User duke leto - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:43:10Z http://mathoverflow.net/feeds/user/1653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7870/stability-analysis-of-a-system-of-2-second-order-nonlinear-differential-equations/8913#8913 Answer by Duke Leto for Stability analysis of a system of 2 second order nonlinear differential equations Duke Leto 2009-12-14T21:36:17Z 2009-12-14T21:36:17Z <p>More generally, nonlinear equations can be linearized along certain "manifolds", i.e. only linear terms are needed to describe the essential dynamics near these manifolds. </p> <p><a href="http://en.wikipedia.org/wiki/Center_manifold" rel="nofollow">http://en.wikipedia.org/wiki/Center_manifold</a></p> http://mathoverflow.net/questions/6325/equations-for-integrable-systems/6604#6604 Answer by Duke Leto for Equations for Integrable Systems Duke Leto 2009-11-23T18:41:36Z 2009-11-23T18:41:36Z <p>Many definitions for "integrable" equations exist, your current definition is quite limiting. I would say what you are describing are a subset of Hamiltonian systems. When most people say "integrable" in PDE's and/or dynamical systems, they usually mean equations for which the Inverse Scattering Transform can be used to construct an analytic solution, which is a much larger class of problems than Hamiltonian systems.</p> http://mathoverflow.net/questions/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur/5141#5141 Answer by Duke Leto for What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? Duke Leto 2009-11-12T01:58:57Z 2009-11-12T01:58:57Z <p>The motivation is to understand how the 4-dimensional dynamics turn into 2-dimensional dynamics (or vice-versa) as the $ \gamma^{(4)}(t)^{i}$ term is "turned on" or "turned off." $ \epsilon $ is the relative amount of that term compared to the others.</p> <p>When the number of dimensions of the dynamics of a system change, that is generally referred to as a 'bifurcation' or 'catastrophe'. Bifurcations usually have the connotations of being low-dimensional $N \le 2$ where catastrophes have the connotation of being higher-dimensional, i.e. $N \gt 2$.</p> http://mathoverflow.net/questions/3699/minimizing-a-functional/4989#4989 Answer by Duke Leto for Minimizing a functional Duke Leto 2009-11-11T06:32:18Z 2009-11-11T06:32:18Z <p>Here is some of your question properly translated.</p> <p>$ T(\theta) = \int \frac{dx}{v_0\cos \theta(x)}dx $</p> <p>$ W = {\theta \in C^1 [0,L]|C(\theta) = y} $</p> <p>The definition of C(\theta) doesn't seem to want to work.</p> http://mathoverflow.net/questions/3234/ansatze-for-solving-pdes-with-wavelets/4828#4828 Answer by Duke Leto for Ansätze for solving PDEs with wavelets Duke Leto 2009-11-10T08:25:57Z 2009-11-11T06:17:49Z <p>The method of choosing a solution Ansatz to an equation and then actually deriving an exact solution is quite common in soliton theory, which is a sub-field of the study of hyperbolic equations. All methods described below, to my knowledge, only work on hyperbolic equations. Sorry, diffusion folks.</p> <p>You must know properties of your equations to know which Ansatz will yield reasonable or good results. If you know that the tails of the solution die off quickly, you may choose a Gaussian $$ A \exp^{-b x^2} $$, or if they die off <strong>very</strong> quickly, a super Gaussian $$ A \exp^{-b a(x)^2} $$, where a(x) can be any polynomial. Also, based on the properties of your equation, you may want to multiply these 'basic' Ansatzen by other functions, to represent behavior that is known to be present. For example, if you know that solutions to the equation are not monotonic and/or 'wiggly', then you might want $$ A \exp^{-b x^2 } \sin(k x) $$ The latter Ansatz is a two-parameter Ansatz and is the most likely to have a chance of working on a real equation. You may think that $ k $ is a third parameter, but actually, it is determined, usually algebraicly, by $A$ and $b$. Single parameter Ansatzen usually only work on very specific coefficients of equations and are too simple to model real equations.</p> <p>There are obviously many, many other good Ansatzen, such as soliton solutions $$ A {sech}^n{\left(k x - \omega t\right)} $$ (where $n$ is a positive even integer, and $\omega=\omega\left(k\right)$ is the dispersion relation) if your equations has symmetry properties. There is a large theory, mostly derived from the work of R. Hirota, of how to derive exact solutions to systems of nonlinear PDE's which have certain symmetry properties or invariants, using the properties of bilinear operators.