Probably More Than You Want to Know About Stability Analysis | FISH 497, Study notes of Agricultural engineering

Download Probably More Than You Want to Know About Stability Analysis | FISH 497 and more Study notes Agricultural engineering in PDF only on Docsity! PROBABLY MORE THAN YOU WANT TO KNOW ABOUT STABILITY ANALYSIS The main objective in stability analysis is to determine whether a system that is pushed slightly from a steady-state (an equilibrium) will return to that steady state. If this is true for small perturbations from equilibrium, then we say that this equilibrium is stable. If a system always returns to that equilibrium, then we say it is globally stable. The difference between a stable and unstable equilibrium can be conceptualized with the "ball and cup" diagrams: A B Figure 1. In this figure, equilibrium A is locally stable, because if the ball is pushed to the left or to the right, it will return back to point A. Point B is unstable because if the ball is pushed in either direction it will move to a new equilibrium. Graphical Stability Analysis There are well-developed mathematical techniques for determining whether a equilibrium is stable or not. If your model has only two state variables, graphical analyses can tell you a lot about local stability. This is usually accomplished by graphing isoclines, which denote the combinations of different values of the state variables for which each equation is at steady state. Consider the simple model of two interacting species, X and Y: ( )(),()( tYtXf dt tdX = ) 1a ( )(),()( tYtXg dt tdY = ) 1b This model simply says the dynamics of X and Y depend on what X and Y are, and this is described by the functions f(X,Y) and g(X,Y). Note that I used a continuous time model here, but I could have just as well used a discrete time model. We begin by drawing an isocline that shows the combination of X and Y for which f(X,Y)=0 (because when f(X,Y)=0, then dX/dt=0 and there is no change in population size in X). We call this the X-isocline, and it might look something like this: X Y Figure 2. X-Isocline, which shows the combinations of X and Y for which dX/dt=0. If X and Y are above the line, then dX/dt is less than zero so X will decrease. When X and Y are below the line, then dX/dt is greater than zero, so X will increase. Remember, the bold line (the X-isocline) shows the combinations of X and Y for which dX/dt=0. If you are above this line, that means that there are too many Y or too many X, so dX/dt will be negative and therefore X will decrease until it hits the isocline. Similarly, if you are below the line, there are too few X and Y, so dX/dt >0 and therefore X will increase until it hits the isocline. The isocline therefore tells us quite a bit about the dynamics of the system. We can now consider an isocline that shows where dY/dt=0 (the Y-isocline). This might look like this: X( t )=X*+x ( t ) 3 where x(t) is the small permutation away from X*. For consistency, lower-case x(t) will refer to the difference between X(t) and X*. Thus, by definition, after the next time step x(t+1) will be: x ( t+1)=X( t+1)-X* 4 What we want to know is whether the x(t) get bigger or smaller over time. If they get smaller, that means the system is moving towards the equilbrium i.e., it is stable. To solve this, we first recognize that X(t+1) can be calculated from Eq. 2 and because X(t)=X*+x(t), we can re-write the right hand side to be: ( ) *)(*)1( XtxXftx −+=+ 5 This equation tells us how the permutations propagate through time. If they get smaller, that means that the system is moving closer to equilibrium. If they are getting bigger, then the system is moving away from equilibrium (because x(t) represents the distance away from the equilibrium). The trick here is figuring out what f(X*+x(t)) is. The solution comes from an approximation that says that any function f can be approximated as a linear function near any point. This is called a Taylor expansion, and yields the following: ( ) )(*)()(* * tx dX dfXftxXf X +=+ 6 where *XdX df is the first derivative of the function f with respect to X, evaluated at X=X*. In other words, we've turned the function f(X(t)) into another function that is linear in x(t), with an intercept f(X*) and slope equal to df /dX. This approximation is only valid near the point at which we are evaluating it, namely at the equilibrium. Substituting this into equation 5, we get )()1( taxtx =+ 7 where a= *XdX df . This