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cloud360
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?How to know if equilibrium points are stable or not. Is my solution correct
solution above
Homework Statement
Homework Equations
The Attempt at a Solution
solution above
HallsofIvy said:For large t, and y close to either -1 or 1, t^2- y- 2 is positive.
For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasing toward -1.
For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasing toward -1 and away from 1.
For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasing away from 1.
Since for starting values close to -1, y then tends to -1, -1 is a stable equilibrium. Since for starting values close to 1, y then tends away from 1, 1 is an unstable equilibrium.
Mark44 said:It would be helpful if you pasted the text into the reply text box instead of uploading a scanned image. A scanned image makes it harder for responders such as myself to identify a particular line. I have to identify the line where there's a problem, and then point out what the problem is.
For c) you said "A fixed point is stable, if when we sub values from either side of the fixed point, we see that the gradient y', are opposite symbols."
This is not true.
From a previous thread, you gave a definition of stability that involved the second derivative. If xc is a fixed point (i.e., f'(xc) = 0), xc is stable if f''(xc) > 0. xc is an unstable fixed point if f''(xc) < 0.
In terms of the first derivative, if xc is a fixed point, xc is a stable fixed point if the derivative f' changes sign from negative to positive, moving left to right. xc is an unstable fixed point if the derivative f' changes sign from positive to negative, moving left to right.
It is not enough to notice that the derivative changes sign. It has to change sign in a certain way for the fixed point to be stable fixed point, and it has to change sign in the opposite way for it to be an unstable fixed point.
HallsofIvy said:For large t, and y close to either -1 or 1, t^2- y- 2 is positive.
For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasing toward -1.
For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasing toward -1 and away from 1.
For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasing away from 1.
Since for starting values close to -1, y then tends to -1, -1 is a stable equilibrium. Since for starting values close to 1, y then tends away from 1, 1 is an unstable equilibrium.
Mark44 said:If I recall, it said x was a stable fixed point if V''(x) > 0, not V(x) > 0.
HallsofIvy said:For large t, and y close to either -1 or 1, t^2- y- 2 is positive.
For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasing toward -1.
For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasing toward -1 and away from 1.
For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasing away from 1.
Since for starting values close to -1, y then tends to -1, -1 is a stable equilibrium. Since for starting values close to 1, y then tends away from 1, 1 is an unstable equilibrium.
You mean 1-y2 = (1-y)(1+y), not (1-y)2, which equals 1-2y+y2.cloud360 said:you said (1-y)^2=(y-1)(y+1)
but that is false
(1-y)^2=-(y-1)(y+1)=(1-y)(y+1)
see here: http://www.wolframalpha.com/input/?i=%281-y^2%29
so doesn't that mean your solution is wrong?
You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.cloud360 said:yes sorry V''(x)>0 is stable , as it is a minimum, but still i don't think this has got anything to do with this question. because the other 1 was related to potential energy.
or am i wrong?
I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).calculating y'' gives -2y+3y^2+4y. this equation contains all y's? not x's? so how can we sub x, it would be V(y)?
Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.cloud360 said:furthermore, when y'=y^2-1,this does not work
as y''=2y
Thanks. I started out intending to say "-(y-1)(y+1)" and lost the negative sign.Mark44 said:HallsOfIvy, you are mistakenly working with y2 - 1 instead of 1 - y2, so some of your results have the wrong sign.
vela said:You mean 1-y2 = (1-y)(1+y), not (1-y)2, which equals 1-2y+y2.
The only thing that does is flip some of the signs in HallsofIvy's analysis. If you understand what he said, you should be able to tell us how that changes the conclusions, if it does at all.
You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.
I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).
Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.
vela said:You mean 1-y2 = (1-y)(1+y), not (1-y)2, which equals 1-2y+y2.
The only thing that does is flip some of the signs in HallsofIvy's analysis. If you understand what he said, you should be able to tell us how that changes the conclusions, if it does at all.
You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.
I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).
Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.
