Steady State
- Thread starter ol98
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The problem explicitly tells you what the two cases are.
When you solved the first differential equation for a steady state, what did you get?
When you solved the second differential equation for a steady state, what did you get?
What variable appears in both results? What does that suggest doing?
When you solved the first differential equation for a steady state, what did you get?
When you solved the second differential equation for a steady state, what did you get?
What variable appears in both results? What does that suggest doing?
Setting the first differential to 0. I got:
Giving 2 solutions of n=0 and bnc+bc-1=0 (Not sure whether to solve for n or c at this stage though)
Setting the 2nd differential to 0. I got:
c=alpha/mu
I might start as you did with
[MATH]\dfrac{dc}{d \tau} = 0 \implies c = \dfrac{\alpha}{\mu}.[/MATH]
Then I might note a special case.
[MATH]c = 0 = \dfrac{dc}{d \tau} \implies \alpha = 0 \implies \dfrac{\alpha \beta}{\mu} = 0 < 1.[/MATH]
[MATH]\text {If also } \dfrac{dn}{d \tau} = 0, \text { then } \dfrac{n}{n + 1} = 0 \implies n = 0.[/MATH]
In other words, if [MATH]\dfrac{\alpha \beta }{\mu} < 1[/MATH], then c = 0 = n is a stable point.
Is that the only one?
If c is not 0, then
[MATH]\dfrac{dn}{d \tau} = 0 = \dfrac{dc}{d \tau} \text { and } \beta = 0 \implies \dfrac{\alpha \beta}{\mu} = 0 < 1 \implies\\ \dfrac{1}{n + 1} = 0 \implies n = 0.[/MATH]In that special case n = 0 and c = [MATH]\dfrac{\alpha}{\mu}[/MATH] is stable.
Now you can think about more general cases.
[MATH]\dfrac{dc}{d \tau} = 0 \implies c = \dfrac{\alpha}{\mu}.[/MATH]
Then I might note a special case.
[MATH]c = 0 = \dfrac{dc}{d \tau} \implies \alpha = 0 \implies \dfrac{\alpha \beta}{\mu} = 0 < 1.[/MATH]
[MATH]\text {If also } \dfrac{dn}{d \tau} = 0, \text { then } \dfrac{n}{n + 1} = 0 \implies n = 0.[/MATH]
In other words, if [MATH]\dfrac{\alpha \beta }{\mu} < 1[/MATH], then c = 0 = n is a stable point.
Is that the only one?
If c is not 0, then
[MATH]\dfrac{dn}{d \tau} = 0 = \dfrac{dc}{d \tau} \text { and } \beta = 0 \implies \dfrac{\alpha \beta}{\mu} = 0 < 1 \implies\\ \dfrac{1}{n + 1} = 0 \implies n = 0.[/MATH]In that special case n = 0 and c = [MATH]\dfrac{\alpha}{\mu}[/MATH] is stable.
Now you can think about more general cases.
So we have a steady state at (0,0) for when (alpha. beta)/mu is less than 1, and it is a stable point.
We have (0, alpha/mu) which is a stable point again when (alpha. beta)/mu is less than 1.
How would I find the cases for when (alpha. beta)/mu is greater than 1?
First, we have found some steady states for when alpha beta / mu < 1. That does not mean we have yet found them all; whether or not there are more is an open question.
Second, I'd be very wary of using the term "stable" when you mean "steady."
Third, it may be helpful to know whether alpha, beta, or mu can take on negative values. (This question may turn out to be irrelevant, but it is prudent to consider issues of sign whenever you have inequalities to deal with.)
So my next step would be to solve for n generally given
[MATH]\dfrac{\alpha \beta}{\mu} \ne 0,\ c = \dfrac{\alpha}{\mu}, \text { and } \dfrac{1}{n + 1} = \dfrac{\alpha \beta}{\mu}.[/MATH]
Do you see where those conditions come from and why they are relevant?
Do you see why we considered the cases of c = 0 = alpha and beta = 0 as special cases?
Yes, I understand. I'm not sure how you would go about doing, (alpha.beta)/mu is not equal to 0 though?
Really, you are making this way harder than it is.
[MATH]0 < \dfrac{1}{n + 1} = \dfrac{\alpha \beta}{\mu} < 1 \implies n + 1 = \dfrac{\mu}{\alpha \beta} > 1 \implies\\ n = \dfrac{\mu}{\alpha \beta} - 1 > 0.[/MATH]I repeat: are there restrictions on alpha, beta, or mu?
I've not been told in the question, but the equations were non-dimensionalised, and was told they were all postive constants
Well, this has wasted some time.
[MATH]\dfrac{dn}{d \tau} = \dfrac{n}{1 + n} - \beta nc \text { and } \dfrac{dc}{d \tau} = \alpha - \mu c.[/MATH]
[MATH]\therefore \dfrac{dc}{d \tau} = 0 \implies \alpha - \mu c \implies c = \dfrac{\alpha}{mu}.[/MATH]
[MATH]\therefore \dfrac{dn}{d \tau} = 0 \implies \dfrac{n}{1 + n} - \beta nc = 0.[/MATH]
Obviously [MATH]n = 0 \implies \dfrac{n}{1 + n} - \beta nc = 0[/MATH].
