numerically, or with perturbation theory I would be extremely appreciative. I have been spinning my wheels for a few weeks now without much to show for it.

Also, if there is a special choice of $\sigma^2(y)/\eta(y)$ or a special choice of $\nu$ that would make solving the above OIDE or infinite dimensional ODE solvable, that would be helpful. The only requirement is that $\sigma>0$ and $\eta>0$.

I think, perhaps the best option at this point would be to try to find a numerical or power series solution to the infinite-dimensional ODE. But, again, I am quite lost, so any suggestions are helpful.

Brownian particle with jump boundary condition

2011-10-23T20:49:00Z

I would like to find a function $f(s)$, which solves the following equation:

$ \int_0^t \int_0^L f(s,x) p(t-s,x,y) dy ds = 1 $

The function $p(\tau,x,y)$ is

$p(\tau,x,y) = \sum_n e^{-\lambda_n \tau} \phi_n(x) \phi_n(y)$

where

$\phi_n(x) = \sqrt{\frac{2}{L}} \sin \left( n \pi x / L \right)$

and

$\lambda_n = \frac{n^2\pi^2}{L^2}$.

i.e. $\psi_n$ and $\lambda_n$ are the eigenfunctions and eigenvalues corresponding to $(-\tfrac{1}{2}\partial^2_{xx})$. Physically, the above equations correspond to the following situation. At time $t=0$ start a Brownian particle at $x \in (0,L)$. Whenever the BM touches a boundary (either $0$ or $L$) immediately send the particle back to $x$, where it begins a new Brownian path. The function $f(s,x)$ represents the probability that a particle found in infinitesimal element $dy$ at time $t$ was started at $x$ at time $s$. The function $p(t,x,y)$ is the transition density of a Brownian particle with a killing boundary condition at $0$ and $L$.

There seems to be a good deal of literature that analyzes the spectrum of BM with a jump boundary. But, as of yet, I have found no papers that specifically say what the transition density of such a process would be. And, that is my interest (i.e. find the transition density of a diffusion in a bounded domain with a jump boundary condition). Any help in solving the top equation or any suggestions for papers to look at would be greatly appreciated.

analytic continuation of a Laplace transform from a countably infinite set of points?

2012-03-16T18:42:07Z

Let $f(\lambda)=\int_0^\infty e^{-\lambda s} F(ds)$, where $F$ is the distribution of a positive random variable. Suppose I know the value of $f(n)$ for $n=0,1,2,\cdots$. Is this enough to uniquely determine $f(\lambda)$ in its domain of convergence (DOC)?

Since the Laplace transform is analytic in its DOC, knowledge of $f(\lambda$) on a line segment uniquely determines $f(\lambda)$ on its domain. But, knowledge of the value of an analytic function at a countably infinite number of points does not, in general, determine the function throughout its domain. I was hoping that, analyticity + the fact that $f(\lambda)$ is a Laplace transform would be enough to analytically continue $f(\lambda)$ from a countably infinite set of points.

It would seem like the fact that $f(\lambda)$ is positive, decreasing and convex might be useful.

question about mixed spectrum of a linear operator $\mathcal{L}$

2011-02-14T22:51:58Z

Suppose $\mathcal{L}$ is a bounded linear operator and I have the solution to Eigenvalue problem

$\mathcal{L} \phi + \lambda \phi = 0$

wish to solve the following PDE

$\left(-\partial_t + \mathcal{L}\right)u = 0$.

If the spectrum of $\mathcal{L}$ is continuous or discrete, then a general solution to the PDE is

$\int C_q e^{- \lambda_q t} \phi_q dq$

$\sum_q C_q e^{- \lambda_q t} \phi_q$,

where the $C$'s are constants.

But, what if the spectrum of $\mathcal{L}$ is mixed and has a continuous part, a discrete part, and a singular part? Is there a general way to write the solution to the above PDE if I do not know the spectrum of $\mathcal{L}$?

