yes,periodic solution

Here is the theorem used in lecture notes

7. Using the appropriate Theorem from lectures, show that the equation T + (x - 1)i +2 +10, has a unique periodic solution if x > 0. 2 marks] 8. For small amplitudes the motion of a soft spring is described by Dufling's equation: where (t) is the displacement, y(t) is the velocity and v is the damping coefficient. Investigate the stability of the equilibrium points for all real valued v. Determine and classify any bifurcation points, and draw, and label, a bifurcation diagram to illustrate your results (6 marks] 7. Using the appropriate Theorem from lectures, show that the equation T + (x - 1)i +2 +10, has a unique periodic solution if x > 0. 2 marks] 8. For small amplitudes the motion of a soft spring is described by Dufling's equation: where (t) is the displacement, y(t) is the velocity and v is the damping coefficient. Investigate the stability of the equilibrium points for all real valued v. Determine and classify any bifurcation points, and draw, and label, a bifurcation diagram to illustrate your results (6 marks] convinetto | Averaging methods Recall x = 0 (25) This has sol's alt) Tacost 7 (25) zlt) = y(t) = -a sint J is ciroles centred on the origin Now we consider * + ghlac, 3) + x = 0, (26) for Elka1 negating Housing) The phase paths will no longer be circles (in general) - we expect unsteady spirals and perhaps a lima't cycle. su Suppose k(0,0)=0 so that (0,0) is an eq. pt, and suppose that (26) has a limit cycle. For small &, we expect a sol" of (24) of the form (25) will be a good approximation to the lima't cycle. * (t) = y(t) = -a sint} (27) some for and the period T = 2TT. Recall from (18) the total s E = 1 (2)' + fx dx. - 1(2)? + 12+ From (186) the change in energy period E(T- E(0) = 1 de dt 1-3 (t) en (x64), 3(t)) dt. (275) Over One at EC) - Edt th), (+) (216) On Lintas tiesak pema) ELTIES) ") kw),160410 Statuting (2) inte. This gives mint hlocest) 4 (2) Ver NL a Calcast.dt. xit) cast and 21 It follows that the energy changes for Ost salt, by (276) (27) - Elo) - f -yehla,y) dt a sint hlacost, -asint) at TE energy gaine SO 1 on the limit cycle let asa. gla.) -0. For a stable limit cycle. energy is gained in the interior e acao, energy is lost in the exterior i asa.. Thus for some so limit eyele g(a) >o for a E(90-8, as) g(a) gbalco for a (@-8.co), rint g(a) so for ae (ao, ats). ext. to In fact, the conditions above are equivalent g(6) & g' (as) co stable, g@) =0 + g(20) > > unistable In fact, the conditions above are equivalent to g(@o) 20 #g'(as) o unstable. Example CD g(a) - sea sint hlacost, -asint) at {a sint [-(acos't-1) asint] dt. -24?]. (a? cos?t-1) sindt at =-{TT (722-1). 11 for aO g() 0 Thus the limit cycle (with a 2) is stable. The converse is true if eco. Alternatively gi'(a) = (-Etrat + ETra?) =- ETa' + 2 Ella = Ella (2-a). So g' (2) = -4 ET go(2) 0 gi (2) >0 for eso frida Writing gla) - safnsint dt, we get gisa) - ef h 2T hsint at tead hsint at U Lecture 13 14.... =eta" + 2 Ela = ETTa(2-a). So g' (2) = -4 ET g'(a) 0 for eso for & O. Writing gla) = ca fa esinedt, get gi(a) -1 2 1 hsint at + ad fhsint at De rute 21 2T - da o parte = 2T ah at ah sint at bring derivati Tea Using an ah ax + ah ay at axat ay y 26 s-asintan -a cost an a ay ah ah ax + ahay da . 