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Equilibrium and Stability of a set of equations

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equilibrium and determine its stability for the model of Sec 1.4. Assume x and y are both
    strictly positive (nonzero).

    2. Relevant equations

    Differential equations of section 1.4

    (1/x)(dx/dt) = a1 - b1y
    (1/y)(dy/dt) = -a2 + b2x



    3. The attempt at a solution

    I solved the two differential equations and got x(t) = e^t(a1 -b1y) and y(t) = e^t(-a2 +b2x) so once I had these I though of what happens to the equations when t approaches infinity but that's where I'm unsure because x and y are apart of each other eqation. Either both of them approach infinity as t approaches infinity and there is no equilibrium or there is something more complex that I can't seem to figure out yet.

    Thanks to anybody who helps.
     
  2. jcsd
  3. Mar 3, 2012 #2

    Dick

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    Those are NOT solutions to your differential equations. You are doing something very wrong. And you don't necessarily have to solve them. What are equilibrium points?
     
  4. Mar 3, 2012 #3
    Well the concept I have so far is that equilibrium is where a state will persist forever but what I think I'm seeing now is that doesn't relate to the solutions but the differential equation where either (dx/dt) or dy/dt) are equal to zero. The equilibrium state would be where the differential equations are constant but I'm not sure how to use that definition in the forms the differential equations are currenly in.
     
  5. Mar 3, 2012 #4

    Dick

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    Yes, the equilibrium points are points where the solutions are constant functions. So dx/dt AND dy/dt are zero. Use that.
     
  6. Mar 4, 2012 #5
    Well if I set the differential equations to zero I get the equilibrium points y=a1/b2 and x = a2/b2 but the question I now have is if these are the equilibrium points I dont see the usefulness in being able to figure out the stability since each differential equation is in respect of the other variable so the graph to find the stability would either always be zero or linear based on it's variable but not the equilibrium point. What is it I need to understand here in order to find the stability?
     
  7. Mar 4, 2012 #6

    Dick

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    You only have one equilibrium point. (x,y)=(a2/b2,a1/b1). You need to figure out what the solutions look like near that point. You must have seen examples of how this is done in your course.
     
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