Mixing
Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
$begingroup$
Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
So if the systems are e.g. notated as:
$$u_t=u(v-1)$$
$$v_t=v(1-u)$$
then would one do stability analysis for $u_t$ and then for $v_t$ or would one consider
$$u_t+v_t=0$$
or
$$u_t=v_t=0 implies u_t-v_t=0$$
and then do the analysis?
simulation
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
$endgroup$
add a comment |
$begingroup$
Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
So if the systems are e.g. notated as:
$$u_t=u(v-1)$$
$$v_t=v(1-u)$$
then would one do stability analysis for $u_t$ and then for $v_t$ or would one consider
$$u_t+v_t=0$$
or
$$u_t=v_t=0 implies u_t-v_t=0$$
and then do the analysis?
simulation
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
$endgroup$
1
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
$endgroup$
– PhysicsMathsLove
Jan 25 at 18:00
1
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
$endgroup$
– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
$endgroup$
– Robert Lewis
Jan 25 at 19:05
1
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
$endgroup$
– PhysicsMathsLove
Jan 25 at 19:24
1
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28
add a comment |
$begingroup$
Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
So if the systems are e.g. notated as:
$$u_t=u(v-1)$$
$$v_t=v(1-u)$$
then would one do stability analysis for $u_t$ and then for $v_t$ or would one consider
$$u_t+v_t=0$$
or
$$u_t=v_t=0 implies u_t-v_t=0$$
and then do the analysis?
simulation
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
$endgroup$
Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
So if the systems are e.g. notated as:
$$u_t=u(v-1)$$
$$v_t=v(1-u)$$
then would one do stability analysis for $u_t$ and then for $v_t$ or would one consider
$$u_t+v_t=0$$
or
$$u_t=v_t=0 implies u_t-v_t=0$$
and then do the analysis?
simulation
simulation
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
edited Jan 25 at 18:31
mavavilj
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
asked Jan 25 at 17:37
mavaviljmavavilj
2,82411137
2,82411137
1
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
$endgroup$
– PhysicsMathsLove
Jan 25 at 18:00
1
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
$endgroup$
– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
$endgroup$
– Robert Lewis
Jan 25 at 19:05
1
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
$endgroup$
– PhysicsMathsLove
Jan 25 at 19:24
1
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28
add a comment |
1
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
$endgroup$
– PhysicsMathsLove
Jan 25 at 18:00
1
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
$endgroup$
– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
$endgroup$
– Robert Lewis
Jan 25 at 19:05
1
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
$endgroup$
– PhysicsMathsLove
Jan 25 at 19:24
1
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28
1
1
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
$endgroup$
– PhysicsMathsLove
Jan 25 at 18:00
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
$endgroup$
– PhysicsMathsLove
Jan 25 at 18:00
1
1
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
$endgroup$
– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
$endgroup$
– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
$endgroup$
– Robert Lewis
Jan 25 at 19:05
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
$endgroup$
– Robert Lewis
Jan 25 at 19:05
1
1
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
$endgroup$
– PhysicsMathsLove
Jan 25 at 19:24
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
$endgroup$
– PhysicsMathsLove
Jan 25 at 19:24
1
1
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), tag 1$
$v_t = v(1 - u). tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other. And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution. Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$mathbf r(u, v) = begin{pmatrix} u \ v end{pmatrix}; tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$mathbf X(u, v) = begin{pmatrix} u(v - 1) \ v(1 - u) end{pmatrix}; tag 4$
the given system is then written in vector form
$dot{mathbf r}(u, v) = mathbf X(u, v). tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, tag 6$
$v_t = v(1 - u) = 0 tag 7$
simultaneously for $u$, $v$; we see that if either
$u = 0 ; text{or} ; v = 0 tag 8$
then
$u = v = 0; tag 9$
also, from (6) and (7),
$u ne 0 Longrightarrow v = 1 Longrightarrow u = 1; tag{10}$
thus another equilibrium point is
