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Numerical Integration

  1. Apr 25, 2007 #1

    kreil

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    1. The problem statement, all variables and given/known data
    Given the system
    [tex]x'(t)=-ax(t)+ky(t)+g[/tex]
    [tex]y'(t)=lx(t)-by(t)+h[/tex]

    If g=h=0,

    a) Find the equilibrium
    b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only solution
    c) choose a,b,l,k such that ab-lk > 0. Find numerically the solution of the system starting in a neighborhood of the equilibrium.


    2. The attempt at a solution
    a) If x'(t)=y'(t)=0, then ax(t)=ky(t) and lx(t)=by(t). This is true for ab = lk, i.e. ab-lk=0.

    and then I run into trouble. I don't know how to explicity show (b), and have even less of an idea on how to start (c)

    Help!
    Josh
     
  2. jcsd
  3. Apr 25, 2007 #2
    Should

    "b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only solution"

    not read

    "b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only equilibrium solution"

    Now, consider that an equilibrium or "steady state" solution correspond to x'=y'=0.
    Finding such a solution here leads you to an algebraic linear system of equations.
    Go back to and reuse the theory for linear systems of equations.
     
  4. Apr 25, 2007 #3

    kreil

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    I don't understand how I'm supposed to get values for x(t) and y(t).

    Can you show me an example?
     
  5. Apr 25, 2007 #4

    HallsofIvy

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    You have, as you said in your first post, the two equations ax= ky and lx= by. Are you saying you do not know how to solve two linear, simultaneous equations?
     
  6. Apr 25, 2007 #5

    kreil

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    The part that confuses is me is that in (b) we prove that if ab-lk does not equal zero then x and y have only one equilibrium point..(0,0). Then part (c) says to solve the system for ab-lk>0, which would imply that (x,y)=(0,0).

    Is this correct or am i missing something?
     
  7. Apr 26, 2007 #6

    HallsofIvy

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    One more time: can you solve the two equations ax= ky and lx= by for x and y? If you can, what happens to that solution if ab- lk DOES equal 0?

    Yes, if ab-lk> 0 then (x,y)= (0,0) is the only EQUILIBRIUM solution. But you are asked to find a solution for an intial value in a neighorborhood of (0,0), not (0,0) itself.
     
  8. May 1, 2007 #7

    kreil

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    I don't think you understand my extreme inability to solve any part of this problem. Here is an attempt at (c):

    a=0.3
    b=0.2
    l=0.1
    k=0.4

    Then ab-lk=0.02 > 0.

    0.3 x(t) = 0.4 y(t)
    0.1 x(t) = 0.2 y(t)

    So, x(t) = 1.333 y(t)
    y(t) = 0.5 x(t)

    And I don't know what to do after this since it doesn't even seem to be true. Should I just plug in values of x(t),y(t) close to 0? I am not given any initial values. I really need help.
     
    Last edited: May 1, 2007
  9. May 2, 2007 #8

    HallsofIvy

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    I have asked you repeatedly to solve the linear equations ax= ky and lx= by. If you cannot do that then you should not be attempting differential equations!
    From ax= ky, y= (a/k)x. Putting that into kx= by we have lx= (ab/k)x so that lkx= abx or (lk-ab)x= 0. If lk- ab is not 0 what is x equal to? What is y equal to? If lk- ab= 0 what can you say about x?

    Yes, it is true: x= (4/3)y and y= (1/2)x so x= (4/3)(1/2)X= (2/3)x. That is, (1/3)x= 0 so x= 0 and then y= 0. The only equilibrium solution is (0,0). Now choose a starting value for (x,y) that is "close to" (0,0) and numerically solve the system of equations.
     
  10. May 2, 2007 #9

    kreil

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    Thanks a lot for your help ivy, I have no other source to turn to on this problem. I have never done differential equations before and my teacher just gave us these projects without covering any of the material in class.

    Did I solve the system correctly for these starting values? (I assume that I am to plug in one value in each equation and solve for the other variable.)

    (x,y) = (0.01, 0.02)

    0.3 * 0.01 = 0.4 * y(t) => y(t) = 0.0075

    0.1 * x(t) = 0.5 * 0.02 => x(t) = 0.1

    edit: Since I am no longer solving for the equilibrium, am I supposed to keep x' and y' in the system?
     
    Last edited: May 2, 2007
  11. May 2, 2007 #10

    kreil

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    This part is my biggest problem. I'm not exactly sure what constitutes a solution here. Am I to use Euler's method and approximate some values of x and y? Or am I trying to come up with a function that satisfies the diff eq? Any help at all here is appreciated!
     
  12. May 4, 2007 #11

    kreil

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    I really need help guys, any takers?
     
  13. May 4, 2007 #12

    J77

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    Yes -- use Eulers method.
     
  14. May 4, 2007 #13

    kreil

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    When I choose values of x,y close to the equilibrium does t=0?
     
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