(4p) See Theorem on pages in the course book. 3. Consider the following system of ODE: z 2. n j 1

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1 MATEMATIK Datum -8- Tid eftermiddag GU, Chalmers Hjälpmedel inga A.Heintz Telefonvakt Aleei Heintz Tel Tenta i ODE och matematisk modellering, MMG, MVE6 Answer rst those questions that look simpler, then take more complicated ones etc. Good luck!. Formulate and give a proof to the Picard-Lindelöf theorem on eistence and uniquness of solutions to initial value problem for sstems of ODE. See Theorem..6 on pages pagest - in the course book.. Formulate and give a proof to the theorem about stabilit of linear sstems of ODE with periodic coe cients. See Theorem.. on pages 6-66 in the course book.. Consider the following sstem of ODE d! r (t) dt 4 = A! r (t), with a constant matri A = 4 = 4 Give general solution to the sstem. bounded solutions to the sstem. 4 4 = 4 Find all those initial vectors! r =! r () that give Solution. General solution is can be given in two was as! r (t) = ep(ta)! r for arbitrar vector! r of initial data at t =, alternativl as a linear combination of columns in an arbitrar fundamental matri solution (t). Practical calculation of ep(ta)! r is based on the idea of considering ep(ta)! r = e jt e t(a j)! r for an eigenvalue j. It is eas to observe that the epression ep(ta)! X t k r = k! (A j) k! r will be nite sum for! r being a linear combination of the eigenvectors and generalized eigenvectors to A corresponding to the eigenvalue j or! r M( j ; A) ep(ta)! r = n j X k= k= t k k! (A j) k! r The set of all eigenvectors and generalized eigenvectors to A build a basis to the space C n and therefore general solution ep(ta)! r can be eplicitel epressed as a sum of the from as above for all eigenvalues of A We remind that generalised eigenvalues v j;(k) satisf the following equations Av j;(k) = j v j;(k ) for k >, and Av j;() = j v j where v j is a eigenvector corresponding to j and v j;(k) are associated generalized eigenvectors. For the given matri A, the characteristic polnomial is + = the characteristic values are = (with multiplicit ) and = (simple). Eigenvectors satisf equations Av = 4 = z, and (A + I)v = 4 4 z =. Or equivalentl.

2 8 + + z = + z = + z = 8, It is eas to see that v = z = + + z = + 4z = and v = 4 There is one generalized eigenvector v ;() corresponding to the eigenvalue = because it has multiplicit. The dimension of the space is and we have alread got two linearl independent eigenvectors corresponding to two di erent eigenvalues and 8 (A I) v ;() = v or + + z = + z = + z = z = + ; and the generalized eigenvector can be chosen (not uniquel!) as v ;() = = + + = ; = 4 = General solution for! r = C v + C v ;() + C v with arbitrar C, C, C is epressed as! r (t) = C v + (I + At) C v ;() e t + C v e t = C v + C v ;() + tv + C v e t or in terms of coordinates (t) = C + C ( + t) + C e t, (t) = C + C ( + t), z(t) = C + C (= + t) + C e t We point out that one can get di erent equivalent epressions of general solution depending on how one chooses eigenvectors and corresponding generalized eigenvectors. The onl initial data giving bounded solutions are vectors are linear combinatoins eigenvectors () = C v + C v. 4. The following sstem of equations describes the evolution of numbers and of two competeng species. = ( ) = ( ) Eplain b analsing their equilibrium points, how these equations make it mathematicall possible but etremel unlikel for both species to survive. Solution. There is onl one possible nonzero equilibrium point =, =. It is the onl possible point for both species to survive. We tr to analse stabilit of this point using linearization. The variational matri is A(; ) = = ; (;)=(;) Characteristic equation + = p It s eigenvectors and eigenvalues in the point (; ) = (; ) are v = $ = p ; v = p $ = p > ; The linearized sstem has a saddle point in the origin, that is hperbolic because both eigenvalues have nonzero real part. General solution to the linearized sstem is r = C e (p+)t v + C e (p )t v +. The onl initial data giving solutions tending to the origin with t! are those on the line r = C v.

