User aaron hoffman - MathOverflow

Answer by Aaron Hoffman for Eigenfunctions of fourth-order differential operator

2013-04-17T14:55:23Z

Choose some $\lambda \ne \lambda_k$ for all $k$. Consider the resolvent $R_\lambda = (\lambda-\partial_x^4)^{-1}$. This is a compact operator on the Hilbert space $H = { \psi \in L^2([0,1]) \; | \; \psi''(0) = \psi''(1) = \psi'''(0) = \psi'''(1) = 0}$. Thus $H$ has a basis of eigenvectors of $R_\lambda$. But $R_\lambda$ and $\partial_x^4$ share eigenvectors, thus $H$ has a basis of eigenvectors of $\partial_x^4$ as desired. The key bit is using compactness to show that a maximizing sequence for the Rayleigh quotient converges to an eigenvector. The argument can be found in Lax's Functional Analysis Ch 28 Thm 3.

Answer by Aaron Hoffman for Spectrum of transition matrix for symmetric random walk

2012-11-30T20:22:52Z

You certainly have enough answers by now, but another name for these are the Neumann eigenvalues for the discrete heat equation. You can compute them as you would compute the Neumann eigenvalues for the heat equation on an interval. Consider the even reflection of your random walk about 0. This generates a symmetric random walk on -n,...n with periodic boundary conditions. The transition matrix is circulant and the eigenvalues can be computed directly. Equivalently, one can diagonalize via discrete Fourier transform to obtain the eigenvalues.

Answer by Aaron Hoffman for Application of inverse function theorem to get short time existence

2012-06-21T16:18:17Z

Perhaps something can be gleaned from the analogous proof of short-time existence for scalar ODEs:

Consider the initial value problem $x' = f(x)$ with $x(0) = x_0$ and define $F(x)(t) = x(t) - x_0 - \int_0^t f(x(s))ds$ so that zeros of F correspond to solutions of the IVP. Regard the function F as acting on some space of functions whose elements obey $x(0) = x_0$. The Derivative of F is $(F'(x)y)(t) = y(t) - \int_0^t f'(x(s))y(s)ds$. To show that F' is an isomorphism, one wants to show that the norm of $y \mapsto \int_0^t f'(x(s))y(s)ds$ is less than one. If you are working on a space of functions from $[-T,T] \to R$, then a cheap estimate is given by T times the maximum value that the absolute value of $f'$ takes. This can be made smaller than one by choosing T sufficiently small. Of course f' has to have a maximum value in the first place. This is dealt with through a short song-and-dance in which one works in an open subset of the function space for which the functions x take a restricted set of values so that on these values f' does take a maximum absolute value.

I suspect that something similar is going on here.

More generally, given an operator A that can be regarded as acting on either a Banach space X or some other Banach space Y, the spectrum of A in general will depend upon the Banach space. That the note produced by a vibrating harp string depends on the length of the string furnishes an example of this phenomenon. (A is the second derivative, X is the set of functions from $[0,L_x]$ to R with Dirichlet boundary conditions and Y is the set of functions from $[0,L_y]$ to R with Dirichlet boundary conditions.)

In your situation again the different function spaces contain points which themselves are functions defined on shorter or longer time intervals.

Answer by Aaron Hoffman for Presenting work in progress

2012-01-12T02:00:52Z

(1) If I've thought carefully enough about a project that I have something to teach my audience by presenting what I know, I'll talk about it (qualifying the parts that I'm unsure of). I prefer to give talks about what I'm most excited by at the moment; this isn't always what is most complete.

(2) The venue doesn't matter, but if there are fancy-pants mathematicians in the audience I will be more nervous about presenting partial results

(3) If I have something that is complete, that I'm excited about, and that will be new to the audience I'm talking to I'll usually choose that over something that is in progress.

Maybe I should be more worried about being scooped, but (a) I'm not presenting a proof of the Riemann Hypothesis (if I were it would be wrong) (b) Its hard to scoop someone based on a talk that they gave (c) Publish-or-perish pressure aside, most of the mathematicians that I've met are people of integrity (d) if someone uses my ideas to do something I might have done but haven't yet done this is a credit to my work (and moreso if they cite me).

decay for spatially discrete parabolic equations with non-constant non-self-adjoint right hand side

2011-08-22T20:39:17Z

Consider the following uniformly parabolic lattice differential equation

$ \begin{array}{ccc} \dot{u}_{n,m} & = & \alpha_{n,m}(u_{n+1,m} - u_{n,m}) + \beta_{n,m}(u_{n-1,m}-u_{n,m}) \\ & & \quad + \gamma_{n,m}(u_{n,m+1}-u_{n,m}) + \delta_{n,m}(u_{n,m-1}-u_{n,m}). \end{array} \quad (1) $

Here the coefficients $\alpha$, $\beta$, $\gamma$ and $\delta$ are bounded uniformly in $n$ and $m$ both above and below by positive constants, thus (1) enjoys a maximum principle.

