Classification of PDE - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:54:04Z http://mathoverflow.net/feeds/question/66917 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66917/classification-of-pde Classification of PDE YBL 2011-06-04T20:00:13Z 2011-06-09T15:56:53Z <p>Recently I have been attending a course on PDE's. I was totally ignorant of the subject and wasn't that motivated to be honest. But I was intrigued and felt I had to take the course seriously both for exams and because I was lead to it through a very circumvoluted path of research (from arithmetics through algebraic geometry to D-modules = holomorphic linear differential equations). </p> <p>Anyway, I was told that PDE were classified into 3 families: hyperbolic, elliptic, parabolic respectively. I saw exemples for each: Wave, Laplace and Heat equations. I also saw lots of different methods to solve these examples: using symmetry to reduce to an ODE, Fourier transform, Distributions, Fourier series (in dimension 1 to deal with boundary conditons). Each time it seemed the answer was "reduce the problem to a polynomial equation or an ODE by any way you can". So my geometric brain kicked in and tried to give a unified geometric interpration to all this.</p> <p>I think I understand the distinction between those equations. We are considering 2nd order PDE with real coefficient. And the idea is to reduce the classification to that of their principal symbols seen as quadratic forms on $\mathbb{R}^n$. </p> <p>So I asked this kind a questions: let's consider a LINEAR PDE $Pf = 0$ where $P$ is a linear differential operator of degree $d$ on $\mathbb{R}^n$.</p> <p><strong>Question 1:</strong> Let's say a "naive elliptic PDE" is any PDE given by a differential operator $P$ whose principal symbol $\sigma(P)$ satisfies $\sigma(P)(x,\xi) \neq 0$ for $\xi\neq 0$. Is this definition any good? </p> <p>If answer to question 1 is yes, </p> <p><strong>Question 2:</strong> What is the analogue of a parabolic or hyperbolic operator? </p> <p>The obvious perfectly nice answer would be: "PDE's are classified by the hypersurfaces defined by their principal symbols". Unfortunatly the answer I got was something like "Don't overthink it, the classification is more heuristic than anything.". Does that mean, "there is a classification along theses lines but it is a bit more subtle" or that "things are much much more complicated in higher dimensions/degrees"? </p> <p>Anyway... </p> <p><strong>Question 3:</strong> Is it at least true that the classification of PDE with constant coefficient is related to the classification of real algebraic projective hypersurfaces ${\sigma(P) = 0 } \subset \mathbb{P}^{n-1}_{\mathbb{R}}$? </p> <p>Let's assume the previous questions aren't completly wrong for trivial reasons. Let's consider a PDE with non-constant coefficients. We should then classify those according to "families of algebraic projective hypersurfaces" in $\mathbb{P}(T^*\mathbb{R}^n)$. </p> <p><strong>Question 4:</strong> Which kind of families can we expect? Is that related to Gabber's theorem of involutivity of the characteristic variety? </p> <p>I am now assuming someone answered all these questions without thinking the words "this is so completly wrong". I have a final question (at least before the next one):</p> <p><strong>Question 5:</strong> Why is this all so hard to learn/teach? </p> <p>PS: Thanks to the people who took time to answer (lots of food for thoughts). I'd be happy to read more especially if you have some references. </p> http://mathoverflow.net/questions/66917/classification-of-pde/66918#66918 Answer by Nilima Nigam for Classification of PDE Nilima Nigam 2011-06-04T20:20:59Z 2011-06-04T20:37:45Z <p>I am unsure of the etiquette surrounding multi-part questions. Here are answers to two sub-parts. Since your Q5. invites opinion, I've addressed that in a comment instead.</p> <p>Q1: yes, the definition of ellipticity via the non-vanishing of the principle symbol is a useful characterization for elliptic PDE, see H&ouml;rmander's book. All manners of existence and regularity properties can be examined from here.</p> <p>Q2: This is murkier. If the principal symbol of a linear PDE, order q, with smooth coefficients is a hyperbolic polynomial, then the PDE is hyperbolic. This doesn't generalize easily to nonlinear cases, and is not an easy condition to check. See an extensive discussion here: <a href="http://math.stackexchange.com/questions/21525/mathematical-precise-definition-of-a-pde-being-elliptic-parabolic-or-hyperbolic" rel="nofollow">http://math.stackexchange.com/questions/21525/mathematical-precise-definition-of-a-pde-being-elliptic-parabolic-or-hyperbolic</a></p> <p>L.C. Evans, in his preface to his AMS text on PDE, mentions that he finds it unsatisfactory to classify PDE, since it creates the false impression that a general classification is available. Several equations change type (eg. Tricomi's equation) and many PDE of interest are highly nonlinear. </p> http://mathoverflow.net/questions/66917/classification-of-pde/66919#66919 Answer by Denis Serre for Classification of PDE Denis Serre 2011-06-04T20:52:08Z 2011-06-04T20:58:48Z <p>It seems you had a course on linear 2nd order scalar PDEs. All these words are meaningful but restrictive. Nowadays, the interesting PDEs are non-linear (for instance the 1M dollars prize for the Navier-Stokes equations). The type of a nonlinear equation is usually obtained by linearizing it around a constant solution. But it can display strange behaviour that linear ones don't. Example: harmonic maps with values in manifolds whose topology is non-trivial. Also, the linear theory is essentially a part of functional analysis (linear operators over topological vector spaces), but the nonlinear theory is much more involved. Sometimes, there is no theory at all! Example: hyperbolic systems of conservation laws in several space variables (my speciality).