Exploring Steady States for Hysteresis in a Nondimensionalized ODE

In summary, the conversation discusses a mathematical model with an ODE that is nondimensionalized and explores the possible nonzero steady states. It is stated that there are three possible steady states if the parameters r and q lie in a specific domain in r-q space. The question of whether the model exhibits hysteresis is also raised. The conversation then delves into solving for the parameters and finding a nose at R = 0.638. The final question asks for guidance on determining the r-q space for the model.
  • #1
Dustinsfl
2,281
5
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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  • #2
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?
 
  • #3
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 $R = 0.638$?
 
  • #4
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
 
  • #5
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.
 
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  • #6
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.
 
  • #7
How would I come up with the r-q space for this model?
 
Last edited:

FAQ: Exploring Steady States for Hysteresis in a Nondimensionalized ODE

What is hysteresis in a nondimensionalized ODE?

Hysteresis in a nondimensionalized ODE refers to a phenomenon where the output of a system lags behind the input. This means that the current state of the system is affected by its past states, causing a delay in the response to changes in the input.

Why is it important to study steady states for hysteresis?

Studying steady states for hysteresis can provide insights into the behavior of complex systems and help predict their response to changes in the input. It is especially important in fields such as physics, engineering, and biology where hysteresis is a common occurrence.

What does it mean to nondimensionalize an ODE?

Nondimensionalizing an ODE means removing the units of measurement from the equation, making it easier to compare and analyze different systems. It is a common technique used in mathematical modeling to simplify the equations and eliminate extraneous variables.

How do you explore steady states for hysteresis in a nondimensionalized ODE?

To explore steady states for hysteresis in a nondimensionalized ODE, one can use mathematical techniques such as phase plane analysis, bifurcation analysis, and numerical simulations. These methods help identify the steady states and their stability, as well as how the system behaves around these states.

What are some real-world applications of studying hysteresis in nondimensionalized ODEs?

Studying hysteresis in nondimensionalized ODEs has many practical applications, such as understanding the behavior of magnetic materials, analyzing the response of electronic circuits, and predicting the behavior of ecological systems. It is also relevant in fields such as economics, where hysteresis can affect the stability of markets and the behavior of financial systems.

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