From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Diff.equation problem... Date: 20 Nov 1994 00:05:24 GMT In article , HONEYMAN ROBLES GEORGE wrote: >Hi! Could someone mail me the solution to the following differential >equation: > > y" = (3/2)*y^2 ( The 2 solutions..) Given a 2nd order DE in which the independent variable does not appear, one finds solutions with the substitution v=dy/dx; then y''=dv/dx can be written as (dv/dy).(dy/dx) by the chain rule; that is, y''=v.(dv/dy). In this way, the DE becomes a first order DE in y and v. In your case, this leads to v.(dv/dy) = (3/2).y^2, which is separable: vdv=(3/2) y^2 dy. Integrate to see v^2/2 = (3/2) y^3/3 + C, or equivalently v^2=y^3+C. Then solve for a relationship between y and x: (dy/dx) = v = sqrt(y^3+C), which is again separable. The solution is x = integral( dy / sqrt(y^3+C) ) (+a different constant of course) Integrate and (if desired) invert to solve for y in terms of x. Oh, and, except for C=0, don't expect too much of this integral in the way of closed form solutions -- this is an elliptic integral. But if you only have to match some initial conditions, you can easily compute numerical values of the integral. dave (who really doesn't think of himself as a DE kind of guy, but is teaching it and feels compelled to turn USENET posts into a chance to keep teaching) ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Diff.equation problem... Date: 21 Nov 1994 06:47:50 GMT In article <3am9f7$8dj@news2.delphi.com>, LAURAHELEN@DELPHI.COM wrote: >if y = 0 then you couldn't turn this into the integral. Is there a >systematic way of looking for trivial solutions? Well, yes, I'm afraid I was a bit hasty. One typically solves equations through mindless manipulation without checking to see if exceptional solutions (your 'trivial' ones) have been missed, or extraneous ones have been introduced (e.g. solving A=B by squaring to get A^2=B^2 would introduce solutions to A=-B). Let's face it, we do it because it works so fast and _usually_ so well. Really one ought to solve and equation not by manipulation but by explanations: "IF y is a solution to [equation] THEN [manipulation], so IF [condition] then [more manipulation] so that y would have to be [form of solution]" This makes it clear that more solutions may occur which don't make [condition] hold, and show as well that y = [form of solution] does not necessarily mean, conversely, that y is a solution to [equation]. Let me take my original solution posted and express it more precisely to look for exceptional or extraneous solutions. I prefer to do it using functions, not variables, since I think the former permits more mathematical precision. >HONEYMAN ROBLES GEORGE wrote: >>Hi! Could someone mail me the solution to the following differential >>equation: >> y" = (3/2)*y^2 ( The 2 solutions..) [my previous solution began:] >Given a 2nd order DE in which the independent variable does not appear, >one finds solutions with the substitution v=dy/dx; then y''=dv/dx can >be written as (dv/dy).(dy/dx) by the chain rule; that is, y''=v.(dv/dy). [formal translation follows] Suppose f is a (twice-differentiable) function defined in a neighborhood of a point x_0 such that y = f(x) satisfies the above DE. Now, IF f'(x_0) <> 0 then by the implicit function theorem, there is a (possibly smaller) neighborhood of x_0 on which f has an inverse g. Then the function h =f'og (that's supposed to be the composition symbol) has hof = f' locally, so by the chain rule f'' =(h' o f). f'. Since we are assuming f'' = (3/2).f^2 for all x, we then have (h' o f). f' = (3/2) . f^2 for all x [near x_0]. [If you're lost, remember we were abbreviating y for f(x) and v for f'(x); h is supposed to be f' (i.e. v) viewed as a function of y, so h' means dv/dy; h'of just means to view this, as well, as a function of x, so we have the equation (dv/dy). v = (3/2). y^2 among several functions of x.] In this way, the DE becomes a first order DE in y and v. From (h' o f). f' = (3/2) . f^2 for all x [near x_0], we compose with g to obtain [remember that f and g are inverses] h' . (f' o g) = (3/2) . id^2 [id is the identity funtion. It might help if I expressed these functions as functions of a variable, say as functions of y. This gives h'(y) . (f'(g(y)) = (3/2) . y^2. ] >In your case, this leads to v.