Predator - Prey Model Trajectories are periodic James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 4, 2013
Outline 1 Showing The PP Trajectories are bounded 2 Trajectories must be periodic 3
Abstract This lecture shows Predator - Prey trajectories that start in Q1 are periodic.
Showing The PP Trajectories are bounded Recall our previous Predator - Prey Model x (t) = 8 x(t) 6 x(t) y(t) y (t) = 7 y(t) + 3 x(t) y(t) For any IC (x 0 >, y 0 > 0), the trajectory (x(t), y(t)) satisfies the NLCL where f (x(t))g(y(t)) = f (x 0 )g(y 0 ) f (x) = x 7 e 3x g(y) = y 8 e 6y
Showing The PP Trajectories are bounded f axis f max The Predator - Prey model f - growth function has the form f (x) = x c /e dx for non negative x. x axis x = c d Figure: The Predator - Prey f growth graph
Showing The PP Trajectories are bounded g axis g max The Predator - Prey model g - growth function has the form g(y) = y a /e by for non negative y. y axis y = a b Figure: The Predator - Prey g growth graph
Showing The PP Trajectories are bounded Figure: Predator - Prey trajectories with initial conditions from Quadrant I are bounded in x and y. The analysis we did in Lecture 39 has shown us the trajectory lives in a bounding box as shown below. y axis No y 2 No Yes 8/6 y 1 No x 1 x 7/3 2 No x axis The x(t) value for the Predator - Prey model lives in the strip x 1 x(t) x 2 and the y(t) value lives in the strip y 1 y(t) y 2. Hence, the trajectory lives in a box. The regions where the trajectory cannot be are labeled as No. The points (x 1, a/b = 8/6), (x 2, a/b = 8/6), (c/d = 7/3, y 1 ) and (c/d = 7/3, y 2 ) are on the trajectory and are marked with a circle.
Trajectories must be periodic We now know that given an initial start (x 0, y 0 ) in Quadrant I of the x y plane, the solution to the Predator - Prey system will not leave Quadrant I. If we piece the various trajectories together for Regions I, II, III and IV, the solution trajectories will either periodic, spiraling in to some center or spiraling out to give unbounded motion. These possibilities are shown in the next three Figures:(periodic), (spiraling out) and (spiraling in). We want to find out which of these possible trajectories is possible.
Trajectories must be periodic y = a b x = 0 x axis y = 0 y axis x = 0 II (, ) y = 0 x = c d I (,+) (x 0, y 0 ) III (+, ) IV (+,+) We show a possible periodic trajectories from a given start (x 0, y 0 ) in Quadrant I. Note that there is a time value, call it T, so that x(0) = x(t ) = x 0 and y(0) = y(t ) = y 0. T is called the period of the trajectory. Note horizontal and vertical lines intersect the trajectory at most two times. Figure: A Periodic Trajectory
Trajectories must be periodic y = a b x = 0 x axis y = 0 y axis x = 0 II (, ) y = 0 x = c d I (,+) (x 0, y 0 ) III (+, ) IV (+,+) We show a trajectory starting from (x 0, y 0 ) in Quadrant I that spirals out. Note horizontal and vertical lines intersect the trajectory more than two times. Figure: A Spiraling Out Trajectory
Trajectories must be periodic y = a b x = 0 x axis y = 0 y axis x = 0 II (, ) y = 0 x = c d I (,+) (x 0, y 0 ) III (+, ) IV (+,+) We show a spiraling in trajectory that starts at (x 0, y 0 ) in Quadrant I. Note horizontal and vertical lines intersect the trajectory more than two times. Figure: A Spiraling In Trajectory
Trajectories must be periodic If the trajectory was not periodic, then there would be horizontal and vertical lines that would intersect the trajectory in more than two places. We will show that we can have at most two intersections which tells us the trajectory must be periodic. We draw the same figure as before for f but we don t need the points x 1 and x 2. This time we add a point x between x 1 and x 2. We ll draw it so that it is between c/d = 7/3 and x 2 but just remember it could have been chosen on the other side.
Trajectories must be periodic f axis f max f (x ) f (x 1 ) f (x 2 ) 0.7 f max We draw the NLCL growth function f for a point x 1 < c/d = 7/3 < x < x 2. x 1 7/3 x x 2 x axis Figure: The f curve with the point x 1 < c/d = 7/3 < x < x 2 added
Trajectories must be periodic At the point x, the NLCL says the corresponding y values satisfy f (x )g(y) = 0.7f max g max. This tells us g(y) = 0.7f max f (x ) g max The biggest the ratio 0.7f max /f (x ) can be is when the bottom f (x ) is the smallest. This occurs when x is chosen to be x 1 or x 2. Then the ratio is 0.7f max /(0.7f max ) = 1. The smallest the ratio 0.7f max /f (x ) can be is when the bottom f (x ) is the largest. This occurs when x is chosen to be c/d = 7/3. Then the ratio is 0.7f max /f max ) = 0.7. So the ratio 0.7f max /f (x ) is between 0.7 and 1. Draw the g now adding in a horizontal line for rg max where r = 0.7f max /f (x ). The lowest this line can be is on the line 0.7g max and the highest it can be is the line of value g max.
