# Runge-Kutta method for 2x2 IVP?

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chrisholt34 el 29 de Abr. de 2015
Respondida: Philip Caplan el 30 de Abr. de 2015
I have a code written to input a vector into my RK4 file, but I keep getting all kinds of errors, including "Undefined function 'mtimes' for input arguments of type 'cell'." my code is as follows
function [Y, t] = RK42d(f, t0, T, y0, N)
%the input of f and y0 must both be vectors
%composed of two fuctions and their respective
%intial conditions
%the vector Y will return a vector as well
h = (T - t0)/(N - 1); %Calculate and store the step size
Y = zeros(2,N); %Initialize the X and Y vector
t = linspace(t0,T,N); % A vector to store the time values
Y(:,1) = y0; % Start Y vector at the intial values.
for i = 1:(N-1)
k1 = f(t(i),Y(i));
k2 = f{t(i)+0.5*h, Y(i)+0.5*h*k1};
k3 = f{t(i)+0.5*h, Y(i)+0.5*h*k2};
k4 = f{t(i)+h, Y(i)+h*k3};
Y{i+1} = Y{i} + (h/6)*(k1+ 2.*k2 + 2*k3 + k4);
%Update approximation
end
%%%END FILE %%%
I've tried running it using another .m file, using
f = {@(t,y)(y(1) - 4.*y(2)); @(t,y)(-y(1) + y(2))};
y0 = [1;0];
t0 = 0;
T = 1;
N = 11;
RK42d(@(t,y)f, t0, T, y0, N)
but for some reason it won't work. Any tips?
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Philip Caplan el 30 de Abr. de 2015
I have made some modifications to the shapes of the arrays in your code and changed the representation of the system of ODEs as a cell array of function handles to an array of function handles. The code below also demonstrates how "ode45" can be used to solve the problem. Both methods give the same result!
function testODE
f = @(t,y) [y(1)-4.*y(2);-y(1)+y(2)];
y0 = [1;0];
t0 = 0;
T = 1;
N = 11;
[Y,t] = RK42d(f, t0, T, y0, N);
figure;
hold on;
plot(t,Y(:,1),'r');
plot(t,Y(:,2),'b');
sol = ode45(f,[0,1],y0);
plot(sol.x,sol.y(1,:),'go');
plot(sol.x,sol.y(2,:),'ko');
legend('RK42d - y1','RK42d - y2','ode45 - y1','ode45 - y2');
end
function [Y, t] = RK42d(f, t0, T, y0, N)
%the input of f and y0 must both be vectors
%composed of two fuctions and their respective
%intial conditions
%the vector Y will return a vector as well
h = (T - t0)/(N - 1); %Calculate and store the step size
Y = zeros(N,2); %Initialize the X and Y vector
t = linspace(t0,T,N); % A vector to store the time values
Y(1,:) = y0; % Start Y vector at the intial values.
for i = 1:(N-1)
y = Y(i,:)';
k1 = f(t(i),y);
k2 = f(t(i) +0.5*h, y +0.5*h*k1);
k3 = f(t(i) +0.5*h, y +0.5*h*k2);
k4 = f(t(i) +h , y +h*k3);
Y(i+1,:) = y + (h/6)*(k1+ 2.*k2 + 2*k3 + k4);
%Update approximation
end
end
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