Making different time scales and limit situation rigorous
$begingroup$
There is a splitting of time scales which I don't understand; it is described below:
We have the following system:
$$
{dot {textbf x}=A({textbf x})textbf catopdot {textbf c}=B({textbf x})textbf c}$$ for $$ {textbf x}={schoose p}$$ and $$textbf c=left( begin{array}{c}
c_0 \ c_1 \ c_2
end{array} right),
$$
with $A({textbf x})$ and $B({textbf x})$ matrices depending on ${textbf x}$.
Let $epsilon = c_0 + c_1 + c_2$ and set
$$
textbf c=epsilon gamma .
$$
This shows that a splitting of time-scales appears when $epsilon$ is small because
$$
dot {textbf x}=epsilon A({textbf x}) gamma;quaddot{gamma}=B({textbf x})gamma.
$$
Introducing a new time variable $tau=epsilon t$, we write
$$
dot {textbf x}=frac{d{textbf x}}{dt}=epsilonfrac{d{textbf x}}{dtau}=epsilon {textbf x}',quad dot{ gamma}=frac{d gamma}{dt}=epsilonfrac{d gamma}{dtau}=epsilon gamma'
$$
and conclude that
$$
{textbf x}'=A({textbf x})gamma,,epsilongamma'=B({textbf x})gamma.
$$
For $epsilon=0$ this reduces to
$$
{textbf x}'=A({textbf x})gamma$$ with $$B({textbf x}) gamma=0 $$ and $$gamma_0+gamma_1+gamma_2=1.
$$
Now my question is: what is exactly happening when we set $epsilon = 0$? I'm looking for a rigorous explanation.
We derive the equation:
$$B({textbf x}) gamma=0 $$
But it seems a bit fishy to me.
Also I don't know how to intepret $gamma$ in that equation, i.e. what is the variable it depends on?
real-analysis ordinary-differential-equations analysis biology
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
$endgroup$
add a comment |
$begingroup$
There is a splitting of time scales which I don't understand; it is described below:
We have the following system:
$$
{dot {textbf x}=A({textbf x})textbf catopdot {textbf c}=B({textbf x})textbf c}$$ for $$ {textbf x}={schoose p}$$ and $$textbf c=left( begin{array}{c}
c_0 \ c_1 \ c_2
end{array} right),
$$
with $A({textbf x})$ and $B({textbf x})$ matrices depending on ${textbf x}$.
Let $epsilon = c_0 + c_1 + c_2$ and set
$$
textbf c=epsilon gamma .
$$
This shows that a splitting of time-scales appears when $epsilon$ is small because
$$
dot {textbf x}=epsilon A({textbf x}) gamma;quaddot{gamma}=B({textbf x})gamma.
$$
Introducing a new time variable $tau=epsilon t$, we write
$$
dot {textbf x}=frac{d{textbf x}}{dt}=epsilonfrac{d{textbf x}}{dtau}=epsilon {textbf x}',quad dot{ gamma}=frac{d gamma}{dt}=epsilonfrac{d gamma}{dtau}=epsilon gamma'
$$
and conclude that
$$
{textbf x}'=A({textbf x})gamma,,epsilongamma'=B({textbf x})gamma.
$$
For $epsilon=0$ this reduces to
$$
{textbf x}'=A({textbf x})gamma$$ with $$B({textbf x}) gamma=0 $$ and $$gamma_0+gamma_1+gamma_2=1.
$$
Now my question is: what is exactly happening when we set $epsilon = 0$? I'm looking for a rigorous explanation.
We derive the equation:
$$B({textbf x}) gamma=0 $$
But it seems a bit fishy to me.
Also I don't know how to intepret $gamma$ in that equation, i.e. what is the variable it depends on?
real-analysis ordinary-differential-equations analysis biology
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
$endgroup$
add a comment |
$begingroup$
There is a splitting of time scales which I don't understand; it is described below:
We have the following system:
$$
{dot {textbf x}=A({textbf x})textbf catopdot {textbf c}=B({textbf x})textbf c}$$ for $$ {textbf x}={schoose p}$$ and $$textbf c=left( begin{array}{c}
c_0 \ c_1 \ c_2
end{array} right),
$$
with $A({textbf x})$ and $B({textbf x})$ matrices depending on ${textbf x}$.
Let $epsilon = c_0 + c_1 + c_2$ and set
$$
textbf c=epsilon gamma .
$$
This shows that a splitting of time-scales appears when $epsilon$ is small because
$$
dot {textbf x}=epsilon A({textbf x}) gamma;quaddot{gamma}=B({textbf x})gamma.
$$
Introducing a new time variable $tau=epsilon t$, we write
$$
dot {textbf x}=frac{d{textbf x}}{dt}=epsilonfrac{d{textbf x}}{dtau}=epsilon {textbf x}',quad dot{ gamma}=frac{d gamma}{dt}=epsilonfrac{d gamma}{dtau}=epsilon gamma'
$$
and conclude that
$$
{textbf x}'=A({textbf x})gamma,,epsilongamma'=B({textbf x})gamma.
$$
For $epsilon=0$ this reduces to
$$
{textbf x}'=A({textbf x})gamma$$ with $$B({textbf x}) gamma=0 $$ and $$gamma_0+gamma_1+gamma_2=1.
$$
Now my question is: what is exactly happening when we set $epsilon = 0$? I'm looking for a rigorous explanation.
