Mixing
Bifurcation of two parameters $lambda$,$mu$
$begingroup$
I'm already familiar with bifurcations of differential equations with only $1$ parameter $lambda$, like:
$$
x'=-lambda x - x^4
$$
but what if we're given a differential equation with $2$ parameters:
$$
x'=lambda-mu x^2 +4x^4 equiv F(x,lambda,mu)
$$
My attempt:
The global minimum of $f(x)$ is the solution $x^*$ to the equation:
$$
frac{partial F(x,lambda,mu)}{partial x}=0, quadfrac{partial^2 F(x^*,lambda,mu)}{partial x^2}>0
$$
which gives:
$$
x=pm sqrt{frac{mu}{8}}
$$
Plugging this solution to $F$'s formula, we yield:
$$
a(lambda,mu)=lambda-frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
which is the function of $F$'s mimima. Therefore, we have a bifurcation when we cross the curve:
$$
lambda=frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
in the $(lambda, mu)$ plane.
From here, how can one determine the number of equilibria and their stability for the different values of $lambda, mu$?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 25 at 12:38
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
I'm already familiar with bifurcations of differential equations with only $1$ parameter $lambda$, like:
$$
x'=-lambda x - x^4
$$
but what if we're given a differential equation with $2$ parameters:
$$
x'=lambda-mu x^2 +4x^4 equiv F(x,lambda,mu)
$$
My attempt:
The global minimum of $f(x)$ is the solution $x^*$ to the equation:
$$
frac{partial F(x,lambda,mu)}{partial x}=0, quadfrac{partial^2 F(x^*,lambda,mu)}{partial x^2}>0
$$
which gives:
$$
x=pm sqrt{frac{mu}{8}}
$$
Plugging this solution to $F$'s formula, we yield:
$$
a(lambda,mu)=lambda-frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
which is the function of $F$'s mimima. Therefore, we have a bifurcation when we cross the curve:
$$
lambda=frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
in the $(lambda, mu)$ plane.
From here, how can one determine the number of equilibria and their stability for the different values of $lambda, mu$?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 25 at 12:38
LoneBoneLoneBone
968
$endgroup$
$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09
add a comment |
$begingroup$
I'm already familiar with bifurcations of differential equations with only $1$ parameter $lambda$, like:
$$
x'=-lambda x - x^4
$$
but what if we're given a differential equation with $2$ parameters:
$$
x'=lambda-mu x^2 +4x^4 equiv F(x,lambda,mu)
$$
My attempt:
The global minimum of $f(x)$ is the solution $x^*$ to the equation:
$$
frac{partial F(x,lambda,mu)}{partial x}=0, quadfrac{partial^2 F(x^*,lambda,mu)}{partial x^2}>0
$$
which gives:
$$
x=pm sqrt{frac{mu}{8}}
$$
Plugging this solution to $F$'s formula, we yield:
$$
a(lambda,mu)=lambda-frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
which is the function of $F$'s mimima. Therefore, we have a bifurcation when we cross the curve:
$$
lambda=frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
in the $(lambda, mu)$ plane.
From here, how can one determine the number of equilibria and their stability for the different values of $lambda, mu$?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 25 at 12:38
LoneBoneLoneBone
968
$endgroup$
I'm already familiar with bifurcations of differential equations with only $1$ parameter $lambda$, like:
$$
x'=-lambda x - x^4
$$
but what if we're given a differential equation with $2$ parameters:
$$
x'=lambda-mu x^2 +4x^4 equiv F(x,lambda,mu)
$$
My attempt:
The global minimum of $f(x)$ is the solution $x^*$ to the equation:
$$
frac{partial F(x,lambda,mu)}{partial x}=0, quadfrac{partial^2 F(x^*,lambda,mu)}{partial x^2}>0
$$
which gives:
$$
x=pm sqrt{frac{mu}{8}}
$$
Plugging this solution to $F$'s formula, we yield:
$$
a(lambda,mu)=lambda-frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
which is the function of $F$'s mimima. Therefore, we have a bifurcation when we cross the curve:
$$
lambda=frac{mu^2}{8}+frac{4mu^4}{8^4}
$$
in the $(lambda, mu)$ plane.
From here, how can one determine the number of equilibria and their stability for the different values of $lambda, mu$?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 25 at 12:38
LoneBoneLoneBone
968
asked Jan 25 at 12:38
LoneBoneLoneBone
968
asked Jan 25 at 12:38
LoneBoneLoneBone
968
asked Jan 25 at 12:38
LoneBoneLoneBone
968
asked Jan 25 at 12:38
LoneBoneLoneBone
968
968
$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09
add a comment |
$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09
$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09
add a comment |
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$begingroup$
We also have a bifurcation when your pass through maximum. It would be much easier if you just factor out $4x^4 - mu x^2 + lambda$ as a biquadratic equation.
$endgroup$
– Evgeny
Jan 25 at 18:50
$begingroup$
@Evgeny Could you please elaborate on this with an answer? I'd be grateful.
$endgroup$
– LoneBone
Jan 25 at 21:09