2 . ayaa = cost asintah . ay costsint terms Car g'(a) = -ca ah dt. ay Example CD hix,y) = (ac-1) y von der Falegn SO ak - 2-1 ay Now g' (2) = - 2 S (2cost) P-7 dt = - 4TE as before 7. Using the appropriate Theorem from lectures, show that the equation T + (x - 1)i +2 +10, has a unique periodic solution if x > 0. 2 marks] 8. For small amplitudes the motion of a soft spring is described by Dufling's equation: where (t) is the displacement, y(t) is the velocity and v is the damping coefficient. Investigate the stability of the equilibrium points for all real valued v. Determine and classify any bifurcation points, and draw, and label, a bifurcation diagram to illustrate your results (6 marks] 7. Using the appropriate Theorem from lectures, show that the equation T + (x - 1)i +2 +10, has a unique periodic solution if x > 0. 2 marks] 8. For small amplitudes the motion of a soft spring is described by Dufling's equation: where (t) is the displacement, y(t) is the velocity and v is the damping coefficient. Investigate the stability of the equilibrium points for all real valued v. Determine and classify any bifurcation points, and draw, and label, a bifurcation diagram to illustrate your results (6 marks] convinetto | Averaging methods Recall x = 0 (25) This has sol's alt) Tacost 7 (25) zlt) = y(t) = -a sint J is ciroles centred on the origin Now we consider * + ghlac, 3) + x = 0, (26) for Elka1 negating Housing) The phase paths will no longer be circles (in general) - we expect unsteady spirals and perhaps a lima't cycle. su Suppose k(0,0)=0 so that (0,0) is an eq. pt, and suppose that (26) has a limit cycle. For small &, we expect a sol" of (24) of the form (25) will be a good approximation to the lima't cycle. * (t) = y(t) = -a sint} (27) some for and the period T = 2TT. Recall from (18) the total s E = 1 (2)' + fx dx. - 1(2)? + 12+ From (186) the change in energy period E(T- E(0) = 1 de dt 1-3 (t) en (x64), 3(t)) dt. (275) Over One at EC) - Edt th), (+) (216) On Lintas tiesak pema) ELTIES) ") kw),160410 Statuting (2) inte. This gives mint hlocest) 4 (2) Ver NL a Calcast.dt. xit) cast and 21 It follows that the energy changes for Ost salt, by (276) (27) - Elo) - f -yehla,y) dt a sint hlacost, -asint) at TE energy gaine SO 1 on the limit cycle let asa. gla.) -0. For a stable limit cycle. energy is gained in the interior e acao, energy is lost in the exterior i asa.. Thus for some so limit eyele g(a) >o for a E(90-8, as) g(a) gbalco for a (@-8.co), rint g(a) so for ae (ao, ats). ext. to In fact, the conditions above are equivalent g(6) & g' (as) co stable, g@) =0 + g(20) > > unistable In fact, the conditions above are equivalent to g(@o) 20 #g'(as) o unstable. Example CD g(a) - sea sint hlacost, -asint) at {a sint [-(acos't-1) asint] dt. -24?]. (a? cos?t-1) sindt at =-{TT (722-1). 11 for aO g() 0 Thus the limit cycle (with a 2) is stable. The converse is true if eco. Alternatively gi'(a) = (-Etrat + ETra?) =- ETa' + 2 Ella = Ella (2-a). So g' (2) = -4 ET go(2) 0 gi (2) >0 for eso frida Writing gla) - safnsint dt, we get gisa) - ef h 2T hsint at tead hsint at U Lecture 13 14.... =eta" + 2 Ela = ETTa(2-a). So g' (2) = -4 ET g'(a) 0 for eso for & O. Writing gla) = ca fa esinedt, get gi(a) -1 2 1 hsint at + ad fhsint at De rute 21 2T - da o parte = 2T ah at ah sint at bring derivati Tea Using an ah ax + ah ay at axat ay y 26 s-asintan -a cost an a ay ah ah ax + ahay da . 2 . ayaa = cost asintah . ay costsint terms Car g'(a) = -ca ah dt. ay Example CD hix,y) = (ac-1) y von der Falegn SO ak - 2-1 ay Now g' (2) = - 2 S (2cost) P-7 dt = - 4TE as before