Outline. Additional Nonlinear Systems. Abstract. Finding Equilibrium Points Numerically. Newton s Method

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1 Outline Finding Equilibrium Points Numerically Additional Nonlinear Systems James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University June 13, 2017 Newton s Method Adding Finite Difference Approximations to the Derivative A Second Nonlinear Model EP One EP Two EP Three The Full Phase Plane Predator Prey Models Abstract We can find the equilibrium points using the root finding methods called bisection and Newton s method. This lecture is going discuss more about nonlinear systems. We need a simple function to find the root of a nice function f of the real variable x using what is called bisection. The method is actually quite simple. We know that if f is a continuous function on the finite interval [a, b] then f must have a zero inside the interval [a, b] if f has a different algebraic sign at the points a and b. This means the product f (a) f (b) is not zero. So we assume we can find an interval [a, b] on which this change in sign satisfies f (a) f (b) 0 (which we can do by switching to f if we have to!) and then if we divide the interval [a, b] into two equal pieces [a, m] and [m, b], f (m) can t have the same sign as both f (a) and f (b) because of the assumed sign difference. So at least one of the two halves has a sign change.

2 Note that if f (a) and f (b) was zero then we still have f (a) f (b) 0 and either a or b could be our chosen root and either half interval works fine. If only one of the point function values is zero, then the bisection of [a, b] into the two halves still finds the one half interval that has the root. So our prototyping Matlab code should use tests like f (x) f (y) 0 rather than f (x) f (y) < 0 to make sure we catch the root. Let s look at a simple Matlab function to perform the Bisection routine. f u n c t i o n r o o t = B i s e c t i o n (fname, a, b, d e l t a ) fname i s the name o f the f u n c t i o n f ( x ) a, b f o r which f (a) f (b) <= 0 d e l t a t h i s i s a non n e g a t i v e r e a l number r o o t t h i s i s the m i d p o i n t o f [ c, d ] with f (c) f (d) <= 0 and d c <= d e l t a + eps max ( c, d ) where eps i s machine z e r o d i s p ( ) d i s p ( k a ( k ) b ( k ) b ( k ) a ( k ) ) k = 1 ; d i s p ( s p r i n t f ( 6d 12.7 f 12.7 f 12.7 f, k, a, b, b a ) ) ; fa = f e v a l (fname, a ) ; f b = f e v a l ( fname, b ) ; w h i l e abs (a b ) > d e l t a + eps max ( abs (a), abs (b) ) mid = ( a+b ) / 2 ; fmid = f e v a l ( fname, mid ) ; i f f a fmid <= 0 t h e r e i s a r o o t i n [ a, mid ] b = mid ; fb = fmid ; t h e r e i s a r o o t i n [ mid, b ] a = mid ; f a = fmid ; k = k +1; d i s p ( s p r i n t f ( 6d 12.7 f 12.7 f 12.7 f, k, a, b, b a ) ) ; r o o t = ( a+b ) / 2 ; We should look at some of these lines more closely. First, to use this routine, we need to write a function definition for the function we want to apply bisection to. We will do this in a file called func.m (Inspired Name, eh?) An example would be the one we wrote for the function Note that the name of our supplied function is passed in as the first argument in single quotes as it is a string. Also, in the Bisection routine, we have added the code to print out what is happening at each iteration of the while loop. Matlab handles prints to the screen a little funny, so do set up a table of printed values we use this syntax: f u n c t i o n y = f u n c ( x ) x r e a l i n p u t y r e a l o u t p u t y = tan ( x /4) 1; f (x) = tan( x 4 ) 1; t h i s p r i n t s a blank l i n e and then a t a b l e heading. note d i s p p r i n t s a s t r i n g o n l y d i s p ( ) d i s p ( k a ( k ) b ( k ) b ( k ) a ( k ) ) now to p r i n t t he k, a, b and b a, we must f i r s t put t h e i r v a l u e s i n t o a s t r i n g u s i n g t h e c l i k e f u n c t i o n s p r i n t f and then use disp to d i s p l y that s t r i n g. so we do t h i s d i s p ( s p r i n t f ( o u t p u t s p e c i f i c a t i o n s h e r e, v a r i a b l e s h e r e ) ) so i n s i d e the w h i l e l o o p we use d i s p ( s p r i n t f ( 6d 12.7 f 12.7 f 12.7 f, k, a, b, b a ) ) ; As mentioned above, we will test this code on the function So to apply bisection to this function on the interval [2, 4] with a stopping tolerance of say 10 4, in Matlab, we would type the command root = Bisection( func,2,4,10^-4). f (x) = tan( x 4 ) 1; on the interval [2, 4] with a stopping tolerance of δ = Our function is the Matlab function func in the file func.m.

