From: Kevin Foltinek Subject: Re: Problem with Partial Differential Equation Date: 26 Jan 2001 00:36:11 -0600 Newsgroups: sci.physics,sci.math Summary: Exterior Differential Systems for solving PDEs ran.dll writes: > Does anyone have any tip on how to approach the following > PDE? > The particular problem involves a PDE which has > partial derivatives with four variables: x,y,z and t. > The sum of x+y+z is one (they are concentrations) and there > is a way of eliminating z from consideration so that the > DE now becomes one involving only three derivatives. The general PDE is of the form F(x,y,z,t,u,u_x,u_y,u_z,u_t,...) = 0 . Let's assume that it is first-order, so we don't have to worry about the "...". This PDE can be written as the system du - p dx - q dy - r dz - s dt = 0 F(x,y,z,t,u,p,q,r,s) = 0 . This method of formulation gives rise to (or generates) an "Exterior Differential System" (EDS). The first equation tells you to interpret p as u_x, q as u_y, etc. (Note I said "interpret"; p, q, etc., are properly treated as new variables.) You can add the constraint x + y + z - 1 = 0 and its differential dx + dy + dz = 0 to the EDS above; as long as you know how to manipulate differential forms (in this case it's not hard; "dx", "dy", etc., behave just like the unit vectors "i", "j", etc., except there is no dot product), you can probably manipulate the system back into a normal-looking PDE. There may be other things to consider in your particular problem, though, which might make this approach not the right one after all. (I would be surprised in this case, but you never know.) > Looking for hints or references in the literature or > a standard text. - ran.dll Look for "Exterior Differential Systems" by Bryant, Chern, Gardner, Goldschmidt, Griffiths. It's not an easy book, but everything you need is in part of the first chapter. Olver wrote a book on the same subject (though in a different direction from BCGGG), whose title I have forgotten, and also wrote "Applications of Lie Groups to Differential Equations", which may have some material of interest to you. I believe Yang wrote a book about EDS, which I think is more readable (but not better in terms of hard content) than BCGGG, better in some ways than Olver's, but I think copies of it are more scarce than the other two. (All of these books contain vastly more information than you actually need.) Kevin. ============================================================================== From: Kevin Foltinek Subject: Re: Problem with Partial Differential Equation Date: 26 Jan 2001 11:55:32 -0600 Newsgroups: sci.physics,sci.math ran.dll writes: > She didn't > describe the specific equation involved but I became interested in > the general problem of reducing the number of independent variables > in a PDE given a restraint of the type described. It turns out that the situation is a little bit more difficult than my previous message implied. Consider, for example, the PDE u_t + u u_x = 0 . As an EDS, this is formulated as 0 = du - p dt - q dx 0 = p + u q . The way to "solve" this is to look at the first equation, 0 = du - p dt - q dx , and assume that u is a function of t and x which solves the system; then (for any function of t and x) du = u_t dt + u_x dx, and (since it solves the system) 0 = du-pdt-qdx; substituting the first into the second, 0 = u_t dt + u_x dx - p dt - q dx = (u_t - p) dt + (u_x - q) dx ; since {dt,dx} are linearly independent, it is true that, for any u(t,x) which is a solution to the EDS, p=u_t and q=u_x. The second equation is then 0 = u_t + u u_x, the original PDE. Now impose the constraint 0 = t + x - 1 0 = dt + dx . Substituting dx=-dt into the original EDS, 0 = du - p dt + q dt = du - (p-q) dt . Now assume that u is a function of t which solves the system; then u_t = p-q or q = p-u_t , and p + u (p-u_t) = 0 . What has happened is that the original PDE, with two independent variables and one dependent, has become a PDE with one independent variable and two dependent (u and p); it is underdetermined. If you choose any function p(t), then you recover a normal determined PDE; or you could look at it another way, that if you choose *any* function u(t), then there is a p(t) so that you have a solution: 0 = p + up - u u_t = p (1+u) - u u_t ; p = u u_t / (1+u) . What is happening here is the following. Given any (smooth) function u(t,x), every point has an associated tangent plane; the original PDE is giving a single condition on the tangent plane to the graph of any solution u(t,x). The constraint, in the independent variables, reduces u to a function of one variable, along the constraint set. We still can think about building a function of one variable, and a set of planes which are tangent to its graph. If two directions defines a plane, then one of these two directions has to be tangent to the graph. The original PDE is *still* a condition on these planes; it tells us what the other direction has to be. (Given any u(t), there is a unique p(t).) On a slightly tangential note (no pun intended), we have p=uu_t/(1+u); what happens when u=-1? Well, when u=-1, the PDE looks like u_t-u_x=0, which is a constraint on the tangent plane in the direction that we already know (the same direction as the constraint set t+x=1); so this is really telling us that, if a solution (with one independent variable) passes through u=-1, then its derivative must be zero at that point. If not, then the original PDE cannot be satisfied; on the other hand, if it is, then p=uu_t/(1+u) = u*0/(1-1) = u*0/0 which is undefined, in other words, p, and the other direction, is undefined (or arbitrary). Returning to your original problem, you mentioned that the constraint arises from the reality condition that (x,y,z) are concentrations. Presumably the PDE, as well, arises from some reality considerations, and is not merely an arbitrary equation; hopefully there is some other reality consideration which can help you make sense of how to choose do the analog of choosing p(t), to recover a determined PDE. (I am hoping that the PDE did not arise from considerations like "vary one concentration while leaving the others fixed".) A possibility which occurs to me is that the PDE involves not concentrations, but quantities, and you are trying to convert it to a PDE involving concentrations. If this is the case, then the PDE must satisfy an extra condition (involving scaling of (x,y,z)), and this, I think, you might be able to exploit to do what you want. We need more information now. :-) Kevin.