This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ECOLOGICAL SYTEMS MODELING SYSTEMS MODEL APPROACH
A) This is a numerical method of examining models which are too large to be fitted
statistically.
1) STATISTICS allows us to fit a certain range of linear models and
non linear models. SYSREG in SAS will even allow to fit several simultaneous linear models.
2) BIOMATH allows us to study the behavior of some models which can be
treated with advanced mathematical techniques
3) SYSTEMS MODELING is a numerical method of examining the behavior of a
virtually limitless number of simultaneous equations, generally not linear. These are not easily examined in their entirety with mathematical analytical
techniques B) SIMPLE MODEL EXPONENTIAL GROWTH
Nt = N! ert
˜ N = r Nt
1) we have discussed the statistical examination of this model
2) it has an exact mathematical solution and can be examined with analytical
mathematical techniques.
3) it can also be modeled numerically, though the solution is less precise.
eg. requires a numerical integration of the formula.
˜N
˜t = rNt or ˜N 1
˜t N> = r Page 1 NUMERICAL INTEGRATION: Rectangular Method (many other methods exist)
let N! = 10 and r = 0.1 per unit time
at each time “t+1", the value of Nt+1 is given by [Nt ˜ Nt ]
EXAMPLE using exponential growth
find Nt where t = 8; exact solution = 22.255 ˜ t=1
r=0.1
t=1
2
3
4
5
6
7
8 ˜ t=2
r=0.2 Nt = 11
12.1
13.3
14.6
16.1
17.7
19.5
21.4 ˜ t=4
r=0.4 ˜ t=8
r=0.8 12
14.4 14
17.3
20.7 19.6 18 We will now discuss problems of the form
˜N
˜t œ N*f(N) Some of these have mathematical solutions (the simpler ones), and we will discuss
a few of these. However, many do not have solutions and must be numerically
integrated such as we did above with the exponential growth curve.
eg. For the fisheries models we have discussed, Beverton and Schaefer have
already described their models as
˜P
˜t œ Pcr(P) g(P) M(P) F(X)d % and
˜P
˜t œ Pcf(P) F(X)d % if we had these rates we could calculate Pt+˜t œ Pt ˜P
˜t Page 2 TWO SYSTEM MODEL the LOTKA VOLTERRA PREDATOR PREY
MODEL
where;
PREY = X
PREDATOR = Y
the Lotka Volterra system of equations for COMPETITION between two species
is
˜ X = r " X ’1 1
K" X !
K" Y“ have you seen this model before?
˜ Y = r # Y ’1 1
K# Y "
K# X“ SIMPLIFIED FOR PREDATOR PREY (where ˜ Y is the predator)
˜ X = r" X [1 ˜ Y = r# Y [1 1
K" X !
K" "
K# X] Y]
œ r# Y " r#
K# XY we will drop the “r" and “K" notation, though it can be carried through and
interpreted Page 3 GENERAL EQUATIONS TWO COMPETING SPECIES
˜ X = X(b!" b"" X b#" Y)
˜ Y = Y(b!# b"# Y b## X)
SECOND EQUATION VERSION FOR PREDATORS
˜ Y = Y( b!# b## X) for predators STATISTICS is not well suited for the examination of this 2 equation system,
individual parameters may be fitted with simpler models
this is true of most of the systems we will examine SYSTEMS REGRESSION may work here since the equations are linear Compare prey model to SCHAEFER'S
˜ N = rN r #
KN C or
˜ N = rN r #
KN qfN compared to ˜ X = r" X ’1 1
K" X !
