Solve a system of three equations with Runge Kutta

2019-02-26 03:02发布

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I want to solve a system of THREE differential equations with the Runge Kutta 4 method in Matlab (Ode45 is not permitted).

After a long time spent looking, all I have been able to find online are either unintelligible examples or general explanations that do not include examples at all. I would like a concrete example on how to implement my solution properly, or the solution to a comparable problem which I can build on.

I have come quite far; my current code spits out a matrix with 2 correct decimals on most of the components, which I am quite happy with.

However, when the step-size is decreased, the errors become enormous. I know the for-loop I have created is not entirely correct. I may have defined the functions incorrectly, but I am quite certain that the problem is solved if some minor changes are made to the for-loop because it appears to be solving the equation-system fairly well already in its current state.

clear all, close all, clc

%{
____________________TASK:______________________
Solve the system of differential equations below 
in the interval 0<t<1, with stepsize h = 0.1.
x'= y                 x(0)=1
y'= -x-2e^t+1         y(0)=0   ,   where x=x(t), y=y(t),  z=z(t)  
z'= -x - e^t + 1      z(0)=1

THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________

%}

h = 0.1;
t    = 0:h:1
N = length(t);

%Defining the functions
x    = zeros(N,1);%I am not entierly sure if x y z are supposed to be defined in this way.
y    = zeros(N,1)
z    = zeros(N,1)

f = @(t, x, y, z) -x-2*exp(t)+1;%Question: Do i need a function for x here as well??
g = @(t, x, y, z) -x - exp(t) + 1;

%Starting conditions
x(1) = 1; 
y(1) = 0;
z(1) = 1;

for i = 1:(N-1)
    K1     = h * ( y(i));%____I think z(i) is supposed to be here, but i dont know in what way. 
    L1     = h * f( t(i)          , x(i)        , y(i) ,      z(i));
    M1     = h * g( t(i)          , x(i)        , y(i) ,      z(i));

    K2     = h *  (y(i) + 1/2*L1 + 1/2*M1);%____Again, z(i) should probably be here somewhere. 
    L2     = h * f(t(i) + 1/2*h,  x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
    M2     = h * g(t(i) + 1/2*h,  x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);

    K3     = h *  (y(i) + 1/2*L2 + 1/2*M2);%____z(i). Should it just be added, like "+z(i)" ? 
    L3     = h * f(t(i) + 1/2*h,  x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
    M3     = h * g(t(i) + 1/2*h,  x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);

    K4     = h *  (y(i) + L3 + M3);%_____z(i) ... ? 
    L4     = h * f( t(i)+h    ,  x(i)+K3     , y(i)+L3,     z(i)+M3);
    M4     = h * g( t(i)+h    ,  x(i)+K3     , y(i)+L3,     z(i)+M3);

    x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);                                                                             
    y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
    z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end

Answer_Matrix = [t' x y z]

2条回答
闹够了就滚
2楼-- · 2019-02-26 03:51

Ok, turns out it was just a minor mistake where the x-variable was not defined as a function of y (as x'(t)=y according to the problem.

So: Below is a concrete example on how to solve a differential equation system using Runge Kutta 4 in matlab:

clear all, close all, clc

%{
____________________TASK:______________________
Solve the system of differential equations below 
in the interval 0<t<1, with stepsize h = 0.1.
x'= y                 x(0)=1
y'= -x-2e^t+1         y(0)=0   ,   where x=x(t), y=y(t),  z=z(t)  
z'= -x - e^t + 1      z(0)=1

THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________

%}

%Step-size
h = 0.1;
t    = 0:h:1
N = length(t);

%Defining the vectors where the answer is stored. 
x    = zeros(N,1);
y    = zeros(N,1)
z    = zeros(N,1)

%Defining the functions
e = @(t, x, y, z) y;
f = @(t, x, y, z) -x-2*exp(t)+1;
g = @(t, x, y, z) -x - exp(t) + 1;

%Starting/initial conditions
x(1) = 1; 
y(1) = 0;
z(1) = 1;

for i = 1:(N-1)
    K1     = h * e( t(i)          , x(i)        , y(i) ,      z(i));
    L1     = h * f( t(i)          , x(i)        , y(i) ,      z(i));
    M1     = h * g( t(i)          , x(i)        , y(i) ,      z(i));

    K2     = h * e(t(i) + 1/2*h,  x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
    L2     = h * f(t(i) + 1/2*h,  x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
    M2     = h * g(t(i) + 1/2*h,  x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);

    K3     = h * e(t(i) + 1/2*h,  x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
    L3     = h * f(t(i) + 1/2*h,  x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
    M3     = h * g(t(i) + 1/2*h,  x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);

    K4     = h * e( t(i)+h    ,  x(i)+K3     , y(i)+L3,     z(i)+M3);
    L4     = h * f( t(i)+h    ,  x(i)+K3     , y(i)+L3,     z(i)+M3);
    M4     = h * g( t(i)+h    ,  x(i)+K3     , y(i)+L3,     z(i)+M3);

    x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);                                                                             
    y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
    z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end

Answer_Matrix = [t' x y z]
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祖国的老花朵
3楼-- · 2019-02-26 03:56

So your main issue was not defining x properly. You were propagating its value using the Runge Kutta 4 (RK4) method, but never actually defined what its derivative was!

