modellen & simulatiefrank011/classes/modsim/handouts/phas… · 4 chapter 1. modelling if n(0)...
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![Page 1: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/1.jpg)
Modellen & SimulatieSlides bij Hoorcollege 3
![Page 2: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/2.jpg)
Programma
• Herhaling stabiliteit van scalaire DVs (incl. Verhulst)
• Slinger model
• Faseruimte
• Lineaire DVs in twee dimensies:
• Eigenwaarde decompositie
• Eigenwaardeprobleem
• Categorizatie van evenwichten
![Page 3: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/3.jpg)
4 CHAPTER 1. MODELLING
If n(0) = n0, then c = n0/(1� n0) and the solution can be written
n(t) =n0e
rt
(1� n0) + n0ert
.
If r > 0 it can be checked thatlim
t!1n(t) = 1
for any initial state n0 > 0. That is, the population eventually saturates (N ! N
⇤) for any initialpopulation. This behavior is shown in the left plot of Figure 1.2.
Next, suppose that in attempt to control the aphid population, a predator is introduced intothe rose farm. One nasty predator is the ladybug (Fig. 1.1b. If you are an aphid, the ladybug isanything but a symbol of nonviolence!). Normally lady bugs will not bother eating aphids unlessthere is su�cient supply. The logistic model can be extended to incorporate predation by ladybugs as follows:
dn
dt
= rn
⇣1� n
s
⌘� n
2
1 + n
2. (1.5)
The last term gradually “turns on” ladybug predation as the population n grows greater than 1.We have re-introduced the saturation level s because there are now two important populationslevels—the natural saturation level, and the level for which ladybugs get “interested”. Somesolutions of this model are shown in the right plot of Figure 1.2. It is interesting to note that fors = 20 and either r = 0.1 or r = 0.6, the aphid population eventually tends to a constant value,independent of the initial population. This is similar to the case of the logistic equation withoutpredation (1.4). For the case s = 20, r = 0.25, however, the final aphid population depends onthe starting population. For a starting population of about n0 < 5, the aphids settle down toa low level of about n = 0.5, whereas for a starting population of n0 > 5, the aphid populationeventually stabilizes at a relatively high level of around n = 14.5. Such a change in the quality ofsolutions with a change in parameter values is called bifurcation.
0 5 10 15 20 25 30 35 400
0.5
1
1.5
t
n
Logistic model, r = 0.25
0 10 20 30 40 50 600
2
4
6
8
10
12
14
16
18
20
t
n
Logistic model with predation, s=20, r = {0.1, 0.25, 0.6}
Figure 1.2: Solutions of the logistic model. On the left, solutions of (1.4) with r = 0.25 for di↵erentinitial conditions. On the right, with predation (1.5), s = 20 and r = 0.6 (green), r = 0.25 (black),and r = 0.1 (red). For r = 0.25, there are two steady states, depending on the initial condition
This model is still rather crude, because it assumes an endless supply of ladybugs. An alterna-tive model might consider an isolated system of aphids and ladybugs, and model the populationsof both. Let p(t) denote the (normalized) population of ladybugs. In the absence of aphids, theladybugs would die out exponentially: dp/dt = �mp for some m > 0. But with an increasingnumber of aphids, the ladybugs will flourish, so the ladybug population should satisfy
dp
dt
= anp�mp
Verhulst model, r=1/4, Pc=1
P(t)
t
![Page 4: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/4.jpg)
Programma
• Herhaling stabiliteit van scalaire DVs (incl. Verhulst)
• Slinger model
• Faseruimte
• Lineaire DVs in twee dimensies:
• Eigenwaarde decompositie
• Eigenwaardeprobleem
• Categorizatie van evenwichten
