From: "Manuel Joseph Loquias" Subject: Help! Date: Fri, 05 Feb 1999 01:48:52 PST Newsgroups: [missing] To: rusin@math.niu.edu Keywords: What does it mean to choose some variables as dependent on others? [deletia--djr] I would like to ask a question on Application of Partial Differentiation to Chain Rule. My professor gave us this exercise: Suppose E,T,V,P are variables related by the equations : F(E,T,V,P) = 0 and G(E,T,V,P) =0. If V and T are independent variables, then the equation (the partial derivative of E with respect to V) minus ( T multiplied to the partial derivative of P with respect to T) plus P = 0 , holds.. Suppose we instead regard P and T as the independent variables, what form does the equation above take now? Our professor solved this and something in the solution bothers me. I do not understand why even if we instead regard P and T as independent variables, the independence of the variables V and T still holds, such that partial derivative of V with respect to T is still 0. PLease explain to me. I consulted with my professor, but I still cannot understand him (He's really good , but he is not good in imparting his ideas). PLease, please, please, reply to me as soon as possible. Any help will be greatly appreciated. Also, I am currently thinking of a thesis topic. Can you send me some possible topics for thesis concerning ODE or PDE? Thanks. [deletia -- djr] ============================================================================== From: Dave Rusin Subject: Re: Help! Date: Sun, 7 Feb 1999 00:45:34 -0600 (CST) Newsgroups: [missing] To: manjo_61@hotmail.com >I visited your Mathematical Atlas, and I find it quite interesting, and >fascinating. I would like to tell you how wonderful your endeavor is. Thank you. But you should be aware that I am hardly an expert in all areas of mathematics! In particular I must confess that I do not feel particularly skilled at PDEs. >Suppose E,T,V,P are variables related by the equations : >F(E,T,V,P) = 0 and G(E,T,V,P) =0. If V and T are independent variables, >then the equation (the partial derivative of E with respect to V) minus >( T multiplied to the partial derivative of P with respect to T) plus P >= 0 , holds.. Suppose we instead regard P and T as the independent >variables, what form does the equation above take now? I'm not sure I understand what you're saying. It is certainly not true that the equations F=G=0 force E_V - (T P_T) + P = 0, as you can see with simple examples e.g. if F and G are linear! So I can only assume you mean, "Suppose E,T,V,P are such that F=0 AND G=0 AND E_V=... Then compute..."; that is, we take the differential expression as an extra hypothesis. >Our professor solved this and something in the solution bothers me. I >do not understand why even if we instead regard P and T as independent >variables, the independence of the variables V and T still holds, such >that partial derivative of V with respect to T is still 0. I think you will find it very confusing to think about PDEs as if there must be some variables which "cause" the other variables to take on certain values. It is much more productive to say, "we only find the variables in the following combinations". Thus I don't know whether "y=x^(2/3)" or "x=y^(3/2)" is the "right" way to distinguish the dependent and independent variables; instead, I would just say, "the variables (x,y) are restricted to the curve y^3-x^2=0". In this setting, both dy/dx and dx/dy make sense, and neither need be preferred. In your setting, what appears to be true is that you are considering a situation in which four variables are measured, but appear to vary together. When you look at the combinations (E,V,T,P) which occur, you find that they do not lie all over four-dimensional space. Rather, you find all the combinations along a certain curve. Certainly your premise that F=0 restricts the combinations to lie in a 3-dimensional hypersurface in R^4; the constraint G=0 does as well, and the intersection can be expected (generically speaking) to give you a 2-dimensional surface in which the points (E,V,T,P) lie. Every point on the surface has four coordinates, but you only need two of them to tell where you are, more or less; and moreover, any two will do. So you can think of E and P as functions of V and T, or vice versa, or use other combinations. The third hypothesis you impose gives a third hypersurface in which the points must lie, so again we expect the dimension of the solution set to drop; the points lie on a curve. Alternatively, you can imagine solveing F=G=0 for P and E in terms of V and T, then substituting these expressions into your third constraint. This leaves you with an ODE involving just V and T. Solve to get, say, V as a function of T, and then substitute in to see everything is a function of T, that is, each solution to your system is a curve (E(T), V(T), T, P(T)) parameterized by T -- or, again, you could use any of the four variables to parameterize the curve. Indeed, it is perhaps most productive to think of the curve as being parameterized by another, independent variable, say z. Let z represent time elapsed as a particle travels along this curve in R^4; then E, V, T, P will all vary with z. An equation like E_V -T P_T + P = 0 then can be expressed (using the Chain Rule) as E'/V' - T P'/T' + P = 0 (the "primes" -- perhaps you call them "dashes"? -- represent differentiation with respect to z). By differentiating the equations F=0 and G=0 with respect to z as well, you then have three equations which involve E', V', T', and P', and so you should expect to be able to solve for any three of them in terms of the other one, and indeed by homogeneity you can then rewrite your answer in a form showing (say) E'/P', V'/P', and T'/P' each to be some function of E, V, T, P. All this says, geometrically, is that the three equations you began with will carry sufficient information to specify the direction the curve will take as it leaves each point (E,V,T,P). If you compare to Euler's method for solving ODE's numerically, you will see this is essentially a solution to the PDE, given enough boundary data. Now, I don't really know what your professor did to derive his or her answer, so I don't know if the answer is justified; you can test the proposed answers with particular solutions to the third constraint, say of the form (E,V,T,P) = ( -y^2 H_y(x,y), x, y, -y H_x(x,y) ) for any sufficiently smooth function H of two variables. But my point is that there is no reason s/he cannot in principle obtain other expressions linking the variables and their derivatives to obtain equations involving, say, dT/dP and dT/dE; since the solutions to the set of equations are curves, these are simply rational expressions in E', V', T', and P' again. I trust this adds to your confusion? :-) (Now that you know the solution set is a family of curves in R^4 you can test some simple examples and see if my explanation, and your teacher's derivations, make any more sense.) >Also, I am currently thinking of a thesis topic. Can you send me some >possible topics for thesis concerning ODE or PDE? Thanks. I am not qualified to do so. But in general you should wait until you have found an advisor; that person can help you formulate a good question, and will be prepared to help you find out how to answer it. dave