(dv/dy) = (3/2).y^2, which is separable: >vdv=(3/2) y^2 dy. Integrate to see v^2/2 = (3/2) y^3/3 + C, or >equivalently v^2=y^3+C. Now, clearly (!) the left hand side is the derivative of h^2/2; indeed the chain rule expresses the derivative of (1/2).h^2 as (1/2).2h^1. h', which is (f' o g).h' from our original definition of h. On the other hand the right-hand side is the derivative of (id^3)/2 (standard calc-1 formula). So (corollary to Mean Value Theorem) the difference h^2/2 - id^3 / 2 is constant, i.e. there is a constant C such that h^2 = id^3 + C on an interval around y_0. [i.e., h(y)^2 = y^3 + C for all y near y_0] >Then solve for a relationship between y and x: >(dy/dx) = v = sqrt(y^3+C), which is again separable. Recalling that h = f'o g we see that f'o g = sqrt(id ^3 + C) on this interval; 'sqrt' could be either square root, although the assumption that f' is (differentiable hence) continuous implies one branch or the other must be chosen across the interval. (You may need to shrink the interval more to exclude spots where f'=0). Composing with f shows f' = sqrt(f^3 + C) on that interval. >The solution is >x = integral( dy / sqrt(y^3+C) ) (+a different constant of course) Let F be a function with F' = 1/sqrt( id^3 + C). There are many such F 's; they differ by constants. Then Fof has derivative (F' o f).f'; given the relationship above for f', this means Fof has derivative of 1, and so must differ from id by a constant. Composing with g shows F = g + D for some constant D. >Integrate and (if desired) invert to solve for y in terms of x. If G is an inverse of F (which must exist since F' is never zero on any interval on which F satisfies the property that defines it), we may compose with f on the left and G on the right to get f = G + f(D), that is, f is the function G up to a constant. Conclusion: the function f we sought is given (up to the addition of a constant) as the inverse G of an integral F of the function 1/sqrt(id^3 + C) for some choice of sqrt and some constant C, all under the assumption that f'(x_0) is not zero. Thus, any solution must be of this form at least on some intervals unless f' vanishes everywhere. That's only possible if f is (locally) constant, of course. So much for incorporating the exceptional solutions. We also ought to eliminate extraneous solutions. For example, all constant functions are extraneous except f(x) ==0. You find this by plugging your purported solution into the original differential equation. In the other family of solutions, note G' = 1/ (F'o G) by the inverse function theorem, so f'= G' = sqrt(id^3+C) o G = sqrt(G^3 + C), so f'' = (1/2)(G^3+C)^(-1/2). 3G^2 . G' = (3/2)(G^3+C)^(-1/2). G^2 . sqrt(G^3 + C) = (3/2) . G^2 so that functions in this family are indeed solutions iff f = G; the other functions f = G + constant are extraneous 'solutions'. (Note that any other choice of F is of the form F + E for some constant E (locally). Then the other possibilities for f = G = F^(-1) are y = G(x-E) for any constant E.) So: yes, you can do this DE carefully and get the complete family of solutions, but no, you probably don't want to bother. (If you think I try to teach a bunch of engineering students differential equations with help of the implicit function theorem, you've misunderstood the teaching concept of 'meeting the students where they come from' :-) ) dave PS - There is more one can say about this differential equation. For example, let F(x) be one fixed antiderivative of 1/sqrt(x^3+1), and G its inverse. Then the integral of 1/sqrt(x^3+C) is C^(-1/6) F(x/C^(1/3)) for C > 0 [please accept complex solutions if C<0 ...] which has inverse C^(1/3) G( x . C^(1/6)). Allowing for the different choices for F and the choices for the square roots, this shows that the general solution to the DE is of the form y = C^(1/3) G( (+ or -) x . C^(1/6) - E) where G is the one given solution. This solution misses the special case C=0 (where F(x) = 3 x^(1/3), G(x) = (x/3)^3) but does include the solution y==0. Also, the functions F and G satisfy certain additivity formulas related to the fact that y^2 = x^3 + C is a Riemann surface homeomorphic to the torus, which is a group. This is the theory of Abelian integrals. The distinctness of the case C=0 results from the fact that in that case, the surface has a singularity but is I believe of genus zero. I'd best not say more here since this is close to all I remember about this fact.