3 root = B i s e c t i o n ( func,2,4,10ˆ 6) Homework 60 k a ( k ) b ( k ) b ( k ) a ( k ) r o o t = Use bisection to find the first five positive solutions of the equation x = tan(x). You can see where this is roughly by graphing tan(x) and x simultaneously. Do this for tolerances {10 1, 10 2, 10 3, 10 4, 10 5, 10 6, 10 7 }. For each root, choose a reasonable bracketing interval [a, b], explain why you chose it, provide a table of the number of iterations to achieve the accuracy and a graph of this number vs. accuracy Use the Bisection Method to find the largest real root of the function f (x) = x 6 x 1. Do this for tolerances {10 1, 10 2, 10 3, 10 4, 10 5, 10 6, 10 7 }. Choose a reasonable bracketing interval [a, b], explain why you chose it, provide a table of the number of iterations to achieve the accuracy and a graph of this number vs. accuracy. Newton s method is based on the tangent line. At a given point x0, we have f (x) = f (x0) + f (x0)(x x0) f (cx)(x x0) 2 where cx is some point between x0 and x. Neglecting the error, we can say f (x) f (x0) + f (x0)(x x0) If we are looking for a zero of f, we would want f (x) = 0. This gives 0 f (x0) + f (x0)(x x0) Now solve for x to find x = x0 f (x0) f (x0) If we think of x0 as being our current guess of the value of the root, x old, we can call x our next guess, x new. So we can rewrite the equation above as x new = x old f (x old ) f (x old ) Of course, if f (x old ) is small this generates a very poor next estimate. Newton s method rapidly converges to a zero of the function f if the original guess is reasonable. Of course, that is the problem. A bad initial guess is a great way to generate random numbers! So usually, we find a good interval where the root might reside by first using bisection. The following code uses a simple test to see which we should do in our zero finding routine.

4 f u n c t i o n ok = S t e p I s I n ( x, fx, fpx, a, b ) x c u r r e n t a p p r o x i m a t e r o o t f x f u n c t i o n f v a l u e a t a p p r o x i m a t e r o o t fpx d e r i v a t i v e f value at approximate root a, b r o o t i s i n t h i s i n t e r v a l ok 1 i f the Newton Step x f x /f p x i s i n [a, b ], 0 i f not i f f p x > 0 ok = ( ( a x ) f p x <= f x ) && ( f x <= ( b x ) f p x ) ; i f f p x < 0 ok = ( ( a x ) f p x >= f x ) && ( f x >= ( b x ) f p x ) ; ok = 0 ; Once the bisection steps have given us an interval where the root might be, we switch to Newton s method. Newton s Method can be rephrased for the scalar function case as xn+1 = xn f (xn) f (xn). f u n c t i o n [ x, fx, n E v a l s, af, bf ] = GlobalNewton ( fname, fpname, a, b, t o l x, t o l f, nevalsmax ) fname name o f f ( x ), f pname name of derivative f ( x ) a, b s e a r c h i n t e r v a l [a, b ] tolx interval size tol, t o l f f ( r o o t approx ) t o l nevalsmax Max Number o f d e r E v a l x s x Approx z e r o o f f, f x i s f ( x ) nevals Num of Deriv E v a l s Needed, af, bf f i n a l i n t e r v a l [af, bf ] T e r m i n [a, b ] has s i z e < t o l x a t i o n o r f ( r o o t ) < t o l f o r num o f e v a l s e x c e e d s nevalsmax where xn denotes x old and xn+1 denotes x new. It is clear the method fails if f (xn) = 0 or is close to 0 at any iteration. MatLab code to implement this method is given next. f u n c t i o n [ x, fx, n E v a l s, af, bf ] = GlobalNewton ( fname, fpname, a, b, t o l x, t o l f, nevalsmax ) f a = f e v a l (fname, a ) ; f b = f e v a l (fname, b ) ; x = a ; f x = f e v a l ( fname, x ) ; f p x = f e v a l ( fpname, x ) ; n E v a l s = 1 ; k = 