makes life very simple! You should be able to see that the value of x(t+n), where n represents the number of time steps into the future, is equal to: )()( txantx n=+ 8 If the absolute value of a is less than 1, then the magnitude of the perturbations (the x(t)) will get smaller with time. In other words, the system will get closer to equilibrium. If the absolute value is greater than 1, then the perturbations will grow over time, meaning that the system is moving away from equilibrium. We now have a condition for stability in a single-species model. Now we can return to a model with 2 state variables. We'll use a similar model as we used in the isocline analysis, but this time represent the model with discrete equations: 9a ( )(),()1( tYtXftX =+ ) )( )(),()1( tYtXgtY =+ 9b Again, we are interested in the stability of the model near some equilibrium, X*,Y*. Like in the single-species model, we can consider how small permutations away from equilibrium propagate through time. Because we have two state variables (X,Y), we represent this permutation as x(t) and y(t), where X( t )=X*+x ( t ) 10a Y( t )=Y*+y( t ) 10b We are going to do a Taylor expansion around the point X*,Y* to derive linear approximations to the functions, so that the time dynamics of x(t) and y(t) are represented by: )()()1( tyatxatx xyxx +=+ 11a )()()1( tyatxaty yyyx +=+ 11b where aij is the effect of species j on species i. These are calculated from the derivatives: **,YX xx X fa δ δ = **,YX yx X ga δ δ = **,YX xy Y fa δ δ = **,YX yy Y ga δ δ = where the derivatives represent the partial derivates (which is why we use the greek δ instead of plain ol' d in the notation) of the functions f and g with respect to X and Y, all evaluated at X*,Y*. How do we assess the stability of the two equations in Eq. 11? To answer this, we first need to revisit what we did in the first model. Remember that we collapsed the function f(x) into a linear one, so that x(t+n)=anx(t). In other words, the dynamics could be expressed as some constant raised to the n power, times some initial value. It turns out that the dynamics of all systems can be represented by a similar equation: x(n )=λnC 13 where λ is some constant, which we call an eigenvalue, and C is some other constant that will be dependent on the initial value of x(t). How does this help us with the two species model? One problem that should be apparent is that Eq. 13 is in only 1 dimension (x), but our model has two dimensions (x and y). But, we can do some simple arithmetic to reduce the system to one dimension. Start with Eq. 11a, and substitute in Eq. 11a for y(t+1) to get: ( ))()()()1()2( tyatxaatxatxatx yyyxxyxxxx ++++=+ 14 and we can get rid of y(t) by using Eq. 11a to obtain: x ( t+2)-(ax x+ay y)x ( t+1)+(ax xay y -ax yay x)x ( t)=0 15 Using Eq 13, we can substitute λnC for x(n) (where n equals 2, 1, or 0) in Eq. 15 to derive a second order polynomial: λ 2- (ax x+ay y) λ +(ax xay y-ax yay x)=0 16 [note that I factored out the common term C]. This equation is known as the characteristic equation, and the roots of the characteristic equation indicate the values of λ for which Eq. 13 is valid. In this case, there will be two roots (λ1, λ2) which are obtained from the quadratic equation (there are always two roots to a second-order polynomial). We can therefore write Eq. 13 for the two species model as: nn AAnx 2211)( λλ += 17 where A1 and A2 represent some constants. These aren't very important, because whether or not x(t+n) gets bigger or smaller over time (n) depends on the values of λ1 or λ2. Consider the case when λ1 > λ2. As n gets really big, λ1n >>> λ2n. In fact, over time the magnitude of the λ2n becomes insignificant compared to λ1n; x(n) will depend almost entirely on λ1. We can therefore approximate Eq. 17 as: x (n )=Am a xλ m a x n 18 where λmax represents the largest (dominant) eigenvalue. This puts is right where we want, because we have an equation that tells us that only the magnitude of the dominant eigenvalue determines the stability of the equilibrium X*Y*. Remember if λmax<1, then the x(n) will get smaller over time i.e., the equilibrium is stable. Keep in mind that largest magnitude means the largest absolute magnitude e.g., |λi|. It turns out that the eigenvalues, λ , can be determined very easily by using matrix algebra. We define the matrix of the coefficients aij as equal to: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = yyyx xyxx aa aa A 19