Is the System with Nonlinear ODEs at Steady State (0,0) Stable?

[582153236]

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

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.

 

[Dick]

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 $z$.

 

[582153236]

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

 

[Dick]

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.