Trajectories must be periodic g axis g max rg max g(y 1 ) g(y 2 ) y 1 z 1 z y 8/6 2 2 0.7 g max y axis We draw the NLCL growth function g along with the line rg max. This line crosses the g two times only except when r = 1 when it just touches g at its maximum point. However, no more than two intersections.
Trajectories must be periodic The figure above shows that there are at most two intersections with the g curve. The case of spiral in or spiral out trajectories implies there are points x with more than two corresponding y values. Hence, spiral in and spiral out trajectories are not possible and the only possibility is that the trajectory is periodic. So there is a smallest postive number T called the period of the trajectory which means (x(0), y(0)) = (x(t ), y(t )) = (x(2t ), y(2t )) = (x(3t ), y(3t )) =...
Trajectories must be periodic Homework 69 For these models, use µ =.8 and prove the trajectories are periodic. 69.1 69.2 69.3 x (t) = 10 x(t) 5 x(t) y(t) y (t) = 9 y(t) + 4 x(t) y(t) x (t) = 6 x(t) 2 x(t) y(t) y (t) = 8 y(t) + 12 x(t) y(t) x (t) = 8 x(t) 5 x(t) y(t) y (t) = 4 y(t) + 3 x(t) y(t)
Now that we know the trajectory is periodic, let s look at the plot more carefully. We know the trajectories must lie within the rectangle [x 1, x 2 ] [y 1, y 2 ]. Mathematically, this means there is a smallest positive number T so that x(0) = x(t ) and y(0) = y(t ). This number T is called the period of the Predator Prey model. We can see the periodicity of the trajectory by doing a more careful analysis of the trajectories.
Now that we know the trajectory is periodic, let s look at the plot more carefully. We know the trajectories must lie within the rectangle [x 1, x 2 ] [y 1, y 2 ]. Mathematically, this means there is a smallest positive number T so that x(0) = x(t ) and y(0) = y(t ). This number T is called the period of the Predator Prey model. We can see the periodicity of the trajectory by doing a more careful analysis of the trajectories. We know the trajectory hits the points (x 1, a b ), (x 2, a b ), ( c d, y 1) and ( c d, y 2) What happens when we look at x points u with x 1 < u < x 2?
Now that we know the trajectory is periodic, let s look at the plot more carefully. We know the trajectories must lie within the rectangle [x 1, x 2 ] [y 1, y 2 ]. Mathematically, this means there is a smallest positive number T so that x(0) = x(t ) and y(0) = y(t ). This number T is called the period of the Predator Prey model. We can see the periodicity of the trajectory by doing a more careful analysis of the trajectories. We know the trajectory hits the points (x 1, a b ), (x 2, a b ), ( c d, y 1) and ( c d, y 2) What happens when we look at x points u with x 1 < u < x 2? For convenience, let s look at the case x 1 < u < c d and the case u = c d separately.
Case 1: u = c d : In this case, the nonlinear conservation law gives f (u) g(v) = µ f max g max. However, we also know f ( c d ) = f max and so we f max g(v) = µ f max g max. or g(v) = µ g max.
Case 1: u = c d : In this case, the nonlinear conservation law gives f (u) g(v) = µ f max g max. However, we also know f ( c d ) = f max and so we f max g(v) = µ f max g max. or g(v) = µ g max. Since µ is less than 1, we draw the µ g max horizontal line on the g graph as usual to obtain the two values of v that give the value µ g max ; namely, v = y 1 and v = y 2. We conclude there are two possible points on the trajectory, ( c d, v = y 1) and ( c d, v = y 2). This gives the usual two points shown as the vertical dots in the figure below.
y axis No y 2 No Yes a/b y 1 No x 1 x c/d 2 No x axis The x(t) value for the Predator - Prey model lives in the strip x 1 x(t) x 2 and the y(t) value lives in the strip y 1 y(t) y 2. Hence, the trajectory lives in a box. The regions where the trajectory cannot be are labeled as No. The points (x 1, a/b), (x 2, a/b), (c/d, y 1 ) and (c/d, y 2 ) are on the trajectory and are marked with a circle. Figure: The points (c/d, y 1 ) and (c/d, y 2 ) are on the trajectory.
