Re: Can I solve this system of nonlinear equations?

*To*: mathgroup at smc.vnet.net*Subject*: [mg125278] Re: Can I solve this system of nonlinear equations?*From*: "Stephen Luttrell" <steve at _removemefirst_stephenluttrell.com>*Date*: Sun, 4 Mar 2012 04:33:06 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <jil5ld$rrm$1@smc.vnet.net> <jit112$e7i$1@smc.vnet.net>

"Ray Koopman" <koopman at sfu.ca> wrote in message news:jit112$e7i$1 at smc.vnet.net... > On Feb 29, 4:28 am, Andy <andy7... at gmail.com> wrote: >> I'm dealing with systems of nonlinear equations that have 8 equations >> and 8 unknowns. Here's an example: >> >> Solve[{(((c - a)/0.002) - (0.995018769272803 + h*b)) == 0, >> (((d - b)/0.002) - (0.990074756047929 + h*c)) == 0, >> (((e - c)/0.002) - (0.985167483257382 + h*d)) == 0, >> (((f - d)/0.002) - (0.980296479563062 + h*e)) == 0, >> (((g - e)/0.002) - (0.975461279165159 + h*f)) == 0, >> (((-1*e + 8*d - 8*b + a)/(12*0.001)) - (0.990074756047929 + h*c)) == >> 0, >> (((-1*f + 8*e - 8*c + b)/(12*0.001)) - (0.985167483257382 + h*d)) == >> 0, >> (((-1*g + 8*f - 8*d + c)/(12*0.001)) - (0.980296479563062 + h*e)) == >> 0}, {a, b, c, d, e, f, g, h}] >> >> Whenever I try this, Mathematica 7 just returns the empty set {}. How >> can I tell if this is unsolvable? Shouldn't I at least be able to get >> a numerical approximation with NSolve? I've tried using stochastic >> optimization to get approximate answers but every method gives poor >> results, and that's why I would like to at least approximately solve >> this if possible. Thanks very much for any help~ > > To see if there is an exact solution to the problem, > make all the coefficients exact and use Solve: > > x = { 500(c - a) - (995018769272803*^-15 + h*b), > 500(d - b) - (990074756047929*^-15 + h*c), > 500(e - c) - (985167483257382*^-15 + h*d), > 500(f - d) - (980296479563062*^-15 + h*e), > 500(g - e) - (975461279165159*^-15 + h*f), > (-1*e + 8*d - 8*b + a)1000/12 - (990074756047929*^-15 + h*c), > (-1*f + 8*e - 8*c + b)1000/12 - (985167483257382*^-15 + h*d), > (-1*g + 8*f - 8*d + c)1000/12 - (980296479563062*^-15 + h*e)}; > > Solve[Thread[x == 0],{a,b,c,d,e,f,g,h}] > > {} > > To get an approximate solution, minimize some measure of the > differences between the two sides of the equations. The sum of > the squared differences is convenient and not obviously improper; > that is, there is nothing to suggest that the equations are on > vastly different scales. > > TimeConstrained[Minimize[x.x,{a,b,c,d,e,f,g,h}],300] aborted. > > Both Minimize[N[x.x],{a,b,c,d,e,f,g,h}] and > NMinimize[x.x,{a,b,c,d,e,f,g,h}] finished in < 2 sec > and gave identical results: > > {3.7850320584543324`*^-11, > {a -> 1.78322069825487`, > b -> 1.7856155567256693`, > c -> 1.7880252569409105`, > d -> 1.79041402958568`, > e -> 1.7928176848807527`, > f -> 1.7952004996719013`, > g -> 1.7975982365756977`, > h -> 0.7881096531282696`}} > > The fit is a little better than Stephen Luttrell got, > but the parameter estimates are far what he got. > > x /. %[[2]] > > {-2.832840262367853`*^-7, > 1.7089251762580915`*^-6, > 3.9068509285478115`*^-6, > 1.6397939603951528`*^-6, > -2.7478474384778906`*^-7, > -1.6998965117753784`*^-6, > -3.348694662008711`*^-6, > -1.6487347667126784`*^-6} > > It looks like only the first 5 digits of the 15-digit constants > are accurate. > When I try to reproduce your solution using NMinimize I get my solution (not yours). I am using version 8.0.1 for Windows (32-bit). I