Skip to main content

differential equations - Runge-Kutta 2nd Order ODE Solver


Suppose I have a 2nd order ODE of the form y''(t) = 1/y with y(0) = 0 and y'(0) = 10, and want to solve it using a Runge-Kutta solver. I've read that we need to convert the 2nd order ODE into two 1st order ODEs, but I'm having trouble doing that at the moment and am hoping someone here might be able to help. This is my code thus far:


Remove["Global`*"]

(*dy/dt=*)f[t_, y_] := 1/y;
(*d^2y/dt^2=*)g[t_, y_, yd_] :=???;
t[0] = 0;
y[0] = 0;
yd[0] = 10;
tmax = 1000;
h = 0.01;

Do[
{t[n] = t[0] + h n,


k1 = h f[t[n], y[n], yd[n]];
l1 = h g[t[n], y[n], yd[n]];

k2 = h f[t[n] + h/2, y[n] + k1/2, yd[n] + l1/2];
l2 = h g[t[n] + h/2, y[n] + k1/2, yd[n] + l1/2];

k3 = h f[t[n] + h/2, y[n] + k2/2, yd[n] + l2/2];
l3 = h g[t[n] + h/2, y[n] + k2/2, yd[n] + l2/2];


k4 = h f[t[n] + h, y[n] + k3, yd[n] + l3];
l4 = h g[t[n] + h, y[n] + k3, yd[n] + l3];

y[n + 1] = y[n] + 1/6 (k1 + 2 k2 + 2 k3 + k4);
yd[n + 1] = yd[n] + 1/6 (l1 + 2 l2 + 2 l3 + l4);

}, {n, 0, tmax}]

As you can see by the question marks for the function g[t_,y_,yd_], I don't know how I can set it in such a way that y''(t) = 1/y. Do I feed the results of y[n+1] into g when running the algorithm? Any help would be appreciated.



Answer




You can specify the functions like this


(*dy/dt=*)f[t_, y_, yd_] := yd;
(*d^2y/dt^2=*)g[t_, y_, yd_] := 1/y;
t[0] = 0;
y[0] = 1;
yd[0] = 10;
tmax = 1000;
h = 0.01;
Do[{t[n] = t[0] + h n, k1 = h f[t[n], y[n], yd[n]];
l1 = h g[t[n], y[n], yd[n]];

k2 = h f[t[n] + h/2, y[n] + k1/2, yd[n] + l1/2];
l2 = h g[t[n] + h/2, y[n] + k1/2, yd[n] + l1/2];
k3 = h f[t[n] + h/2, y[n] + k2/2, yd[n] + l2/2];
l3 = h g[t[n] + h/2, y[n] + k2/2, yd[n] + l2/2];
k4 = h f[t[n] + h, y[n] + k3, yd[n] + l3];
l4 = h g[t[n] + h, y[n] + k3, yd[n] + l3];
y[n + 1] = y[n] + 1/6 (k1 + 2 k2 + 2 k3 + k4);
yd[n + 1] = yd[n] + 1/6 (l1 + 2 l2 + 2 l3 + l4);}, {n, 0, tmax}]

The function f is the derivative of y and therefore equal to yd. The function g is the derivative of yd which means it is the second derivative of the function you are looking for. Here you can specify the right hand side of the equation y''=1/y.



Also, you shouldn't specify y(0) = 0 in your initial conditions, because then 1/y is not defined.


Comments

Popular posts from this blog

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 - 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 - 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.