Runge-Kutta Numerical Method Bad Aproximation
I´m attempting to use Runge-Kutta method to compare it to the lsode function. But it is performing rather poorly, every other method I used (fowards and backwards Euler, Heun) to compare to lsode do a way better job to the point they are almost indistinguishable from lsode.
This is what my code returns
https://i.stack.imgur.com/vJ6Yi.png
If anyone can pointout a way to improve it or if I doing something wrong I´d appreciate it.
The following is what I use for the Runge-Kutta method
%Initial conditions
u(1) = 1;
v(1) = 2;
p(1) = -1/sqrt(3);
q(1) = 1/sqrt(3);
%Graf interval / step size
s0 = 0;
sf = 50;
h = 0.25;
n=(sf-s0)/h;
s(1) = s0;
%-----------------------------------------------------------------------%
for j = 2:n
i = j-1;
k1_u(j) = p(i);
k1_v(j) = q(i);
k1_p(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
k1_q(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
u1(j) = p(i) + (1/2)*k1_u(j)*h;
v1(j) = q(i) + (1/2)*k1_v(j)*h;
p1(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_p(j)*h;
q1(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_q(j)*h;
k2_u(j) = p1(j);
k2_v(j) = q1(j);
k2_p(j) = (-2*v1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
k2_q(j) = (-2*u1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
u2(j) = p(i) + (1/2)*k2_u(j)*h;
v2(j) = q(i) + (1/2)*k2_v(j)*h;
p2(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_p(j)*h;
q2(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_q(j)*h;
k3_u(j) = p2(j);
k3_v(j) = q2(j);
k3_p(j) = (-2*v2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
k3_q(j) = (-2*u2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
u3(j) = p(i) + k3_u(j)*h;
v3(j) = q(i) + k3_v(j)*h;
p3(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_p(j)*h;
q3(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_q(j)*h;
k4_u(j) = p3(j);
k4_v(j) = q3(j);
k4_p(j) = (-2*v3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
k4_q(j) = (-2*u3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
s(j) = s(j-1) + h;
u(j) = u(j-1) + (h/6)*(k1_u(j) + 2*k2_u(j) + 2*k3_u(j) + k4_u(j));
v(j) = v(j-1) + (h/6)*(k1_v(j) + 2*k2_v(j) + 2*k3_v(j) + k4_v(j));
p(j) = p(j-1) + (h/6)*(k1_p(j) + 2*k2_p(j) + 2*k3_p(j) + k4_p(j));
q(j) = q(j-1) + (h/6)*(k1_q(j) + 2*k2_q(j) + 2*k3_q(j) + k4_q(j));
endfor
subplot(2,3,1), plot(s,u);
hold on; plot(s,v); hold off;
title ("Runge-Kutta");
h = legend ("u(s)", "v(s)");
legend (h, "location", "northwestoutside");
set (h, "fontsize", 10);
matlab octave numerical-methods ode runge-kutta
add a comment |
I´m attempting to use Runge-Kutta method to compare it to the lsode function. But it is performing rather poorly, every other method I used (fowards and backwards Euler, Heun) to compare to lsode do a way better job to the point they are almost indistinguishable from lsode.
This is what my code returns
https://i.stack.imgur.com/vJ6Yi.png
If anyone can pointout a way to improve it or if I doing something wrong I´d appreciate it.