1 ; d i s p ( ) d i s p ( Step k a ( k ) x ( k ) b ( k ) ) d i s p ( s p r i n t f ( S t a r t 6d 12.7 f 12.7 f 12.7 f, k, a, x, b ) ) ; w h i l e ( abs (a b )>t o l x ) && ( abs ( f x )> t o l f ) && ( nevals<nevalsmax ) ( n E v a l s ==1) check = S t e p I s I n ( x, fx, fpx, a, b ) ; [a, b ] b r a c k e t s a r o o t and x=a o r x=b i f check x = x fx / fpx ; Take Newton Step x = ( a+b ) / 2 ; Take a B i s e c t i o n Step : f x = f e v a l ( fname, x ) ; f p x = f e v a l ( fpname, x ) ; n E v a l s = n E v a l s +1; i f f a fx <=0 b = x ; f b = f x ; t h e r e i s a r o o t i n [a, x ]. Use r i g h t e n d p o i n t. a = x ; f a = f x ; t h e r e i s a r o o t i n [ x, b ]. B r i n g i n l e f t e n d p o i n t. k = k +1; i f ( check ) d i s p ( s p r i n t f ( Newton 6d 12.7 f 12.7 f 12.7 f, k, a, x, b ) ) ; d i s p ( s p r i n t f ( B i s e c t i o n 6d 12.7 f 12.7 f 12.7 f, k, a, x, b ) ) ; af = a ; bf = b ; We will apply our global newton method root finding code to a simple example: find a root for f (x) = sin(x) in the interval [ 7π 2, 15π + 0.1]. We code the function and its derivative in two simple Matlab files f u n c t i o n y = f 1 ( x ) y = s i n ( x ) ; and f u n c t i o n y = f 1 p ( x ) y = c o s ( x ) ; To run this code on this example, we would then type a phrase like the one below: [ x, fx, n E v a l s, alast, b L a s t ] = GlobalNewton ( f1, f1p, 7 p i /2,15 p i +.1,10ˆ 6,10ˆ 8,200) Here we set the interval tolerance to 10 6 so if the width of the interval [a, b] < 10 6 the algorithm stops. We also set the root tolerance to 10 8 which means the algorithm terminates once the value of the approximate root f (x) < 10 8.

5 Homework 61 Here is the runtime output: [ x, fx, n E v a l s, alast, b L a s t ] = GlobalNewton ( f1, f1p, 7 p i /2,15 p i +.1,10ˆ 6,10ˆ 8,200) Step k a ( k ) x ( k ) b ( k ) S t a r t B i s e c t i o n B i s e c t i o n Newton Newton Newton x = f x = e 16 n E v a l s = 6 a L a s t = b L a s t = Use the Global Newton Method to find the first five positive solutions of the equation x = tan(x). Do this for root tolerances {10 1, 10 2, 10 3, 10 4, 10 5, 10 6, 10 7 } and always use the interval tolerance First make up the two functions f and fp which are f (x) = x tan(x) and f (x) = 1 sec 2 (x). These will be in file with names like myfunc.m and myfuncp.m. Sketch tan(x) and x together. The positive roots are clearly close to 3π/2, 5π/2 and so on. So for each one, pick a starting interval [a, b] so the x tan(x) changes size at the points. This requires a bit of trial and error. The function call for the first root tolerance is [x,fx,nevals,alast,blast] = GlobalNewton( myfunc, myfuncp,a,b,10^-6,10^-1,200) where a and b are your starting guesses for the first interval. Keep track of how many iterations you use for each choice of root tolerance and display this as a table for each root. Homework 61 We can also choose to replace the derivative function for f with a finite difference approximation. We will use f (x) f (xc + δc) f (xc) δc 61.2 Follow the guidelines of Problem 61.1 and use the Global Newton Method to find the largest real root of the function f (x) = x 6 x 1. Do this for the root tolerances {10 1, 10 2, 10 3, 10 4, 10 5, 10 6, 10 7 } and always use the interval tolerance to approximate the value of the derivative at the point xc. Some care is required to pick a size for δc so that round-off errors do not destroy the accuracy of our finite difference approximation to f. The simple Matlab code to implement this is given below: f v a l = f e v a l ( fname, x ) ; f p v a l = ( f e v a l (fname, x+d e l t a ) f v a l ) / d e l t a ; We can also use a secant approximation as follows: f v a l = f e v a l ( fname, x ) ; f v a l c = f e v a l ( fname, c ) ; f p c = ( f v a l c f v a l ) /( x c ) ; We add the finite difference routines into our Global Newton s Method as follows:

