Stable and unstable fixed points of $y' = 500y^2(1 - y)$
$begingroup$
$y' = 500y^2(1 - y)$ has fixed points at $y^*=0$ and $y^*=1$
$y(0) =0$
There are 2 definitions;
1 says that:
A fixed point $y^*$ is asymptotically stable if
$$
sigma(Df(y^*)) subset Bbb C^- ~~text{ where }~~ Bbb C^- := {z in Bbb C : Re(z) < 0} ;
$$
$sigma(Df(y^*))$ is the set of eigenvalues of the matrix $Df(y^*)$.
2 says that:
A fixed point $y^*$ is asymptotically stable (or attractive) if there exists a ball
$B_delta(y^*)$ (of radius $delta> 0$ and centered at $y^*$) such that, whenever $y_0 in B_delta(y^*)$,
the solution to $y' = f (y)$, $y(0) = y_0$ satisfies the limit $y(t) to y^*$ for $t rightarrow infty$.
I am convinced the first definition cannot be applied and so I need to show that for the fixed point $y^*=1$ there exists a $delta$ such that whenever $|y_0 - 1|< delta $, the solution to $y' = 500y^2(1-y), y(0)=0$ satisfies $y(t) = 1$ as $t rightarrow infty$.
How would i go about doing this
ordinary-differential-equations eigenvalues-eigenvectors stability-in-odes
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
$y' = 500y^2(1 - y)$ has fixed points at $y^*=0$ and $y^*=1$
$y(0) =0$
There are 2 definitions;
1 says that:
A fixed point $y^*$ is asymptotically stable if
$$
sigma(Df(y^*)) subset Bbb C^- ~~text{ where }~~ Bbb C^- := {z in Bbb C : Re(z) < 0} ;
$$
$sigma(Df(y^*))$ is the set of eigenvalues of the matrix $Df(y^*)$.
2 says that:
A fixed point $y^*$ is asymptotically stable (or attractive) if there exists a ball
$B_delta(y^*)$ (of radius $delta> 0$ and centered at $y^*$) such that, whenever $y_0 in B_delta(y^*)$,
the solution to $y' = f (y)$, $y(0) = y_0$ satisfies the limit $y(t) to y^*$ for $t rightarrow infty$.
I am convinced the first definition cannot be applied and so I need to show that for the fixed point $y^*=1$ there exists a $delta$ such that whenever $|y_0 - 1|< delta $, the solution to $y' = 500y^2(1-y), y(0)=0$ satisfies $y(t) = 1$ as $t rightarrow infty$.
How would i go about doing this
ordinary-differential-equations eigenvalues-eigenvectors stability-in-odes
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04
add a comment |
$begingroup$
$y' = 500y^2(1 - y)$ has fixed points at $y^*=0$ and $y^*=1$
$y(0) =0$
There are 2 definitions;
1 says that:
A fixed point $y^*$ is asymptotically stable if
$$
sigma(Df(y^*)) subset Bbb C^- ~~text{ where }~~ Bbb C^- := {z in Bbb C : Re(z) < 0} ;
$$
$sigma(Df(y^*))$ is the set of eigenvalues of the matrix $Df(y^*)$.
2 says that:
A fixed point $y^*$ is asymptotically stable (or attractive) if there exists a ball
$B_delta(y^*)$ (of radius $delta> 0$ and centered at $y^*$) such that, whenever $y_0 in B_delta(y^*)$,
the solution to $y' = f (y)$, $y(0) = y_0$ satisfies the limit $y(t) to y^*$ for $t rightarrow infty$.
I am convinced the first definition cannot be applied and so I need to show that for the fixed point $y^*=1$ there exists a $delta$ such that whenever $|y_0 - 1|< delta $, the solution to $y' = 500y^2(1-y), y(0)=0$ satisfies $y(t) = 1$ as $t rightarrow infty$.
