Phase Plane. Ex. 6.4.2 “Rabbits & Sheep” John F. Mateus R.* Universidad Pedag´ ogica y Tecnol´ ogica de Colombia–UPTC 13 de octubre de 2017
Consider the following “rabbits vs sheep” problem, whe-re x, y 0. Find the fixed points, investigate their stability, draw the nullclines, and sketch plausible phase portraits. Indicate the basins of attraction of any stable fixed points.
≥
= x(3 (3 x˙ = x = y(2 (2 y˙ = y
− 2x − y ) , − x − y) .
(1)
Solution Fixed points:
First, we find the fixed points ((x (0, 0). We have that the Eqs. (1) x∗ , y∗ ) for which (˙x, y˙ ) = (0, become in:
0 = x ∗ (3 0 = y ∗ (2
− 2x∗ − y∗) , − x∗ − y∗) .
(2)
−→ − → 2 − y∗ = 0 − −→ → y ∗ = 2 .
(3)
−→ − → 3 − 2x∗ = 0 − −→ → x∗ = 32 .
(4)
Thus, the first fixed point is obviously (0, (0 , 0). ∗ ∗ If y = 0 but x = 0 we have:
(2
− x∗ − y∗)|
x∗ =0
= 0
So, the second fixed point is (0, (0, 2). ∗ ∗ Now, if y butt x = 0 we have: y = 0 bu
(3 − 2x∗ − y∗ )|
y ∗ =0
= 0
So, the third fixed point is (3/ (3 /2, 0). ∗ ∗ Finally, if (x (x , y ) = (0, (0, 0) we have:
3 *
− 2x∗ − y∗ = 0 ,
∧
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1
2
− x∗ − y∗ = 0 .
(5)
Substract from the left side Eq. the right side Eq. in (5) we have: 1
− x∗ = 0 −→ x∗ = 1 .
(6)
As we have y ∗ = 2 x∗ from right side Eq. in (5), then y ∗ = 1, and therefore, the fourth fixed point is (1, 1). Thus, we have the next fixed points for this exercise:
−
P 1 = (0, 0)
P 2 = (0, 2)
P 3 = (3/2, 0)
P 4 = (1, 1)
(7)
Nullclines:
We find now the nullclines, that is, the points at the phase plane where the flux is only vertical or only horizontal. We started with horizontal flux, that is, y ˙ = 0. If y = ˙ 0 we have y = 0 or 2 x y = 0. If y = 0 the nullcline is the x-axis, for all points at this axis the flux is only horizontal. To know the direction of this flux we must analized x under ˙ the condition y = 0. For this we have:
− −
˙ x = x(3
− 2x − y)| = x(3 − 2x) . (8) Hence, x˙ > 0 in x ∈ [0, 3/2) and x˙ < 0 in x ∈ (3/2, ∞), that is, flux goes rigth for 0 < x < 3/2 and goes left for 3/2 < x < ∞. If y = 0 we must have 2 − x − y = 0, so y = 2 − x is the second nullcline for only horizontal − 2x − (2 − x)) = x(1 − x). The flux going to the rigth flux. Therefore, for x we ˙ have x = x(3 ˙ y =0
for 0 < x < 1 and going to the left for 1 < x < 2. This situations are shown in the Fig. 1 in blue color. y
NV2
P 2
P 4
NV1 NH4
P 1
NH3
P 3
x
Figura 1: Nullclines and direction of the flux for this exercise. In blue we show only horizontal
flux and in red we show only vertical flux. Now, we analize the vertical flux only. The same manner like before, we have ˙x = 0 that implies two situations: first consider that x = 0 or 3 2x y = 0. If we take first x = 0 the y-axis
− −
become in a nullcline. For this line we have that the flux will be up for 0 < y < 2 and down for 2 < y < . If x = 0, we need that 3 2x y = 0, so x = (3 y)/2 is the second vertical flux nullcline. This situation implies that y is ˙ upward for 1 < x < 3/2 and downward for 0 < x < 1. This is shown in Fig. 1.
∞
− −
−
Stability of fixed points:
To stablished the stability of the fixed points, we need the linearized system. To make this, we use the Jacobian matrix for the system (1) and then we evaluate this matrix at each fixed point. The Jacobian matrix for (1) is:
J
J =
− 3
4x
−y
−y
−x 2 − x − 2y
(9)
.
At P 1 :
J (P ) = 1
3 0
.
0 2
(10)
For this point we have τ = 5, ∆ = 6 and τ 2 4∆ = 1, so, P 1 is an unstable node. Furthermore the eigenvalues are λ1 = 3 and λ2 = 2 with eigenvectors v1 = (1, 0) and v2 = (0, 1). We can see that λ1 > λ2 hence v1 is the fast eigendirection and v2 is the slow eigendirection.
−
| | | |
At P 2 :
J (P ) = −12 −02 . (11) For this point we have τ = −1, ∆ = −2 and τ − 4∆ = 9, so, P is a saddle node. Furthermore the eigenvalues are λ = 1 and λ = −2 with eigenvectors v = (3, −2) and v = (0, 1). We can see that |λ | < |λ | hence v is the slow eigendirection and v is the fast eigendirection. 2
2
1
1
2
2
2
1
1
2
2
At P 3 :
J (P ) = 3
−
3
0
−3/2 1/2
.
(12)
For this point we have τ = 5/2, ∆ = 3/2 and τ 2 4∆ = 49/4, so, P 3 is a saddle node. Furthermore the eigenvalues are λ1 = 1/2 and λ2 = 3 with eigenvectors v1 = ( 3, 7) and v2 = (1, 0). We can see that λ1 < λ2 hence v1 is the slow eigendirection and v2 is the fast eigendirection.
− − | | | |
− −
−
Finally, at P 4 :
− −
J (P ) = −21 −11 . (13) For this point we have τ = −3,√ ∆ = 1 and τ − 4∆ √ = 5, so, P is a stable node. Furthermore √ − − the eigenvalues are = and = with eigenvectors v = (1 − 5, 2) and λ λ √ v = (1 + 5, 2). We can see that |λ | < |λ | hence v is the slow eigendirection and v is the 4
2
3+ 2
1
5
2
2
4
3+ 5 2
1
2
1
1
2
fast eigendirection. This situations are sketeched in Fig. 2. y
f
P 2
f s P 4
s
s
s P 1
f
P 3
f
x
Figura 2: Stability of all fixed points. We show the eigendirections (fast in red, slow in blu) for
each fixed point and the behavior near to the fixed points