RENDICONTI DEL SEMINARIO MATEMATICO

104 H. Koçak - K. Palmer - B. Coomesassumption of no floating point errors, we obtain(2)u k+1 − A k u k = S ∗ k+1 g k for k = 0,..., N − 1u 0 − A N z N = S ∗ 0 g Nands k = −‖ f (y k+1 )‖ −2 f (y k+1 ) ∗ {Dφ h k(y k )S k u k + g k } for k = 0,..., N − 1s N = −‖ f (y 0 )‖ −2 f (y 0 ) ∗ {Dφ h N(y N )S N u N + g N },where A k = Sk+1 ∗ Dφh k(y k )S k for k = 0,..., N − 1 and A N = S0 ∗ Dφh N(y N )S N . Sothe main problem is to solve Eq. (2). This is solved by exploiting the local hyperbolicityalong the pseudo orbit which implies the existence of contracting and expanding directions.We use a triangularization procedure which enables us to solve forward firstalong the contracting directions and then backwards along the expanding directions.Note that it is numerically impossible to solve the whole system forwards because ofthe expanding directions.Once the new pseudo periodic orbit {y k + z k }k=0 N with associated times {h k +s k }k=0 N is found, we check if its delta is small enough. If it is not, we repeat the procedurefor further refinement. For complete details see [19]. A similar method to refine acrude pseudo homoclinic orbit is given in [22].(ii) Verifying the invertibility of the operator (or finding a suitable right inverse)and calculating an upper bound on the norm of the inverse: Again we just look atthe periodic case. A similar procedure is used in the finite-time and homoclinic casesbut in the homoclinic case it is rather more complicated since the sequence spaces areinfinite-dimensional; however, we can handle it due to the periodicity at both ends.First we outline the procedure which would be used in the case of exact computations.To construct L −1y , we need to find the unique solution z k ∈ Y k ofz k+1 = P k+1 Dφ h k(y k )z k + g k , for k = 0,..., N − 1z 0 = P N Dφ h k(y k )z N + g N .whenever g k is in Y k+1 for k = 0,..., N−1 and in Y 0 for k = N. We use the n×(n−1)matrices S k as defined in (i) and make the transformationz k = S k u k ,where u k is in IR n−1 . Making this transformation, our equations becomeu k+1 − A k u k = S ∗ k+1 g k, for k = 0,..., N − 1u 0 − A N u N = S ∗ 0 g N,where A k is the (n − 1)×(n − 1) matrix A k = S ∗ k+1 Dφh k(y k )S k for k = 0,..., N − 1and A N = S ∗ 0 Dφh N(y N )S N .As in (i), these equations are solved by exploiting the hyperbolicity which impliesthe existence of contracting and expanding directions.