Therefore (0, alpha/mu) is a steady point regardless of the the values of alpha, beta, and mu.
[MATH]n \ne 0 \text { and } \dfrac{dn}{d \tau} = 0 \implies \dfrac{1}{n + 1} - \beta c = 0 \implies \\ \dfrac{1}{n + 1} = \beta c \implies n + 1 = \dfrac{1}{\beta c} \implies n = \dfrac{1}{\beta c} - 1.[/MATH]Any questions to here?
[MATH]\therefore \dfrac{dc}{d \tau} = 0,\ n \ne 0 \text { and } \dfrac{dn}{d \tau} = 0 \implies \\ c = \dfrac{\alpha}{\mu} \text { and } n = \dfrac{1}{\beta c} - 1 = \dfrac{\mu}{\alpha \beta} - 1.[/MATH]
So what do the inequalities tell you?
[MATH]\dfrac{dn}{d \tau} = \dfrac{n}{1 + n} - \beta nc \text { and } \dfrac{dc}{d \tau} = \alpha - \mu c.[/MATH]
[MATH]\therefore \dfrac{dc}{d \tau} = 0 \implies \alpha - \mu c \implies c = \dfrac{\alpha}{mu}.[/MATH]
[MATH]\therefore \dfrac{dn}{d \tau} = 0 \implies \dfrac{n}{1 + n} - \beta nc = 0.[/MATH]
Obviously [MATH]n = 0 \implies \dfrac{n}{1 + n} - \beta nc = 0[/MATH].
Therefore (0, alpha/mu) is a steady point regardless of the the values of alpha, beta, and mu.
[MATH]n \ne 0 \text { and } \dfrac{dn}{d \tau} = 0 \implies \dfrac{1}{n + 1} - \beta c = 0 \implies \\ \dfrac{1}{n + 1} = \beta c \implies n + 1 = \dfrac{1}{\beta c} \implies n = \dfrac{1}{\beta c} - 1.[/MATH]Any questions to here?
[MATH]\therefore \dfrac{dc}{d \tau} = 0,\ n \ne 0 \text { and } \dfrac{dn}{d \tau} = 0 \implies \\ c = \dfrac{\alpha}{\mu} \text { and } n = \dfrac{1}{\beta c} - 1 = \dfrac{\mu}{\alpha \beta} - 1.[/MATH]
So what do the inequalities tell you?
Thanks again, I've got one question about the method, I was just wondering how you did this step:
View attachment 20911
When you set dn/dT to 0, can you just cancel out the n's like you did?
and so if n= (mu/alpha. beta)- 1 is that for the case when (alpha.beta)/mu is greater than 1.
and if n= (1/beta. c) - 1 than thats for the case when it's less than 1.
If you go back to the post that you are questioning, I first considered the case of n = 0. And no, you cannot do cancelling in that case. Then I explicitly considered the case if n is not zero.
[MATH]n \ne 0 \text { and } \dfrac{n}{n + 1} - \beta nc = 0 \implies \\ \dfrac{1}{n} * \left ( \dfrac{n}{n + 1} - \beta nc \right ) = \dfrac{1}{n} * 0 \implies\\ \dfrac{1}{n + 1} - \beta c = 0.[/MATH]So, yes, once it is clear that n is not zero, you can cancel like that.
This is friendly and hopefully constructive criticism. As I told you in a private message, my knowledge of differential equations is superficial, but the questions you have been raising in this thread have been about elementary algebra. I understand that you have uncertainty about concepts introduced by differential equations, but algebra has not been repealed. Everything you learned in algebra and calculus is still true: you do not need to feel perplexed about everything.
We are back to basic algebra.
[MATH]c = \dfrac{\alpha}{\mu} \text { and } n = \dfrac{1}{\beta c} - 1 \implies \\ n = \dfrac{1}{\beta * \dfrac{\alpha}{\mu}} - 1 = \dfrac{1}{\dfrac{\alpha \beta}{\mu}} - 1 = \dfrac{\mu}{\alpha \beta} - 1.[/MATH]This should be absolutely straight forward.
Now we know that alpha, beta, and mu are all positive, and we are asked to consider two cases.
Case I
[MATH]0 < \dfrac{\alpha \beta}{\mu} < 1 \implies \dfrac{\mu}{\alpha \beta} > 1.[/MATH]
Do you agree?
[MATH]\therefore \dfrac{\mu}{\alpha \beta} > 1 \implies \dfrac{\mu}{\alpha \beta} - 1 > 0 \implies n > 0.[/MATH]
That presumably has some material significance in the model.
Case II
[MATH]\dfrac{\alpha \beta}{\mu} > 1 \implies 0 < \dfrac{\mu}{\alpha \beta} < 1.[/MATH]
Do you agree? Do you see why it is necessarily positive?
[MATH]\therefore 0 < \dfrac{\mu}{\alpha \beta} < 1 \implies - 1 < \dfrac{\mu}{\alpha \beta} - 1 < 0 \implies - 1 < n < 0.[/MATH]
That presumably has some material significance in the model.