• Support PF! Buy your school textbooks, materials and every day products Here!

Stability of nonlinear ODE's

  • Thread starter 582153236
  • Start date
  • #1
14
0

Homework Statement


Find whether this system is stable or unstable at the steady state (x1,x2)=(0,0)
dx1/dt = -x1+2sin(x1)+x2
dx2/dt=2sin(x2)

Homework Equations


upload_2014-12-15_23-57-6.png

The Attempt at a Solution


z1=x1-0
z2=x2-0
dz1/dt=-z1+z2+2z1
dz2/dt=2z2

Jacobian =
[ 1 1 ]
[ 0 2 ]
so the system is unstable.

This problem is from my notes from class; I'm not 100% certain that it is written correctly. I am very confused about the part where z1 is set equal to x1 (I believe this is called linearization)? Could someone please clarify this step for me and how on earth
dx2/dt=2sin(x2) becomes dz2/dt=2z2 ?

The rest is pretty straightforward.
 

Attachments

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


Find whether this system is stable or unstable at the steady state (x1,x2)=(0,0)
dx1/dt = -x1+2sin(x1)+x2
dx2/dt=2sin(x2)

Homework Equations


View attachment 76651

The Attempt at a Solution


z1=x1-0
z2=x2-0
dz1/dt=-z1+z2+2z1
dz2/dt=2z2

Jacobian =
[ 1 1 ]
[ 0 2 ]
so the system is unstable.

This problem is from my notes from class; I'm not 100% certain that it is written correctly. I am very confused about the part where z1 is set equal to x1 (I believe this is called linearization)? Could someone please clarify this step for me and how on earth
dx2/dt=2sin(x2) becomes dz2/dt=2z2 ?

The rest is pretty straightforward.
If ##z_1=x_1-0## then ##x_1=z_1##. I don't see any problem with that. And the series expansion of ##sin(z)## around ##z=0## is ##z-z^3/3!+z^5/5!+...##. The linearization part is where you ignore the higher powers of ##z## and just replace ##sin(z)## with ##z##, since if ##z## is very close to 0 then the higher powers of ##z## are much smaller than ##z##.
 
  • #3
14
0
If ##z_1=x_1-0## then ##x_1=z_1##. I don't see any problem with that. And the series expansion of ##sin(z)## around ##z=0## is ##z-z^3/3!+z^5/5!+...##. The linearization part is where you ignore the higher powers of ##z## and just replace ##sin(z)## with ##z##, since if ##z## is very close to 0 then the higher powers of ##z## are much smaller than ##z##.
Thanks!
If instead I had dx1/dt=-x13+x2, would it be valid to keep z1 as it is defined now (z1=x1-0) and would dz1/dt=-z13+x2 be true? Since the first term of the series expansion of x3 is x3
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
Thanks!
If instead I had dx1/dt=-x13+x2, would it be valid to keep z1 as it is defined now (z1=x1-0) and would dz1/dt=-z13+x2 be true? Since the first term of the series expansion of x3 is x3
No, 'linearization' mean you only keep first powers of variables around the steady state. You drop the higher powers.
 

Related Threads on Stability of nonlinear ODE's

  • Last Post
Replies
5
Views
919
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
794
  • Last Post
Replies
1
Views
563
  • Last Post
Replies
4
Views
951
Replies
2
Views
2K
Top