![Page 5: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/5.jpg)
Vector Field (✓̇, v̇)✓(t) v(t)
✓̇ = v
v̇ = � sin ✓Pendulum
![Page 6: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/6.jpg)
Phase plot✓(t) v(t) (✓(t), v(t))
✓̇ = v
v̇ = � sin ✓Pendulum
![Page 7: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/7.jpg)
Phase plot✓(t) v(t) (✓(t), v(t))
✓̇ = v
v̇ = � sin ✓Pendulum
![Page 8: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/8.jpg)
Phase plot✓(t) v(t) (✓(t), v(t))
✓̇ = v
v̇ = � sin ✓Pendulum
![Page 9: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/9.jpg)
Phase plot (✓(t), v(t))
✓̇ = v
v̇ = � sin ✓Pendulum
![Page 10: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/10.jpg)
d = det
s = spoor
d = s2/4
�1,2 2 R
zadelpunt
knoop
spiraal�1 = �̄2 2 C
cent
rum
punt
![Page 11: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/11.jpg)
Phase portraitVector Field
�1 = �1.5, �2 = �0.8 v1 =
✓10
◆, v2 =
✓01
◆
Stable node
![Page 12: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/12.jpg)
Phase portraitVector Field
v1 =
✓10
◆, v2 =
✓01
◆
Unstable node
�1 = 1.5, �2 = 0.8
![Page 13: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/13.jpg)
v1 =
✓10
◆, v2 =
✓11
◆
Phase portraitVector Field
�1 = �1.5, �2 = �0.8
Stable node
![Page 14: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/14.jpg)
d = det
s = spoor
d = s2/4
�1,2 2 R
zadelpunt
knoop
spiraal�1 = �̄2 2 C
cent
rum
punt
![Page 15: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/15.jpg)
Phase portraitVector Field
v1 =
✓10
◆, v2 =
✓01
◆�1 = �1.5, �2 = 0.8
Saddle point
![Page 16: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/16.jpg)
v1 =
✓10
◆, v2 =
✓11
◆
Phase portraitVector Field
�1 = �1.5, �2 = 0.8
Saddle point
![Page 17: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/17.jpg)
d = det
s = spoor
d = s2/4
�1,2 2 R
zadelpunt
knoop
spiraal�1 = �̄2 2 C
cent
rum
punt
![Page 18: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/18.jpg)
Phase portraitVector Field
�1,2 = �0.5± 1.0 i
Stable spiral (focus)
![Page 19: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/19.jpg)
Phase portraitVector Field
�1,2 = �0.5± 1.0 i
Stable spiral (focus)
![Page 20: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/20.jpg)
d = det
s = spoor
d = s2/4
�1,2 2 R
zadelpunt
knoop
spiraal�1 = �̄2 2 C
cent
rum
punt
![Page 21: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/21.jpg)
Phase portraitVector Field
�1,2 = ±2i
Center point
![Page 22: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/22.jpg)
Phase portraitVector Field
�1,2 = ±2i
Center point
![Page 23: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/23.jpg)
d = det
s = spoor
d = s2/4
�1,2 2 R
zadelpunt
knoop
spiraal�1 = �̄2 2 C
cent
rum
punt
![Page 24: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/24.jpg)
Phase portraitVector Field
v1 =
✓10
◆, v2 =
✓01
◆�1 = �1, �2 = 0
Singular case
![Page 25: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/25.jpg)
Phase portraitVector Field
�1 = �1, �2 = 0
Singular matrix A =
�1 01 0
�
![Page 26: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/26.jpg)
Phase portraitVector Field
v1 =
✓10
◆, v2 =
✓01
◆�1 = �2 = �1
Stable singular node A =
�1 00 �1
�
![Page 27: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/27.jpg)
Phase portraitVector Field
�1 = �2 = �1 v1 =
✓01
◆
Stable degenerate node A =
�1 10 �1
�
![Page 28: Modellen & Simulatiefrank011/Classes/modsim/Handouts/Phas… · 4 CHAPTER 1. MODELLING If n(0) = n 0, then c = n 0/(1 n 0) and the solution can be written n(t)= n 0ert (1 n 0)+n 0ert](https://reader034.vdocuments.site/reader034/viewer/2022051913/600460fbaf19b20cf875f4b6/html5/thumbnails/28.jpg)
Phase portraitVector Field
�1 = �2 = �1
Stable degenerate node