Re: FindRoot / Jacobian

*To*: mathgroup at smc.vnet.net*Subject*: [mg84410] Re: FindRoot / Jacobian*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Fri, 21 Dec 2007 20:42:23 -0500 (EST)*References*: <fk73ub$6el$1@smc.vnet.net> <fk87t2$4f6$1@smc.vnet.net> <fkcst0$1fv$1@smc.vnet.net> <fkfstl$3n$1@smc.vnet.net>

I apologise. That wasn't a very clear (or even correct!) way of explaining it. Try this: Evaluate Plot3D[solU, {p, 0, 10 \[Pi]}, {q, 0, 10 \[Pi]}] Plot3D[solV, {p, 0, 10 \[Pi]}, {q, 0, 10 \[Pi]}] which shows the periodicity (modulo 2Pi) of solU and solV in the (p,q) plane, which means that each solution of FindRoot[solU == N[u] && solV == N[v], {{p, 0}, {q, 0}}] is also modulo 2Pi. A quick fix to my earlier solution is to constrain the solutions to lie in [-Pi,Pi] solutions=Table[FindRoot[solU==N[u]&&solV==N[v],{{p,0,-\[Pi],\[Pi]},{q,0,-\[Pi],\[Pi]}}],{u,-\[Pi],\[Pi],(2\[Pi])/70},{v,-\[Pi],\[Pi],(2\[Pi])/70}]; then the quick graphic to visualise the solutions is Graphics[{Map[Point,#,{2}],Map[Line,#],Map[Line,Transpose[#]]}&[{p,q}/.solutions//Chop]] which looks much smoother than the graphic I obtained before without constraining the solutions, though there are still places where the lines in the graphic jump about. I presume that by using a more intelligent choice of initial search point and constraint range in FindRoot you can make the solution vary smoothly everywhere. Stephen Luttrell West Malvern, UK "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk> wrote in message news:fkfstl$3n$1 at 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 >> >> > >