Runge-Kutta for sovling 2nd ODE

[hsong9]

Hi,

Could someone please show me how to solve the following simple problem using the Runge-Kutta (RK4) integration method?

(tw')' + tw = 0 with w(0) = 1, w'(0) = 0 on the interval [0,1]
by introducing the new variable v=tw' and considering the resulting first order differential system involving w and v
computed solution (w_h(1),v_h(1)) for h=1/10.


I'm trying to understand how to apply Runge-Kutta to this problem.


Thanks for your attention.

 

[HallsofIvy]

If you let v= tw' then your problem becomes v'+ v= 0 or v'= -v with v(0)= 0. Now, what exactly is your question? What do you know about "Runge-Kutta"?

 

[D H]

If you let v= tw' then your problem becomes v'+ v= 0 or v'= -v with v(0)= 0.

Not quite. The problem become v'+tw=0. v is not tw. It is tw'.

 

[HallsofIvy]

Ah, I misread the problem!

hSong9, the standard way of applying Runge-Kutta, or any numerical method for first order de's to second order is to rewrite the problem as two first order problems.

The differential equation is tw''+ w'+ tw= 0. If you let u= w' then w''= u' so the equation becomes tu'+ u+ tw= 0 or tu'= -u- tw. We also, of course, have w'= u so we have two first order equations for u and w. Run two simultaneous Runge-Kutta solvers for the two first order equations, at each step using the currrent values you have for both u and w in the formulas.

Your initial values, of course, will be w(0)= 1, w'(0)= u(0)= 0.

 

[blue_raver22]

Sure, I'd be happy to explain how to use Runge-Kutta (RK4) to solve this 2nd order differential equation. First, let's rewrite the equation in its standard form:

w'' + (1/t)w = 0

Now, we can introduce the new variable v = tw', as suggested. This gives us the following system of two first order differential equations:

w' = v/t
v' = -w/t

Next, we can use RK4 to numerically solve this system over the interval [0,1] with a step size of h = 1/10. The general RK4 formula for solving a first order differential equation is:

y_n+1 = y_n + (1/6)*(k1 + 2*k2 + 2*k3 + k4)

where k1, k2, k3, and k4 are calculated as follows:

k1 = hf(t_n, y_n)
k2 = hf(t_n + h/2, y_n + k1/2)
k3 = hf(t_n + h/2, y_n + k2/2)
k4 = hf(t_n + h, y_n + k3)

In our case, we have two equations, so we will use this formula twice, once for w and once for v. Here is the specific implementation for our problem:

w_n+1 = w_n + (1/6)*(k1w + 2*k2w + 2*k3w + k4w)
v_n+1 = v_n + (1/6)*(k1v + 2*k2v + 2*k3v + k4v)

where:

k1w = h*v_n/t_n
k2w = h*(v_n + k1v/2)/(t_n + h/2)
k3w = h*(v_n + k2v/2)/(t_n + h/2)
k4w = h*(v_n + k3v)/(t_n + h)

k1v = -h*w_n/t_n
k2v = -h*(w_n + k1w/2)/(t_n + h/2)
k3v = -h*(w_n + k2w/2)/(t_n + h/2)
k4v = -h*(w_n + k3w)/(t_n +