At the bottom of this answer is a function which can take any given number of equations and their initial conditions. This has been included to address your need for a clear example for three (or more) equations.

For reference, the equations can be directly lifted from the standard RK4 method described here.

Working Script

This is comparable to yours, but uses slightly clearer naming conventions and structure.

% Initialise step-size variables
h = 0.1;
t = (0:h:1)';
N = length(t);

% Initialise vectors
x = zeros(N,1);    y = zeros(N,1);    z = zeros(N,1);
% Starting conditions
x(1) = 1;     y(1) = 0;    z(1) = 1;

% Initialise derivative functions
dx = @(t, x, y, z) y;                  % dx = x' = dx/dt
dy = @(t, x, y, z) - x -2*exp(t) + 1;  % dy = y' = dy/dt
dz = @(t, x, y, z) - x -  exp(t) + 1;  % dz = z' = dz/dt

% Initialise K vectors
kx = zeros(1,4); % to store K values for x
ky = zeros(1,4); % to store K values for y
kz = zeros(1,4); % to store K values for z
b = [1 2 2 1];   % RK4 coefficients

% Iterate, computing each K value in turn, then the i+1 step values
for i = 1:(N-1)        
    kx(1) = dx(t(i), x(i), y(i), z(i));
    ky(1) = dy(t(i), x(i), y(i), z(i));
    kz(1) = dz(t(i), x(i), y(i), z(i));

    kx(2) = dx(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
    ky(2) = dy(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
    kz(2) = dz(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));

    kx(3) = dx(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
    ky(3) = dy(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
    kz(3) = dz(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));

    kx(4) = dx(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
    ky(4) = dy(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
    kz(4) = dz(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));

    x(i+1) = x(i) + (h/6)*sum(b.*kx);       
    y(i+1) = y(i) + (h/6)*sum(b.*ky);       
    z(i+1) = z(i) + (h/6)*sum(b.*kz);        
end    

% Group together in one solution matrix
txyz = [t,x,y,z];

Implemented as function

You wanted code which can "be applied to any equation system". To make your script more usable, let's take advantage of vector inputs, where each variable is on its own row, and then make it into a function. The result is something comparable (in how it is called) to Matlab's own ode45.

% setup
odefun = @(t, y) [y(2); -y(1) - 2*exp(t) + 1; -y(1) - exp(t) + 1];
y0 = [1;0;1];
% ODE45 solution
[T, Y] = ode45(odefun, [0,1], y0);
% Custom RK4 solution
t = 0:0.1:1;
y = RK4(odefun, t, y0);
% Compare results
figure; hold on;
plot(T, Y); plot(t, y, '--', 'linewidth', 2)

You can see that the RK4 function (below) gives the same result of the ode45 function.

results comparison

The function RK4 is simply a "condensed" version of the above script, it will work for however many equations you want to use. For broad use, you would want to include input-checking in the function. I have left this out for clarity.

function y = RK4(odefun, tspan, y0)
% ODEFUN contains the ode functions of the system
% TSPAN  is a 1D vector of equally spaced t values
% Y0     contains the intial conditions for the system variables

    % Initialise step-size variables
    t = tspan(:); % ensure column vector = (0:h:1)';
    h = t(2)-t(1);% define h from t
    N = length(t);

    % Initialise y vector, with a column for each equation in odefun
    y = zeros(N, numel(y0));
    % Starting conditions
    y(1, :) = y0(:)';  % Set intial conditions using row vector of y0

    k = zeros(4, numel(y0));              % Initialise K vectors
    b = repmat([1 2 2 1]', 1, numel(y0)); % RK4 coefficients

    % Iterate, computing each K value in turn, then the i+1 step values
    for i = 1:(N-1)        
        k(1, :) = odefun(t(i), y(i,:));        
        k(2, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(1,:));        
        k(3, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(2,:));        
        k(4, :) = odefun(t(i) + h, y(i,:) + h*k(3,:));

        y(i+1, :) = y(i, :) + (h/6)*sum(b.*k);    
    end    
end
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