Case 2: x 1 < u < c d : The analysis is is very similar to the one we just did for u = c d. First, for this choice of u, we can draw a new graph as shown below f axis f (u) f max f (x 1 ) f (x 2 ) µ f max c d x 1 u x 2 x axis The horizontal line µ f max intersects the f curve in two points, (x 1, f (x 1 )) and (x 2, f (x 2 )). The choice x 1 < u < c d gives the vertical line shown which intersects the f curve in the point (u, f (u)) with f (x 1 ) < f (u) < f ( c d ).
Here, the conservation law gives f (u) g(v) = µ f max g max. Dividing through by f (u), we seek v values satisfying g(v) = µ fmax f (u) g max.
Here, the conservation law gives f (u) g(v) = µ f max g max. Dividing through by f (u), we seek v values satisfying g(v) = µ fmax f (u) g max. Here the ratio f max /f (u) is larger than 1 (just look at the previous figure to see this). Call this ratio r. Hence, µ < µ (f max /f (u)) and so µg max < µ (f max /f (u)) g max.
Here, the conservation law gives f (u) g(v) = µ f max g max. Dividing through by f (u), we seek v values satisfying g(v) = µ fmax f (u) g max. Here the ratio f max /f (u) is larger than 1 (just look at the previous figure to see this). Call this ratio r. Hence, µ < µ (f max /f (u)) and so µg max < µ (f max /f (u)) g max. Also from the previous figure, we see the ratio µ f max < f (u) which tells us (µ f max )/f (u) g max < g max. Now look at the figure below. The inequalities above show us we must draw the horizontal line µ r g max above the line µ g max and below the line g max.
So we seek v values that satisfy µ g max < g(v) = µ fmax f (u) g max = µ r g max < g max.
So we seek v values that satisfy µ g max < g(v) = µ fmax f (u) g max = µ r g max < g max. We already know the values of v that satisfy g(v) = µ g max which are labeled in the previous figure as v = y 1 and v = y 2. Since the number µ r is larger than µ, we see from the figure below there are two values of v, v = z 1 and v = z 2 for which g(v) = µ r g max and y 1 < z 1 < a b < z 2 < y 2.
So we seek v values that satisfy µ g max < g(v) = µ fmax f (u) g max = µ r g max < g max. We already know the values of v that satisfy g(v) = µ g max which are labeled in the previous figure as v = y 1 and v = y 2. Since the number µ r is larger than µ, we see from the figure below there are two values of v, v = z 1 and v = z 2 for which g(v) = µ r g max and y 1 < z 1 < a b < z 2 < y 2. From the figure, we see that in the case x 1 < u < c d, there are always 2 and only 2 possible v values on the trajectory. These points are (u, z 1 ) and (u, z 2 ).
g axis g max g(z 1 ) g(z 2 ) g(y 1 ) g(y 2 ) µ r g max µ g max Analyzing the Predator - Prey trajectories for x 1 < u < c d ; the g growth dynamics. a b y 1 z 1 z 2 y 2 y axis
Case 3: x 1 < u 1 < u 2 < c d : What happens if we pick two points, x 1 < u 1 < u 2 < c d? The f curve analysis is essentially the same but now there are two vertical lines that we draw as shown in the figure below. f axis f (u f 1 ) (u 2)f max f (x 1 ) f (x 2 ) µ f max The horizontal line µ f max intersects the f curve in two points, (x 1, f (x 1 )) and (x 2, f (x 2 )). The choices x 1 < u 1 < u 2 < c d give vertical lines as shown which intersect the f curve in the points (u 1, f (u 1 )) and (u 2, f (u 2 )) with f (x 1 ) < f (u 1 ) < f (u 2 ) < f ( c d ). c d x 1 u 1 u 2 x 2 x axis
Now, applying conservation law gives two equations f (u 1 ) g(v) = µ f max g max, f (u 2 ) g(v) = µ f max g max This implies we are searching for v values in the following two cases: g(v) = µ fmax f (u 1 ) g max and g(v) = µ fmax f (u 2 ) g max.
Now, applying conservation law gives two equations f (u 1 ) g(v) = µ f max g max, f (u 2 ) g(v) = µ f max g max This implies we are searching for v values in the following two cases: g(v) = µ fmax f (u 1 ) g max and g(v) = µ fmax f (u 2 ) g max. Since f (u 1 ) is smaller than f (u 2 ), we see the ratio f max /f (u 1 ) is larger than f max /f (u 2 ) and both ratios are larger than 1 (just look at the figure to see this). Call these ratios r 1 (the one for u 1 ) and r 2 (for u 2 ). It is easy to see r 2 < r 1 from the figure.
