Runge-Kutta for solving differential equation with final value (backward integration)

Hello,

I have the following simple differential equaiotn: x' = -x-2 with a final value x(2)=0.

I solved it anlytically and got: x(t)=2(exp(2-t) -1). It satisfies the final condition x(2)=0. Also, x(0)=12.77.

Then I tried to solve it numerically using RK4 and plot the numerical solution with the analytical solution togehter

but I can see a big difference at t=0. Here is my code:

% x_f = 0

function y = Test(x_f)

test = -1;

epsn = 0.001;

N = 1000;

t = linspace(0,2,N+1);

h = 1/N;

h2 = h/2;

x = zeros(1,N+1);

x(N+1)=x_f;

while(test < 0)

oldx = x;

% Backward sweep

for i = 1:N

j = N + 2 - i;

k1 = -2-x(j);

k2 = -2-(x(j)-h2*k1);

k3 = -2-(x(j)-h2*k2);

k4 = -2-(x(j)-h*k3);

x(j-1) = x(j)-(h/6)*(k1 + 2*k2 + 2*k3 + k4);

end

temp = epsn*sum(abs(x)) - sum(abs(oldx - x));

test = temp;

end

% Analytical solution

x_cal = 2*(exp(2-t)-1);

y = x;

figure

plot(t,x,t,x_cal)

I'd appreciate any feedback on this issue.

Saleh

Answer 1

[Divija Aleti]

The huge difference in the initial values is because the value of ‘h’ should be 2/N.

This value is calculated as follows:

h = (t_final – t_initial)/(Number of Points - 1) = (2-0)/((N+1)-1) = 2/N

[Saleh Alshamali]

Many thanks. It's working now.