Equation (1) is related to diffusion on a bidirectional graph as well as a semi-discretization of a parabolic PDE. In my own work it arose in the study of stability of planar fronts in lattice differential equations. I don't understand the behavior of solutions to (1) very well. I am curious to know whether or not their qualitative asymptotic behavior is well-known and/or somewhat trivial or if there is some deepish mathematics lurking here. I suspect that there is deep mathematics in the quantitative asymptotics because heat kernels for graphs have been a topic of study for a long time and to the best of my knowledge there are no results on decay estimates for heat kernels in the non-self-adjoint case.

In the case that the coefficients $\alpha$, $\beta$, $\gamma$ and $\delta$ are independent of space ($n$ and $m$), solutions to (1) can be obtained via Fourier transform. One can show for example that $u \to 0$ (e.g. in $\ell^2$ for initial data in $\ell^2$).

In the case that the right hand side is self-adjoint ( $\alpha_{n,m} = \beta_{n,m}$ and $\gamma_{n,m} = \delta_{n,m}$ ) an energy estimate together with a Nash inequality gives algebraic decay in e.g. $\ell^2$.

In the case that the right hand side is neither self-adjoint nor constant-coefficient, $\ell^1$ norm need not be preserved and $\ell^2$ norm need not be non-increasing. Arguments involving the Nash inequality and/or concentration compactness seem not to have traction.

Question: What can be said about the asymptotic behavior of solutions to (1). Do solutions to (1) with initial data in $\ell^\infty$ necessarily approach a constant? Do solutions with initial data in $\ell^1$ necessarily approach zero? Can anything be said about the rate of decay?

Answer by Aaron Hoffman for Question on PDE

2011-04-25T11:06:10Z

This answer makes some assumptions about what the OP is asking. In particular I am using Scott's interpretation that the boundaries of interest correspond to level sets of the fundamental solution.

I suppose you have to specify which level set you are talking about. Since the heat equation has infinite propagation speed, you can make the $t_i$ as close to zero as you like by looking at level sets $a_i = \epsilon$ for $\epsilon$ sufficiently small. If you fix $\epsilon$ (or $\epsilon_i$) then you are looking at a collection of three circles in the plane with radius $r_i(t) = \left(4kt \log(1/\epsilon) + 2kt \log(4 \pi k t)\right)^\frac{1}{2}$ (assuming the standard heat equation with diffusion constant k) and you are asking when these circles intersect. The circle about $A_i$ will intersect the circle about $A_j$ when $r_i + r_j \ge |A_i - A_j|$.

Regularity of reflection coefficients (or more generally the scattering transform)

2010-05-06T22:19:08Z

Consider the Schrodinger operator $L(q) = -\partial_x^2 + q(x)$ where the potential $q$ is a real-valued function of a real variable which decays sufficiently rapidly at $\pm \infty$.

We define the scattering data in the usual way, as follows:

The essential spectrum of $L(q)$ is the positive real axis $[0,\infty)$ and it has multiplicity two. The Jost functions $f_\pm(\cdot,k)$ corresponding to $L(q)$ solve $L(q)f_\pm = k^2f_\pm$ with $f_\pm(x,k) \sim e^{ikx}$ as $x \to \pm \infty$.

The reflection coefficients $R_\pm(k)$ are defined so that $f_\pm = \bar{f_\mp} + R_\pm(k)f_\mp$ where here overbar denotes complex conjugate. The intuition is that $R$ measures the amount of energy which is reflected back to spatial $\infty$ when a wave with spatial frequency $k$ that originates at spatial $\infty$ interacts with the potential $q$.

The scattering transform (the map from $q$ to the scattering data, of which $R_+$ is a part) and its inverse are important in the theory of integrable PDE.

My question is the following: What is the regularity of the map $q \mapsto R_+$? Is it continuously differentiable?

To answer this question, we first must specify the spaces that $q$ and $R_+$ live in. I don't really care so long as they are reasonable spaces, for example take $q$ in weighted $H^1$ where the weight enforces a rapid decay at $\pm \infty$.

Answer by Aaron Hoffman for Fundamental Examples

2009-11-11T12:56:05Z

The KdV equation in integrable systems. It was through a numerical study of KdV that the word soliton was coined. This numerical study lead to much analytical work, including the development of Lax Pairs. (Answer by Aaron Hoffman)

Answer by Aaron Hoffman for Stability analysis of a system of 2 second order nonlinear differential equations

2009-12-06T12:46:28Z

This is an answer to Charles' restatement of the question.

