?How to know if equilibrium points are stable or not. Is my solution correct

[cloud360]

?How to know if equilibrium points are stable or not. Is my solution correct

Homework Statement

Homework Equations

The Attempt at a Solution

solution above

 

[HallsofIvy]

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.

 

[cloud360]

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.

ok i followed what your saying. can you tell me if my solution below is correct? for part c and d

i said C is stable, and so is D

 

[Mark44]

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 x_c is a fixed point (i.e., f'(x_c) = 0), x_c is stable if f''(x_c) > 0. x_c is an unstable fixed point if f''(x_c) < 0.

In terms of the first derivative, if x_c is a fixed point, x_c is a stable fixed point if the derivative f' changes sign from negative to positive, moving left to right. x_c 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.

 

[cloud360]

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 x_c is a fixed point (i.e., f'(x_c) = 0), x_c is stable if f''(x_c) > 0. x_c is an unstable fixed point if f''(x_c) < 0.

In terms of the first derivative, if x_c is a fixed point, x_c is a stable fixed point if the derivative f' changes sign from negative to positive, moving left to right. x_c 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.

actually, the other time i was dealing with potential energy, and the notes said V(x)>0 is stable (minimum)

where V=potential energy equation

 

[Mark44]

If I recall, it said x was a stable fixed point if V''(x) > 0, not V(x) > 0.

 

[cloud360]

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 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?

 

[cloud360]

If I recall, it said x was a stable fixed point if V''(x) > 0, not V(x) > 0.

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?

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)?

if i sub y=1 i get -2+3+4=5

since V''(1)=5>0, means x=1 is stable

V''(-1) gives 2+3+4=7

V''(-1)=7>0, means x=-1 is stable

so agrees with my solution

 

[cloud360]

furthermore, when y'=y^2-1,this does not work

as y''=2y

we have fixed points -1 and 1

V''(-1)=-2<0, so unstable

but -1 is actually stable. according to website below. see last question.

http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf

but my method of showing symbols change, shows -1 is stable (just like above website). the above website has error. it says x=1 is stable. but the graph they sketched clearly shows its unstable !

again the fixed point 1 gives V''(1)=2>0, so is stable. but is actually unstable !

 

[Mark44]

HallsOfIvy, you are mistakenly working with y² - 1 instead of 1 - y², so some of your results have the wrong sign.

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.

 

[vela]

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 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.

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?

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.

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)?

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).

furthermore, when y'=y^2-1,this does not work

as y''=2y

Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.

 

[HallsofIvy]

HallsOfIvy, you are mistakenly working with y² - 1 instead of 1 - y², so some of your results have the wrong sign.

Thanks. I started out intending to say "-(y-1)(y+1)" and lost the negative sign.

 

[cloud360]

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.

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?

 

[cloud360]

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.

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

 

[Mark44]

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

I don't see how you're getting y'' = 0.

If y' = y² - 1 (where y' means dy/dt)
then y'' = 2y*y', by the chain rule.

 

[vela]

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?

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.

 

[cloud360]

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.

thanks again for response. when differentiating. do i have to do implicit differentiation?

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]

when differentiating. do i have to do implicit differentiation?

You need to use the chain rule.

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).
"

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.

 

[cloud360]

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.

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

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]

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

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.

Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.

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!

 

[cloud360]

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.

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

but isn't this just the same as my method in post 3, except a small error.

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

I think of a negative y’ as <== and positive y’ as==>. Then if we have ==><== at either side. Is stable. is this correct?

 

[vela]

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

Right. In your solution, you should probably explain how you determined the sign of y' in the three regions.

but isn't this just the same as my method in post 3, except a small error.

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.

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

Lower and higher values of what?

I think of a negative y’ as <== and positive y’ as==>. Then if we have ==><== at either side. Is stable. is this correct?

Yes, it's correct. Can you explain why you think of y'<0 as <== and y'>0 as ==>?

 

[cloud360]

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 ==>?

no. but i guess i have to explain that in the exam. why is it?

i just thinking of positive as moving forwards, thus ==>

and negative as moving backwards, thus <==

 

[vela]

no. but i guess i have to explain that in the exam. why is it?

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]

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."

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]

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)??

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 f_y have to be defined and continuous on some rectangle around (1, 2). The part that you seem to have skipped is finding
$$f_y = \frac{\partial }{\partial y}f(t, y)$$

 

[cloud360]

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 f_y have to be defined and continuous on some rectangle around (1, 2). The part that you seem to have skipped is finding
$$f_y = \frac{\partial }{\partial y}f(t, y)$$

First y’’=-2(t^2-2)y+3y^2-1,

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

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]

First y’’=-2(t^2-2)y+3y^2-1,

No, this isn't even very close.
y' = f(t, y) = (t² - y - 2)(1 - y²). Here, y' means dy/dt.
To calculate f_y, which in different notation is
$$\frac{\partial f}{\partial y}$$
you need to use the product rule. I think you might have done this, but what you got is incorrect.

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

This isn't about solutions or unique solutions. All you need to do is confirm that f(t, y) and f_y(t, y) are continuous in some rectangle around the point (1, 2).

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?

No. See above.

 

[cloud360]

No, this isn't even very close.
y' = f(t, y) = (t² - y - 2)(1 - y²). Here, y' means dy/dt.
To calculate f_y, which in different notation is
$$\frac{\partial f}{\partial y}$$
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 f_y(t, y) are continuous in some rectangle around the point (1, 2).


No. See above.

I did differentaite w.r.t y, see wolfram

http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29

then i sub t=1 y=2 and got 15

 

[Mark44]

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 f^y 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 f_y(1, 2). You need to show that f(t, y) and f_y(t, y) are continuous in some rectangle around (1, 2). Can you do that?

 

[cloud360]

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 f^y 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 f^y(1, 2). You need to show that f(t, y) and f^y(t, y) are continuous in some rectangle around (1, 2). Can you do that?

Fy=-2(t^2-2)y+3y^2-1

i have no idea how to show this is continuous?

is there a method. if so i don't know what it is, cany you give me a link to a website which possibly teaches you how to know if somethig is continous.

i know how to prove a function is 1-1 though

 

[Mark44]

Both functions are defined at all point (t, y) in the plane. Since the only operations involved in their formulas are multiplication and addition, both functions are also continuous at every point (t, y) in the plane.

The basis ideas involved here are these: if g(x) and h(x) are continuous functions, the the sum (g + h)(x) and product gh(x) are continous. both f(t, y) and f_y(t, y) can be split up into the sum or product of simple linear or quadratic functions, which is enough to show that f(t, y) and f_y(t, y) are continuous. This is enough to show that the differential equation y' = f(t, y) has a unique solution that contains the point (1, 2). That's what you wanted to show in part e, so you're done.