Mixing
Use Poincaré-Bendixson to show that a limit cycle exists in the first quadrant.
Clash Royale CLAN TAG#URR8PPP
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The entire problem reads:
Consider the predator-prey model $dot x = left( 4-x-frac2y1+x right)$, $dot y = y(x-1)$. 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 already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = frac(4-x)(1+x)2$ because that's where $dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 31 at 0:05
John B
11.9k51639
asked Jul 30 at 1:42
obewanjacobi
310110
add a comment |Â
up vote
0
down vote
favorite
The entire problem reads:
Consider the predator-prey model $dot x = left( 4-x-frac2y1+x right)$, $dot y = y(x-1)$. 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 already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = frac(4-x)(1+x)2$ because that's where $dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 31 at 0:05
John B
11.9k51639
asked Jul 30 at 1:42
obewanjacobi
310110
3
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
1
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
1
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The entire problem reads:
Consider the predator-prey model $dot x = left( 4-x-frac2y1+x right)$, $dot y = y(x-1)$. 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 already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = frac(4-x)(1+x)2$ because that's where $dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 31 at 0:05
John B
11.9k51639
asked Jul 30 at 1:42
obewanjacobi
310110
The entire problem reads:
Consider the predator-prey model $dot x = left( 4-x-frac2y1+x right)$, $dot y = y(x-1)$. 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 already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = frac(4-x)(1+x)2$ because that's where $dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 31 at 0:05
John B
11.9k51639
asked Jul 30 at 1:42
obewanjacobi
310110
edited Jul 31 at 0:05
John B
11.9k51639
edited Jul 31 at 0:05
John B
11.9k51639
edited Jul 31 at 0:05
John B
11.9k51639
11.9k51639
asked Jul 30 at 1:42
obewanjacobi
310110
asked Jul 30 at 1:42
obewanjacobi
310110
asked Jul 30 at 1:42
obewanjacobi
310110
310110
3
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
1
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
1
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33
add a comment |Â
3
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
1
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
1
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33
3
3
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
1
1
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
1
1
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,infty)$ and its $omega$-limit set is compact and nonempty.
Let $L$ stand for the $É$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $(4,0)$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix
$$
beginbmatrix
-1 & -frac25
\
0 & 3
endbmatrix.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold
is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $dotx = 4 - x$, $doty equiv 0$, so for any $(x_1,0)$ we have $omega((x_1,0)) = (4,0)$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $omega$-limit point is the saddle. Hence, if $L = (4,0)$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
answered Jul 31 at 8:53
user539887
1,4361313
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
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
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,infty)$ and its $omega$-limit set is compact and nonempty.
Let $L$ stand for the $É$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $(4,0)$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix
$$
beginbmatrix
-1 & -frac25
\
0 & 3
endbmatrix.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold
is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $dotx = 4 - x$, $doty equiv 0$, so for any $(x_1,0)$ we have $omega((x_1,0)) = (4,0)$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $omega$-limit point is the saddle. Hence, if $L = (4,0)$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
answered Jul 31 at 8:53
user539887
1,4361313
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
add a comment |Â
up vote
1
down vote
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,infty)$ and its $omega$-limit set is compact and nonempty.
Let $L$ stand for the $É$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $(4,0)$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix
$$
beginbmatrix
-1 & -frac25
\
0 & 3
endbmatrix.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold
is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $dotx = 4 - x$, $doty equiv 0$, so for any $(x_1,0)$ we have $omega((x_1,0)) = (4,0)$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $omega$-limit point is the saddle. Hence, if $L = (4,0)$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
answered Jul 31 at 8:53
user539887
1,4361313
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,infty)$ and its $omega$-limit set is compact and nonempty.
Let $L$ stand for the $É$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $(4,0)$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix
$$
beginbmatrix
-1 & -frac25
\
0 & 3
endbmatrix.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold
is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $dotx = 4 - x$, $doty equiv 0$, so for any $(x_1,0)$ we have $omega((x_1,0)) = (4,0)$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $omega$-limit point is the saddle. Hence, if $L = (4,0)$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
answered Jul 31 at 8:53
user539887
1,4361313
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $mathbbR^2_+ := , (x, y) : x ge 0, y ge 0 ,$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,infty)$ and its $omega$-limit set is compact and nonempty.
Let $L$ stand for the $É$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $(4,0)$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix
$$
beginbmatrix
-1 & -frac25
\
0 & 3
endbmatrix.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold
is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $dotx = 4 - x$, $doty equiv 0$, so for any $(x_1,0)$ we have $omega((x_1,0)) = (4,0)$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $omega$-limit point is the saddle. Hence, if $L = (4,0)$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
answered Jul 31 at 8:53
user539887
1,4361313
edited Aug 2 at 11:06
answered Jul 31 at 8:53
user539887
1,4361313
answered Jul 31 at 8:53
user539887
1,4361313
answered Jul 31 at 8:53
user539887
1,4361313
1,4361313
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
add a comment |Â
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
Just a minor addition: due to Poincaré–Bendixson theorem, a homoclinic loop can also be a limit set. This option is however trivially excluded here.
– Evgeny
Aug 2 at 9:31
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
@Evgeny Thank you. Just corrected.
– user539887
Aug 2 at 11:10
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
In the OP's question, as well in my solution, it is assumed that all positive solutions are bounded. To me, proving that would be the real problem!
– user539887
Aug 2 at 11:18
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
@Evgeny Now, I have serious doubts if that assumption can be satisfied at all! At $(0,y)$ with $y>2$ the $x$-coordinate of the field is negative. Perhaps the OP forgot to put $x$ before the parentheses in the formula for $dotx$?
– user539887
Aug 2 at 11:32
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
You are right about observation that positive quadrant is not invariant in this system. I'd agree that it is very likely that there is a typo in equations. The system with additional $x$ as multiplier also wouldn't have all the nasty stuff near the $x=-1$. At best the original system might have a limit cycle in the upper half-plane, but I really don't like the singularity at $x = -1$.
– Evgeny
Aug 2 at 15:21
add a comment |Â
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3
The system has some similarities to that posed by you 6 days ago Show that this system has a limit cycle in the first quadrant.. Did you see my answer there? Do you understand it? If not, why didn't you ask me for clarification? Or show that it is wrong?
– user539887
Jul 30 at 7:47
I am trying to solve this one in a different way (without the use of software). I was curious to see if there would be any input.
– obewanjacobi
Jul 31 at 0:44
1
In my answer the sketch was only for illustrative purposes, and had very little to do with the reasoning. I repeat: do you understand my reasoning there?
– user539887
Jul 31 at 7:48
1
Are you sure that you have $dotx=left(4-x-frac2y1+xright)$? Not $dotx=xleft(4-x-frac2y1+xright)$?
– user539887
Aug 2 at 11:33