Question

Consider the nonlinear system: - x + (x – 1) y y + 4x° (1 – x). (a) Show that the system has a unique fixed point at the origin (0, 0). (b) Use a linear approximation to determine the stability of the fixed point. (c) Apply the Liapunov direct method to determine the stability of the fixed point. Is your conclusion different form that of Part (a)? Why? (d) Can the system have closed orbits (trajectories)? Explain.

Answer 1

Given syatem is: • @ fised þorinto e the 8olufion of X-1 + 4x° CI-x)=0 X-1 → x=0, 42" (x~2x+) +1=0 has no real voote. The only fired point is (0, 0). •O clape do (a.4): (40), we linerine the eguation ) and O onithe ariumption that tal<<t, 19!<< L 80 Hhat =- x-Ye. (2a 2b she 80lution f y6)>ge-t Uaing e in O, we get Bhi may be intaradel immediately to x()= G-G ¢}et° this 36)

thuo the DD me sigerveluen of the liner syetem A,=-1 omd d,= -J. Mareouet - 25) 091e Lim (at), y )= (0,0), the fied point, Hence the fined print (0,0) is stable nade. • © Liafunov divect methad of stability. Let 'L(a.) =2x'+", then L(a) = 0 and L(1.9) >0 fore fie) f (o9). %3D Now, - gx- 8ay (-)-ayt8x (1*) = - 82-ay co Honce dy Liabinov dinect method of stabiltity conclude fhat the ifired þrint (0,0) is stable mode. Hence the conelusion is same as in we port@0). •Q Since the Liepumov meathend of stability showe that Ahe fined porint (0.0) is stable and 'n porl (), the eigenvaluer d, ko aned dg <o (lie, '(0,))is stable node)! homee the syetem orbit. On order to heve a clooed orbit, the system munt Aaa han purely imazinery hos no clooed linear rooto.

Note: In part (c), by Liapunov direct method to determine the stability of fixed point, we can say that the fixed point (0,0) is globally asymptotically stable.

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