By Paul H. Rabinowitz

The papers during this quantity signify the complaints of the complex Seminar on purposes of Bifurcation conception held in Madison on October 27-29, 1976. aside from the survey by means of M. G. Crandall, the papers are released within the order during which they're awarded.

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**Example text**

Let t0 ∈ I, since z(t0 ) is an n × 1 constant vector and φ1 (t0 ), φ2 (t0 ), · · · , φn (t0 ) are linearly independent n × 1 constant vectors, there are constants a1 , a2 , · · · , an such that a1 φ1 (t0 ) + a2 φ2 (t0 ) + · · · + an φn (t0 ) = z(t0 ). 3). for t ∈ I. 2. The Vector Equation x′ = A(t)x 31 First we will see how to solve the vector differential equation x′ = Ax, where A is a constant n × n matrix. We recall the definitions of eigenvalues and eigenvectors for a square matrix A. 12 Let A be a given n × n constant matrix and let x be a column unknown n-vector.

8) is given by x(t) = c1 cos(2t) −3 cos(2t) − 2 sin(2t) + c2 sin(2t) 2 cos(2t) − 3 sin(2t) , for t ∈ R. 2 involving two masses attached to springs for the special case that all the parameters are equal to one. In this case we have 0 1 0 0 −2 −1 1 0 . A= 0 0 0 1 1 0 −2 −1 By expanding det(A − λI) along the first row, we get the characteristic equation 0 = det(A − λI) = = = λ(λ + 1)(λ2 + λ + 2) + 2(λ2 + λ + 2) − 1 (λ2 + λ + 2)2 − 1 (λ2 + λ + 1)(λ2 + λ + 3). Hence the eigenvalues of A are √ √ 3 11 1 1 i, − ± i.

Proof Let ψ1 , ψ2 , · · · , ψn be n linearly independent constant n × 1 vectors and let t0 ∈ I. Then let φi be the solution of the IVP x′ = A(t)x, x(t0 ) = ψi , for 1 ≤ i ≤ n. Assume c1 , c2 , · · · , cn are constants such that c1 φ1 (t) + c2 φ2 (t) + · · · + cn φn (t) = 0, for all t ∈ I. Letting t = t0 we have c1 φ1 (t0 ) + c2 φ2 (t0 ) + · · · + cn φn (t0 ) = 0 30 2. Linear Systems or, equivalently, c1 ψ1 + c2 ψ2 + · · · + cn ψn = 0. Since ψ1 , ψ2 , · · · , ψn are n linearly independent constant vectors, we have that c1 = c2 = · · · = cn = 0.