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Show that this system has a limit cycle in the first quadrant.
Clash Royale CLAN TAG#URR8PPP
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The full question reads as follows:
Consider the predator-prey model
$$dotx = x left(6-x-frac3y1+x right), doty = y(x-2)$$
Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have finished part (a) and found that $(0,0)$ is a saddle, $(6,0)$ is a saddle, and $(2,4)$ is an unstable focus. I have attempted to do part (b) using some phase plane analysis, but I am looking for a more definite way to prove this. Thanks in advance for any help!
dynamical-systems mathematical-modeling nonlinear-system
asked Jul 24 at 1:22
obewanjacobi
310110
add a comment |Â
up vote
2
down vote
favorite
The full question reads as follows:
Consider the predator-prey model
$$dotx = x left(6-x-frac3y1+x right), doty = y(x-2)$$
Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have finished part (a) and found that $(0,0)$ is a saddle, $(6,0)$ is a saddle, and $(2,4)$ is an unstable focus. I have attempted to do part (b) using some phase plane analysis, but I am looking for a more definite way to prove this. Thanks in advance for any help!
dynamical-systems mathematical-modeling nonlinear-system
asked Jul 24 at 1:22
obewanjacobi
310110
It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
1
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
1
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The full question reads as follows:
Consider the predator-prey model
$$dotx = x left(6-x-frac3y1+x right), doty = y(x-2)$$
Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have finished part (a) and found that $(0,0)$ is a saddle, $(6,0)$ is a saddle, and $(2,4)$ is an unstable focus. I have attempted to do part (b) using some phase plane analysis, but I am looking for a more definite way to prove this. Thanks in advance for any help!
dynamical-systems mathematical-modeling nonlinear-system
asked Jul 24 at 1:22
obewanjacobi
310110
The full question reads as follows:
Consider the predator-prey model
$$dotx = x left(6-x-frac3y1+x right), doty = y(x-2)$$
Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have finished part (a) and found that $(0,0)$ is a saddle, $(6,0)$ is a saddle, and $(2,4)$ is an unstable focus. I have attempted to do part (b) using some phase plane analysis, but I am looking for a more definite way to prove this. Thanks in advance for any help!
dynamical-systems mathematical-modeling nonlinear-system
asked Jul 24 at 1:22
obewanjacobi
310110
edited Jul 25 at 1:46
asked Jul 24 at 1:22
obewanjacobi
310110
asked Jul 24 at 1:22
obewanjacobi
310110
asked Jul 24 at 1:22
obewanjacobi
310110
310110
It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
1
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
1
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40
add a comment |Â
It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
1
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
1
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40
It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
1
1
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
1
1
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40
add a comment |Â
1 Answer
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Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$. It follows then that the domain of any such solution contains $[0, infty)$, and that the $omega$-limit set is compact and nonempty.
Let $L$ stand for the $omega$-limit set of some point, $(x_0, y_0)$, sufficiently close to the unstable focus $(2,4)$. By the Poincaré–Bendixson theorem, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
$L$ cannot be an equilibrium. Indeed, there are three possibilities: either $L = (0, 0)$, or $L = (6,0)$, or else $L = (2, 4)$. In the first two cases, the equilibria are saddles, so $(x_0, y_0)$ must belong to the stable manifold, which means either $, (0, y) : y > 0 ,$ or $, (x, 0) : x in (0, 6) cup (6, infty) ,$. In the third case, the only point whose $omega$-limit set is $(2,4)$ is $(2,4)$ itself.
We proceed now to excluding the case of heteroclinic cycle (EDIT: or homoclinic loop). But the only possible connection is from $(0, 0)$ to $(6,0)$, so there are no heteroclinic cycles.
Consequently, $L$ is a limit cycle surrounding $(2,4)$.
Below is a sketch of the phase portrait:
Incidentally, it is not so easy to show that all solutions starting in $mathbbR^2_+$ are bounded for positive times. The OP gave that as an assumption, but it should follow somehow from the form of the system.
answered Jul 24 at 20:36
user539887
1,4611313
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$. It follows then that the domain of any such solution contains $[0, infty)$, and that the $omega$-limit set is compact and nonempty.
Let $L$ stand for the $omega$-limit set of some point, $(x_0, y_0)$, sufficiently close to the unstable focus $(2,4)$. By the Poincaré–Bendixson theorem, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
$L$ cannot be an equilibrium. Indeed, there are three possibilities: either $L = (0, 0)$, or $L = (6,0)$, or else $L = (2, 4)$. In the first two cases, the equilibria are saddles, so $(x_0, y_0)$ must belong to the stable manifold, which means either $, (0, y) : y > 0 ,$ or $, (x, 0) : x in (0, 6) cup (6, infty) ,$. In the third case, the only point whose $omega$-limit set is $(2,4)$ is $(2,4)$ itself.
We proceed now to excluding the case of heteroclinic cycle (EDIT: or homoclinic loop). But the only possible connection is from $(0, 0)$ to $(6,0)$, so there are no heteroclinic cycles.
Consequently, $L$ is a limit cycle surrounding $(2,4)$.