K" Y“ ˜ X = X(b!" b"" X b#" Y)
the equations are the same if predator = effort
ie. where fish (N) = prey (X) and effort (E) = predator (Y)
the coefficients may be considered equal Page 4 We will use a BIOMATHEMATICAL APPROACH to examine the behavior of this
simple system, and numerical integration for bigger systems later
examine prey first
˜ X = X(b!" b"" X b#" Y)
the first term is (after multiplying out)
= b!" X
exponential growth (when X is “small" and X# is even smaller)
the first two terms are
= b!" X b"" X#
logistic growth in the absence of a predator
the three terms together give
X(b!" b"" X b#" Y)
NOTE: negative interspecific competition logistic growth curve when predators = 0
negative intraspecific competition, this can be two competing species or it can be a
prey with a predator or a parasite
The coefficients b#" and b## are constants which describe the frequency of
encounter between X and Y resulting in predations of X as either X and /or Y increases the value of these terms get larger Page 5 The predator model
˜ Y = Y( b!# b## X) when there is no prey then ˜ Y = b!# Y note that b!# is negative this is an exponential decay model, so we have only predator mortality in the
absence of prey the only positive factor comes from the encounter of predators and prey there is no intraspecific competition for the predator in this model Page 6 BIOMATHEMATICAL APPROACH (one of many) quick and dirty ARROW and ISOCLINE PHASE PLANE
uses some calculus
PHASE PLANE plot of Y on X; Y t1 t2
t3 t10
t9 t4 t8
t5 t7 t6 X
where the line within the plane describes the systems behavior over time Page 7 if the two variables are examined independently over time the behavior is Y time X time
these lines, combined on to a phase plane give the line above Page 8 There are a number of aspects of the behavior of a line describing changes over
time that we will try to define, particularly
1) the direction of movement in various sectors of the phase plane
2) steady state points
To accomplish this we must first define an isocline lines in the phase plane move over time to find the direction of movement of a track in the phase plane at any particular time,
we could find the tangent to the line at that time
eg. Y X Page 9 if the line is to change direction, then the tangent must pass either the horizontal or
the vertical at these points
either fX = 0 (tangent horizontal)
or fY = 0 (tangent vertical) Y ΔY=0 ΔY=0 ΔX=0 X Page 10 note that when the line changes direction, the tangent is either horizontal or vertical
the direction doesn't have to change when the tangent is horizontal or vertical
(inflection), but it will in the models we examine Y
ΔY=0 X Page 11 we will describe a line along which the tangents are either horizontal or vertical,
THESE ARE ISOCLINES
since our equations are already in a form describing ˜ X and ˜ Y, this is simple Y ΔY=0 ΔY=0 ΔX=0 X Page 12 1) ISOCLINE DETERMINATION lines connecting points where ˜ X=0 or ˜ Y=0
a) the VERTICAL ISOCLINE occurs when
˜ X = 0 = b!" X b"" Xr b#" XY
this is true when X = 0 (occurs along the Y axis, not as important) and when
0 = b!" b"" X b#" Y
Y= b!"
b#" b""
b#" X which is an EQUATION FOR A LINE this is a simple linear equation where the intercept is
to Y on the X axis the isocline touches when Y=0 b!"
b#" on the axis corresponding so;
0 = b!" b"" X b#" Y
0 = b!" b"" X b#" (0)
0 = b"! b"" X
and
X= b!"
b"" THESE TWO POINTS ARE THE ENDPOINTS (ON THE X AND Y AXES) of
the isocline for X (ie. when ˜ X = 0) so the phase plane line must cross this line
moving VERTICALLY b01
b21
ΔY=0 Y
X b01
b11
Page 13 b) THE HORIZONTAL (Y) ISOCLINE IS DESCRIBED BY
˜ Y = 0 = b!# Y b## XY
which occurs when Y = 0 (not as important) or when
0 = b!# b## X
X= b!#
b## which describes the Y ISOCLINE ( ˜ Y = 0) so the phase plane line must move
HORIZONTALLY across this line
note that this isocline has no "slope", it is a vertical line Y ΔX=0 X Page 14 Plot the two isoclines together, there are 2 cases
CASE 1: b!#
b## b!"
b"" b01
b21 Y
b02
b22
CASE 2: b!"
b#" b01
b11 X b#!
b## b01
b21 Y
X b01 b02
b11 b22 Page 15 2) Now, we examine behavior of the line over time The isoclines show where the line (over time) is either
HORIZONTAL
˜ Y = 0 or
VERTICAL
˜X=0 b01
b21 4
1 Pred
2 3
b02
b22 Prey b01
b11 Page 16 all that remains is to place arrows in the quadrants, indicating direction of movement
examine the equations
˜ X = b!" X b"" Xr b#" YX
˜ Y = b!# Y b## XY
1) when X and Y are both VERY large [if desired Ä _]
then ˜ X is ( ) since the two negative terms predominate in the model, so X
decreases
˜ Y is ( )
since the only term with X and Y is positive Y increases 2) when X and Y are both VERY small then ˜ X is ( )
since the only term without X and Y is positive X increases
˜ Y is ( )
since the only term without X and Y is negative Y decreases 3) when X is MODERATELY LARGE and Y is SMALL , then
˜ X is ( )
the term with X still predominates X increases ˜ Y is ( )
the term with X is sufficiently large to change the balance Y increases 4) when X is SMALL and Y is VERY LARGE , then
˜ X is ( )
since the term with Y is negative X decreases fY is ( )
since X is very small Y decreases Page 17 Now the completed arrow and isocline PHASE PLANES
CASE 1: b!#
b## b!"
