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

In summary, the student is trying to find the function values for the x and y variables at each time step in a Runge-Kutta method.
  • #1

Homework Statement

Consider the system of coupled second-order differential equations



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, [tex]\alpha=\beta=3/2[/tex] to find u, u', v, v' at t=0.2

Homework Equations

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



[tex]k_2=hf(x_n+\alpha h,y_n+\beta k_1)[/tex]

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

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


If we now let




and we have


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




so that





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

[tex]k_{2u}=hf(x_0+\alpha h)[/tex]

[tex]k_{2v}=hm(y_0+\alpha h)[/tex]

or is it supposed to be

[tex]k_{2u}=hf(x_0+\alpha k_{1u})[/tex]

[tex]k_{2v}=hm(y_0+\alpha k_{1v})[/tex]


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

Thanks for your help!
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  • #2
Try thinking of your Runge-Kutta equations as a vector equation, with yk having four components (u,v,x,y), and with the dependent variable, xx = tk. 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 yk = (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.
  • #3
Right, let's rewrite the Runge-Kutta equations in terms of z and w then:



[tex]k_2=hf(w_n+\alpha h,z_n+\beta k_1)[/tex]

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

[tex]k_{2u}=hf(x_0+\alpha h)[/tex]

[tex]k_{2v}=hm(y_0+\alpha h)[/tex]

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... :grumpy:

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

  • #4
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.

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