This has come up in my research because I need to work with the $e^{- \lambda_q t}$ but I do not know what the spectrum of $\mathcal{L}$?

Comment by psyduck

2013-01-05T16:00:27Z

Concerning Mathematica, try: FourierTransform[Exp[cx],{x,w}]

Comment by psyduck

2012-07-15T01:04:42Z

Right. Yes, I should check that there is an integrable function that bounds the $|g_n|^2$

Comment by psyduck

2012-06-29T01:45:09Z

I think what often happens when I post on this board is that people give me exactly the answer that I need, but it takes me a long time to realize it. In any case, I do learn a lot by posting my research question here. And there is no doubt that my papers are more rigorous for the help the MathOverflow community provides.

Comment by psyduck

2012-06-28T01:45:57Z

Thanks Emilio (and Robert). This is quite helpful. Might either of you have a reference for Robert's statement? Citing "MathOverflow" probably won't be acceptable to whichever journal I eventually submit to.

Comment by psyduck

2012-06-28T01:19:10Z

Ah...I see. That is unfortunate. Any suggestions on how I might find conditions under which g(x) DOES have a power series expansion? I phrased my question rather generally. But, in fact, my question refers to a very specific problem. So, I still have hope that my logic, which is generally incorrect, may still be correct for my particular problem.

Comment by psyduck

2012-06-28T00:09:58Z

I study financial mathematics. The paper is about deriving the implied volatility smile for exponential Levy models...not sure if that makes any sense to somebody that doesn't study financial math.

Comment by psyduck

2012-06-25T02:20:52Z

Oh...I just noticed the comment by Pietro Majer. Thank you. That is helpful.

Comment by psyduck

2012-06-25T00:03:54Z

Davide, This is incredibly helpful to me! If it is okay with you, I would like to add an acknowledgement of your help in my paper. I can also send you a draft the paper before I submit it if you like.

Comment by psyduck

2012-05-15T20:56:01Z

It seems that choosing $\nu(dz) = \mathbb{I}_{(-\infty,0]}e^{az} dz$ enables turning the OIDE into a finite order ODE. Once the transormation is complete, it is fairly easy to look up solutions in ODE literature.

Comment by psyduck

2012-04-05T03:09:35Z

Thanks Mike. To be honest, I have really looked into this question much since I posted it a few months ago, since this question isn't really a major area of research for me. But, I did find the problem interesting enough to look for an answer. Now that I know where to look, I think I will revisit this problem. Thanks.

Comment by psyduck

2012-03-16T19:31:41Z

Perfect! I can't tell you how much this helps my paper.

Comment by psyduck

2011-07-29T21:32:09Z

Yes, I believe when I was learning about scattering in quantum mechanics we found the green's function by adding a small imaginary component to the resolvant and took a limit $g^{\pm}(E) = \lim_{\epsilon \to 0}\frac{1}{H - E \pm i \epsilon}$ I did try that in the above scenario and still ended up with a solution that blew up. But, to be honest, I didn't do things as carefully as I should have. I'll try this again.

Comment by psyduck

2011-07-29T20:45:30Z

By "blow up" I mean that if I try to evaluate an integral of the form $\int \frac{e^{ipx} h(p) }{p^2 - q^2} \delta'(p-q) dp = - \int \delta(p-q) \frac{d}{dp} ( \frac{e^{ipx} h(p) }{p^2 - q^2} ) dp$ the result is infinity.

Comment by psyduck

2011-07-29T20:21:56Z

sorry. $\psi_p^0$ and $\psi_0^p$ were supposed to read $\phi_p^0$ above.

Comment by psyduck

2011-07-29T20:19:34Z

Until I read you post, I had not heard of the "trace class" before. But, after a google search, if my understanding is correct, unless I can show $\int < |(L^0 + i)^{-1} L^1 | \psi_p^0, \psi^p_0 > dp < \infty$ Then my search for a solution is hopeless.