differential equations - Strange DAE solution

I have a system, consisting of one differential equation and one algebraic equation

req = (1 - xD)/τD - δxD xD ν;
ν = νmax (2/(1 + Exp[(-2 (u - u0))/kv]) - 1);
eqh = u == Istim + gE ν (xD - 0.5);
τD = 2; δxD = 0.01; νmax = 100; u0 = 150; kv = 20; \
Istim = 126; gE = 5;
eqs1 = xD'[t] == (req /. {xD -> xD[t], u -> u[t]});
eqs2 = eqh /. {xD -> xD[t], u -> u[t]}

ics = xD[0] == 0.99;

I try to solve it with the next command:

eqsol = NDSolve[{eqs1, eqs2, ics}, {xD[t], u[t]}, {t, 0, 1}];

Then I want to make sure that this is correct solution, so I plot both variable xD[t] and u[t] and expect to find them on the curve defined by implicit equation from the system.

ParametricPlot[Evaluate[{u[t], xD[t]} /. eqsol], {t, 0, 1}, 
 PlotRange -> Full]

I also can plot that curve by expressing the variable xD[t] as a function of the variable u: xD=xD(u).

xDimp = Solve[eqh, xD];
Plot[Evaluate[xD /. xDimp], {u, u0, 300}]

But I found that these curves don't coincide with each other. I suspect that the integration was executed wrong, but no errors were displayed during the integration. By observing the apt derivative I investigated that the function u(t) can be explicitly found from the implicit equation only for xD>0.54. This solution satisfies me. Due to this constraint I set the initial value as xD=0.99. But shall I define this constraint explicitly? Or else I don't know where the error can be.

Answer

Your equation set actually has 3 solutions. Just enlarge the domain of definition of Plot a bit:

curve = Plot[Evaluate[xD /. xDimp], {u, -300, 400}, 
  Epilog -> {Dashed, Line[{{-300, 0.99}, {400, 0.99}}]}]

Mathematica graphics

Clearly there're three possible values of u that satisfys the i.c. xD==0.99. What's found by NDSolve automatically is the leftmost one:

solplot1 = ParametricPlot[Evaluate[{u[t], xD[t]} /. eqsol], {t, 0, 1}, PlotStyle -> Red];

curve~Show~solplot1

Mathematica graphics

Why does NDSolve find the leftmost one only? The answer is hidden in the obscure tutorial tutorial/NDSolveDAE#1966826857. In short, since the i.c. for u isn't explicitly given in the code, NDSolve will try to determine it from an initial guess u[0] == 1, which converges to the leftmost solution in this case.

To make NDSolve converge to other solutions, we need to use other initial guesses:

eqsol2 = NDSolve[{eqs1, eqs2, ics}, {xD[t], u[t]}, {t, -1, 1/2}, 
     Method -> {"DAEInitialization" -> {"Collocation", 

         "DefaultStartingValue" -> #}}][[1]] & /@ {-120, 150, 400}

solplot2 = ParametricPlot[Evaluate[{u[t], xD[t]} /. eqsol2], {t, -1, 1/2}, 
   PlotStyle -> {Green, Orange, Red}];

curve~Show~solplot2

Mathematica graphics

If you don't want to play with those options, explicitly giving i.c. of u to NDSolve is OK, of course:

ustart = u /. FindRoot[eqh /. xD -> 0.99, {u, #}] & /@ {0, 160, 300}

(* {-119., 152.142, 371.} *)    
eqsol3 = NDSolve[{eqs1, eqs2, ics, u[0] == #}, {xD[t], u[t]}, {t, -1, 1/2}][[1]] & /@ 
  ustart

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

Blog

Search This Blog