In different areas, steady state has slightly different meanings, so please be aware of that.
We want a theory to study the qualitative properties of solutions of differential equations, without solving the equations explicitly.
Moreover, we often want to know whether a certain property of these solutions remains unchanged if the system is subjected to various changes (often called perturbations).
It is very important to be able to study how sensitive the particular model is to small perturbations or changes of initial conditions and of various paramters.
This leads us to an area of DEQ called Stability Analysis using phase space methods and we would consider this for both autonomous and nonautonomous systems under the umbrella of the term equilibrium.
Autonomous
Definition: The equilibrium solution ${y}0$ of an autonomous system $y' = f(y)$ is said to be stable if for each number $\varepsilon$ $>0$ we can find a number $\delta$ $>0$ (depending on $\varepsilon$) such that if $\psi(t)$ is any solution of $y' = f(y)$ having $\Vert$ $\psi(t)$ $- {y_0}$ $\Vert$ $<$ $\delta$, then the solution $\psi(t)$ exists for all $t \geq {t_0}$ and $\Vert$ $\psi(t)$ $- {y_0}$ $\Vert$ $<$ $\varepsilon$ for $t \geq {t_0}$ (where for convenience the norm is the Euclidean distance that makes neighborhoods spherical).
Definition: The equilibrium solution ${y_0}$ is said to be asymptotically stable if it is stable and if there exists a number ${\delta_0}$ $> 0$ such that if $\psi(t)$ is any solution of $y' = f(y)$ having $\Vert$ $\psi(t)$ $- {y_0}$ $\Vert$ $<$ ${\delta_0}$, then $\lim_{t\rightarrow+\infty}$ $\psi(t)$ = ${y_0}$.
The equilibrium solution ${y_0}$ is said to be unstable if it is not stable.
Equivalent definitions can be written for the nonautonomous system $y' = f(t, y)$.
Now we can add notions of globally asymptoctically stable, regions of asymptotic stability and so forth.
From all of these definitions, we can write nice theorems about Linear and Almost Linear system by looking at eigenvalues and we can add notions of conditional stability.
Update
You might also want to peruse the web for notes that deal with the above. For example DEQ.
Regards
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How to find the steady state solution
- Thread starter davedave
- Start date Apr 18, 2011
consider and determine the steady state solution of the differential equation below. dy/dx = y(y-1)(y+1) We can separate the variables, break the integrand into partial fractions, and integrate the fractions easily. Solving gives y = the square root of 1 / (1 - e^(2t)). as t goes to infinity, y goes to zero which the steady state solution. But, the
actual wording of the problem goes like this. Find the steady state solution of the differential equation WITHOUT determining the exact solution and taking t to infinity. How can that be done? Please help. Thanks.
Answers and Replies
- Apr 18, 2011
- #2
When the system reaches the steady state solution, what do you know about y(x) (in general)? How does the steady-state solution change as x changes? (hint: that's a trick question). Think about what that means in terms of your DE.
- Apr 19, 2011
- #3
To find the steady state solution, I set dy/dx=0 in the differential equation dy/dx=y(y-1)(y+1) So, y=0, y=1 or y=-1. The book says the only answer is y=0. Why are y=-1 and y=+1 rejected as steady state solutions?
- Apr 19, 2011
- #4
You give us an equation in y for dy/dx and a solution in terms of t. Quite often when students give us answers we are able to work out what the questions were but generally we find it easier the other way round. Steady state may be - guessing - dy/dt = 0, and dx/dt = 0? Maybe your other solutions do not work for that? Like dy/dx = 0 , when dy/dt = 0 but dx/dt ≠ = 0,
which is possible but not a steady state. ? Nor really checked but doubtful about the solution you have given.
- Apr 20, 2011
- #5
There is a typo in the solution that I write down on the posting. It should be given in terms of x NOT t. sorry about that. ie, y = the square root of 1/(1-e^(2x)) Back to my issue. By setting dy/dx = 0, we should get three steady state solutions. y=0, y=1 or y=-1 Is there a way to verify that ONLY y=0 is correct which is the book's answer?
Thanks.
- Apr 20, 2011
- #6
OK, so not what I thought. I trust you have redifferentiated your solution to check it is correct. I did and agree it is - but it is only one solution, not the general solution. You have forgotten to put in a constant of integration. I think the solution is y = (1 - Ke^2x)^-(1/2) with K arbitrary constant. Then you can get solutions through any
point in the x,y plane. Including through those y values you mentioned. I will put up a plot of the solutions if I can and you will see they are just what you can expect qualitatively directly from the d.e. Just from the language used, "steady state" really suggests x represents time. In that case I see nothing wrong with what you say and all three y's correspond to steady states. Just that y = 0 is a stable steady state to which anything starting between y = -1 and y = 1
converges, while the other two are unstable steady states from which anything off those lines however slightly goes away from them as time goes on. But they are called steady states nonetheless. I don't know what trick Mute has in mind. Are you sure you reproduced the question verbatim? Worry about the fact that some values of K in the solution, as also in your original solution, seem to give you infinite y.
Last edited: Apr 20, 2011
- Apr 20, 2011
- #7
What I mean. x → , y ↑
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