Constructing a differential equation with a bifurcation
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I am trying to construct a differential equation $x' = f_a(x)$ where the number of equilibrium solutions depends on $a$ in this fashion: if $a<0$, no equilibrium; if $a=0$, one equilibrium; if $a>0$, four equilibria.
I have started by making $f$ a quartic function where $-a$ is a $y$ intercept (vertical shift). But this makes the order of equilibria solutions at best $0, 1, 3, 4$ or $0, 2, 4$. I am stuck on how to make the $a = 0$ equation have only $1$ zero?
differential-equations
edited Nov 18 at 6:03
Rócherz
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
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up vote
0
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I am trying to construct a differential equation $x' = f_a(x)$ where the number of equilibrium solutions depends on $a$ in this fashion: if $a<0$, no equilibrium; if $a=0$, one equilibrium; if $a>0$, four equilibria.
I have started by making $f$ a quartic function where $-a$ is a $y$ intercept (vertical shift). But this makes the order of equilibria solutions at best $0, 1, 3, 4$ or $0, 2, 4$. I am stuck on how to make the $a = 0$ equation have only $1$ zero?
differential-equations
edited Nov 18 at 6:03
Rócherz
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to construct a differential equation $x' = f_a(x)$ where the number of equilibrium solutions depends on $a$ in this fashion: if $a<0$, no equilibrium; if $a=0$, one equilibrium; if $a>0$, four equilibria.
I have started by making $f$ a quartic function where $-a$ is a $y$ intercept (vertical shift). But this makes the order of equilibria solutions at best $0, 1, 3, 4$ or $0, 2, 4$. I am stuck on how to make the $a = 0$ equation have only $1$ zero?
differential-equations
edited Nov 18 at 6:03
Rócherz
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
I am trying to construct a differential equation $x' = f_a(x)$ where the number of equilibrium solutions depends on $a$ in this fashion: if $a<0$, no equilibrium; if $a=0$, one equilibrium; if $a>0$, four equilibria.
I have started by making $f$ a quartic function where $-a$ is a $y$ intercept (vertical shift). But this makes the order of equilibria solutions at best $0, 1, 3, 4$ or $0, 2, 4$. I am stuck on how to make the $a = 0$ equation have only $1$ zero?
differential-equations
differential-equations
edited Nov 18 at 6:03
Rócherz
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
edited Nov 18 at 6:03
Rócherz
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
edited Nov 18 at 6:03
Rócherz
2,6612721
edited Nov 18 at 6:03
Rócherz
2,6612721
edited Nov 18 at 6:03
Rócherz
2,6612721
2,6612721
asked Nov 18 at 3:45
MathGuyForLife
746
asked Nov 18 at 3:45
MathGuyForLife
746
asked Nov 18 at 3:45
MathGuyForLife
746
746
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Consider
begin{align}
x' =(x^2-a)(x^2-2a).
end{align}
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Consider
begin{align}
x' =(x^2-a)(x^2-2a).
end{align}
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
add a comment |
up vote
2
down vote
accepted
Consider
begin{align}
x' =(x^2-a)(x^2-2a).
end{align}
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Consider
begin{align}
x' =(x^2-a)(x^2-2a).
end{align}
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
Consider
begin{align}
x' =(x^2-a)(x^2-2a).
end{align}
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
answered Nov 18 at 4:09
Jacky Chong
17.3k21027
17.3k21027
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
add a comment |
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
Oh wow, I did not think of something that simple! Thank you.
– MathGuyForLife
Nov 18 at 4:34
add a comment |
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