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Equilibrium Solutions

  1. Sep 21, 2013 #1
    Hello,

    I want to make certain of my understanding of equilibrium solutions. Are equilibrium solutions the value(s) of y such that [itex]\frac{dy}{dt} = 0[/itex]? So, suppose y = y0 is one of those solutions. Graphically, does this mean that one the horizontal line y = y0, little slope lines are too horizontal; and does this correspond to where the solution curves attain a maximum or a minimum?
     
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  3. Sep 21, 2013 #2
    The equilibrium solutions are when dy/dx=0. Graphically it means the all along the line y=y0, there are little horizontal lines (slope of 0). Just because the equilibrium solution is y=y0 does not mean that the function obtains a minimum or a maximum. You must plot points below and above the equilibrium solution to check the direction of where the solutions are going, hence the term to describe the graph as a direction field. Let's say you set y=1+y0(assuming y>0) and plug it into the differential equation(dy/dx) and the number you get is positive. Then you plug in y=y0-1 into the differential equation and get a positive munber. This would mean the there would not be a max or a min obtained. And it would mean that the equilibrium solution y=y0 is semistable. If you were to get opposite slope values above and below the equilibrium solution, then you would obtain either a max or a min.
     
  4. Sep 21, 2013 #3

    Office_Shredder

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    Typically equilibrium solutions will not correspond to actual extrema of non-equilibrium solutions. If your differential equation is reasonably well behaved then given a point there is a unique solution passing through it - if that solution is the equilibrium solution then it means no other solution is passing through it. Typically solutions will asymptotically approach the equilibrium either as t goes to infinity or minus infinity (or both, depending on the differential equation)
     
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