The following is what I use for the Runge-Kutta method
%Initial conditions
u(1) = 1;
v(1) = 2;
p(1) = -1/sqrt(3);
q(1) = 1/sqrt(3);
%Graf interval / step size
s0 = 0;
sf = 50;
h = 0.25;
n=(sf-s0)/h;
s(1) = s0;
%-----------------------------------------------------------------------%
for j = 2:n
i = j-1;
k1_u(j) = p(i);
k1_v(j) = q(i);
k1_p(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
k1_q(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
u1(j) = p(i) + (1/2)*k1_u(j)*h;
v1(j) = q(i) + (1/2)*k1_v(j)*h;
p1(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_p(j)*h;
q1(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_q(j)*h;
k2_u(j) = p1(j);
k2_v(j) = q1(j);
k2_p(j) = (-2*v1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
k2_q(j) = (-2*u1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
u2(j) = p(i) + (1/2)*k2_u(j)*h;
v2(j) = q(i) + (1/2)*k2_v(j)*h;
p2(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_p(j)*h;
q2(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_q(j)*h;
k3_u(j) = p2(j);
k3_v(j) = q2(j);
k3_p(j) = (-2*v2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
k3_q(j) = (-2*u2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
u3(j) = p(i) + k3_u(j)*h;
v3(j) = q(i) + k3_v(j)*h;
p3(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_p(j)*h;
q3(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_q(j)*h;
k4_u(j) = p3(j);
k4_v(j) = q3(j);
k4_p(j) = (-2*v3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
k4_q(j) = (-2*u3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
s(j) = s(j-1) + h;
u(j) = u(j-1) + (h/6)*(k1_u(j) + 2*k2_u(j) + 2*k3_u(j) + k4_u(j));
v(j) = v(j-1) + (h/6)*(k1_v(j) + 2*k2_v(j) + 2*k3_v(j) + k4_v(j));
p(j) = p(j-1) + (h/6)*(k1_p(j) + 2*k2_p(j) + 2*k3_p(j) + k4_p(j));
q(j) = q(j-1) + (h/6)*(k1_q(j) + 2*k2_q(j) + 2*k3_q(j) + k4_q(j));
endfor
subplot(2,3,1), plot(s,u);
hold on; plot(s,v); hold off;
title ("Runge-Kutta");
h = legend ("u(s)", "v(s)");
legend (h, "location", "northwestoutside");
set (h, "fontsize", 10);
matlab octave numerical-methods ode runge-kutta
I'd writeu(i).^2
rather thenu(i)*u(i)
. Makes your code a bit more readable :)
– Pablo Jeken
Nov 12 at 8:33
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08
add a comment |
I´m attempting to use Runge-Kutta method to compare it to the lsode function. But it is performing rather poorly, every other method I used (fowards and backwards Euler, Heun) to compare to lsode do a way better job to the point they are almost indistinguishable from lsode.
This is what my code returns
https://i.stack.imgur.com/vJ6Yi.png
If anyone can pointout a way to improve it or if I doing something wrong I´d appreciate it.
The following is what I use for the Runge-Kutta method
%Initial conditions
u(1) = 1;
v(1) = 2;
p(1) = -1/sqrt(3);
q(1) = 1/sqrt(3);
%Graf interval / step size
s0 = 0;
sf = 50;
h = 0.25;
n=(sf-s0)/h;
s(1) = s0;
%-----------------------------------------------------------------------%
for j = 2:n
i = j-1;
k1_u(j) = p(i);
k1_v(j) = q(i);
k1_p(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
k1_q(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
u1(j) = p(i) + (1/2)*k1_u(j)*h;
v1(j) = q(i) + (1/2)*k1_v(j)*h;
p1(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_p(j)*h;
q1(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_q(j)*h;
k2_u(j) = p1(j);
k2_v(j) = q1(j);
k2_p(j) = (-2*v1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
k2_q(j) = (-2*u1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
u2(j) = p(i) + (1/2)*k2_u(j)*h;
v2(j) = q(i) + (1/2)*k2_v(j)*h;
p2(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_p(j)*h;
q2(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_q(j)*h;
k3_u(j) = p2(j);
k3_v(j) = q2(j);
k3_p(j) = (-2*v2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
k3_q(j) = (-2*u2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
u3(j) = p(i) + k3_u(j)*h;
v3(j) = q(i) + k3_v(j)*h;
p3(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_p(j)*h;
q3(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_q(j)*h;
k4_u(j) = p3(j);
k4_v(j) = q3(j);
k4_p(j) = (-2*v3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
k4_q(j) = (-2*u3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
s(j) = s(j-1) + h;
u(j) = u(j-1) + (h/6)*(k1_u(j) + 2*k2_u(j) + 2*k3_u(j) + k4_u(j));
v(j) = v(j-1) + (h/6)*(k1_v(j) + 2*k2_v(j) + 2*k3_v(j) + k4_v(j));
p(j) = p(j-1) + (h/6)*(k1_p(j) + 2*k2_p(j) + 2*k3_p(j) + k4_p(j));
q(j) = q(j-1) + (h/6)*(k1_q(j) + 2*k2_q(j) + 2*k3_q(j) + k4_q(j));
endfor
subplot(2,3,1), plot(s,u);
hold on; plot(s,v); hold off;
title ("Runge-Kutta");
h = legend ("u(s)", "v(s)");
legend (h, "location", "northwestoutside");
set (h, "fontsize", 10);
matlab octave numerical-methods ode runge-kutta
I´m attempting to use Runge-Kutta method to compare it to the lsode function. But it is performing rather poorly, every other method I used (fowards and backwards Euler, Heun) to compare to lsode do a way better job to the point they are almost indistinguishable from lsode.
