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# SPEC TOPICS AFS FISH 497

UW

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This 10 page Class Notes was uploaded by Percy Wintheiser on Wednesday September 9, 2015. The Class Notes belongs to FISH 497 at University of Washington taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/192251/fish-497-university-of-washington in Aquatic And Fishery Sciences at University of Washington.

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Date Created: 09/09/15

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 steadystate 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 quotball and cupquot diagrams B A Figure 1 In this gure 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 welldeveloped 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 fX 010 1a dm dt gXrYr 1b This model simply says the dynamics of X and Y depend on whatX and Y are and this is described by the functions fXY and gXY 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 fXY0 because when fXY0 then dXdt0 and there is no change in population size in X We call this the Xisocline and it might look something like this X Figure 2 XIsocline which shows the combinations of X and Y for which dXdt0 If X and Y are above the line then dXdt is less than zero so X will decrease When X and Y are below the line then dXdt is greater than zero so X will increase Remember the bold line the Xisocline shows the combinations of X and Y for which dXdFO If you are above this line that means that there are too many Y or too manyX so dXdt will be negative and thereforeX will decrease until it hits the isocline Similarly if you are below the line there are too few X and Y so dXdt gt0 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 dYdFO the Y isocline This might look like this X Figure 3 Ylsocline which shows the combinations of X and Y for which dYdt0 If X and Y are above the line then dYdt is less than zero so Y will decrease When X and Y are below the line then dYdt is greater than zero so Y will increase Remember that the isocline shows the combination of X and Y for which dYdt is zero Thus for points above the line there are too manyX and Y so that dYdt is less than zero and Y will therefore decrease The opposite holds for points below the line We can now combine these two graphs together as follows Figure 4 X and Y isoclines plotted on the same graph Equilibrium points are indicated by the letters a b c and d Black arrows indicate the direction of the dynamics in each region Wherever the two lines intersect there is an equilibrium because both dXdt and dYdt are both zero In the gure above this happens at point b and point c Also where each isocline crosses it39s own axis represents an equilibrium because we assume that dYdt0 when Y 0 and dXdt0 when X 0 so these represent equilibrium conditions where only 1 population is present The question is which if any of these equilibrium are stable in the way depicted in Figure 1 We can use the arrows on the gure to help us sort that out In several regions of the graph I ve shown the same arrows that I showed in Figures 2 and 3 which represent the direction the system will want to move towards each isocline these are shown in gray So in the bottom left hand comer the point is below the X isocline so X will want to increase arrow points to right and the point is also below the Y isocline so Y will want to increase arrow points up The net direction of the XY system will be the 2 dimensional representation of these two vectors indicated by the black arrow In other words if you started in that lower left hand comer initially you would see an increase inX and Y leading to movement in XY space that was upwards and to the right Let s take a look at point b imagining what would happen if there was a small perturbation off of this point in any direction No matter which direction we move we end up in a region where the arrows point us back to point b For example if we move to a point that is slightly to the right of point b then we are in a region that wants to push us back to the left The same holds for any possible direction that we move Thus we say that this point is locally stable because any small perturbation will return back to the same point Now let s look at point c Suppose we move to a point that is slightly to the right of this equilibrium The initial direction will be to move back towards the left but also down But once we move down below the X isocline we see that we are in a totally different region where the system wants to move down and to the Lig In other words it wants to move away from the equilibrium point Thus point c is an unstable equilibrium The same exercise will indicate that point a is an unstable equilibrium while point d is a stable equilibrium This system therefore has two stable states one of which has a high abundance of Y and a low abundance of X point b and the other where Y is extirpated andX is at a high abundance Mathematical Analysis of Stability Graphical analysis provides a quick way to evaluate the stability of steady states in simple 2 dimensional systems When there are more than 2 state variables graphical methods are not very helpful Also graphical methods don39t tell us whether a system will cycle or show damped oscillations