6 f u n c t i o n [ x, fx, n E v a l s, af, bf ] =... In Matlab the smallest number we can uniquely have is eps which is e 16. We use this to set the δ for our finite difference approximations. in the line delta = sqrt(eps)* abs(x);. GlobalNewtonFD (fname, a, b, t o l x, t o l f, nevalsmax ) fname a s t r i n g t h a t i s the name o f the f u n c t i o n f ( x ) a, b we l o o k f o r the r o o t i n the i n t e r v a l [a, b ] t o l x t o l e r a n c e on t h e s i z e o f t h e i n t e r v a l t o l f t o l e r a n c e o f f ( c u r r e n t a p p r o x i m a t i o n to r o o t ) nevalsmax Maximum Number o f d e r i v a t i v e E v a l u a t i o n s x Approximate z e r o o f f f x The v a l u e o f f a t t h e a p p r o x i m a t e z e r o n E v a l s The Number o f D e r i v a t i v e E v a l u a t i o n s Needed af, bf the f i n a l i n t e r v a l the a p p r o x i m a t e r o o t l i e s in, [ af, bf ] T e r m i n a t i o n I n t e r v a l [a, b ] has s i z e < t o l x f ( a p p r o x i m a t e r o o t ) < t o l f Have exceeded nevalsmax d e r i v a t i v e E v a l u a t i o n s f u n c t i o n [ x, fx, n E v a l s, af, bf ] = GlobalNewtonFD ( fname, a, b, t o l x, t o l f, nevalsmax ) f a = f e v a l (fname, a ) ; f b = f e v a l (fname, b ) ; x = a ; f x = f e v a l (fname, x ) ; d e l t a = s q r t ( eps ) abs (x) ; f p v a l = f e v a l (fname, x+d e l t a ) ; f p x = ( f p v a l f x ) / d e l t a ; n E v a l s = 1 ; k = 1 ; w h i l e ( abs (a b )>t o l x ) && ( abs ( f x )> t o l f ) && ( nevals<nevalsmax ) ( n E v a l s ==1) check = S t e p I s I n ( x, fx, fpx, a, b ) ; [a, b ] b r a c k e t s a r o o t and x=a o r x=b i f check x = x fx / fpx ; Take Newton Step x = ( a+b ) / 2 ; Take a B i s e c t i o n Step : f x = f e v a l ( fname, x ) ; f p v a l = f e v a l ( fname, x+d e l t a ) ; f p x = ( f p v a l f x ) / d e l t a ; n E v a l s = n E v a l s +1; i f f a fx <=0 b = x ; t h e r e i s a r o o t i n [a, x ]. Use r i g h t e n d p o i n t. fb = fx ; a = x ; t h e r e i s a r o o t i n [ x, b ]. B r i n g i n l e f t e n d p o i n t. fa = fx ; k = k +1; af = a ; bf = b ; We will apply our finite difference global newton method root finding code to the same simple example: find a root for f (x) = sin(x) in the interval [ 7π 2, 15π + 0.1]. Homework 62 To run this code on this example, we would then type [ x, fx, n E v a l s, alast, b L a s t ] = GlobalNewtonFD ( f1, 7 p i /2,15 p i +.1,10ˆ 6,10ˆ 8,200) This generates [ x, fx, n E v a l s, alast, b L a s t ] = GlobalNewtonFD ( f1, 7 p i /2,15 p i +. 1,... 10ˆ 6,10ˆ 8,200) Step k a ( k ) x ( k ) b ( k ) S t a r t B i s e c t i o n B i s e c t i o n Newton Newton Newton x = f x = e 15 n E v a l s = 6 a L a s t = b L a s t = Use the Finite Difference Global Newton Method to find the second positive solution of the equation x = tan(x). Do this for the interval tolerance 10 6 and the root tolerance For this problem, we will look at what happens as we vary the δ we use to find the finite difference approximations to the derivative at each step. So for the δ choices {10 4, 10 6, 10 8, } change the line delta = sqrt(eps)* abs(x); in the code to be delta = 10^-4; etc and run the code. Provide a table of the root accuracy versus the δ choice.