Recall that equation F(x,x',x'') = 0 (e.g. x'' + sin x = 0) can be written as a system

X' = f(X), where X = (x,x')^T (e.g. f(x',x) = (-sin x, x')^T) and that system can be linearized about an equilibrium E = (x_*,x'_*)^T to obtain a linear equation X' = AX where A is the 2 x 2 matrix given by the derivative of f at E.

So too a larger system F(x,x',x'',y,y',y'') = 0 can be written as a system

X' = f(X) where X = (x,x',y,y')

Given an equilibrium E = (x_*,x'_*,y_*,y'_*), the linearization of X' = f(X) about E is again X' = AX where X = (x',x,y',y) and A is the derivative of f evaluated at E. A is a 4 by 4 matrix. If all eigenvalues of A have negative real part, the system is stable. If one eigenvalue has positive real part, the system is unstable. If there are no eigenvalues with positive real part, and there are eigenvalues which lie on the imaginary axis, then the equilibrium is "spectrally stable," and further analysis is required to determine the nonlinear stability.

With regard to the energy of the system, you are looking for a function of the form V(x,x',y,y') whose time-derivative along solutions is constant. A good place to start is with the guess 1/2((x')^2 + (y')^2) + F(x,y). You should be able to figure out what F needs to be in this particular example.

For future reference, this forum is (I believe) intended primarily for questions from students and practitioners who are a little further along in their study. I decided to answer the question because I can imagine being very frustrated at not having some information that it would take an expert five minutes to explain and the question didn't strike me as the kind which would encourage others to attempt to turn this forum into a homework help site. If the moderators disagree, I apologize.

Answer by Aaron Hoffman for What is an integrable system

2009-12-04T13:31:07Z

This is soft -- but I think of an integrable system as one whose dynamics are dominated by algebra. For finite dimensional integrable systems, the symmetries (related to conserved quantities by Noether's theorem) force the trajectories to live on half-dimensional tori. For infinite dimensional integrable systems, where the flow on the scattering data is isospectral the symmetries force solutions to be n-soliton solutions plus dispersive modes.

There is a blog post of Terry Tao's (apologies for not having the link) which talks about how algebra is the right tool to understand structure while analysis is the right tool to understand randomness. The claim is that one mark of an good problem is the presence of an interesting relationship between structure and randomness and hence the requirement that both algebra and analysis be used -- to some degree -- in order to get a good answer to the problem. The soliton resolution conjecture is by this standard a good problem because the asymptotic n-soliton solutions are fundamentally algebraic while the dispersive modes are fundamentally analytic objects.

I agree with Dmitri that there isn't a dichotomy. The symmetries can have a large or small role in the dynamics as can the ergodicity.

Answer by Aaron Hoffman for CAS for finding closed form solutions to PDEs and SDEs?

2009-10-30T12:08:05Z

Most ODEs, nevermind PDEs and SDEs don't have what one would usually call "closed form" solutions.

Are you interested in a special class of equations (e.g. linear, constant coefficient, on flat manifolds without boundary) for which closed form solutions are likely to exist?

Comment by Aaron Hoffman

2013-04-18T17:53:34Z

The standard Duhamel integral representation is \[ F(x) = G_F(x,0)F(0) + \int_0^x G_F(x,x')[S_{13}[F(x') + S_2(F(x')]dx' \] Of course this is not a closed form as the unknown F appears on the right hand side. My claim (not at all carefully checked) is that in the generality that you've stated the question the terms $f_{13}$ and $g_{13}$ don't help you.

Comment by Aaron Hoffman

2013-04-18T15:40:04Z

Rewriting in vector notation your equation is $F' = A(x)F + S_{13}(F) + S_2(F)$. You want to update \[F(x+\delta) \approx F(x) + \int_x^{x+\delta} G_F(x,x')\left[S_{13}(F(x')) + S_2(F(x'))\right]dx'\] and iterate. Its tempting to write the first two terms as $F_{13}(x+\delta)$ but this of course depends on the initial condition $F_{13}(x)$. The $S_2$ term kicks the $F_{13}$ term away from a true solution. Upon iteration error will accrue.

Comment by Aaron Hoffman

2013-04-18T14:42:10Z

(1) You are absolutely right about the space. $H$ makes no sense. Instead of $L^2$ I should have $H^4$. (2) I was confused about where you were stuck. The last two sentences are probably not relevant. (3) Perhaps I'm being quite sloppy here, but: given $R_\lambda f = g$ we have $\|g\|_{L^2} \le C\|f\|_{L^2}$ and $\|g''''\|_{L^2} \le \lambda\|g\|_{L^2} + \|f\|_{L^2}$ which implies \|g\|_{H^4} \le C'\|f\|_2$. Since bounded sets in $H^4([0,1])$ are compact in $L^2([0,1])$, the resolvent is compact.

