Re: Table NDSolve Plot
- To: mathgroup at smc.vnet.net
- Subject: [mg121951] Re: Table NDSolve Plot
- From: Peter Pein <petsie at dordos.net>
- Date: Fri, 7 Oct 2011 04:46:52 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <j6joh8$m3g$1@smc.vnet.net>
Am 06.10.2011 10:23, schrieb Howie: > I know this is a basic question, but I am having a hard time figuring > this out on my own. > > I would like to numerically solve an ordinary differential equation > with a parameter which varies over a range of numbers and plot all the > solutions with different values of the parameter on the same plot. > Also, I would like to be able to then over-plot another solution on > the same plot. > > For instance: > > h=0 > n=0 > k=0.25 > s = NDSolve[{(h - k^2 n (1 + n) z^2) w[z] - > z (1 + k^2 - 2 k^2 z^2) Derivative[1][w][ > z] + (1 - z^2) (1 - k^2 z^2) (w^\[Prime]\[Prime])[z] == 0, > w[2] == 1, w'[2] == -1}, w, {z, 1.1, 3.0}] > Plot[Evaluate[w[z] /. s], {z, 1.1, 3.0}, PlotStyle -> Automatic] > Plot[1/2 Log[(z + 1)/(z - 1)], {z, 1.1, 3.0}, PlotStyle -> Automatic] > I would like to do this for instance 10 increments of k from 0 to 1, > and also the Log plot, all on the same plot. > > Thanks! > Hi, why not solving the differential equation exactly? Then create all plots you want and display these together with the log-plot using Show. something like: In[1]:= f[z_] = FullSimplify[With[{h = 0, n = 0}, w[z] /. First[DSolve[{(1 - z^2)*(1 - k^2*z^2)*Derivative[2][w][z] - z*(1 + k^2 - 2*k^2*z^2)* Derivative[1][w][z] + (h - n*(n + 1)*k^2*z^2)*w[z] == 0, w[2] == Log[3]/2, Derivative[1][w][2] == -(1/3)}, w, z]]], 0 < k < 1 && 1/k < z] Out[1]= (1/6)*(2*I*Sqrt[3 - 12*k^2]*(EllipticF[ArcSin[2], k^2] - EllipticF[ArcSin[z], k^2]) + Log[27]) In[2]:= Show[Join[{Plot[(1/2)*Log[(z + 1)/(z - 1)], {z, 1.1, 3.}, PlotStyle -> Thick]}, Table[Plot[f[z], {z, 1, 3}, PlotStyle -> Hue[k]], {k, Range[9]/10}]], PlotRange -> {All, All}, Axes -> None, Frame -> True, GridLines -> {Automatic, (1/4)*Log[3]*Range[5]}, GridLinesStyle -> LightGray] Peter