For these applications we need to rely on some mathematical formulations It is easiest to begin with a simple system that has only 1 state variable Let s consider a discrete time model where Xt1fXt 2 where fXt is some function of X t that describes population growth Assume that there is some equilibrium point X that we are interested in By definition at equilibrium XtlXtX Therefore fXX Now let39s consider what happens if we move slightly away from equilibrium XtXxt 3 where xt is the small permutation away from X For consistency lowercase xt will refer to the difference between Xt and X Thus by de nition after the next time step xtl will be xtlXtlX 4 What we want to know is whether the xt get bigger or smaller over time If they get smaller that means the system is moving towards the equilbrium ie it is stable To solve this we first recognize that X H1 can be calculated from Eq 2 and because XtX xt we can rewrite the right hand side to be xtl fXxt X 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 xt represents the distance away from the equilibrium The trick here is figuring out what fX xt 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 df fX xt fX E W xt 6 is the first derivative of the function f with respect to X evaluated at XX In M other words we ve turned the function fXt into another function that is linear in xt with an intercept fX and slope equal to dfdX 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 where i dX xt 1 axt 7 where a i This makes life very simple You should be able to see that the value of X xtn where n represents the number of time steps into the future is equal to xt n aquotxt 8 If the absolute value of a is less than 1 then the magnitude of the perturbations the xt 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 singlespecies model Now we can return to a model with 2 state variables We39ll use a similar model as we used in the isocline analysis but this time represent the model with discrete equations Xt 1 fXtYt 9a Yt 1 gXtYt 9b Again we are interested in the stability of the model near some equilibrium XY Like in the singlespecies model we can consider how small permutations away from equilibrium propagate through time Because we have two state variables XY we represent this permutation as xt and yt where XtXxt 10a YtYyt 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 xt and yt are represented by xt l amxa ay yt 1 la yt l ayxt ayyyt 1 lb where 61 is the effect of species j on species 139 These are calculated from the derivatives 03 5g an 2 ayr 6X XV 6X XV a i a g 0 yy 61 XV Y XV where the derivatives represent the partial derivates which is why we use the greek 5 instead of plain 01 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 x into a linear one so that xtnd1xt In other words the dynamics could be expressed as some constant raised to the 71 power times some initial value It turns out that the dynamics of all systems can be represented by a similar equation xn7tquotC 13 where 7 is some constant which we call an eigenvalue and C is some other constant that will be dependent on the initial value of xt 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 foryt1 to get xt 2 axt 1 axt ay a xa ayy yt 14 and we can get rid ofyt by using Eq 11a to obtain xt2amayyxt1amayyaxyayxxt0 15 Using Eq 13 we can substitute MC for xn where 7 equals 2 1 or 0 in Eq 15 to derive a second order polynomial kzax rayy A amayyaxyayx0 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 7 for which Eq 13 is valid In this case there will be two roots M M which are obtained from the quadratic equation there are always two roots to a secondorder polynomial We can therefore write Eq 13 for the two species model as xnA111quot 1421 17 where A1 and A2 represent some constants These aren t very important because whether or not xtn gets bigger or smaller over time 7 depends on the values of X1 or M Consider the case when M gt M As 71 gets really big 7L1 gtgtgt Mquot In fact over time the magnitude of the Mquot becomes insigni cant compared to 7L1 xn will depend almost entirely on M We can therefore approximate Eq 17 as xquot Amax7Imaxn 18 where 7cm 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 XY Remember if Xmaxlt1 then the xn will get smaller over time ie the equilibrium is stable Keep in mind that largest magnitude means the largest absolute magnitude e g ml It turns out that the eigenvalues 7t can be determined very easily by using matrix algebra We define the matrix of the coefficients 61 as equal to so that Equation 11 can be expressed as ztl Azt 20 where z is a vector containing xt and yt The matrix A has several names but is often called the Jacobian matrix but I39ve also heard it affectionately called quotTonyquot Because A is a square matrix then there exists 2 eigenvalues 7t and eigenvectors V Recall from linear algebra that for any square matrix m x m there exists m eigenvalues 7t and m m x l eigenvectors v that satisfy the relation AV7LV 21 It turns out that the eigenvalues of the Jacobian matrix A equal the roots of the characteristic equation Thus if we had three state variables we would have three eigenvalues and if the magnitude of the largest eigenvalue was less than 1 then the equilibrium would be stable Thus only the Jacobian matrix needs to be known to evaluate