7 Homework 62 Now let s look at this model 62.2 Use the Finite Difference Global Newton Method to find the largest real root of the function f (x) = x 6 x 1. Do this for the interval tolerance 10 6 and the root tolerance For this problem, we will look at what happens as we vary the δ we use to find the finite difference approximations to the derivative at each step. So for the δ choices {10 4, 10 6, 10 8, } change the line delta = sqrt(eps)* abs(x); in the code to be delta = 10^-4; etc and run the code. Provide a table of the root accuracy versus the δ choice. x =.5( h(x) + y) y =.2( x 1.5y + 1.2) for h(x) = 17.76x x x x x 5, This is model of how an electrical component called a diode behaves called the trigger model and the details are not really important as we are just investigating how to use our code and theoretical ideas. The equilibrium points are the solutions to the simultaneous equations y = 17.76x x x x x 5 y = 2 3 x We can see these solutions graphically by plotting the two curves simultaneously and using the cursor to locate the roots and read off the (x, y) values from the plot. This is not quite accurate so a better way is to find the roots numerically. The plot which shows the equilibrium points graphically is shown below: g x ) (12 10 x ) / 1 5 ; h x x. ˆ x. ˆ x. ˆ x. ˆ 5 ; X = l i n s p a c e ( 0, 1, ) ; Y1 = h (X) ; Y2 = g (X) ; p l o t (X, Y1, X, Y2 ) ; q x ) h ( x ) g ( x ) ; [ x1, fx1, Nevals1, af1, bf1 ] =... GlobalNewtonFD ( q,0,.1,10ˆ 6,10ˆ 8,10) ; [ x2, fx2, Nevals2, af2, bf2 ] =... GlobalNewtonFD ( q,.2,.4,10ˆ 6,10ˆ 8,10) ; [ x3, fx3, Nevals3, af3, bf3 ] =... GlobalNewtonFD ( q,.4,.9,10ˆ 6,10ˆ 8,10) ; y1 = h ( x1 ) ; y2 = h ( x2 ) ; y3 = h ( x3 ) ; EP = {[ x1, y1 ], [ x2, y2 ], [ x3, y3 ] } ; [ x3, fx3, Nevals3, af3, bf3 ] =... GlobalNewtonFD ( q,.4,.9,10ˆ 6,10ˆ 8,10) ; y3 = h ( x3 ) ; EP = {[ x1, y1 ], [ x2, y2 ], [ x3, y3 ] } ; EP EP = { [ 1, 1 ] = [ 1, 2 ] = [ 1, 3 ] = The graphical and numerical solutions are shown in the following MatLab/ Octave session.

8 For the equilibrium point Q1, we have J ( x1, y1 ) ans = There are now three equilibrium points Q1 = ( , ) Q2 = ( , ) Q3 = ( , ) [V1,D1 ] = e i g (J(x1, y1 ) ) V1 = D1 = The Jacobian is 0.5h J(x, y) = (x) D i a g o n a l M a t r i x Hence, there are two real distinct eigenvalues, r1 = and r2 = The eigenvectors are E1 = and E2 = The dominant eigenvector is E1 and it is easy to plot the resulting trajectories. We modify our function linearsystem to the new function linearsystemep so that we can plot the trajectories centered at the equilibrium point Q1. A1 = J ( x1, y1 ) ; p1 = [ A1 ( 1, 1 ), A1 ( 1, 2 ), A1 ( 2, 1 ), A1 ( 2, 2 ), x1, y1 ] p1 = AutoPhasePlanePlotLinearSystemRKF5 ( x a x i s, y a x i s, Q1 Plot,1.0 e 6,1.0 e 6,. 0 1,. 2, l i n e a r s y s t e m e p, p1,. 0 1, 0,. 4, 1 2, 1 2, 0, 1, 0, 1 ) ; f u n c t i o n f = l i n e a r s y s t e m e p ( p, t, x ) a = p ( 1 ) ; b = p ( 2 ) ; c = p ( 3 ) ; d = p ( 4 ) ; ex = p ( 5 ) ; ey = p ( 6 ) ; u = x ( 1 ) ex ; v = x ( 2 ) ey ; f = [ a u+b v ; c u+d v ] ; We set up the plot as follows. We define the coefficient matrix A1 of our linearization and set up the parameter p1 by listing all the entries of A1 followed by the coordinates of Q1. We then generate the automatic phase plane plot.

