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PEME5220M Adv. Math. Tech. ACM Coursework-Non linear Stability

Peishin Tan 200305149

Question: Investigate stability of given system: dx =y dt dy = −x + x 3 − y dt

Solution: dx dy and must equal to zero. Hence we have y = 0 and dt dt x ( x 2 −1) = 0 thus x = 0 or x = ±1 .

To find stationary points,

Therefore we have fixed points at (0,0) , (1,0) and (-1,0) . 

0

1 

 Jacobian of the system, J =  2  3 x − 1 − 1 2 Thus Tr ( J ) = −1 and Det ( J ) =1 − 3x 2 , λ1, 2 = − 1 ± 1 − 4(1 − 3x )

2

2

⇒ At point (0,0) 0 J ( 0, 0 ) =   −1 

1   ; Tr ( J ) = −1 ; Det ( J ) =1 ; λ1, 2 = − 1 ± −1 2 

−3 2

With conditions: Tr ( J ) < 0 , Det ( J ) > 0 , Tr ( J ) 2 − 4 Det ( J ) < 0 indicated stable focus behaviour.

⇒ At point (1,0) 0 J (1,0 ) =   2

1  1 3  ; Tr ( J ) = −1 ; Det ( J ) = −2 ; λ1, 2 = − ± = 1 or − 2  −1 2 2

With conditions: Tr ( J ) < 0 , Det ( J ) < 0 , Tr ( J ) 2 − 4 Det ( J ) > 0 indicated saddle point behaviour.  −1 J − λ1 I =  2 

1   −2 

2

1

 ; J − λ2 I =    2 1

Eigenvectors at point (1,0) must satisfy ( J − λI ) X = 0  − 1 1  x1   2 1 x2    = 0 and    = 0 Thus we have   2 − 2  y1   2 1 y 2  ∴x1 = y1 and y 2 = −2x 2

1

PEME5220M Adv. Math. Tech. ACM Coursework-Non linear Stability

Peishin Tan 200305149

⇒ At point (-1,0) 0 J ( −1,0 ) =  2 

1  1 3  ; Tr ( J ) = −1 ; Det ( J ) = −2 ; λ1, 2 = − ± = 1 or − 2 −1 2 2 

With conditions: Tr ( J ) < 0 , Det ( J ) < 0 , Tr ( J ) 2 − 4 Det ( J ) > 0 indicated saddle point behaviour.  −1 J − λ1 I =  2 

1   −2 

2

1

 ; J − λ2 I =    2 1

Eigenvectors at point (1,0) must satisfy ( J − λI ) X = 0  − 1 1  x1   2 1 x2    = 0 and    = 0 Thus we have   2 − 2  y1   2 1 y 2  ∴x1 = y1 and y 2 = −2x 2

Therefore the phase plane diagram is sketched as below:

-1

0

1

2