We derive the equation:
$$B({textbf x}) gamma=0 $$
But it seems a bit fishy to me.
Also I don't know how to intepret $gamma$ in that equation, i.e. what is the variable it depends on?
real-analysis ordinary-differential-equations analysis biology
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
$endgroup$
There is a splitting of time scales which I don't understand; it is described below:
We have the following system:
$$
{dot {textbf x}=A({textbf x})textbf catopdot {textbf c}=B({textbf x})textbf c}$$ for $$ {textbf x}={schoose p}$$ and $$textbf c=left( begin{array}{c}
c_0 \ c_1 \ c_2
end{array} right),
$$
with $A({textbf x})$ and $B({textbf x})$ matrices depending on ${textbf x}$.
Let $epsilon = c_0 + c_1 + c_2$ and set
$$
textbf c=epsilon gamma .
$$
This shows that a splitting of time-scales appears when $epsilon$ is small because
$$
dot {textbf x}=epsilon A({textbf x}) gamma;quaddot{gamma}=B({textbf x})gamma.
$$
Introducing a new time variable $tau=epsilon t$, we write
$$
dot {textbf x}=frac{d{textbf x}}{dt}=epsilonfrac{d{textbf x}}{dtau}=epsilon {textbf x}',quad dot{ gamma}=frac{d gamma}{dt}=epsilonfrac{d gamma}{dtau}=epsilon gamma'
$$
and conclude that
$$
{textbf x}'=A({textbf x})gamma,,epsilongamma'=B({textbf x})gamma.
$$
For $epsilon=0$ this reduces to
$$
{textbf x}'=A({textbf x})gamma$$ with $$B({textbf x}) gamma=0 $$ and $$gamma_0+gamma_1+gamma_2=1.
$$
Now my question is: what is exactly happening when we set $epsilon = 0$? I'm looking for a rigorous explanation.
We derive the equation:
$$B({textbf x}) gamma=0 $$
But it seems a bit fishy to me.
Also I don't know how to intepret $gamma$ in that equation, i.e. what is the variable it depends on?
real-analysis ordinary-differential-equations analysis biology
real-analysis ordinary-differential-equations analysis biology
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
asked Jan 15 at 15:53
Jens WagemakerJens Wagemaker
550312
550312
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $epsilon = 0$, $x' = 0$, or $x$ stays constant.
Usually you want to take $epsilon$ very small, but not $0$.
The point of multiple time scale analysis is to simplify complex dynamical system that has multiple components acting on multiple time scale. By isolating the slow/fast time scale component, you either assume that the component changing on a slower time scale to be approximately constant, or the component changing on a fast time scale to reach a (quasi) steady state first. Setting $epsilon = 0$ simply means one component is not changing at all, which is not the point of multiple time scale analysis.
answered Feb 1 at 22:22
PaichuPaichu
761616
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
If $epsilon = 0$, $x' = 0$, or $x$ stays constant.
Usually you want to take $epsilon$ very small, but not $0$.
The point of multiple time scale analysis is to simplify complex dynamical system that has multiple components acting on multiple time scale. By isolating the slow/fast time scale component, you either assume that the component changing on a slower time scale to be approximately constant, or the component changing on a fast time scale to reach a (quasi) steady state first. Setting $epsilon = 0$ simply means one component is not changing at all, which is not the point of multiple time scale analysis.
answered Feb 1 at 22:22
PaichuPaichu
761616
$endgroup$
add a comment |
$begingroup$
If $epsilon = 0$, $x' = 0$, or $x$ stays constant.
Usually you want to take $epsilon$ very small, but not $0$.
The point of multiple time scale analysis is to simplify complex dynamical system that has multiple components acting on multiple time scale. By isolating the slow/fast time scale component, you either assume that the component changing on a slower time scale to be approximately constant, or the component changing on a fast time scale to reach a (quasi) steady state first. Setting $epsilon = 0$ simply means one component is not changing at all, which is not the point of multiple time scale analysis.
answered Feb 1 at 22:22
PaichuPaichu
761616
$endgroup$
add a comment |
$begingroup$
If $epsilon = 0$, $x' = 0$, or $x$ stays constant.
Usually you want to take $epsilon$ very small, but not $0$.
The point of multiple time scale analysis is to simplify complex dynamical system that has multiple components acting on multiple time scale. By isolating the slow/fast time scale component, you either assume that the component changing on a slower time scale to be approximately constant, or the component changing on a fast time scale to reach a (quasi) steady state first. Setting $epsilon = 0$ simply means one component is not changing at all, which is not the point of multiple time scale analysis.
answered Feb 1 at 22:22
PaichuPaichu
761616
$endgroup$
If $epsilon = 0$, $x' = 0$, or $x$ stays constant.
Usually you want to take $epsilon$ very small, but not $0$.
The point of multiple time scale analysis is to simplify complex dynamical system that has multiple components acting on multiple time scale. By isolating the slow/fast time scale component, you either assume that the component changing on a slower time scale to be approximately constant, or the component changing on a fast time scale to reach a (quasi) steady state first. Setting $epsilon = 0$ simply means one component is not changing at all, which is not the point of multiple time scale analysis.
answered Feb 1 at 22:22
PaichuPaichu
761616
answered Feb 1 at 22:22
PaichuPaichu
761616
answered Feb 1 at 22:22
PaichuPaichu
761616
answered Feb 1 at 22:22
PaichuPaichu
761616
761616
add a comment |
add a comment |
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