Comment by Aaron Hoffman

2013-04-11T14:13:16Z

Yes and No. No: There is no point equidistant to every point on an ellipsoid so there is no point whose value will be given by the mean of the boundary values. Yes: If you weight the boundary values properly you can recover any interior value you like.

Comment by Aaron Hoffman

2013-04-10T12:15:29Z

Why not look at $G: \mathbb{R}^3 \to \mathbb{R}$ given by $G(x,y,z) = F(x,y)-z$ and seek $\theta(x,z)$ so that $G(x,\theta(x,z),z) = 0$ describes all roots of $G$ in $\mathbb{R}^3$? You now at least are only worrying about a singly global IFT.

Comment by Aaron Hoffman

2013-03-30T21:47:12Z

The keyword singular value decomposition (SVD) might be of interest to the OP

Comment by Aaron Hoffman

2013-03-26T20:21:09Z

To be concrete let's take $\partial_x^2$ defined on $H^2(\mathbb{R}) \subset L^2(\mathbb{R})$. What do you mean by eigenvectors? The solutions to the eigenvector equation $u_{xx} = \lambda u$ don't live in $L^2$. If we think of an orthonormal system of eigenvectors as forming the columns of an orthogonal matrix which diagonalizes the operator, perhaps the appropriate generalization is to seek a unitary transformation (Fourier Transform) which conjugates the operator with multiplication (by $-k^2$).

Comment by Aaron Hoffman

2013-03-22T21:14:32Z

How well does $\left(\frac{1}{\mathrm{vol}(S_p)}\int_{S_p}(v \cdot w)^m dw\right)^\frac{1}{m}$ approximate $\mu^m(f)$? here $S_p = \{w \in \R^n \; | \; \sum_k w_k^p = 1\}$ is the unit $\ell^p$ sphere in $\R^n$ and the functional $f$ is given by $f(e) = v \cdot e$.

Comment by Aaron Hoffman

2013-03-22T01:42:11Z

In this vein, convolution against a function which vanishes at zero or more generally integration against a kernel $K(x,y)$ for which $K(x,x)$ vanishes identically also work.

Comment by Aaron Hoffman

2013-03-07T19:49:39Z

Hi Bryan, Your question is likely to be closed for being too localized. Please see the FAQ, noting that this is not a math help forum but rather a site for research level math questions. math.stackexchange.com might better serve your needs, but there too your question should be carefully edited.

Comment by Aaron Hoffman

2013-02-08T16:04:39Z

Since solution trajectories are level sets of the Hamiltonian $H(x,\dot{x}) = \frac{1}{2}\dot{x}^2 + \frac{1}{2}x^2 + \frac{1}{3}x^3$, are you asking about the level set of $H$ running through (.5,0)?

Comment by Aaron Hoffman

2013-02-07T18:52:06Z

If my memory serves me, The Nash inequality was developed in this paper. This inequality controls $\|u'\|_{L^2}$ in terms of $\|u\|_{L^2}$ and $\|u\|_{L^1}$ and hence allows the basic energy estimate for the heat equation to close. (Note that we are on the whole line here so there is no Poincare inequality). In this paper I also so for the first time the trick of obtaining $L^\infty$ bounds on Green's functions from $L^2$ bounds on the same by writing the semigroup from 0 to t as the product of semigroups from 0 to $t/2$ and from $t/2$ to $t$.

Comment by Aaron Hoffman

2012-12-27T16:43:24Z

(1) Its not clear to me that $fg \in \Omega_t$ unless by $fg$ you mean $f p_s^t g$. (2) Have you restricted attention to the case $n = 1$, $\Omega_t = [0,t]$ and worked out the answer in coordinates there?

Comment by Aaron Hoffman

2012-12-01T18:20:02Z

This does look like homework. But ... You know that the eigenfunctions are going to be of the form $e^{cx}$, $xe^{cx}$, and $e^{dx}$ and the differential operator will be $p(\partial_x)$ where $p(z) = (z-c)^2(z-d)-\lambda$. You can now reverse-engineer your boundary conditions so that all three eigensolutions to the linear homogeneous ODE $p(\partial_x)u = 0$ (including the generalized eigensolution) satisfy them.

Comment by Aaron Hoffman

2012-11-30T23:22:02Z

For the record, I am saying nothing that Pablo Lessa had not already said earlier. Had I read more carefully before posting I would have realized this.