Now, applying conservation law gives two equations f (u 1 ) g(v) = µ f max g max, f (u 2 ) g(v) = µ f max g max This implies we are searching for v values in the following two cases: g(v) = µ fmax f (u 1 ) g max and g(v) = µ fmax f (u 2 ) g max. Since f (u 1 ) is smaller than f (u 2 ), we see the ratio f max /f (u 1 ) is larger than f max /f (u 2 ) and both ratios are larger than 1 (just look at the figure to see this). Call these ratios r 1 (the one for u 1 ) and r 2 (for u 2 ). It is easy to see r 2 < r 1 from the figure. We also still have (as in our analysis of the case of one point u) that both µ r 1 and µ r 2 are less than 1. We conclude µ < µ f max f (u 2 ) = µ r 1 < µ f max f (u 1 ) = µ r 2 < 1.
Now look at the next figure. The inequalities above show us we must draw the horizontal line µ r 1 g max above the line µ r 2 g max which is above the line µ g max. We already know the values of v that satisfy g(v) = µ g max which are labeled previously as v = y 1 and v = y 2.
Now look at the next figure. The inequalities above show us we must draw the horizontal line µ r 1 g max above the line µ r 2 g max which is above the line µ g max. We already know the values of v that satisfy g(v) = µ g max which are labeled previously as v = y 1 and v = y 2. Since the number µ r 2 is larger than µ, we see from the next figure there are two values of v, v = z 21 and v = z 22 for which g(v) = µ r 2 g max and y 1 < z 21 < a b < z 22 < y 2 as shown. But we can also do this for the line µ r 1 g max to find two more points z 11 and z 12 satisfying y 1 < z 21 < z 11 < a b < z 12 < z 22 < y 2 as seen in the next figure also.
We also see that the largest spread in the y direction is at x = c d giving the two points ( c d, y 1) and ( c d, y 2) which corresponds to the line segment [y 1, y 2 ] drawn at the x = c d location. If we pick the point x 1 < u 2 < c d, the two points on the trajectory give a line segment [z 21, z 22 ] drawn at the x = u 2 location.
We also see that the largest spread in the y direction is at x = c d giving the two points ( c d, y 1) and ( c d, y 2) which corresponds to the line segment [y 1, y 2 ] drawn at the x = c d location. If we pick the point x 1 < u 2 < c d, the two points on the trajectory give a line segment [z 21, z 22 ] drawn at the x = u 2 location. Note this line segment is smaller and contained in the largest one [y 1, y 2 ]. The corresponding line segment for the point u 1 is [z 11, z 12 ] which is smaller yet.
y axis y 2 a b y 1 u 1 u 2 c d x 1 x 2 x axis The Predator - Prey model trajectory is shown for the points (u 1, z 11 ), (u 1, z 12 ), (u 2, z 21 ), (u 2, z 22 ) which live in the bounding box [x 1, x 2 ] [y 1, y 2 ]. Note the length of the line segments in the vertical direction is decreasing as we move away from the center line through x = c d.
If you think about it a bit, if we picked three points as follows, x 1 < u 1 < u 2 < u 3 < c d and three more points < u 4 < u 5 < u 6 < x 2, we would find line segments as follows: c d Point Spread x 1 One point (x 1, a b ) u 1 [z 11, z 12 ] u 2 [z 21, z 22 ] contains [z 11, z 12 ] u 3 [z 31, z 32 ] contains [z 21, z 22 ] c d [y 1, y 2 ] contains [z 21, z 22 ] u 4 [z 41, z 42 ] inside [y 1, y 2 ] u 2 [z 51, z 52 ] inside [z 41, z 42 ] u 1 [z 61, z 62 ] inside [z 51, z 52 ] x 2 One point (x 2, a b ) inside [z 51, z 52 ]
y axis y 2 a b y 1 u 1 u 2 u3 u4u5 u 6 c d x 1 x 2 x axis The Predator - Prey model trajectory is shown for the points (u 1, z 11 ), (u 1, z 12 ), (u 2, z 21 ), (u 2, z 22 ), (u 3, z 31 ), (u 4, z 42 ) and (u 5, z 51 ), (u 6, z 62 ) which all live in the bounding box [x 1, x 2 ] [y 1, y 2 ]. Note the length of the line segments in the vertical direction is decreasing as we move away from the center line through x = c d.
Again, note since the trajectory is periodic is there is a smallest positive number T so that x(t + T ) = x(t) and y(t + T ) = y(t) for all values of t.
Again, note since the trajectory is periodic is there is a smallest positive number T so that x(t + T ) = x(t) and y(t + T ) = y(t) for all values of t. Note the value of this period is really determined by the initial values (x 0, y 0 ) as they determine the bounding box [x 1, x 2 ] [y 1, y 2 ] since the initial condition determines µ.