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

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

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}]
Post a Comment
Read more