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

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the system of coupled second-order differential equations

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

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

    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

    2. Relevant equations

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

    [tex]y_{n+1}=y_n+ak_1+bk_2[/tex]

    [tex]k_1=hf(x_n,y_n)[/tex]

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

    3. 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]

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

    If we now let

    [tex]u'=x[/tex]
    [tex]v'=y[/tex]

    then

    [tex]u''=x'[/tex]
    [tex]v''=y'[/tex]

    and we have

    [tex]u'=f(x)=x[/tex]
    [tex]v'=m(y)=y[/tex]

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

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

    Furthermore

    [tex]u(0)=1[/tex]
    [tex]u'(0)=x_0=2[/tex]
    [tex]v(0)=3[/tex]
    [tex]v'(0)=y_0=4[/tex]

    so that

    [tex]k_{1u}=hf(x_0)
    =(0.2)f(2)
    =(0.2)(2)
    =0.4[/tex]

    [tex]k_{1v}=hm(y_0)
    =(0.2)m(4)
    =(0.2)(4)
    =0.8[/tex]

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

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

    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!
    phyz
     
    Last edited: Mar 13, 2010
  2. jcsd
  3. Mar 13, 2010 #2

    phyzguy

    User Avatar
    Science Advisor

    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.
     
  4. Mar 13, 2010 #3
    Right, let's rewrite the Runge-Kutta equations in terms of z and w then:

    [tex]z_{n+1}=z_n+ak_1+bk_2[/tex]

    [tex]k_1=hf(w_n,z_n)[/tex]

    [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.

    Cheers!
     
  5. Aug 4, 2011 #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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