How would i go about doing this
ordinary-differential-equations eigenvalues-eigenvectors stability-in-odes
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$y' = 500y^2(1 - y)$ has fixed points at $y^*=0$ and $y^*=1$
$y(0) =0$
There are 2 definitions;
1 says that:
A fixed point $y^*$ is asymptotically stable if
$$
sigma(Df(y^*)) subset Bbb C^- ~~text{ where }~~ Bbb C^- := {z in Bbb C : Re(z) < 0} ;
$$
$sigma(Df(y^*))$ is the set of eigenvalues of the matrix $Df(y^*)$.
2 says that:
A fixed point $y^*$ is asymptotically stable (or attractive) if there exists a ball
$B_delta(y^*)$ (of radius $delta> 0$ and centered at $y^*$) such that, whenever $y_0 in B_delta(y^*)$,
the solution to $y' = f (y)$, $y(0) = y_0$ satisfies the limit $y(t) to y^*$ for $t rightarrow infty$.
I am convinced the first definition cannot be applied and so I need to show that for the fixed point $y^*=1$ there exists a $delta$ such that whenever $|y_0 - 1|< delta $, the solution to $y' = 500y^2(1-y), y(0)=0$ satisfies $y(t) = 1$ as $t rightarrow infty$.
How would i go about doing this
ordinary-differential-equations eigenvalues-eigenvectors stability-in-odes
ordinary-differential-equations eigenvalues-eigenvectors stability-in-odes
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
edited Dec 5 '18 at 15:49
Dylan
12.9k31027
12.9k31027
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
asked Dec 1 '18 at 14:12
pablo_mathscobarpablo_mathscobar
996
996
$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04
add a comment |
$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04
$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04
$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As the equation is scalar, you can more directly explore the behavior near the stationary solutions.
For $yapprox y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.
For $yapprox y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.
For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,infty)$ converge towards $1$, so that you could take $δ=1$.
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
$endgroup$
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
As the equation is scalar, you can more directly explore the behavior near the stationary solutions.
For $yapprox y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.
For $yapprox y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.
For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,infty)$ converge towards $1$, so that you could take $δ=1$.
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
$endgroup$
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
add a comment |
$begingroup$
As the equation is scalar, you can more directly explore the behavior near the stationary solutions.
For $yapprox y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.
For $yapprox y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.
For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,infty)$ converge towards $1$, so that you could take $δ=1$.
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
$endgroup$
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
add a comment |
$begingroup$
As the equation is scalar, you can more directly explore the behavior near the stationary solutions.
For $yapprox y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.
For $yapprox y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.
For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,infty)$ converge towards $1$, so that you could take $δ=1$.
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
$endgroup$
As the equation is scalar, you can more directly explore the behavior near the stationary solutions.
For $yapprox y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.
For $yapprox y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.
For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,infty)$ converge towards $1$, so that you could take $δ=1$.
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
edited Dec 9 '18 at 17:12
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
answered Dec 1 '18 at 14:47
LutzLLutzL
58.6k42055
58.6k42055
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
add a comment |
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
thanks i see what you are saying but i have updated the question with an exact definition of asymptotically stable points. How would I adapt your answer to solve the problem in terms of the definition
$endgroup$
– pablo_mathscobar
Dec 1 '18 at 15:00
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Your point 1) is not a definition, it is a theorem. And it does not apply at $y=0$ as there you have $f'(0)=0$. However $f'(1)=-500$. The second is the usual definition. You would have to add bounds to the statements I made, but they will not change the picture.
$endgroup$
– LutzL
Dec 1 '18 at 15:22
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
$begingroup$
Explicit solutions of approximating differential equations are not needed here. It might be more fruitful to point the OP to the rigorous, standard (and efficient) way to solve this question (and quite a few similar ones).
$endgroup$
– Did
Dec 1 '18 at 15:57
add a comment |
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$begingroup$
The phase diagram of the differential equation is quite simple and it fully and completely solves this: for every $y_0>0$, $y(t)to1$ while, for every $y_0<0$, $y(t)to0$, hence the fixed point $0$ is ... and the fixed point $1$ is ...
$endgroup$
– Did
Dec 1 '18 at 15:04