Runge-Kutta for solving differential equation with final value (backward integration)
7 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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
0 commentaires
Réponses (1)
Divija Aleti
le 27 Août 2020
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
Voir également
Catégories
En savoir plus sur Numerical Integration and Differential Equations dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!