Re: Re: Solving simple equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg83248] Re: [mg83206] Re: Solving simple equations*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Fri, 16 Nov 2007 05:24:22 -0500 (EST)*References*: <fheh0u$mm7$1@smc.vnet.net> <18907167.1195135705102.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

Things are not hopeless, even without numerical values for everything. meqn = {x'[t] == beta (x[t] + (Pf - p[t])), p'[t] == (Tanh[a1 (x[t] + (Pf - p[t])) + a2 x[t]] - x[t])}; eqns = meqn /. {p'[t] -> 0, x'[t] -> 0} {0 == beta (Pf - p[t] + x[t]), 0 == Tanh[a2 x[t] + a1 (Pf - p[t] + x[t])] - x[t]} Notice the first equation has a trivial solution (beta == 0) and a second solution that simplifies everything: Quiet@Solve[eqns[[1]]] {{beta -> 0}, {Pf -> p[t] - x[t]}} {first, simple} = Flatten@DeleteCases[eqns /. %, True, {2}] /. x[t] -> x // Simplify {x == Tanh[a2 x + a1 (Pf + x - p[t])], x == Tanh[a2 x]} Either of these can be solved numerically, but I'll just take the second. Solving for x is no help: Solve[simple, {x}] Solve::tdep: The equations appear to involve the variables to be \ solved for in an essentially non-algebraic way. >> Solve[x == Tanh[a2 x], {x}] but solving for a2 gives us some idea where to start: Quiet@Solve[simple, {a2}] {{a2 -> ArcTanh[x]/x}} NMinimize[ArcTanh[x]/x, {x}] {1., {x -> -1.45306*10^-8}} Plot[ArcTanh[x]/x, {x, -5, 5}] So we see that a2 >= 1. The ratio is flat at x == 0, so inverting it numerically can be unstable there. I deal with this using Series: seriesSoln = x /. Solve[x == Normal@Series[Tanh[a x], {x, 0, 9}], x]; seriesSoln /. a -> 1.1 {0, -0.981186 + 1.01603 \[ImaginaryI], 0.981186- 1.01603 \[ImaginaryI], -0.981186 - 1.01603 \[ImaginaryI], 0.981186+ 1.01603 \[ImaginaryI], -0.503006, 0.503006, -1.38767, 1.38767} 0.503006 is the correct root; the others are spurious: Plot[{x, Tanh[1.1 x]}, {x, 0, 2}] Finally, x in terms of a can be calculated efficiently by Clear[xnum] xnum[a_] /; 1 <= a < 1.1 = seriesSoln[[-3]]; xnum[a_] /; a >= 1.1 := x /. FindRoot[x == Tanh[a x], {x, 1}] xnum[a_] /; a < 1 := xnum[2 - a] xint = Quiet@FunctionInterpolation[xnum[a], {a, -5, 5}]; Plot[xint[a], {a, -5, 5}, PlotRange -> All] If the beta = 0 case is important, I'd start by simplifying first // ExpandAll x == Tanh[a1 Pf + a1 x + a2 x - a1 p[t]] to just x == Tanh[a x + b] and applying similar techniques to those above, first finding allowable ranges for a and b, then using FindRoot and Series to get x values. I hope that helps! Bobby On Thu, 15 Nov 2007 04:32:48 -0600, Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote: > Holger wrote: > >> I'm trying to solve these two simple equations but it doesn't work. >> >> meqn = { >> x'[t]==beta(x[t]+(Subscript[P,f]-p[t])), >> p'[t]==(Tanh[Subscript[a,1](x[t]+(Subscript[P,f]-p[t]))+ >> Subscript[a,2]x[t]]-x[t]) >> }; >> >> eqp = NSolve[ >> meqn/.{ >> p'[t]->0,x'[t]->0},{p[t], x[t] >> } >> ] >> >> I guess I'm doing a mistake somewhere. Does anyone have idea where the >> mistake is? > > So you want to solve *numerically* the system meqn (note that I have > modified the subscripted stuff). > > In[1]:= meqn = {x'[t] == beta (x[t] + (Pf - p[t])), > p'[t] == (Tanh[a1 (x[t] + (Pf - p[t])) + a2 x[t]] - x[t])}; > meqn /. {p'[t] -> 0, x'[t] -> 0} > > Out[2]= {0 == beta (Pf - p[t] + x[t]), > 0 == Tanh[a2 x[t] + a1 (Pf - p[t] + x[t])] - x[t]} > > To do so, you must provide some numeric values for *all* the parameter= s, > that is beta, Pf, a1, and a2, (though in the above example the value o= f > beta does not matter and beta can be discarded before solving). > > Of course some (many?) other things might be wrong, but since you did > tell us neither what you mean by "not working" nor reported any error > messages returned by by Mathematica, it is hard to tell more (perhaps > DSolve or NDSolve might be more appropriate). > > Regards, -- = DrMajorBob at bigfoot.com