Runge-Kutta for solving differential equation with final value (backward integration)
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Antworten (1)
Divija Aleti
am 27 Aug. 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
Siehe auch
Kategorien
Mehr zu Numerical Integration and Differential Equations finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!