presume that the discrepency is because the solution space is not a single point, so different versions and/or methods will arrive at different points in the solution space. Anyway, I derived the Hessian matrix to have a close look at the neighbourhood of my solution point to see how the objective function behaved, where the solution space seems to be 2-dimensional. This doesn't tell us what it might look like further away. More work would be needed to characterise it all. Equations. eqns = {(((c - a)/0.002) - (0.995018769272803 + h*b)) == 0, (((d - b)/0.002) - (0.990074756047929 + h*c)) == 0, (((e - c)/0.002) - (0.985167483257382 + h*d)) == 0, (((f - d)/0.002) - (0.980296479563062 + h*e)) == 0, (((g - e)/0.002) - (0.975461279165159 + h*f)) == 0, (((-1*e + 8*d - 8*b + a)/(12*0.001)) - (0.990074756047929 + h*c)) == 0, (((-1*f + 8*e - 8*c + b)/(12*0.001)) - (0.985167483257382 + h*d)) == 0, (((-1*g + 8*f - 8*d + c)/(12*0.001)) - (0.980296479563062 + h*e)) == 0}; Objective function. obj = Total[eqns /. (x : _) == 0 :> x^2]; Minimise. vars = {a, b, c, d, e, f, g, h}; soln = NMinimize[obj, vars] giving what I got before ... {3.94032*10^-11, {a -> 0.314882, b -> 0.315979, c -> 0.317109, d -> 0.318197, e -> 0.319318, f -> 0.320396, g -> 0.321509, h -> 0.374577}} Now compute the matrix of second derivatives (Hessian) of the objective function at this solution point. hessian = Outer[D[obj, #1, #2] &, vars, vars] /. soln[[2]]; Then compute the eigenvalues of the Hessian. evals = Eigenvalues[hessian] {4.19009*10^6, 3.42241*10^6, 1.39642*10^6, 1.00435*10^6, 403393., 0.0159205, -2.57089*10^-10, 7.18138*10^-12} Eyeballing this (or using ListPlot[evals, Filling -> Axis]) you see 5 large (order 10^6), 1 very small (order 10^-2), and 2 noise level (order 10^(-10 or -11)) eigenvalues, so the the objective function stays at its minimum value as you move away from the solution point in 2 (or 3, almost) independent directions. The solution space is 2-dimensional (at least) around the solution point. You can use EigenSystem[hessian] to look at this 2D solution space in more detail. {evals, evects} = Eigensystem[hessian]; The last 2 eigenvectors (i.e. with noise level eigenvalues) span the 2-dimensional solution space using coefficients (u,v). solnoffset = Thread[vars -> vars + {u, v}.Take[evects, -2]] {a -> a + 0.37754 u - 0.000830755 v, b -> b + 0.377681 u - 0.000462092 v, c -> c + 0.377823 u - 0.000199144 v, d -> d + 0.377964 u + 0.000171975 v, e -> e + 0.378106 u + 0.000437376 v, f -> f + 0.378247 u + 0.000810937 v, g -> g + 0.378389 u + 0.00107878 v, h -> h - 0.000381899 u + 0.999999 v} Check what the objective function now looks like in the 2-dimensional neighbourhood of the solution point. objsoln[u_, v_] = obj /. solnoffset /. soln[[2]] // Expand // Chop 1.66682*10^-7 u^4 - 0.000872909 u^3 v + 1.14285 u^2 v^2 + 0.000884953 u v^3 + 1.39221*10^-6 v^4 The coefficients are not all minuscule, but all of the terms are quartic which helps the objective function to stay close to zero over a sizeable 2-dimensional region around the solution point. Check what the objective function looks like in this region by generating a contour plot. ContourPlot[objsoln[u, v], {u, -0.1, 0.1}, {v, -0.1, 0.1}] which shows that the (u,0) and (0,v) directions are the "flattest". Anyway, those are the sorts of techniques that I would use to get a "toe hold" on the solution space around a particular found solution point. -- Stephen Luttrell West Malvern, UK