Below is a sketch of the phase portrait:
Incidentally, it is not so easy to show that all solutions starting in $mathbbR^2_+$ are bounded for positive times. The OP gave that as an assumption, but it should follow somehow from the form of the system.
answered Jul 24 at 20:36
user539887
1,4611313
add a comment |Â
up vote
1
down vote
accepted
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$. It follows then that the domain of any such solution contains $[0, infty)$, and that the $omega$-limit set is compact and nonempty.
Let $L$ stand for the $omega$-limit set of some point, $(x_0, y_0)$, sufficiently close to the unstable focus $(2,4)$. By the Poincaré–Bendixson theorem, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
$L$ cannot be an equilibrium. Indeed, there are three possibilities: either $L = (0, 0)$, or $L = (6,0)$, or else $L = (2, 4)$. In the first two cases, the equilibria are saddles, so $(x_0, y_0)$ must belong to the stable manifold, which means either $, (0, y) : y > 0 ,$ or $, (x, 0) : x in (0, 6) cup (6, infty) ,$. In the third case, the only point whose $omega$-limit set is $(2,4)$ is $(2,4)$ itself.
We proceed now to excluding the case of heteroclinic cycle (EDIT: or homoclinic loop). But the only possible connection is from $(0, 0)$ to $(6,0)$, so there are no heteroclinic cycles.
Consequently, $L$ is a limit cycle surrounding $(2,4)$.
Below is a sketch of the phase portrait:
Incidentally, it is not so easy to show that all solutions starting in $mathbbR^2_+$ are bounded for positive times. The OP gave that as an assumption, but it should follow somehow from the form of the system.
answered Jul 24 at 20:36
user539887
1,4611313
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$. It follows then that the domain of any such solution contains $[0, infty)$, and that the $omega$-limit set is compact and nonempty.
Let $L$ stand for the $omega$-limit set of some point, $(x_0, y_0)$, sufficiently close to the unstable focus $(2,4)$. By the Poincaré–Bendixson theorem, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
$L$ cannot be an equilibrium. Indeed, there are three possibilities: either $L = (0, 0)$, or $L = (6,0)$, or else $L = (2, 4)$. In the first two cases, the equilibria are saddles, so $(x_0, y_0)$ must belong to the stable manifold, which means either $, (0, y) : y > 0 ,$ or $, (x, 0) : x in (0, 6) cup (6, infty) ,$. In the third case, the only point whose $omega$-limit set is $(2,4)$ is $(2,4)$ itself.
We proceed now to excluding the case of heteroclinic cycle (EDIT: or homoclinic loop). But the only possible connection is from $(0, 0)$ to $(6,0)$, so there are no heteroclinic cycles.
Consequently, $L$ is a limit cycle surrounding $(2,4)$.
Below is a sketch of the phase portrait:
Incidentally, it is not so easy to show that all solutions starting in $mathbbR^2_+$ are bounded for positive times. The OP gave that as an assumption, but it should follow somehow from the form of the system.
answered Jul 24 at 20:36
user539887
1,4611313
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$. It follows then that the domain of any such solution contains $[0, infty)$, and that the $omega$-limit set is compact and nonempty.
Let $L$ stand for the $omega$-limit set of some point, $(x_0, y_0)$, sufficiently close to the unstable focus $(2,4)$. By the Poincaré–Bendixson theorem, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
$L$ cannot be an equilibrium. Indeed, there are three possibilities: either $L = (0, 0)$, or $L = (6,0)$, or else $L = (2, 4)$. In the first two cases, the equilibria are saddles, so $(x_0, y_0)$ must belong to the stable manifold, which means either $, (0, y) : y > 0 ,$ or $, (x, 0) : x in (0, 6) cup (6, infty) ,$. In the third case, the only point whose $omega$-limit set is $(2,4)$ is $(2,4)$ itself.
We proceed now to excluding the case of heteroclinic cycle (EDIT: or homoclinic loop). But the only possible connection is from $(0, 0)$ to $(6,0)$, so there are no heteroclinic cycles.
Consequently, $L$ is a limit cycle surrounding $(2,4)$.
Below is a sketch of the phase portrait:
Incidentally, it is not so easy to show that all solutions starting in $mathbbR^2_+$ are bounded for positive times. The OP gave that as an assumption, but it should follow somehow from the form of the system.
answered Jul 24 at 20:36
user539887
1,4611313
edited Aug 2 at 11:09
answered Jul 24 at 20:36
user539887
1,4611313
answered Jul 24 at 20:36
user539887
1,4611313
answered Jul 24 at 20:36
user539887
1,4611313
1,4611313
add a comment |Â
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It appears to me that the "focus"--if it is a focus (I say this because I haven't checked, I'm going with what you wrote)--is at $(2, 4)$, not $(2, 3)$. This is why I'm not sure what type of critical point this is, since I'm not sure we got the coordinates right!
– Robert Lewis
Jul 24 at 2:00
1
Just edited, but my question still stands. Thanks Robert!
– obewanjacobi
Jul 24 at 3:47
Which tools do you know to prove the existence of a cycle?
– Did
Jul 24 at 17:32
By the way: (2,4) a stable focus, are you sure about this?
– Did
Jul 24 at 17:34
1
According to Mathematica, the eigenvalues at $(2,4)$ are $frac13(1pm isqrt71)$, so $(2,4)$ is an unstable focus.
– user539887
Jul 24 at 19:40