Skip to main content

equation solving - 4th-order Runge-Kutta method to solve a system of coupled ODEs




I am a beginner at Mathematica programming and with the Runge-Kutta method as well. I'm trying to solve a system of coupled ODEs using a 4th-order Runge-Kutta method for my project work.


I have solved it by NDSolve, but I want to solve this by 4th-order Runge-Kutta method. Here is my problem:


Γ = 1.4    
k = 0
z = 0
β = 0.166667

k1 = (d[η] v[η] η (1 - z d[η]) (v[η] - η) - 2 p[η] η (1 - z d[η]) - ϕ[η]^2 d[η]
(1 - z d[η]) - Γ p[η] v[η])/((Γ p[η] - (v[η] - η)^2 d[η] (1 - z d[η])) η)


k2 = (d[η] (1 - z d[η]) (v[η] d[η] (v[η] - 2 η) (v[η] - η) + 2 p[η] η + ϕ[η]^2
d[η]))/((Γ p[η] - (v[η] - η)^2 d[η] (1 - z d[η])) (v[η] - η) η)

k3 = (p[η] d[η] (2 η (v[η] - η)^2 (1 - z d[η]) + Γ v[η] (v[η] - 2 η) (v[η] - η) +
ϕ[η]^2 Γ))/((Γ p[η] - (v[η] - η)^2 d[η] (1 - z d[η])) (v[η] - η) η)

k4 = -((ϕ[η] (v[η] + η))/(η (v[η] - η)))

k5 = -(w[η]/(η (v[η] - η)))


sol = NDSolve[{v'[η] == k1, d'[η] == k2, p'[η] == k3, ϕ'[η] == k4, w'[η] == k5,
v[1] == (1 - β), d[1] == 1/β, p [1] == (1 - β), ϕ[1] == 0.01, w[1] == 0.02},
{v, d, p, ϕ, w}, {η, 0, 1}, MaxSteps -> 30000]

Please guide me how can I solve the above problem with 4th-order Runge-Kutta method, thanks.


code for RK4 method are given in Solving a system of ODEs with the Runge-Kutta method


but how can I apply those codes to my problem...please guide me...



Answer



According to your statement, I think what you need is just 4th-order Runge-Kutta method, and a completely self-made implementation of 4th-order Runge-Kutta method isn't necessary, then the answer from J.M. has shown you the optimal direction:


(* Unchanged part omitted. *)


ClassicalRungeKuttaCoefficients[4, prec_] :=With[{amat = {{1/2}, {0, 1/2}, {0, 0, 1}},
bvec = {1/6, 1/3, 1/3, 1/6}, cvec = {1/2, 1/2, 1}}, N[{amat, bvec, cvec}, prec]]

sol = NDSolve[{v'[η] == k1, d'[η] == k2, p'[η] == k3, ϕ'[η] == k4, w'[η] == k5,
v[1] == (1 - β), d[1] == 1/β, p[1] == (1 - β), ϕ[1] == 0.01, w[1] == 0.02},
{v, d, p, ϕ, w}, {η, 0, 1},
Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 4,
"Coefficients" -> ClassicalRungeKuttaCoefficients}, StartingStepSize -> 1/10000]


However, what I really want to point out is, despite the above code seems to solve your ODE set up to η = 0.0001, I'm afraid it's not reliable at all:


 {{nl, nr}} = (v /. sol)[[1]]["Domain"];
Plot[{v@n, d@n, p@n, ϕ@n, w@n} /. sol // Evaluate, {n, nl, nr}]

enter image description here


NDSolve by default setting doesn't manage to solve this set of equation, too. It stopped at about η = 0.9576. (I'm not sure what do you mean by saying you have solved it by NDSolve.) I'm not surprised though, your ODEs are non-linear. As for how to solve the ODEs, it's another question. I vote to close this question as a duplicate.


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.