b"" b01
b21 Pred b02
b22 Prey b01
b11 The model then describes an inward spiral, the model will go to a steady state at
PEACEFUL COEXISTANCE where the lines intersect
THIS IS A “STEADY STATE" POINT
this can be evaluated by employing the equations to solve for the point where the two
linear equations intersect, or, evaluate
Y = b!"
b#" b""
b#" X when
X= b!#
b## then at steady state, the point is
(X,Y) = ’ b!# ,
b## b!"
b#" b"" b!#
b#" b## “ Page 18 If we look at only the behavior of X or Y over time we have + =29<> =:3<+6
numerically, as long as we don't start too near the axes the system is stable (recall
that the axis was also an isocline)
Using the same approach, the directions of movement can be determined, giving
the model below, b01
b21 Pred b02
b22 Prey b01
b11 Page 19 CASE 2: b!"
b#" b#!
b## b01
b21 Y
X b01 b02
b11 b22 This model describes a situation where the PREDATOR WILL GO TO EXTINCTION,
and the PREY will go to its steady state (carrying capacity) at b!!
b#"
steady state occurs at the point (X,Y) = Š b!" , 0‹
b#" Page 20 3) Another possibilities
a) a change in the sign of either b"" or b"# gives the angled line a positive slope
OUTWARD SPIRAL
so the system is unstable, and will "blow up". There is no steady state point
b) if b"" = 0 then the system describes a stable circle or oval
this is a stable oscillating system
All of the stable models tend to a steady state over time for these DETERMINISTIC
MODELS
DETERMINISTIC for a given set of initial conditions, the track followed
through the phase plane is always the same Page 21 COMPETITION 2 species both with negative interspecific and intraspecific
interaction
˜ X = X(b!" b"" X b#" Y)
˜ Y = Y(b!# b"# Y b## X)
both behave as the previous PREY model, so that both have linear equations with
slopes for ISOCLINES b01
b21 Y
X b01
b11 b02
b22 Y
X b02
b12 Page 22 This gives rise to 4 situations depending on which combination of the following
conditions exist
b!"
b#" or b!#
b## on the Y axis b!"
b"" or b!#
b"# on the X axis and Page 23 COMPETITION MODELS 4 SITUATIONS
EQUATIONS
˜ X = X(b!" b"" X b#" Y)
˜ Y = Y(b!# b"# Y b## X)
SITUATION 1: b!"
b#" b!#
b## and b!"
b"" b!#
b"# b02
b22
b01
b21 Y
X b01
b11 b02
b12 Under these conditions only one species (Y) remains, regardless of initial conditions Page 24 SITUATION 2: b!"
b#" b!#
b## and b!"
b"" b!#
b"# b01
b21
b02
b22 Y
X b02
b12 b01
b11 Under these conditions, again only one species (X) remains, regardless of initial
conditions, but not the same species as above Page 25 SITUATION 3: b!"
b#" b!#
b## b!"
b"" and b!#
b"# b01
b21
b02
b22 Y
X b02
b12 b01
b11 CASE 3: PEACEFUL COEXISTANCE regardless of the starting point (INCON)
Steady state at:
X= Y= b!"
b#"
b""
b#" b!"
b#" b!#
b"#
b##
b"# = b!" b"# b#" b!#
b"" b"# b#" b## b"" b!" b"# b#" b!#
b#" ’ b"" b"# b#" b## “ = b"" b!# b!" b##
b"" b"# b#" b## Page 26 SITUATION 4: b!"
b#" b!#
b## and b!"
b"" b!#
b"# b02
b22
b01
b21 Y
X b02
b12 b01
b11 CASE 4: This is the only model where the final result (baring a blow up) varies with the
initial conditions Page 27 B0
B1
B2
Initial
dt X
0.1
0.01
0.001
1
1 Y
0.001
0
0.005
65 140 1.8
1.6 120 1.4 100
1.2
1 80 0.8 60 0.6 40
0.4 20 0.2 0
0 500 1000 1500 2000 2500 3000 3500 4000 0
0 500 1000 1500 2000 2500 3000 3500 4000 130 Predator Prey 120
110 Y
100
90
80
70
60
0.00 0.20 0.40 0.60 0.80 1.00 X 1.20 1.40 1.60 1.80 ...
View
Full
Document
This note was uploaded on 12/29/2011 for the course EXST 7025 taught by Professor Geaghan,j during the Spring '08 term at LSU.
 Spring '08
 Geaghan,J

Click to edit the document details