This is what my code returns
https://i.stack.imgur.com/vJ6Yi.png
If anyone can pointout a way to improve it or if I doing something wrong I´d appreciate it.
The following is what I use for the Runge-Kutta method
%Initial conditions
u(1) = 1;
v(1) = 2;
p(1) = -1/sqrt(3);
q(1) = 1/sqrt(3);
%Graf interval / step size
s0 = 0;
sf = 50;
h = 0.25;
n=(sf-s0)/h;
s(1) = s0;
%-----------------------------------------------------------------------%
for j = 2:n
i = j-1;
k1_u(j) = p(i);
k1_v(j) = q(i);
k1_p(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
k1_q(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1);
u1(j) = p(i) + (1/2)*k1_u(j)*h;
v1(j) = q(i) + (1/2)*k1_v(j)*h;
p1(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_p(j)*h;
q1(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k1_q(j)*h;
k2_u(j) = p1(j);
k2_v(j) = q1(j);
k2_p(j) = (-2*v1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
k2_q(j) = (-2*u1(j)*p1(j)*q1(j)) / (u1(j)*u1(j) + v1(j)*v1(j) + 1);
u2(j) = p(i) + (1/2)*k2_u(j)*h;
v2(j) = q(i) + (1/2)*k2_v(j)*h;
p2(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_p(j)*h;
q2(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + (1/2)*k2_q(j)*h;
k3_u(j) = p2(j);
k3_v(j) = q2(j);
k3_p(j) = (-2*v2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
k3_q(j) = (-2*u2(j)*p2(j)*q2(j)) / (u2(j)*u2(j) + v2(j)*v2(j) + 1);
u3(j) = p(i) + k3_u(j)*h;
v3(j) = q(i) + k3_v(j)*h;
p3(j) = (-2*v(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_p(j)*h;
q3(j) = (-2*u(i)*p(i)*q(i)) / (u(i)*u(i) + v(i)*v(i) + 1) + k3_q(j)*h;
k4_u(j) = p3(j);
k4_v(j) = q3(j);
k4_p(j) = (-2*v3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
k4_q(j) = (-2*u3(j)*p3(j)*q3(j)) / (u3(j)*u3(j) + v3(j)*v3(j) + 1);
s(j) = s(j-1) + h;
u(j) = u(j-1) + (h/6)*(k1_u(j) + 2*k2_u(j) + 2*k3_u(j) + k4_u(j));
v(j) = v(j-1) + (h/6)*(k1_v(j) + 2*k2_v(j) + 2*k3_v(j) + k4_v(j));
p(j) = p(j-1) + (h/6)*(k1_p(j) + 2*k2_p(j) + 2*k3_p(j) + k4_p(j));
q(j) = q(j-1) + (h/6)*(k1_q(j) + 2*k2_q(j) + 2*k3_q(j) + k4_q(j));
endfor
subplot(2,3,1), plot(s,u);
hold on; plot(s,v); hold off;
title ("Runge-Kutta");
h = legend ("u(s)", "v(s)");
legend (h, "location", "northwestoutside");
set (h, "fontsize", 10);
matlab octave numerical-methods ode runge-kutta
matlab octave numerical-methods ode runge-kutta
edited Nov 12 at 0:58
asked Nov 12 at 0:18
R.FALLEN
11
11
I'd writeu(i).^2
rather thenu(i)*u(i)
. Makes your code a bit more readable :)
– Pablo Jeken
Nov 12 at 8:33
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08
add a comment |
I'd writeu(i).^2
rather thenu(i)*u(i)
. Makes your code a bit more readable :)
– Pablo Jeken
Nov 12 at 8:33
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08
I'd write
u(i).^2
rather then u(i)*u(i)
. Makes your code a bit more readable :)– Pablo Jeken
Nov 12 at 8:33
I'd write
u(i).^2
rather then u(i)*u(i)
. Makes your code a bit more readable :)– Pablo Jeken
Nov 12 at 8:33
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08
add a comment |
1 Answer
1
active
oldest
votes
You misunderstood something in the method. The intermediate values for p,q
are computed the same way as the intermediate values for u,v
, and both are "Euler steps" with the last computed slopes, not separate slope computations. For the first ones that is
u1(j) = u(i) + (1/2)*k1_u(j)*h;
v1(j) = v(i) + (1/2)*k1_v(j)*h;
p1(j) = p(i) + (1/2)*k1_p(j)*h;
q1(j) = q(i) + (1/2)*k1_q(j)*h;
The computation for the k2
values then is correct, the next midpoints need to be computed correctly via "Euler steps", etc.
