Could this nondimensionalized ODE reveal hysteresis through its steady states?

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SUMMARY

The discussion centers on a nondimensionalized ordinary differential equation (ODE) that aims to demonstrate the existence of three nonzero steady states under the condition \(rq > 4\). The ODE is defined as \(\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right) = 0\), with functions \(h(u) = ru\left(1 - \frac{u}{q}\right)\) and \(f(u) = 1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\). The derivatives \(h'(u)\) and \(f'(u)\) are also provided, leading to the equation \(r = \frac{2uq}{\varepsilon(q - 2u)}\exp\left[-\frac{u^2}{\varepsilon}\right]\). The discussion seeks guidance on demonstrating hysteresis and the implications of substituting \(r = \frac{R}{\sqrt{\varepsilon}}\) and \(u = U\sqrt{\varepsilon}\).

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Dustinsfl
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I need to demonstrate that there are 3 possible nonzero steady states if r and q lie in a domain in r,q space given approximately by rq>4. Could this model exhibit hysteresis?
The below ODE is nondimensionalized.
$0<\varepsilon\ll 1$

$\displaystyle
\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right)=0
$
From that we get these two equations,
$\displaystyle
h(u) = ru\left(1 - \frac{u}{q}\right), \quad f(u) = 1 - \exp\left[-\frac{u^2}{\varepsilon}\right].
$
The derivatives of $h$ and $f$ are
$\displaystyle
h'(u) = r - \frac{2ru}{q}, \quad f'(u) = \frac{2u}{\varepsilon}\exp\left[-\frac{u^2}{\varepsilon}\right].
$
Solving for $\displaystyle r = \frac{2uq}{\varepsilon(q - 2u)}\exp\left[-\frac{u^2}{\varepsilon}\right]$.

I tried substitution r into the original equation and solving for q but that is ridiculous. What can I do here?
 
Last edited:
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dwsmith said:
I need to demonstrate that there are 3 possible nonzero steady states if r and q lie in a domain in r,q space given approximately by rq>4. Could this model exhibit hysteresis?
The below ODE is nondimensionalized.
$0<\varepsilon\ll 1$

$\displaystyle
\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right)=0
$
From that we get these two equations,
$\displaystyle
h(u) = ru\left(1 - \frac{u}{q}\right), \quad f(u) = 1 - \exp\left[-\frac{u^2}{\varepsilon}\right].
$
The derivatives of $h$ and $f$ are
$\displaystyle
h'(u) = r - \frac{2ru}{q}, \quad f'(u) = \frac{2u}{\varepsilon}\exp\left[-\frac{u^2}{\varepsilon}\right].
$
Solving for $\displaystyle r = \frac{2uq}{\varepsilon(q - 2u)}\exp\left[-\frac{u^2}{\varepsilon}\right]$.

I tried substitution r into the original equation and solving for q but that is ridiculous. What can I do here?
Can anyone offer any guidance? I tried using Mathematica but the solution is unmanageable.

What should I do since it says rq > 4? I don't really understand how that affects. Does epsilon being really small change anything?
 
In order to determine if the model has hysteresis, I have to make the substitution

$r=\frac{R}{\sqrt{\varepsilon}}$ and $u=U\sqrt{\varepsilon}$.

And show that there is a nose at $R = 0.638$.

After I make the substitutions, what do I do to show the nose at $R = 0.638$?
 
dwsmith said:
I need to demonstrate that there are 3 possible nonzero steady states if r and q lie in a domain in r,q space given approximately by rq>4. Could this model exhibit hysteresis?
The below ODE is nondimensionalized.
$0<\varepsilon\ll 1$

$\displaystyle
\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right)=0
$
From that we get these two equations,
$\displaystyle
h(u) = ru\left(1 - \frac{u}{q}\right), \quad f(u) = 1 - \exp\left[-\frac{u^2}{\varepsilon}\right].
$
The derivatives of $h$ and $f$ are
$\displaystyle
h'(u) = r - \frac{2ru}{q}, \quad f'(u) = \frac{2u}{\varepsilon}\exp\left[-\frac{u^2}{\varepsilon}\right].
$
Solving for $\displaystyle r = \frac{2uq}{\varepsilon(q - 2u)}\exp\left[-\frac{u^2}{\varepsilon}\right]$.

I tried substitution r into the original equation and solving for q but that is ridiculous. What can I do here?

Can you explain your notation, in particular what are constants and what are functions of time and state?

Also, when you have:

\[ \displaystyle
\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right)=0 \]

you do not proceed by looking for solutions of:

\[ \displaystyle
ru\left(1 - \frac{u}{q}\right) = \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right) \]

CB
 
CaptainBlack said:
Can you explain your notation, in particular what are constants and what are functions of time and state?

Also, when you have:

\[ \displaystyle
\frac{du}{d\tau} = ru\left(1 - \frac{u}{q}\right) - \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right)=0 \]

you do not proceed by looking for solutions of:

\[ \displaystyle
ru\left(1 - \frac{u}{q}\right) = \left(1 - \exp\left[-\frac{u^2}{\varepsilon}\right]\right) \]

CB

u is a parameter and everything else is a constant.

I know one steady state is u = 0 but the others are rather difficult to solve for which is what I need some aid with.

As well as parameterizing the r and q.
 
Last edited:
dwsmith said:
In order to determine if the model has hysteresis, I have to make the substitution

$r=\frac{R}{\sqrt{\varepsilon}}$ and $u=U\sqrt{\varepsilon}$.

And show that there is a nose at $R = 0.638$.

After I make the substitutions, what do I do to show the nose at $R = 0.638$?
I have everything solved now except the hysteresis. If you decide to respond to this question, ignore everything else.

Thanks.
 
How would I come up with the r-q space for this model?
 
Last edited:

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