9 For equilibrium point Q2, we find the linearization just like we did for Q1. First, we find the Jacobian at this point. J ( x2, y2 ) ans = A2 = J ( x2, y2 ) ; p2 = [ A2 ( 1, 1 ), A2 ( 1, 2 ), A2 ( 2, 1 ), A2 ( 2, 2 ), x2, y2 ] p2 = AutoPhasePlanePlotLinearSystemRKF5 ( x a x i s, y a x i s, Q1 Plot,1.0 e 6,1.0 e 6,. 0 1,. 2, l i n e a r s y s t e m e p, p2,. 0 1, 0,. 4, 1 2, 1 2, 0. 5, 1, 0, 1 ) ; [V2,D2 ] = e i g (J(x2, y2 ) ) V2 = D2 = Hence, there are two real distinct eigenvalues, r1 = and r2 = The eigenvectors are E1 = and E2 = The dominant eigenvector is again E1. We set the coefficient matrix A2 of our linearization and the parameter p2 of our model and generate the automatic phase plane plot. Finally, we analyze the model near the point Q3. The Jacobian is now ( x3, y3 ) ans = [V3,D3 ] = e i g (J(x3, y3 ) ) V3 = D3 = A3 = J ( x3, y3 ) ; p3 = [ A3 ( 1, 1 ), A3 ( 1, 2 ), A3 ( 2, 1 ), A3 ( 2, 2 ), x3, y3 ] p3 = AutoPhasePlanePlotLinearSystemRKF5 ( x a x i s, y a x i s, Q1 Plot,1.0 e 6,1.0 e 6,. 0 1,. 2, l i n e a r s y s t e m e p, p3,. 0 1, 0,. 4, 1 2, 1 2, 0, 1, 0, 1 ) ; Hence, there are two real distinct eigenvalues, r1 = and r2 = The eigenvectors are E1 = and E2 = The dominant eigenvector is now E2. We set the coefficient matrix A3 of our linearization and the parameter p3 of our model and generate the automatic phase plane plot.

10 We can also generate the full plot of the original system using the function triggermodel. f u n c t i o n y = t r i g g e r m o d e l ( t, x ) u = x ( 1 ) ; v = x ( 2 ) ; y = [ 0.5 (17.76 u u.ˆ u.ˆ u.ˆ u.ˆ5 v ) ; ( u 1.5 v + 1.2) ] ; We then plot the trajectories using AutoPhasePlanePlotRKF5NoP. You can see in this plot how trajectories do not stay near Q2; instead, they move to Q1 or to Q3. Hence, you can a trigger a move from Q1 to Q3 or vice versa by choosing the right Initial condition! So this gives us a way to implement computer memory. Choose say Q1 to represent a binary 1 and Q3 to represent a binary 0 or to implement a model of how emotional states can switch quickly. AutoPhasePlanePlotRKF5NoP ( x axis, y axis, Trigger Model Phase Plane,1.0 e 6,1.0 e 6,. 0 1,. 0 5, t r i g g e r m o d e l,. 0 1, 0, 2, 2 0, 2 0, 0. 2, 1, 0, 1 ) ; For the equilibrium point Q1, we have The next nonlinear system is the familiar Predator - Prey model. Consider the following example x (t) = 3 x(t) 4 x(t) y(t) y (t) = 5 y(t) + 7 x(t) y(t) The equilibrium points (x, y) solve the equations x (3 4y) = 0 y ( 5 + 7x) = 0 which has the familiar solutions Q1 = (0, 0) and Q2 = ( 5 7, 3 4 ). The Jacobian here is 3 4y 4x J(x, y) = 7y 5 + 7x A1 = J ( 0, 0 ) A1 = [ V1, D1 ] = e i g ( A1 ) V1 = D1 = Hence, there are two real distinct eigenvalues, r1 = 5 and r2 = 3. The eigenvectors are 0 1 E1 = and E2 = 1 0 The dominant eigenvector is thus E2 and it is easy to plot the resulting trajectories. Using the function linearsystemep we set up the plot. Using the coefficient matrix A1 of our linearization and the coordinates of Q1, we set the parameter p1 by listing all the entries of A1 followed by the coordinates of Q1. We then generate the automatic phase plane plot.