</p> <p>Note: Directly translated, the word der Ansatz in German has many meanings, but it most usually is translated as 'approach' or 'basic approach', but it really just means: an educated guess of a solution, with enough degrees of freedom (in the form of parameters) such that the Ansatz is able to solve the equation.</p> <p>Also, in the above equations, $A$ can be constant, or only a function of $ t $ or a function of both $x$ and $t$, depending on which behavior is being modeled.</p> http://mathoverflow.net/questions/3471/are-the-asymptotics-of-fourier-coefficients-to-periodic-solutions-of-ode-known/4987#4987 Answer by Duke Leto for Are the asymptotics of Fourier coefficients to periodic solutions of ODE known? Duke Leto 2009-11-11T06:14:10Z 2009-11-11T06:14:10Z <p>I am quite sure I found a perturbative solution to this during graduate school, what you are asking for is the Fourier representation of that. Are you familiar with the Method of Dominant Balance ?</p> http://mathoverflow.net/questions/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur/4926#4926 Answer by Duke Leto for What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? Duke Leto 2009-11-10T21:13:08Z 2009-11-10T21:19:35Z <p>Bender and Orszag is probably the most approachable book and is what I learned from as well. The references given in Bender and Orszag are where I would go from here, unless you have very specific knowledge of what properties your system has, in which I may be able to suggest more specific references.</p> <p>Are you looking for more specific refs on bifurcation theory? If you tell me more about your specific problem, I can do that. </p> http://mathoverflow.net/questions/2969/what-is-a-rigorous-statement-for-linear-time-invariant-systems-can-be-represente/4832#4832 Answer by Duke Leto for What is a rigorous statement for "linear time-invariant systems can be represented as convolutions"? Duke Leto 2009-11-10T08:58:54Z 2009-11-10T08:58:54Z <p>As a more "down to earth" answer, I would say that linear systems have linear solutions, and convolution is a linear operator (or possibly bi-linear, based on the type of convolution) and as such the solutions of these equations can be represented as convolutions. The time-invariant property probably imposes some additional restrictions on the properties of the convolution.</p> http://mathoverflow.net/questions/4812/in-an-n-dimensional-linear-2nd-order-ode-why-is-the-transpose-inverse-to-a-syste/4830#4830 Answer by Duke Leto for In an n-dimensional linear 2nd-order ODE, why is the transpose-inverse to a system of solutions also a solution? Duke Leto 2009-11-10T08:42:22Z 2009-11-10T08:42:22Z <p>It seems like when you get to the variation-of-parameters step, everything gets fuzzy. Have you tried doing variation-of-parameters and finding the form of that solution? Your solution is probably not the fastest way, but it may still be correct.</p> http://mathoverflow.net/questions/3471/are-the-asymptotics-of-fourier-coefficients-to-periodic-solutions-of-ode-known/4826#4826 Answer by Duke Leto for Are the asymptotics of Fourier coefficients to periodic solutions of ODE known? Duke Leto 2009-11-10T07:27:34Z 2009-11-10T07:27:34Z <p>This sounds like a homework problem :) By which means are we allowed to derived the form of $ a_n(g) $ ? What is the asymptotic formula to be used for?</p> http://mathoverflow.net/questions/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur/4814#4814 Answer by Duke Leto for What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? Duke Leto 2009-11-10T04:58:05Z 2009-11-10T06:45:54Z <p>I believe that the method of solution to your problem is called the method of "dominant balance", and in this case, "singular dominant balance." If you do a web search for that, you should be able to find the information you need.</p> <p>This method will you give a perturbative solution to as high a degree as you have the propensity to calculate. You can analyze this solution to answer various questions that you implied in your original question, such as what the decay behavior of the solution is, which continuity and smoothness properties it has, etc...</p> <p>If you want to study the solutions of a large class of coefficient functions, not just a specific set, you can leave arbitrary constants in a solution "ansatz" and then develop a parameterized family of solutions. Note that the algebraic expressions involved in finding the simple-looking solutions grow exponentially in the number terms which end up simplifying in the end. Computer algebra is needed to find the simplified form of these solutions, lest you go mad and kill many trees.</p> <p>You may also want to search for "catastrophe theory", which catalogs the types of bifurcations that happen in systems such as you have described. This is a one-dimensional bifurcation problem, which are well-studied. </p>