I don't see how you're getting y'' = 0.cloud360 said:also, for the last question on this page
http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf
it says to determine wether the fixed points of y'=y^2-1 is stable or not
where y'=y dot=dy/dt
but if we differentiate again we get y''=0? does this mean all points are unstable?
or to determine stability, do we look at the last non 0, ODE
This isn't the approach you want to take. For one thing, you've differentiated incorrectly. Also, you have to keep track of which variable you're talking about. Your equilibrium points are at y=±1. That's completely different than saying t=±1. This complication makes the problem a lot harder.cloud360 said:ok, for the question in the opening. the one about stability.
if i differentiate with respect to t. i get
-2ty^2+2t. and t=1 gives
-2y^2+2=0
and y is imaginery. does this mean at t=1, point is unstable?
also at t=-1, i get 2y^2-2==>y^2=1
i got y=+/- 1?? how do i know if it is stable or not?
vela said:This isn't the approach you want to take. For one thing, you've differentiated incorrectly. Also, you have to keep track of which variable you're talking about. Your equilibrium points are at y=±1. That's completely different than saying t=±1. This complication makes the problem a lot harder.
Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.
You need to use the chain rule.cloud360 said:when differentiating. do i have to do implicit differentiation?
You have dy/dt = y2 - 1cloud360 said:i differentiated with respect to t, because you said "I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).
"
Mark44 said:You need to use the chain rule.
You have dy/dt = y2 - 1
So, d2y/dt2 = d/dt(y2 - 1) = d/dy(y2 - 1) * dy/dt. Of course, this should be simplified further.
No, this is wrong. Wolframalpha is calculating the partial derivative with respect to t, and the assumption is that y and t are independent variables. They are not, since y is a function of t. I already showed you what you needed to do in using the chain rule.cloud360 said:but if i differentiate w.r.t t i get 0
http://www.wolframalpha.com/input/?i=differentiate+with+respect+to+t+dy/dt+=+y2+-+1
vela said:Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.
cloud360 said:this is so confusing. because i have not seen this kind of differentiation. the answer is definitely 0. if it isnt, then i have no idea how to get the solution
i put what you said into wolfram
http://www.wolframalpha.com/input/?i=+d/dy(y2+-+1)+*+dy/dt
but how is that findinh y''. this is differentiating the multiply of (^2-1)y', which is just (y')^2. so we are differentiating y'^2!
Mark44 said:No, this is wrong. Wolframalpha is calculating the partial derivative with respect to t, and the assumption is that y and t are independent variables. They are not, since y is a function of t. I already showed you what you needed to do in using the chain rule.
In any case, and to repeat what vela said, this is NOT THE RIGHT APPROACH.
Right. In your solution, you should probably explain how you determined the sign of y' in the three regions.cloud360 said:y'=-(t^2- y- 2)(y- 1)(y+ 1)
-1<y<1 for large t, y' is positive
y<-1 for large t,y' is negative
y>1 for large y, y' is negative
<==-1==>1<==
so y=1 is a stable, y=-1 is unstable
I wouldn't say you made a small error. You had a conceptual error about the difference between a stable and unstable fixed point. Considering the problem was about identifying exactly this characteristic of the fixed points, I'd say it was a major error.but isn't this just the same as my method in post 3, except a small error.
Lower and higher values of what?all i have learned here is that lower values than the fixed point must give a positive gradient. and higher values must be negative gradient
Yes, it's correct. Can you explain why you think of y'<0 as <== and y'>0 as ==>?I think of a negative y’ as <== and positive y’ as==>. Then if we have ==><== at either side. Is stable. is this correct?
vela said:Right. In your solution, you should probably explain how you determined the sign of y' in the three regions.
I wouldn't say you made a small error. You had a conceptual error about the difference between a stable and unstable fixed point. Considering the problem was about identifying exactly this characteristic of the fixed points, I'd say it was a major error.
Lower and higher values of what?
Yes, it's correct. Can you explain why you think of y'<0 as <== and y'>0 as ==>?