$u = v = 1; tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{mathbf X}(u, v) = begin{bmatrix} u_{t, u} & u_{t, v} \ v_{t, u} & v_{t, v} end{bmatrix}, tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = dfrac{partial u_t}{partial u}, tag{13}$
and so forth; then
$J_{mathbf X}(u, v) = begin{bmatrix} v - 1 & u \ -v & 1 - u end{bmatrix}; tag{14}$
at $(0, 0)$,
$J_{mathbf X}(0, 0) = begin{bmatrix} -1 & 0 \ 0 & 1 end{bmatrix}; tag{15}$
the eigenvalues of this matrix are clearly $pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{mathbf X}(1, 1) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, tag{16}$
the eigenvalues of which are easily seen to be $pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5). At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it. But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
$endgroup$
add a comment |
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$begingroup$
Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), tag 1$
$v_t = v(1 - u). tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other. And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution. Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$mathbf r(u, v) = begin{pmatrix} u \ v end{pmatrix}; tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$mathbf X(u, v) = begin{pmatrix} u(v - 1) \ v(1 - u) end{pmatrix}; tag 4$
the given system is then written in vector form
$dot{mathbf r}(u, v) = mathbf X(u, v). tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, tag 6$
$v_t = v(1 - u) = 0 tag 7$
simultaneously for $u$, $v$; we see that if either
$u = 0 ; text{or} ; v = 0 tag 8$
then
$u = v = 0; tag 9$
also, from (6) and (7),
$u ne 0 Longrightarrow v = 1 Longrightarrow u = 1; tag{10}$
thus another equilibrium point is
$u = v = 1; tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{mathbf X}(u, v) = begin{bmatrix} u_{t, u} & u_{t, v} \ v_{t, u} & v_{t, v} end{bmatrix}, tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = dfrac{partial u_t}{partial u}, tag{13}$
and so forth; then
$J_{mathbf X}(u, v) = begin{bmatrix} v - 1 & u \ -v & 1 - u end{bmatrix}; tag{14}$
at $(0, 0)$,
$J_{mathbf X}(0, 0) = begin{bmatrix} -1 & 0 \ 0 & 1 end{bmatrix}; tag{15}$
the eigenvalues of this matrix are clearly $pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{mathbf X}(1, 1) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, tag{16}$
the eigenvalues of which are easily seen to be $pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5). At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it. But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
$endgroup$
add a comment |
$begingroup$
Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), tag 1$
$v_t = v(1 - u). tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other. And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution. Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$mathbf r(u, v) = begin{pmatrix} u \ v end{pmatrix}; tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$mathbf X(u, v) = begin{pmatrix} u(v - 1) \ v(1 - u) end{pmatrix}; tag 4$
the given system is then written in vector form
$dot{mathbf r}(u, v) = mathbf X(u, v). tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, tag 6$
$v_t = v(1 - u) = 0 tag 7$
simultaneously for $u$, $v$; we see that if either
$u = 0 ; text{or} ; v = 0 tag 8$
then
$u = v = 0; tag 9$
also, from (6) and (7),
$u ne 0 Longrightarrow v = 1 Longrightarrow u = 1; tag{10}$
thus another equilibrium point is
$u = v = 1; tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{mathbf X}(u, v) = begin{bmatrix} u_{t, u} & u_{t, v} \ v_{t, u} & v_{t, v} end{bmatrix}, tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = dfrac{partial u_t}{partial u}, tag{13}$
and so forth; then
$J_{mathbf X}(u, v) = begin{bmatrix} v - 1 & u \ -v & 1 - u end{bmatrix}; tag{14}$
at $(0, 0)$,
$J_{mathbf X}(0, 0) = begin{bmatrix} -1 & 0 \ 0 & 1 end{bmatrix}; tag{15}$
the eigenvalues of this matrix are clearly $pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{mathbf X}(1, 1) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, tag{16}$
the eigenvalues of which are easily seen to be $pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5). At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it. But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
$endgroup$
add a comment |
$begingroup$
Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), tag 1$
