MHB How Do I Solve this 2nd Order Non-Linear ODE and Find Its Equilibrium Points?

[metstandard16]

Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y²y'+y-y⁵=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u²v-u+u⁵

The second equation can be factored out to = (u-1)(u+1)(u³-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes. Somehow I think I have 3 eq points at (0,0) , (1,0) , & (-1,0) but I wasn't completely sure how that equation panned out.

Thanks for your help!

 

[Prove It]

Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y²y'+y-y⁵=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u²v-u+u⁵

The second equation can be factored out to = (u-1)(u+1)(u³-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes. Somehow I think I have 3 eq points at (0,0) , (1,0) , & (-1,0) but I wasn't completely sure how that equation panned out.

Thanks for your help!

Are you trying to solve this exactly, or approximately with a numerical scheme?

 

[metstandard16]

Are you trying to solve this exactly, or approximately with a numerical scheme?

Thanks for your reply. I'm trying to solve it numerically, eventually with the aid of MATLAB and with a few initial conditions based on the phase portraits.

I guess my confusion lies around the equilibrium points and the jacobian matrix system.

To expand on my work in the post above:

v' = y'' = v-u²v-u+u⁵
v' = v(1-u²)+u(u⁴-1)
v = u(1-u⁴)/(1-u²)
= u(1+u²)

v=0
Eq Point: (0,0) seems to be the only real solution, and perhaps a +/- i on the non-real part because of the 1+u^2.

Am I concerned with the imaginary part at this point? Just needed help getting this so I could classify them as (saddle, center, etc) and work up some approximate phase portraits.

Thanks!

 

[HallsofIvy]

Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y²y'+y-y⁵=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u²v-u+u⁵

The second equation can be factored out to = (u-1)(u+1)(u³-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes.

And there is no "(v)/y'" here! Your two equations are u'= v and v'= v- u^2v- u+ u^5.

The equilibrium points are where u'= 0 and v'= 0. That is, u'= v= 0 and v'= v- u^2v- u+ u^5= 0. Setting v= 0 in the second equation, we have -u+ u^5= u(u^2+ 1)(u- 1)(u+ 1)=0 so u= 0, u= 1, and u= -1 are the real roots. Those give equilibrium points, (u, v)= (0, 0), (1, 0), and (-1, 0). The simplest way to find the phase diagram is to linearize about each equilibrium point.