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
You misunderstood something in the method. The intermediate values for p,q
are computed the same way as the intermediate values for u,v
, and both are "Euler steps" with the last computed slopes, not separate slope computations. For the first ones that is
u1(j) = u(i) + (1/2)*k1_u(j)*h;
v1(j) = v(i) + (1/2)*k1_v(j)*h;
p1(j) = p(i) + (1/2)*k1_p(j)*h;
q1(j) = q(i) + (1/2)*k1_q(j)*h;
The computation for the k2
values then is correct, the next midpoints need to be computed correctly via "Euler steps", etc.
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
add a comment |
You misunderstood something in the method. The intermediate values for p,q
are computed the same way as the intermediate values for u,v
, and both are "Euler steps" with the last computed slopes, not separate slope computations. For the first ones that is
u1(j) = u(i) + (1/2)*k1_u(j)*h;
v1(j) = v(i) + (1/2)*k1_v(j)*h;
p1(j) = p(i) + (1/2)*k1_p(j)*h;
q1(j) = q(i) + (1/2)*k1_q(j)*h;
The computation for the k2
values then is correct, the next midpoints need to be computed correctly via "Euler steps", etc.
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
add a comment |
You misunderstood something in the method. The intermediate values for p,q
are computed the same way as the intermediate values for u,v
, and both are "Euler steps" with the last computed slopes, not separate slope computations. For the first ones that is
u1(j) = u(i) + (1/2)*k1_u(j)*h;
v1(j) = v(i) + (1/2)*k1_v(j)*h;
p1(j) = p(i) + (1/2)*k1_p(j)*h;
q1(j) = q(i) + (1/2)*k1_q(j)*h;
The computation for the k2
values then is correct, the next midpoints need to be computed correctly via "Euler steps", etc.
You misunderstood something in the method. The intermediate values for p,q
are computed the same way as the intermediate values for u,v
, and both are "Euler steps" with the last computed slopes, not separate slope computations. For the first ones that is
u1(j) = u(i) + (1/2)*k1_u(j)*h;
v1(j) = v(i) + (1/2)*k1_v(j)*h;
p1(j) = p(i) + (1/2)*k1_p(j)*h;
q1(j) = q(i) + (1/2)*k1_q(j)*h;
The computation for the k2
values then is correct, the next midpoints need to be computed correctly via "Euler steps", etc.
answered Nov 12 at 8:51
LutzL
13.4k21326
13.4k21326
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
add a comment |
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
Thanks a lot, that fixed it. :)
– R.FALLEN
Nov 12 at 18:08
1
1
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
Then please mark the question as answered @R.FALLEN
– Pablo Jeken
Nov 13 at 10:09
add a comment |
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I'd write
u(i).^2
rather thenu(i)*u(i)
. Makes your code a bit more readable :)– Pablo Jeken
Nov 12 at 8:33
Thanks for the advice, I will change it in my code.
– R.FALLEN
Nov 12 at 18:08