the stability Most computer packages will calculate eigenvalues for you in Matlab use the quoteigquot command because the math is very messy when m gt3 Therefore for any number of state variables the stability of any equilibrium can be found from the magnitude of the dominant eigenvalue of A the Jacobian matrix Pretty slick eh The eigenvalues also tell us one more thing about the equilibrium The roots to the characteristic equation can be real or imaginary numbers If the roots eigenvalues are real then the system will monotonically move towards or away from the equilibrium If they are complex it turns out that imaginary numbers raised to a power in this case n are oscillatory with over time This comes from the fact that any complex number can be represented as a b i which is a point in the complex plain with coordinates ab An alternative way to represent this number is to consider the angle formed by the vector a b from the origin call it 1 and r is the length of the vector a2b20395 Thus arcos and brsin I This means that 61biquotrquotcosn rquotsinn i The magnitude of a complex eigenvalue is r and if this is less than 1 then the system will show damped oscillations towards equilibrium and if they are greater than 1 the system will show oscillations with increasing amplitude away from equilibrium For the special case where rl there will be stable oscillations around the equilibrium In summary the dominant eigenvalue tells us 2 things about an equilibrium It tells us whether the system will return to an equilibrium following a perturbation and it tells us whether it moves towards or away from an equilibrium monotonically LIKELIHOOD Recall our unicorn example from the last class period In that exercise we pretended as if we knew the true unicorn density and we calculated the probabilities of seeing different numbers of unicorns More realistically and I use that term loosely given that we re talking about unicoms here we wouldn t know what the unicorn density is In fact the entire reason for conducting the survey was to estimate unicorn density How might we use our data to decide what the true unicorn density is Suppose in a survey of 20 kmz you spotted 4 unicorns From this you want an estimate of unicorn density r that is in good agreement with the data What constitutes a good agreement One reasonable criteria is that the observed data should be not be very unusual given the estimate of r that you ve chosen Better yet you might want to identify the value of r for which the probability of observing 4 unicoms in 20 km2 is maximized With respect to the parameters ofthe Poisson distribution k 4 t 20 kmz so that we would choose r based on maxPF4lrt How can we do this We de ne the likelihood of r as the probability of an observed outcome given a potential value of r In other words we calculate the likelihood for any value of r as Likelihood Lr Pk 4lrt What s the difference between probability and likelihood Mathematically they are directly related to each other However there is a difference in what they represent For probabilities the parameters are known and the data are unknown while for likelihoods the data are known and the parameter values are unknown To summarize Probability Likelihood Know the parameters of the random Don t know the parameters of the variable s probability distribution random variable s pdf function pdf Want to calculate the probability of Have a known outcome different future events So if we saw 4 unicoms after searching 20 km2 we can use the PDF for a Poisson distribution and substitute F20 and 64 to calculate likelihood for any value of r exp 20r20r4 LrPk4lrt 439 Use Excel to calculate likelihoods for values of r ranging from 002 to 06 For what value of r are the data most likely Congratulations You ve just found the maximum likelihood estimate for r Please pause a moment to pat yourself on the back For what range of r are the likelihoods gt50 of the maximum likelihood gt10 of the maximum value Feedback this one is tricky The example that we considered above considered unicoms to be completely randomly distributed Suppose you new that unicoms are actually very patchily distributed so that you might expect to see very few or very many unicoms on any given sample date We often represent this type of distribution with a negative binomial pdf This is very similar to the Poisson except that in addition to the parameters r and t we have a new parameter n which describes the over dispersion or patchiness in unicorn abundances Low values of 71 indicate high patchiness and as 71 gets very large you return to random unicorn encounters Use the spreadsheet called Patchy Unicorn to see a negative binomial distribution of unicorn sightings with a dispersion parameter set at n 125 How is the probability distribution shown here different from that generated by the Poisson function Suppose you knew that n equaled 125 a fairly patch distribution and you spotted 4 unicoms in 20 km2 of searching Use this spreadsheet by incrementally changing the value of r until you find the r that makes the observation of 4 unicoms most likely Is your maximum likelihood estimate of r different than it was under the Poisson model If so offer an explanation for why this might be the case

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