Solving Second Order Differential Equations using Runge Kutta

  1. 1. The problem statement, all variables and given/known data
    In aerodynamics, one encounters the following initial value problem for Airy’s equations:

    y''(x) + xy = 0, y(0) = 1, y'(0) = 0

    Using the Runge-Kutta method with h=0.005 and determine values between x=0 and x=10 sufficient to sketch the relationship.



    2. Relevant equations
    y''(x) + xy = 0, y(0) = 1, y'(0) = 0
    I think,
    k1 = h*f(Xn, Yn)
    k2 = h*f(Xn+h/2, Yn+k1/2)
    k3 = h*f(Xn+h/2, Yn+k2/2)
    k4 = h*f(Xn + h, Yn + k3)

    3. The attempt at a solution
    From what I have read you cant do second order ODE using runge kutta without breaking it into a system of first order ODEs so thats what I tried.

    I tried:
    d2y/dx2 + xy = 0

    dy/dx = z, y(0) = 1

    dz/dx + xy = 0

    dz/dx = -xy, z(0) = 0

    I dont know if that is right or not and if it is I have no idea where to go from here.

    Thanks for any help.
     
  2. jcsd
  3. Have you ever used mathcad? It has a very easy to use RK function. First, however, you do need to reduce the order and define your system of DE's.
     
  4. HallsofIvy

    HallsofIvy 40,308
    Staff Emeritus
    Science Advisor

    Yes, that's correct. Now use two simultaneous Runge-Kutta calculations where, at each step, you calculate the values for both y and z and use both in the calculations for the next step.

     
  5. D H

    Staff: Mentor

    There is nothing in the Runge Kutta formalism that precludes using on a vector as opposed to a scalar. For example (note well: This is not your problem), suppose

    [tex]\frac{d\vec x}{dt} = A \vec x\,, \quad\vec x(0) = \vec x_0[/tex]

    where [itex]\vec x[/itex] is some n-vector, A is a constant n-by-n matrix, and x0 is the initial (vectorial) value.

    The Runge Kutta equations can be used to numerically this problem.

    Back to your problem: Create a 2-vector, I'll call it u, that comprises your y and its derivative:

    [tex]\begin{aligned}
    z &\equiv \frac{dy}{dx} \\[6 pt]
    \vec u &= \bmatrix y \\ z\endbmatrix
    \end{aligned}[/tex]

    You have already derived that [itex]dy/dx = z, y(0) = 1[/itex] and [itex]dz/dx = -xy, z(0) = 0[/itex]. From these, can you come up with an expression for the initial value of the vector u and the derivative of the vector u with respect to x?
     
  6. I am doing this all in MATLAB, This is what I have come up with but I am not sure if this is correct or not.

    Attached are my MATLAB code, a Graph of the results, and an excel sheet of all results.
    It is zipped because it was to big otherwise. the matlab code can be opened in notepad.

    Thanks for any help.
     

    Attached Files:

  7. Yes, your approach is right.

    d2y/dx2 + xy = 0

    dy/dx = z -------------------(1)

    dz/dx = -xy -----------------(2)


    y' = z, then define
    y(n+1) = y(n) + (h/6)*(k1 + 2*k2 + 2*k3 +k4)

    and for,
    z' = -xy, define,
    z(n+1) = z(n) + (h/6)*(j1 + 2*j2 + 2*j3 +j4)

    now define k1 , k2 , k3 , k4 and j1 , j2 , j3 , j4

    k1 = z
    j1 = -x*y

    k2 = z + 0.5 * j1
    j2 = -( x + 0 .5 * h ) *(y + 0.5 * k1)

    k3 = z + 0.5 * j2
    j3 = -( x + 0 .5 * h ) *(y + 0.5 * k2)

    k4 = z + j2
    j4 = -( x + h ) *(y + k2)
     
  8. I have a similar problem with Runge Kutta with y''+2xy'=0 given boundary conditions are y=1 at x=0 and y=0 at x=infinity. I am not able to understand how to transform the second boundary condition (i.e.y=0 at x=infinity) to find a definitive value of y' at x=0.
     
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