Re: solve f(x)=0, where f:Rn+1 -> Rn

*To*: mathgroup at smc.vnet.net*Subject*: [mg26212] Re: [mg26189] solve f(x)=0, where f:Rn+1 -> Rn*From*: "Carl K. Woll" <carlw at u.washington.edu>*Date*: Sat, 2 Dec 2000 02:10:39 -0500 (EST)*References*: <200012010302.WAA11186@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Pavel, A while back I put forth a function which I believe does exactly what you want. The function is called ImplicitSolve, and it has the usage message: ?ImplicitSolve ImplicitSolve[eqns, {x->x0,y->y0,...}, {x, xmin, xmax}, opts] finds a solution to the implicit equations eqns for the functions y, ... with the independent variable x in the range xmin to xmax. The root {x0,y0,...} should satisfy the equations, or should provide a good starting point for finding a solution when using FindRoot. Currently, the only available option is AccuracyGoal, but a better ImplicitSolve would include the possibility of supplying options for both the FindRoot and NDSolve function calls. I will give the function definition at the end of this message (modified slightly from my previous post). For your circle example, you would use ImplicitSolve in the following way: ImplicitSolve[{x^2 + y^2 == 1}, {x -> 0, y -> 1}, {x, -1, 1}, AccuracyGoal -> 10] and Mathematica would return an interpolating function for y as a function of x, with an error of about 10^-8. Carl Woll Physics Dept U of Washington Here is the definition of ImplicitSolve: Clear[ImplicitSolve] Options[ImplicitSolve]={AccuracyGoal->6}; ImplicitSolve[eqn_List,rt_,{x_,min_,max_},opts___?OptionQ]:=Module[{x0,y,y0, acc}, (* options *) acc=AccuracyGoal/.{opts}/.Options[ImplicitSolve]; (* root *) x0=Cases[rt,(x->a_)->a]; If[x0=={}, Message[ImplicitSolve::badroot,x];Return[], x0=x0[[1]]]; {y,y0}=Transpose[DeleteCases[rt,x->x0]/.Rule->List]; (* check root *) If[!(And@@Thread[Abs[eqn/.Equal->Subtract/.rt]<10^-acc]), Message[ImplicitSolve::inaccurate]; y0=FindRoot[ Evaluate[eqn/.x->x0], Evaluate[Sequence@@(Transpose[{y,y0}])],opts][[All,2]], Null, Message[ImplicitSolve::incomplete];Return[] ]; (* get interpolating function *) NDSolve[ Flatten[{D[eqn/.Thread[Rule[y,#[x]&/@y]],x],Thread[#[x0]&/@y==y0]}], y, {x,min,max}, AccuracyGoal->acc PrecisionGoal->acc] ] ImplicitSolve::usage="ImplicitSolve[eqns, {x->x0,y->y0,...}, {x, xmin, xmax}, opts] finds a solution to the implicit equations eqns for the functions y, ... with the independent variable x in the range xmin to xmax. The root {x0,y0,...} should satisfy the equations, or should provide a good starting point for finding a solution when using FindRoot. Currently, the only available option is AccuracyGoal, but a better ImplicitSolve would include the possibility of supplying options for both the FindRoot and NDSolve function calls."; ImplicitSolve::badroot="Supplied root is missing value for `1`"; ImplicitSolve::incomplete="Supplied root is incomplete"; ImplicitSolve::inaccurate="Supplied root is inaccurate, using FindRoot to improve accuracy"; ----- Original Message ----- From: <Pavel.Pokorny at vscht.cz> To: mathgroup at smc.vnet.net Subject: [mg26212] [mg26189] solve f(x)=0, where f:Rn+1 -> Rn > Dear Mathematica friends > > Is there a way in Mathematica 4.0 to solve (numerically) the problem > f(x) = 0 > where f: R^{n+1} -> R^n, > i.e. f has n+1 real arguments and n real results ? > > The solution is (under certain conditions on f) > a curve in (n+1)-dim space. > > Example > x^2 + y^2 - 1 = 0 > is a unit circle. > > This problem is called "continuation" in nonlinear system analysis > see > Seydel: Tutorial on Continuation > Int.J.Bif.Chaos, Vol.1 No.1 (1991) pp 3-11. > > -- > Pavel Pokorny > Math Dept, Prague Institute of Chemical Technology > http://staff.vscht.cz/mat/Pavel.Pokorny >