3 The Grobman-Hartmann theorem states that in the neighbourhood of the point (; ) the phase portrait of the original nonlinear sstem is homeomor c to one of the linear sstem =. Therefore there is a curve through the equilibrium point (; ) such that evolution of the original sstem starting on this curve leads to the equilibrium point. All other trajectories escape this equilibrium point as the do for the linearized equation. Another possibilit for both species to survive could be rising in nitel. One can see that it is impossible from the following analsis. It is eas to observe that coordinate aises consist of trajectories to the sstem = ( ) and = ( ) that have for > and > the onl attracting equilibrium points (; ) and (; ). The right hand of the ODE side is a smooth function, therefore solutions are unique and trajectories starting in the rst quadrant cannot cross coordinate aises and will sta in the rst quadrant, (t); (t) >. It implies that ( ) and ( ) everwhere in the rst quadrant and implies that (t); (t) are bounded for t!.. Consider the sstem of ODE = = 6 i) Using the test function V (; ) = + + show that the origin is asmptoticall stable and ii) Estimate the domain of attraction for the equilibrium point. Solution. The sstem has two equilibrium points ( ; ) and (; ). V f (; ) = V ((t); (t)) = =. The epression + for ( Therefore V is a Lapunov function in half plane > point. ; because + = ( + ), see the picture. and the origin is a stable equilibrium graph of the polnomial + To show that the origin is also an asmptoticall stable equilibrium we appl the La Salle theorem. On the set V f () where V f (; ) = is the -ais, we have =, = 6 = (+). All trajectories starting at points of the ais with > eept the equilibrium point (; ) cross it and cannot belong to it. Therefore this hal ine > includes no! - invariant sets eept the equilibrium point (; ). Therefore the origin is asmptoticall stable. The largest of the level set of the Lapunov function V including onl one equilibrium is one going through the second equilibrium point ( ; ) namel V ( ; ) = 4. This domain is

4 bounded b the closed curve + + = 4. This curve is smmetric with respect to the - ais and can be represented b two graphs () = p 4 p with [ ; ], ( ) = () = ; () = p, see picture. The domain inside this curve is positivel invariant set because of V f (; ) = and evidentl V decreases inside the level set with equation + + = 4 Therefore La Salle invariance principle implies that the origin is asmptoticall stable. Trajectories starting inside the identi ed! - invariant set are attracted to the origin with t! Domain of attraction i) Formulate de nition of an! -limit set!(p) of a point p for a continuous dnamical sstem. ii) What general properties have! -limit sets of continuous dnamical sstems? iii) Formulate the Poincare- Bendion theorem and use it to show that the following sstem of ODE has an! - limit set that is a periodic orbit. = + + = + Solution. i)!(p) is an! -limit set of a point p for a continuous dnamical sstem (t; ) if for an point z!(p) there is a sequence of times ft n g n= depending on z, such that t n! and n! (t n ; p)! z. n! ii) If the trajector ([; ]; ) of the dnamical sstem is bounded, then the! -limit set!(p) is not empt, closed, connected and positivel (or!) - invariant set. iii) Poincare Bendison theorem states that if an ODE in plane has a positivel invariant set U without equilibrium points then it must include a periodic orbit that is an! - limit set for all points in thes set U. We tr to use a simple test function V (; ) = + to localize a positivel invariant set without equilibrium points. V ((t); (t)) = = + + It implies that for large enough + + 6= we have V ((t); (t)) >. we have V ((t); (t)) and for small enough 4

5 Therefore there are and R; R such that the ring p + R is a positivel invariant set and therefore must include at least one periodic orbit that is an! - limit set for all points in this ring = + + ( ) + + () = Ma 4 points; Thresholding for marks for GU VG 9 points; G points. For Chalmers points; 4 7 points; points; One must pass both the home assignments and the eam to pass the course. Total points for the course will be the average of the points for the home assignments (%) and for this eam (7%). The same thresholding is valid for the eam, for the home assignments, and for the total points.