MATLAB: Initial condition of y(0) = 1

differential equestionsinitial conditionrunge kuttay=0

I am solving a differential equation (dy/dt = (-1/2*y*sin^2(t)) where y(0)=1) using different methods for my homework. For the Runge-Kutta method I keep getting the error: "Subscript indices must either be real positive integers or logicals."

I know that the y(0) = 1 initial condition is the problem, and that there just can't be a 0 in the parentheses, but I have know idea how to correct this? How can I put in the initial condition and make the program run. Below is the code that I am using if you need it.

clear;
clc;
h = input('What would you like the time step to be? >> ');
x = 0:h:10*pi;
y = zeros(1,length(x));
y(0) = 1;
F_xy = @(t,y) ((-1/2)*y*(sin(t)*sin(t)));
for i = 1:(length(x)-1)
  k_1 = F_xy(x(i),y(i));
  k_2 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_1));
  k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
  k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
    y(i+1) = y(i)+(1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
end

Best Answer

Just put it

y(1) = 1.0;

The '0' does not refer to the spatial x-value, but is an array index. Keep in mind that x(1)=0, and y(1) is the solution value at x(1).

Best wishes

Torsten.

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MATLAB: Runge-Kutta 4th order method

First up, you will need a much smaller step size to get an accurate solution using this explicit RK4 (with no error control). I suggest h = 0.05. Validate using say ode45 (which does have error control).

Then you will need to run your ode above three separate times, once starting from y(1) = 0.5, again with y(1) = 0.3, etc.

Then finally plot the result with plot(x,y,'o-').

h=0.5;                                             % step size
x = 0:h:100;                                         % Calculates upto y(3)
Y = zeros(1,length(x)); 
%y(1) = [-0.5;0.3;0.2];
y(1) = -0.5; % redo with other choices here. 
    % initial condition
F_xy = @(t,r) 3.*exp(-t)-0.4*r;                    % change the function as you desire
for i=1:(length(x)-1)                              % calculation loop
    k_1 = F_xy(x(i),y(i));
    k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
    k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
    k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
  
end
% validate using a decent ODE integrator  
tspan = [0,100]; y0 = -0.5;
[tx, yx] = ode45(F_xy, tspan, y0)
plot(x,y,'o-', tx, yx, '--')

MATLAB: Merge three initial values into one code by runge-kutta

h=0.5;                                             % step size
x = 0:h:100;                                         % Calculates upto y(3)
ibc = [0.2;0.3;0.2];
Y = zeros(length(ibc),length(x));
for j = 1:length(ibc)
    y = zeros(1,length(x)) ;
    y(1) = ibc(j) ;
    % redo with other choices here.
    % initial condition
    F_xy = @(t,r) 3.*exp(-t)-0.4*r;                    % change the function as you desire
    for i=1:(length(x)-1)                              % calculation loop
        k_1 = F_xy(x(i),y(i));
        k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
        k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
        k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
        y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
        
    end
    Y(j,:) = y ;
    
    % validate using a decent ODE integrator
    tspan = [0,100]; y0 = -0.5;
    [tx, yx] = ode45(F_xy, tspan, y0) ; 
    plot(x,y,'o-', tx, yx, '--')
    hold on
end

Best Answer

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