</p> <p>Going back to your questions:</p> <blockquote> <p><em>Question 1</em>. You are correct as long as your unkown is scalar. When it is vector valued, it is slightly more involved. A system of PDE writes $A(\nabla)\vec u=0$, where $A$ is a matrix of differential operators. The principal symbol $A_0(\xi)$ has some homogeneity: $a_{ij}(\xi)$ is a homogeneous polynomial of degree $\mu_i-\nu_j$. The system is elliptic if $A_0(\xi)$ is non-singular for every $\xi\ne0$.</p> <p><em>Question 2</em>. A parabolic operator is more delicate to define because its principal part is not of homogeneous order. I am not sure that there is a general enough definition of it. Hyperbolicity is very interesting. You have to distinguish a direction $\vec e$ which you call time-like. Let me focus on scalar equations and consider the principal part, say of order $n$. Its symbol $P(X)$ is a homogeneous polynomial of degree $n$. It is hyperbolic if $P(\vec e)\ne0$ and for every $\xi$, the univariate polynomial $t\mapsto P(t\vec e+\xi)$ has its $n$ roots real. Therefore hyperbolicity has a lot to do with real algebraic geometry. The Petrowsky school (in particular O. A. Oleinik) is famous in this field. </p> <p><em>Question 3</em>. This is correct.</p> <p><em>Question 4</em>. The $x$-dependence of the coefficients makes the analysis more difficult, but it does not change the classification above. The situation is different when you consider operators that do not fall in these three categories. The case of <em>sub-elliptic</em> operators is especially interesting. Hormander remarked that an operator of the form $X_0+X_1^2+\cdots+X_r^2$, where the $X_j$'s are smooth vector fields, has ''elliptic properties'' if the Lie algebra spanned by the vector fields has full rank at every point $x$. This applies for instance to the Fokker-Planck operator $\partial_t-\Delta_v-v\cdot\nabla_x$.</p> <p><em>Question 5</em>. It is like medicine in a hospital. One theory is not enough. You need to know a lot of different branches of mathematics. As much as you can. Recently, I worked on a problem from gas dynamics and I reproved the existence of complete minimal surfaces in a Riemann space of negative curvature... without knowing it. Until a colleague pointed it out to me.</p> </blockquote> http://mathoverflow.net/questions/66917/classification-of-pde/66920#66920 Answer by Ben McKay for Classification of PDE Ben McKay 2011-06-04T20:57:30Z 2011-06-04T21:02:44Z <p>PDE books often discuss classification, but they always restrict attention to the case of second order equations, especially for one function of several variables, with good reason. The point of a classification is to find categories of PDE whose analysis has many common features, but there really isn't any general classification in that sense, since the world of PDE is a huge zoo (once you leave the 3 familiar families of elliptic, hyperbolic and parabolic). Think about how you would define parabolic PDEs, even in second order. You already need to look beyond the symbol to distinguish $\partial_t u=\partial_{xx} u$ from $0=\partial_{xx} u$. As the OP points out, the symbol is certainly an important part of the classification''. The symbol is only a part of the tableau, which gives a little more information in an algebraic format; see the book of Bryant, et. al, <em>Exterior Differential Systems</em>. But systems of differential equations with the same tableau often have different analysis. Think about the famous Lewy counterexample. There are so many very different genera of animals in the zoo, and broad classifications don't give us much insight. Also look at Gromov, <em>Partial Differential Relations</em>, for lots of examples of PDEs that are locally the same, but globally very different, and are nothing like elliptic, hyperbolic or parabolic. So question 1: yes, question 2: hyperbolic is tricky to define beyond second order, because already for second order, hyperbolic is very different from ultrahyperbolic, so you really need something to distinguish a Lorentzian geometry from a more general pseudo-Riemannian geometry. On the other hand, your definition of ellipticity is perfect, and does give us some tools to carry out analysis. question 3: a little bit like yes, in that each PDE system gives rise to an algebraic variety, but finally no in that the classification of constant coefficient PDE systems is much finer than the classification of their symbols (it is in fact exactly the classification of their tableau), question 4: yes, you prolong until you hit involution, and so the classification of involutive tableau is not known, a huge messy algebra problem, question 5: like biology, it is messy because there are too many very different animals. </p> http://mathoverflow.net/questions/66917/classification-of-pde/67088#67088 Answer by Deane Yang for Classification of PDE Deane Yang 2011-06-06T21:32:45Z 2011-06-06T21:32:45Z <p>If you are interested in a real scalar linear PDE with constant coefficients, then it has all been worked out, primarily by Ehrenpreis, using the Fourier transform. Unfortunately, I don't know of a decent reference.</p> <p>If you are interested in variable coefficient complex scalar linear PDE's, then there is extensive work by many people, including Hormander, Nirenberg, Treves.</p> <p>If you're interested in analytic or hyperfunction (dual to analytic functions, analogous to distributions for smooth functions) solutions to constant coefficient systems, then I believe that leads to D-modules and therefore territory unknown to me. I believe there are ties to algebraic geometry.</p> <p>But as other have mentioned, PDE's, whether a scalar equation or a system of equations, do not succumb to any kind of neat classification or general approach. Only a few special types (the ones you cite) are useful and tractable. Most are not useful and more or less impossible to understand.</p> <p>Also, many if not most of the most important and interesting are nonlinear. And of first or (more often) second order. There are higher order PDE's that are worth studying but relatively few.</p>