$v_t = v(1 - u). tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other. And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution. Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$mathbf r(u, v) = begin{pmatrix} u \ v end{pmatrix}; tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$mathbf X(u, v) = begin{pmatrix} u(v - 1) \ v(1 - u) end{pmatrix}; tag 4$
the given system is then written in vector form
$dot{mathbf r}(u, v) = mathbf X(u, v). tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, tag 6$
$v_t = v(1 - u) = 0 tag 7$
simultaneously for $u$, $v$; we see that if either
$u = 0 ; text{or} ; v = 0 tag 8$
then
$u = v = 0; tag 9$
also, from (6) and (7),
$u ne 0 Longrightarrow v = 1 Longrightarrow u = 1; tag{10}$
thus another equilibrium point is
$u = v = 1; tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{mathbf X}(u, v) = begin{bmatrix} u_{t, u} & u_{t, v} \ v_{t, u} & v_{t, v} end{bmatrix}, tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = dfrac{partial u_t}{partial u}, tag{13}$
and so forth; then
$J_{mathbf X}(u, v) = begin{bmatrix} v - 1 & u \ -v & 1 - u end{bmatrix}; tag{14}$
at $(0, 0)$,
$J_{mathbf X}(0, 0) = begin{bmatrix} -1 & 0 \ 0 & 1 end{bmatrix}; tag{15}$
the eigenvalues of this matrix are clearly $pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{mathbf X}(1, 1) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, tag{16}$
the eigenvalues of which are easily seen to be $pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5). At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it. But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
$endgroup$
Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), tag 1$
$v_t = v(1 - u). tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other. And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution. Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$mathbf r(u, v) = begin{pmatrix} u \ v end{pmatrix}; tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$mathbf X(u, v) = begin{pmatrix} u(v - 1) \ v(1 - u) end{pmatrix}; tag 4$
the given system is then written in vector form
$dot{mathbf r}(u, v) = mathbf X(u, v). tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, tag 6$
$v_t = v(1 - u) = 0 tag 7$
simultaneously for $u$, $v$; we see that if either
$u = 0 ; text{or} ; v = 0 tag 8$
then
$u = v = 0; tag 9$
also, from (6) and (7),
$u ne 0 Longrightarrow v = 1 Longrightarrow u = 1; tag{10}$
thus another equilibrium point is
$u = v = 1; tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{mathbf X}(u, v) = begin{bmatrix} u_{t, u} & u_{t, v} \ v_{t, u} & v_{t, v} end{bmatrix}, tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = dfrac{partial u_t}{partial u}, tag{13}$
and so forth; then
$J_{mathbf X}(u, v) = begin{bmatrix} v - 1 & u \ -v & 1 - u end{bmatrix}; tag{14}$
at $(0, 0)$,
$J_{mathbf X}(0, 0) = begin{bmatrix} -1 & 0 \ 0 & 1 end{bmatrix}; tag{15}$
the eigenvalues of this matrix are clearly $pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{mathbf X}(1, 1) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, tag{16}$
the eigenvalues of which are easily seen to be $pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5). At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it. But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
edited Jan 25 at 19:35
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
answered Jan 25 at 19:01
Robert LewisRobert Lewis
48.1k23167
48.1k23167
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1
$begingroup$
I would do it for both at the same time. The fixed points satisfy $u(v-1) = 0$ and $v(1-u) = 0$ so either $u=0$ or $v=1$ or $v= 0$ or $u = 1$, giving 4 points $(0, 0)$, $(0, 1)$, $(1, 0)$ and $(1, 1)$. Then you can calculate the Jacobean and evaluate the stability. However, I am a student so there may be other ways...
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– PhysicsMathsLove
Jan 25 at 18:00
1
$begingroup$
@PhysicsMathsLove This www2.hawaii.edu/~taylor/z652/PredatorPreyModels.pdf has an example where they calculate the stationary points so that each equation produces the other point. Then they form the Jacobian based on these points.
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– mavavilj
Jan 25 at 18:30
$begingroup$
@PhysicsMathsLove: $(0, 1)$ and $(1, 0$ are not equilibria. Cheers!
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– Robert Lewis
Jan 25 at 19:05
1
$begingroup$
@RobertLewis why not? It causes both u_t and v_t = 0 which is the definition of a fixed point, no?
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– PhysicsMathsLove
Jan 25 at 19:24
1
$begingroup$
@PhysicsMathsLove: check out $u = 1$, $v = 0$; $u_t = u(v - 1) = -1 ne 0$!
$endgroup$
– Robert Lewis
Jan 25 at 19:28