Steady State

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Subhotosh Khan said:
What is the characteristic of "steady-state"?

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sorry, I wasn't aware of the rules. So it's the wording of "as a function of E" that I'm not entirely sure with. I know for a steady state you want to set du/dt to 0. So my working out is as follows:

0=u*(1-u*)(1+u*)-Eu*
Eu*=u*(1-u*)(1+u*)
E=(1-u*)(1+u*)
Giving u*=1-E or u*=E-1

Any feedback on this would be great, thanks
 
ol98 said:
sorry, I wasn't aware of the rules. So it's the wording of "as a function of E" that I'm not entirely sure with. I know for a steady state you want to set du/dt to 0. So my working out is as follows:

0=u*(1-u*)(1+u*)-Eu*
Eu*=u*(1-u*)(1+u*)
E=(1-u*)(1+u*)
Giving u*=1-E or u*=E-1 .............................................How are you getting these?!

Any feedback on this would be great, thanks
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No!

You have:

E=(1-u*)(1+u*)

This is a quadratic equation in u* \(\displaystyle \ \ \to \ \ \) solve for u* accordingly.

However, if you "play" with the equation - you could solve it in another (more intuitive) way.
 
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ol98 said:
u*=0, u*= sqrt(1-E), u*=-sqrt(1-E) ?
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Correct - however

Why does your question use the term "biologically relevant"?

What is the significance of that term?
 
that was what I was going to go onto next. The differential is referring to an animal population. If the steady state is 0 is this indicating that there is no change in the system population, and the negative square root term is indicating a decline in the population?
 
ol98 said:
that was what I was going to go onto next. The differential is referring to an animal population. If the steady state is 0 is this indicating that there is no change in the system population, and the negative square root term is indicating a decline in the population?
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No, do you remember what u denotes? can a population be negative?
 
yoscar04 said:
No, do you remember what u denotes? can a population be negative?
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u is denoting an animal population. So the steady state that is -sqrt(1-E) is biologically irrelevant as a population cannot be negative. The steady state 0 indicates the population of the animal species is extinct, and the positive sqrt steady state indicates an increase population?
 
ol98 said:
u is denoting an animal population. So the steady state that is -sqrt(1-E) is biologically irrelevant as a population cannot be negative. The steady state 0 indicates the population of the animal species is extinct, and the positive sqrt steady state indicates an increase population?
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It sounds good to me besides the last part. Drop the "increase", this is just a steady state solution (positive constant value which is biological relevant).
 
yoscar04 said:
It sounds good to me besides the last part. Drop the "increase", this is just a steady state solution (positive constant value which is biological relevant).
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to determine the stability, would you look at a phase line plot?
 
ol98 said:
to determine the stability, would you look at a phase line plot?
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You need to write the equation for a perturbation solution based on the steady state that you just found: u(t)=uss+upert. At the end you should get something like dup/dt=constant[MATH]\cdot[/MATH]up. The stability will be determined by the sign of the constant. What textbook are you using? I recommend you Lee Segel and Leah Edelstein-Keshet: A primer on mathematical models in Biology (2013). I used an older version of this book many years ago in my class.
 
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