How can I use Runge Kutta to solve coupled first order DE's in Matlab?

Click For Summary
To solve coupled first-order differential equations using the Runge Kutta method in MATLAB, the discussion focuses on applying Heun's method, a specific variant of the RK2 approach. The equations provided are dp/dt = aq - bp and dq/dt = -aq + bp, with constants a=3 and b=4, and a conservation relationship nt = p(t) + q(t). Participants clarify the formulation of the algorithm, emphasizing the need to define separate variables for p and q in the numerical solution. The conversation also touches on the correct application of the RK2 method, confirming that Heun's method can be adapted for these equations.
HoosierDaddy
Messages
4
Reaction score
0
Hello All,

I need to utilize a Runge Kutta second order approach to solve two coupled first order DE's simultaneously given some initial conditions and a conservation relationship.

The DE's are as follows:

\frac{dp}{dt} = aq - bp

\frac{dq}{dt} = -aq + bp

Where a and b are constants, a=3, b=4, and the conservation relationship is nt = p(t) + q(t). Here, nt = 3.

I'm having trouble getting the ball rolling with writing the R.K. algorithm for this, will be programming in matlab. I have solved the equations analytically by hand and have found the following:

p(t) = -0.055972e-3t + 3.055972e-4
q(t) = -0.055972e-7t + 3.055972

Any help on how to get started with the Runge Kutta algorithm to solve numerically would be appreciated!
 
Physics news on Phys.org
You have to do a little work before we can help you. What do you know about Runge Kutta methods? About RK2 in particular? Which particular variant of RK2 (there are more than one) are you supposed to use?
 
Not sure exactly whick RK approach it is, we will be incrementing time such that ...

yj+1 = y*j+1 + \frac{h}{2}(k2 - k1)

Where y*j+1 = yj + hk1

and k1 = f(xj, yj) and k2 = f(xj+1, y*j+1)

Does this identify which RK2 method I am attempting to use?
 
That is Heun's method.

An alternative is the midpoint method:

\aligned<br /> y_{j+1} &amp;= y_j + hk_{j+1/2} \\<br /> x_{j+1/2} &amp;= x_j +h/2 \\<br /> y_{j+1/2} &amp;= y_j + hk_j/2 \\<br /> k_j &amp;= f(x_j,y_j) \\<br /> k_{j+1/2} &amp;= f(x_{j+1/2},y_{j+1/2})<br /> \endaligned
 
Thanks D H, I believe we are required to use Heun's method. Can I just draw a parallel to Heun's method with my values? i.e. xj = tj, xj+1 = tj+1 = tj + \Deltat?

If this is the case, what would my yj be? Would I have a separate yj for my p and q equations?
 
Consider this set of six differential equations:

\aligned<br /> \frac {d\mathbf r}{dt} &amp;= \mathbf v \\<br /> \frac {d\mathbf v}{dt} &amp;= \mathbf a_{\text{ext}} \equiv \mathbf F_{\text{ext}}/m<br /> \endaligned

In other words, Newton's laws of motion. Another way to look at this is as a six vector:

\aligned<br /> \mathbf u &amp;= \left[ r_x \, r_y \,r_z \,\,v_x \,v_y \,v_z \right]^T \\<br /> \frac{\mathbf u}{dt} &amp;= \left[ v_x \,v_y \,v_z \,\,a_x \,a_y \,a_z \right]^T<br /> \endaligned

Integration techniques such as Runge Kutta can be used on vectors as well as on scalars. Molecular simulations, for example, essentially work on vectors with thousands of elements or more.
 
Last edited:
HoosierDaddy said:
Not sure exactly whick RK approach it is, we will be incrementing time such that ...

yj+1 = y*j+1 + \frac{h}{2}(k2 - k1)

Where y*j+1 = yj + hk1

and k1 = f(xj, yj) and k2 = f(xj+1, y*j+1)

Does this identify which RK2 method I am attempting to use?

Are you sure it is k2 - k1 and not k2 + k1?

The full Heun is

k0 = f(xi, yi)
k1 = f(xi + h, yi + hk0)

yi+1 = yi + (1/2 k0 + 1/2 k1)h
 
That is exactly what he is doing, just written a little differently.
 

Similar threads

  • · Replies 65 ·
3
Replies
65
Views
7K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
8K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K