11 A1 = J ( x1, y1 ) ; p1 = [ A1 ( 1, 1 ), A1 ( 1, 2 ), A1 ( 2, 1 ), A1 ( 2, 2 ), x1, y1 ] p1 = AutoPhasePlanePlotLinearSystemRKF5 ( x a x i s, y a x i s, Q1 Plot,1.0 e 6,1.0 e 6,. 0 1,. 2, l i n e a r s y s t e m e p, p1,. 0 1, 0,. 4, 1 2, 1 2, 1, 1, 1, 1 ) ; Note the linearization shows a set of trajectories that moves out from the origin along the x axis deping on the quadrant the initial condition comes from. We also know that no trajectory of the predator - prey model that starts in Quadrant One can cross the x and y axis, so the lesson that the linearization is an approximation to the true trajectories really comes home here. For equilibrium point Q2, we find the Jacobian at this point. A2 = J ( x2, y2 ) A2 = [ V2, D2 ] = e i g ( A2 ) V2 = i i i i D2 = i i Hence, there is now a complex conjugate eigenvalue pair which has zero real part. Thus, these trajectories will be circles about Q2. The eigenvalues are r1 = i and r2 = i. We set coefficient matrix A2 and p2 and generate the automatic phase plane plot. A2 = J ( x2, y2 ) ; p2 = [ A2 ( 1, 1 ), A2 ( 1, 2 ), A2 ( 2, 1 ), A2 ( 2, 2 ), x2, y2 ] p2 = AutoPhasePlanePlotLinearSystemRKF5 ( x a x i s, y a x i s, Q2 Plot,1.0 e 6,1.0 e 6,. 0 1,. 2, l i n e a r s y s t e m e p, p2,. 0 1, 0, 2. 4, 1 2, 1 2, 0. 5,. 1, 0,. 1 ) ; The full plot of the original system uses the standard function PredPrey modified for the model at hand. f u n c t i o n f = PredPrey ( t, x ) f = [3 x (1) 4 x (1) x (2) ; 5 x (2)+7 x (1) x (2) ] ; We then plot the trajectories using AutoPhasePlanePlotRKF5NoPMultiple. We had to write a new function to generate the plot as the equilibrium point Q1 generates trajectories with large x and y values which complicate the calculations. So the new function manually builds four separate plots, one for each quadrant, and stays carefully away from the trajectories on the x and y axis. It is pretty hard to automate this kind of plot! In each quadrant, the number of trajectories and the size of the box from which the initial conditions come must be chosen inside the code. Also, the length of time the calculations run in a given quadrant must be handpicked. Hence, there is a lot of tweaking here!