At this point in your education, your instructor probably assumes you understand this concept, so you don't need to explain it on an exam. You should be able to explain it, however. If someone were to ask you why y'>0 means ==> as opposed to <==, you should be able to give a reason better than "it gives me the right answer when I assume that" or "that's what someone else told me."cloud360 said:no. but i guess i have to explain that in the exam. why is it?
vela said:At this point in your education, your instructor probably assumes you understand this concept, so you don't need to explain it on an exam. You should be able to explain it, however. If someone were to ask you why y'>0 means ==> as opposed to <==, you should be able to give a reason better than "it gives me the right answer when I assume that" or "that's what someone else told me."
cloud360 said:ok, i am very grateful for your help. i have my exam in 3 days.
can you please kindly tell me how do i answer part e. wether there is a unique solution through the point (1,2).
i know that to prove uniqueness, we must have y'=f(t,y), and differentiate w.r.t to y. i.e find (f(t,y))'
and then show that (f(t,y))' is continuous.
to show whether or not it has a unique solution at (1,2), should i sub t=1 and y=2 into (f(t,y))', and if i get a unique solution. does that mean there is a unique solution through the point (1,2)??
Mark44 said:Aren't you skipping a step? You have y' = f(t, y) = (t2 - y - 2)(1 - y2).
For there to be a unique solution at (1, 2), both f and fy have to be defined and continuous on some rectangle around (1, 2). The part that you seem to have skipped is finding
[tex]f_y = \frac{\partial }{\partial y}f(t, y)[/tex]
No, this isn't even very close.cloud360 said:First y’’=-2(t^2-2)y+3y^2-1,
This isn't about solutions or unique solutions. All you need to do is confirm that f(t, y) and fy(t, y) are continuous in some rectangle around the point (1, 2).cloud360 said:http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29
at t=1 and y=2, we get that the only solution is 15. So the point (1,2) gives a unique solution, as required
No. See above.cloud360 said:http://www.wolframalpha.com/input/?i=-2+%28t^2-2%29+y%2B3+y^2-1%2C+t%3D1%2Cy%3D2
is my solution correct?
Mark44 said:No, this isn't even very close.
y' = f(t, y) = (t2 - y - 2)(1 - y2). Here, y' means dy/dt.
To calculate fy, which in different notation is
[tex]\frac{\partial f}{\partial y}[/tex]
you need to use the product rule. I think you might have done this, but what you got is incorrect.
This isn't about solutions or unique solutions. All you need to do is confirm that f(t, y) and fy(t, y) are continuous in some rectangle around the point (1, 2).
No. See above.
Mark44 said:Apologies, your derivative was correct. Your notation was not so good, though.
If y' means dy/dt, then y'' means d2y/dt2, and that's not what is called for here. Instead of y'', you should use fy or other notation to indicate that you're taking the partial with respect to y.
The object isn't merely to evaluate f(1, 2) and fy(1, 2). You need to show that f(t, y) and fy(t, y) are continuous in some rectangle around (1, 2). Can you do that?
To determine if an equilibrium point is stable, you need to analyze the behavior of the system around the point. If the system returns to the equilibrium point after being disturbed, it is stable. If the system moves away from the equilibrium point after being disturbed, it is unstable.
A stable equilibrium point is one where the system returns to the same point after being disturbed. An unstable equilibrium point is one where the system moves away from the point after being disturbed.
To prove that an equilibrium point is stable, you can use the linearization method. This involves linearizing the system around the equilibrium point and analyzing the eigenvalues of the resulting matrix. If the eigenvalues have negative real parts, the equilibrium point is stable.
No, an equilibrium point can only be either stable or unstable. However, it is possible for a system to have multiple equilibrium points, some of which may be stable and others unstable.
To ensure the correctness of your solution, you can compare it with other methods such as phase plane analysis or Lyapunov stability analysis. Additionally, you can check your calculations and assumptions to make sure they are accurate.