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How to find the steady state solution

• Context: Undergrad
• Thread starter Thread starter davedave
• Start date Start date Apr 18, 2011
• Tags Tags State Steady Steady state

Click For Summary

Discussion Overview

The discussion revolves around finding the steady state solution of the differential equation dy/dx = y(y-1)(y+1). Participants explore different methods to identify steady state solutions without deriving the exact solution or taking limits as t approaches infinity. The conversation includes theoretical considerations and interpretations of steady state in the context of differential equations.

Discussion Character

• Exploratory
• Technical explanation
• Debate/contested
• Mathematical reasoning

Main Points Raised

• One participant proposes that the steady state solution can be found by setting dy/dx = 0, leading to potential solutions y = 0, y = 1, and y = -1.
• Another participant questions why y = 1 and y = -1 are rejected as steady state solutions, suggesting that the book's assertion of y = 0 being the only solution needs verification.
• A later reply suggests that steady state may imply conditions such as dy/dt = 0 and dx/dt = 0, indicating that not all solutions may represent steady states.
• One participant mentions the importance of including a constant of integration in the general solution and discusses the stability of the steady states, noting that y = 0 is stable while y = 1 and y = -1 are unstable steady states.
• There is a correction regarding the variable used in the solution, with one participant clarifying that the solution should be expressed in terms of x instead of t.
• Another participant expresses uncertainty about the interpretation of the question and the nature of the steady states, indicating that all three y values could correspond to steady states under certain conditions.

Areas of Agreement / Disagreement

Participants express differing views on the validity of y = 1 and y = -1 as steady state solutions, with some arguing for their inclusion and others supporting the book's claim that only y = 0 is valid. The discussion remains unresolved regarding the acceptance of these solutions.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the nature of steady states and the definitions used. The conversation also highlights the potential for infinite values in the solutions depending on the constant of integration.

Who May Find This Useful

This discussion may be of interest to students and educators in mathematics and physics, particularly those studying differential equations and steady state analysis.

davedave Messages 50 Reaction score 0 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. Physics news on Phys.org

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Mute Homework Helper Messages 1,388 Reaction score 12 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. davedave Messages 50 Reaction score 0 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? epenguin Science Advisor Messages 3,637 Reaction score 1,014 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. davedave Messages 50 Reaction score 0 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. epenguin Science Advisor Messages 3,637 Reaction score 1,014 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 epenguin Science Advisor Messages 3,637 Reaction score 1,014 What I mean. x → , y ↑

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