What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:13:51Z http://mathoverflow.net/feeds/question/4459 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? Theo Johnson-Freyd 2009-11-06T23:10:43Z 2010-08-21T10:33:25Z <h3>Background</h3> <p>Lagrangian mechanics on $\mathbb R^n$ is usually defined by picking a <strong>Lagrangian</strong> function $L: {\rm T}\mathbb R^n \to \mathbb R$, where ${\rm T}\mathbb R^n = \mathbb R^{2n}$ is the tangent bundle of the configuration space $\mathbb R^n$. Such a function determines the <strong>Euler-Lagrange</strong> equations: $$\frac{d}{dt}\left[ \frac{\partial L}{\partial v^i}\bigl( \dot\gamma(t), \gamma(t)\bigr) \right] - \frac{\partial L}{\partial q^i} \bigl( \dot\gamma(t), \gamma(t)\bigr) = 0$$ Here $(v^i,q^i)$ for $i=1,\dots,n$ are the standard coordinates on ${\rm T}\mathbb R^n$, $\gamma: [0,T] \to \mathbb R^n$ is a smooth function, and $\dot\gamma^i(t) = \frac{d\gamma^i}{dt}$. Suppose that the matrix $\frac{\partial^2 L}{\partial v^i\partial v^j}(v,q)$ is invertible for any $(v,q) \in {\rm T}\mathbb R^n$. Then the Euler-Lagrange equations are a nondegenerate second-order differential equation on $\mathbb R^n$. I am interested in the <strong>boundary-value problem</strong> for $L$. Namely, fix $T > 0$ and <img src="http://latex.mathoverflow.net/png?%24q%5F1%2Cq%5F2%20%5Cin%20%5Cmathbb%20R%5En%24" alt="$q\sb 1,q\sb 2 \in \mathbb R^n$" title="" />; the BVP asks to find the set <img src="http://latex.mathoverflow.net/png?%24C%28q%5F1%2Cq%5F2%2CT%29%24" alt="$C(q\sb 1,q\sb 2,T)$" title="" /> of all paths $\gamma: [0,T] \to \mathbb R^n$ with <img src="http://latex.mathoverflow.net/png?%24%5Cgamma%280%29%20%3D%20q%5F1%24" alt="$\gamma(0) = q\sb 1$" title="" /> and <img src="http://latex.mathoverflow.net/png?%24%5Cgamma%28t%29%20%3D%20q%5F2%24" alt="$\gamma(t) = q\sb 2$" title="" />. Generically, this is a discrete set.</p> <h3>My question</h3> <p>Suppose that if instead of the Euler-Lagrange equations above, I pick some small parameter $\epsilon$ and consider the differential equation $$\frac{d}{dt}\left[ \frac{\partial L}{\partial v^i}\bigl( \dot\gamma(t), \gamma(t)\bigr) \right] - \frac{\partial L}{\partial q^i} \bigl( \dot\gamma(t), \gamma(t)\bigr) = \epsilon\gamma^{(4)}(t)^i$$ where $\gamma^{(4)}(t)^i$ is the $i$th component of the fourth derivative of $\gamma$ with respect to $t$ (the "jounce", a word I just learned from <a href="http://en.wikipedia.org/wiki/Jounce" rel="nofollow">Wikipedia</a>). For $\epsilon \neq 0$, the EL equations are a nondegenerate fourth-order differential equation, and so generically solutions to the boundary-value-problem above form a two-dimensional family. To restrict to a discrete set, we should fix more boundary values. Pick <img src="http://latex.mathoverflow.net/png?%24%28v%5F1%2Cq%5F1%29%2C%20%28v%5F2%2Cq%5F2%29%20%5Cin%20%7B%5Crm%20T%7D%5Cmathbb%20R%5En%24" alt="$(v\sb 1,q\sb 1), (v\sb 2,q\sb 2) \in {\rm T}\mathbb R^n$" title="" /> and $T > 0$, and define <img src="http://latex.mathoverflow.net/png?%24C%5F%5Cepsilon%28v%5F1%2Cq%5F1%2Cv%5F2%2Cq%5F2%2CT%29%24" alt="$C\sb \epsilon(v\sb 1,q\sb 1,v\sb 2,q\sb 2,T)$" title="" /> to be the set of solutions $\gamma$ to the $\epsilon$-dependent EL equations with <img src="http://latex.mathoverflow.net/png?%28%5Cdot%5Cgamma%280%29%2C%5Cgamma%280%29%29%20%3D%20%28v%5F1%2Cq%5F1%29" alt="(\dot\gamma(0),\gamma(0)) = (v\sb 1,q\sb 1)" title="" /> and <img src="http://latex.mathoverflow.net/png?%28%5Cdot%5Cgamma%28T%29%2C%5Cgamma%28T%29%29%20%3D%20%28v%5F2%2Cq%5F2%29" alt="(\dot\gamma(T),\gamma(T)) = (v\sb 2,q\sb 2)" title="" />.</p> <p>My question is: as $\epsilon \to 0$, in what sense do we have <img src="http://latex.mathoverflow.net/png?C%5F%5Cepsilon%28v%5F1%2Cq%5F1%2Cv%5F2%2Cq%5F2%2CT%29%20%5Cto%20C%28q%5F1%2Cq%5F2%2CT%29" alt="C\sb \epsilon(v\sb 1,q\sb 1,v\sb 2,q\sb 2,T) \to C(q\sb 1,q\sb 2,T)" title="" />?</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> 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/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur/5115#5115 Answer by timur for What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? timur 2009-11-11T22:46:44Z 2009-11-11T22:46:44Z <p>What is the motivation behind adding epsilon? Why can't you solve the second order BVP directly?</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/4459/what-happens-to-the-solutions-of-a-fourth-order-boundary-value-problem-as-you-tur/36270#36270 Answer by Aleksey Pichugin for What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? Aleksey Pichugin 2010-08-21T10:33:25Z 2010-08-21T10:33:25Z <p>I do not understand your equations in detail, but I think I can say what can/will happen from the point of view of formal asymptotics. After modifying your PDE in the described way you have to face two kinds of issues.</p> <p>Firstly, you can potentially change the type of PDE that you were solving originally. Suppose, for example, your original PDE is hyperbolic and you are interested in the Cauchy problem for it. Then you need to ensure that your additional terms do not turn your problem into elliptic, which would lead to an ill-posed problem. Hence, your perturbations cannot be completely arbitrary.</p> <p>Secondly, by introducing higher order terms into your equation you added new integrals to the general solution. In the linear case your new general solution becomes the sum of the (slightly modified) solution of the problem with $\epsilon=0$ and the new integral. These new integrals are spurious and have absolutely no physical significance. How do you ensure that your new solution is "close" in some sense to the solution of the original problem? by selecting the additional boundary conditions in such way as to minimize the contribution of these spurious integrals.</p> <p>Variants of this problem are long known (although not widely appreciated) in the long-wave asymptotics of thin elastic structures. I have a brief paper that applies this mode of thinking to an elementary mechanics problem. I am not aware of these issues discussed anywhere outside of mechanics, but am sure that physicists must have thought about this at some point. If anyone can point to papers that construct additional boundary conditions reducing the contribution of the spurious integrals in singularly perturbed PDEs, I would really love to learn about them.</p> <p>I hope this helps.</p>