Re: Iteration of Initial Conditions for required Boundary Value with NDSolve

*To*: mathgroup at smc.vnet.net*Subject*: [mg52811] Re: Iteration of Initial Conditions for required Boundary Value with NDSolve*From*: bghiggins at ucdavis.edu*Date*: Mon, 13 Dec 2004 04:23:54 -0500 (EST)*References*: <cpejtd$6vr$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

If I understand your problem correctly, believe the problem as stated is overspecified. You cannot specify smax and then search for a value of sz such that x(smax)=3 and y(smax)=2. What you need to do is search for both sz and smax . One way to do this is to use a shooting method. The following code does the task automatically given initial guesses for sz and smax. I should point out that the solution space has a turning point(limit point) so there are likely two solutions. depending on what your initial guesses are for sz and smax. I have a web site that has a notebook that discusses the shooting method in some detail: http://www.higgins.ucdavis.edu/chemmath.php Here is the code for your problem kgi = 3 10^-3; soln[sz1_, smax1_] := NDSolve[{ph''[ s] == ph'[s]/s, ph[sz1] == 10^-7, ph'[sz1] == kgi, x'[s] == Cos[ph[s]], x[sz1] == 0, y'[s] == Sin[ph[s]], y[sz1] == 0}, {ph, x, y}, {s, sz1, smax1}]; yend[sz1_?NumericQ, smax1_?NumericQ] := First[y[smax1] /. soln[sz1, smax1]] xend[sz1_?NumericQ, smax1_?NumericQ] := First[x[smax1] /. soln[sz1, smax1]] val = {sz1, smax1} /. FindRoot[Evaluate[{yend[sz1, smax1] == 2, xend[ sz1, smax1] == 3}], {sz1, 0.01, 0.02}, {smax1, 3.5, 4}]; xyplot[{sz1_, smax1_}] := ParametricPlot[Evaluate[{x[s], y[s]} /. soln[sz1, smax1]], {s, sz1, smax1}] xyplot[val]