2nd Order Runge-Kutta: 2nd Order Coupled Differential Equations

[phyzmatix]

Homework Statement

Consider the system of coupled second-order differential equations

u''-(t+1)(u')^2+2uv-u^3=\cos{t}

2v''+(\sin{t})u'v'-6u=2t+3

with initial conditions u(0)=1, u'(0)=2, v(0)=3, v'(0)=4. Use the second order Runge-Kutta method with h=0.2 and a=2/3, b=1/3, \alpha=\beta=3/2 to find u, u', v, v' at t=0.2

Homework Equations

For the equation dy/dx=f(x,y)

y_{n+1}=y_n+ak_1+bk_2

k_1=hf(x_n,y_n)

k_2=hf(x_n+\alpha h,y_n+\beta k_1)

The Attempt at a Solution

I have no background in Differential Equations but some searching around on the net gave me a little bit of insight and I believe I'm supposed to change the two given equations into 4 separate first-order ODE's.

So what I have is as follows:

Rearranging the two 2nd order equations gives

u''=\cos{t}+(t+1)(u')^2-2uv+u^3

v''=t+\frac{3}{2}-\frac{1}{2}(\sin{t})u'v'+3u

If we now let

u'=x
v'=y

then

u''=x'
v''=y'

and we have

u'=f(x)=x
v'=m(y)=y

x'=g(t,x,u,v)=\cos{t}+(t+1) (x)^2-2uv+u^3

y'=n(t,x,y,u)=t+\frac{3}{2}-\frac{1}{2}(\sin{t})xy+3u

Furthermore

u(0)=1
u'(0)=x_0=2
v(0)=3
v'(0)=y_0=4

so that

k_{1u}=hf(x_0)<br /> =(0.2)f(2)<br /> =(0.2)(2)<br /> =0.4

k_{1v}=hm(y_0)<br /> =(0.2)m(4)<br /> =(0.2)(4)<br /> =0.8

k_{1x}=hg(t,x_0,u,v)<br /> =(0.2)g(0,2,1,3)<br /> =(0.2)[1+(2)^2-2(1)(3)+(1)^3]<br /> =0

k_{1y}=hn(t,x_0,y_0,u)<br /> =(0.2)n(0,2,4,1)<br /> =(0.2)[(0)+\frac{3}{2}-(0)+3(1)]<br /> =0.9

Assuming all of the above is correct then, well, what now? Do I say, e.g.

k_{2u}=hf(x_0+\alpha h)

k_{2v}=hm(y_0+\alpha h)

or is it supposed to be

k_{2u}=hf(x_0+\alpha k_{1u})

k_{2v}=hm(y_0+\alpha k_{1v})

?

And what about k_{2x} and k_{2y}, each of which has four variables?

Thanks for your help!
phyz

 

[phyzguy]

Try thinking of your Runge-Kutta equations as a vector equation, with y_k having four components (u,v,x,y), and with the dependent variable, x_x = t_k. So at each time step, you need to calculate all four components of the vector, and plug them into the Runge-Kutta formula. You might want to rename x and y as z and w so you don't get confused so then you'll have y_k = (u,v,z,w) ), since the x and y in your diff eqs are different from the x and y in the Runge-Kutta formula.

 

[phyzmatix]

Right, let's rewrite the Runge-Kutta equations in terms of z and w then:

z_{n+1}=z_n+ak_1+bk_2

k_1=hf(w_n,z_n)

k_2=hf(w_n+\alpha h,z_n+\beta k_1)

If we have to increment the value of the independent variable with h and that of the function values with those of the previous k values, then we should have

k_{2u}=hf(x_0+\alpha h)

k_{2v}=hm(y_0+\alpha h)

since from f(x) and m(y) we have x and y as independent variables (?)

Which would then mean that, in the functions g(t,x,u,v) and n(t,x,y,u) the variables t,x,y are independent and u,v are function values?

I really need an example of a similar problem to help me figure this out...

Please tell me which bits of what I've done so far are correct.

Cheers!

 

[physneil]

Have anyone been able to solve this. I have been stuck on the same question now for days at about the same place.

Help would really be appreciated.