12 function AutoPhasePlanePlotRKF5NoPMultiple ( Hlabel, Vlabel, Tlabel, errortol, s t e p t o l, minstep, maxstep, fname, h i n i t,q1,q2,q3,q4) fname i s t h e name o f t h e model dynamics e r r o r t o l i s t o l e r a n c e i n RKF5 a l g o r i t h m s t e p t o l i s a n o t h e r t o l e r a n c e i n RKF5 a l g o r i t h m m i n s t e p i s t h e minimum s t e p s i z e a l l o w e d maxstep i s t h e maximum s t e p s i z e a l l o w e d h i n i t i s the i n i t i a l s t e p s i z e Q1 i s 1 i f we want to g e n e r a t e t h e p l o t i n q u a d r a n t 1 Q2 i s 1 i f we want to g e n e r a t e t h e p l o t i n q u a d r a n t 2 Q3 i s 1 i f we want to g e n e r a t e t h e p l o t i n q u a d r a n t 3 Q4 i s 1 i f we want to g e n e r a t e t h e p l o t i n q u a d r a n t 4 t i s t he f i n a l time function AutoPhasePlanePlotRKF5NoPMultiple ( Hlabel, Vlabel, Tlabel, errortol, s t e p t o l, minstep, maxstep, fname, h i n i t,q1,q2,q3,q4) newplot ; hold a l l ; i f (Q3 == 1) LL = 2 ; ull = l i n s p a c e (.2,.1, LL ) ; vll = l i n s p a c e (.2,.1, LL ) ; u = ull ; v = vll ; f o r i =1: LL f o r j =1: LL y0 = [ u ( i ) ; v ( j ) ] ; [ t v a l s, yvals, f v a l s, h v a l s ] = RKF5NoP( e r r o r t o l, s t e p t o l, minstep, maxstep,... fname, h i n i t, 0, 0. 5, y0 ) ; p l o t ( y v a l s ( 1, : ), y v a l s ( 2, : ) ) ; i f (Q2 == 1) UL = 2 ; uul = l i n s p a c e (.2,.1,UL ) ; vul = l i n s p a c e (.1,.2, UL ) ; u = uul ; v = vul ; f o r i =1:UL f o r j =1:UL y0 = [ u ( i ) ; v ( j ) ] ; [ t v a l s, yvals, f v a l s, h v a l s ] = RKF5NoP( e r r o r t o l, s t e p t o l, minstep, maxstep,... fname, h i n i t, 0, 0. 6, y0 ) ; p l o t ( y v a l s ( 1, : ), y v a l s ( 2, : ) ) ; C o n t i n u e t h e s o u r c e l i s t i n g i f (Q4 == 1) LR = 2 ; ulr = l i n s p a c e (. 1,. 2, LR ) ; vlr = l i n s p a c e (.2,.1, LR ) ; u = ulr v = vlr f o r i =1: LR f o r j =1: LR y0 = [ u ( i ) ; v ( j ) ] ; [ t v a l s, yvals, f v a l s, h v a l s ] = RKF5NoP( e r r o r t o l, s t e p t o l, minstep, maxstep,... fname, h i n i t, 0, 0. 5, y0 ) ; p l o t ( y v a l s ( 1, : ), y v a l s ( 2, : ) ) ; i f (Q1 == 1) UR = 4 ; uur = l i n s p a c e (.1,2.0, UR ) ; vur = l i n s p a c e (.1,2.0, UR ) ; u = uur ; v = vur ; f o r i =1:UR f o r j =1:UR y0 = [ u ( i ) ; v ( j ) ] ; [ t v a l s, yvals, f v a l s, h v a l s ] = RKF5NoP( e r r o r t o l, s t e p t o l, minstep, maxstep,... fname, h i n i t, 0, 2. 4, y0 ) ; p l o t ( y v a l s ( 1, : ), y v a l s ( 2, : ) ) ; x l a b e l ( H l a b e l ) ; y l a b e l ( V l a b e l ) ; t i t l e ( T l a b e l ) ; hold o f f ; AutoPhasePlanePlotRKF5NoPMultiple ( x a x i s, y a x i s, P r e d a t o r Prey Phase Plane, 1. 0 e 6,1.0 e 6,. 0 1,. 0 5, PredPrey,. 0 1, 1, 1, 1, 1 ) ;

13 Homework 63 Homework 63 For each of the following systems, do this: 1. Graph the nullclines x = 0 and y = 0 and show on the x y plane the regions where x and y take on their various algebraic signs. 2. Find the equilibrium points. 3. At each equilibrium point, find the Jacobian of the system and analyze the linearized system we have discussed in class. This means: find eigenvalues and eigenvectors if the system has real eigenvalues. You don t need the eigenvectors if the eigenvalues are complex. sketch a graph of the linearized solutions near the equilibrium point. 4. Using 1 and 3 to combine all this information into full graph of the system x = y y = x + x 3 6 y x = x + y y =.1x 2y x 2.1x 3 x = y y = x + y (1 3x 2 2y 2 )