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
phyzmatix
313
0

Homework Statement



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

Homework 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]

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:
Physics news on Phys.org
  • #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]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!
 
  • #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.
 

1. What is 2nd Order Runge-Kutta?

2nd Order Runge-Kutta is a numerical method used to solve second order differential equations. It is an improvement upon the original Runge-Kutta method, providing greater accuracy and stability.

2. How does 2nd Order Runge-Kutta work?

2nd Order Runge-Kutta works by using a weighted average of two slopes at different points to approximate the solution to a second order differential equation. This method is often used when the equation cannot be solved analytically.

3. What are some applications of 2nd Order Runge-Kutta?

2nd Order Runge-Kutta is commonly used in physics, engineering, and other sciences to model and solve problems involving second order differential equations. It is also used in computer simulations and mathematical modeling.

4. What are the advantages of using 2nd Order Runge-Kutta over other numerical methods?

Compared to other numerical methods, 2nd Order Runge-Kutta is relatively simple to implement and provides a good balance between accuracy and computational efficiency. It is also more stable than the original Runge-Kutta method.

5. Are there any limitations to using 2nd Order Runge-Kutta?

Like any numerical method, 2nd Order Runge-Kutta is not exact and relies on approximations. It may also have difficulty handling stiff equations or problems with rapidly changing solutions. In these cases, other numerical methods may be more appropriate.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
114
  • Calculus and Beyond Homework Help
Replies
2
Views
220
  • Calculus and Beyond Homework Help
Replies
1
Views
175
  • Calculus and Beyond Homework Help
Replies
2
Views
465
  • Calculus and Beyond Homework Help
Replies
14
Views
232
  • Calculus and Beyond Homework Help
Replies
7
Views
128
  • Calculus and Beyond Homework Help
Replies
19
Views
673
  • Calculus and Beyond Homework Help
Replies
5
Views
195
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
663
Back
Top