Re: FindRoot / Jacobian

*To*: mathgroup at smc.vnet.net*Subject*: [mg84389] Re: FindRoot / Jacobian*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Fri, 21 Dec 2007 03:18:17 -0500 (EST)*References*: <fk73ub$6el$1@smc.vnet.net> <fk87t2$4f6$1@smc.vnet.net> <fkcst0$1fv$1@smc.vnet.net>

I don't think the problem is to do with precision issues. If you look at the plots Plot3D[solU, {p, 0, 2 \[Pi]}, {q, 0, 2 \[Pi]}] Plot3D[solV, {p, 0, 2 \[Pi]}, {q, 0, 2 \[Pi]}] it is clear that there is a continuum of solutions from which FindRoot[solU == N[u] && solV == N[v], {{p, 0}, {q, 0}}] picks a single solution, whilst warning you that it is having difficulty finding a unique solution by outputting the following message: "The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function." I seem to have I overlooked this warning message earlier. Stephen Luttrell West Malvern, UK <sigmundv at gmail.com> wrote in message news:fkcst0$1fv$1 at smc.vnet.net... > Thank you for the suggestion, Steve. It sure helped me clean up the > code and get rid of the complex values (which was expected since I > used complex starting points in FindRoot before). > > Do you know how I can increase the precision in FindRoot, without > increasing the computing time dramatically? If you plot p/.solutions > and q/.solutions, respectively, using e.g. the ListSurfacePlot3D > command, you see some peaks. Ideally those peaks should not be there. > Do you have any clever suggestion on how to get rid of those peaks? > > Sigmund Vestergaard > Kongens Lyngby, Denmark > > On Dec 18, 11:37 am, "Steve Luttrell" > <steve at _removemefirst_luttrell.org.uk> wrote: >> Here is how you can solve your problem: >> >> solU=(E^((-I)*p-(2*ArcTan[Tan[q/2]/Sqrt[3]])/Sqrt[3])*(1+E^((2*I)*p)))/2 >> solV=Sin[p]/E^((2*ArcTan[Tan[q/2]/Sqrt[3]])/Sqrt[3]) >> >> solutions=Table[FindRoot[solU==N[u]&&solV==N[v],{{p,0},{q,0}}],{u,-\[Pi],\[Pi],(2\[Pi])/70},{v,-\[Pi],\[Pi],(2\[Pi])/70}]; >> >> You can do a quick visualisation of the solutions with the following >> graphics: >> >> Graphics[{Map[Point,#,{2}],Map[Line,#],Map[Line,Transpose[#]]}&[{p,q}/.solutions//Chop]] >> >> Stephen Luttrell >> West Malvern, UK >> >> <sigmu... at gmail.com> wrote in messagenews:fk73ub$6el$1 at smc.vnet.net... >> > Dear readers, >> >> > Consider the following expressions: >> >> > solU = (E^((-I)*p - (2*ArcTan[Tan[q/2]/Sqrt[3]])/Sqrt[3])*(1 + >> > E^((2*I)*p)))/2 >> >> > solV = Sin[p]/E^((2*ArcTan[Tan[q/2]/Sqrt[3]])/Sqrt[3]) >> >> > I then want to solve the system of equations (u = solU, v = solV) >> > numerically on a grid (u,v) \in (-pi,pi). For that purpose I use >> > FindRoot, and execute the following: >> >> > u1 = Pi; du = 2*Pi/70; >> >> > uvPts1 = Flatten[Table[{u, v}, {u, u1 + du, u2 - du, du}, {v, v1 + >> > du, >> > v2 - du, dv}], 1]; >> >> > uvNew = {p, q} /. Table[FindRoot[{solU == uvPts1[[i, 1]], solV == >> > uvPts1[[i, 2]]}, {p, >> > -2}, {q, 1+I}], {i, Length[uvPts1]}]; >> >> > This gives me a p-value and a q-value for each point on the grid, no >> > problem. However, I get the wrong solution at some of the grid points. >> > Since I get several of the following type of messages, this would be >> > expected: >> >> > FindRoot::jsing:Encountered a singular Jacobian at the point {p,q} = \ >> > {-10.4516+0.784848 \[ImaginaryI],108.092+104.013 \[ImaginaryI]}. Try \ >> > perturbing the initial point(s). >> >> >> > Then I would like to know if there is any 'nice' way to handle these >> > 'singular Jacobian' errors? Does there exist any kind of 'event >> > handler', as in NDSolve? >> >> > Kind regards, >> > Sigmund Vestergaard > >