Linear Algebra

D.6 General Geometric Theory

Here I will consider the case where the matrix A has both positive and negative eigenvalues. First here is a useful lemma.

Proof: I want an estimate on the solutions to the system

Φ′(t) = AΦ (t),Φ (0) = I.

for t < 0. Let s = −t and let Ψ

= Φ. Then writing this in terms of Ψ,

Ψ′(s) = − A Ψ (s),Ψ (0) = I.

Now the eigenvalues of −A have real parts less than −δ because these eigenvalues are obtained from the eigenvalues of A by multiplying by −1. Then by Theorem D.5.3 there exists a constant, C such that for any x,

|Ψ (s)x| ≤ Ce− δs|x|.

Therefore, from the definition of Ψ,

|Φ (t) x| ≤ Ceδt|x |.■

Here is another essential lemma which is found in Coddington and Levinson [6]

Proof: Suppose the largest exponent of any of the p_j is M and let λ_j = a_j + ib_j. First assume each a_j = 0. This is convenient because σ = 0 in this case and the largest of the Re

occurs in every λ_j.

Then arranging the above sum as a sum of decreasing powers of t,

f (t) = tM fM (t)+ ⋅⋅⋅+ tf1(t)+ f0(t).

Then

( ) t−Mf (t) = fM (t)+ O 1 t

where the last term means that tO

is bounded. Then

∑m ibjt fM (t) = cje j=1

It can’t be the case that all the c_j are equal to 0 because then M would not be the highest power exponent. Suppose c_k≠0. Then

1 ∫ T ∑m 1 ∫ T lim -- t−M f (t)e−ibktdt = cj-- ei(bj−bk)tdt = ck ⁄= 0. T→∞ T 0 j=1 T 0

Letting r =

, it follows > r for arbitrarily large values of t. Thus it is also true that > r for arbitrarily large values of t.

Next consider the general case in which σ is given above. Thus

−σt ∑ bjt e f (t) = pj(t)e + g(t) j:aj=σ

where lim_t→∞g

= 0, g being of the form ∑ _sp_se

(as−σ+ibs)

t where a_s − σ < 0. Then this reduces to the case above in which σ = 0. Therefore, there exists r > 0 such that

||−σt || e f (t) > r

for arbitrarily large values of t. ■

Next here is a Banach space which will be useful.

Proof: Let

be a Cauchy sequence in E_γ. Then since BC is a Banach space, there exists y ∈ BC such that e^γtx_k converges uniformly on [0,∞) to y. Therefore e^−γte^γtx_k = x_k converges uniformly to e^−γty on [0,∞). Define x ≡ e^−γty. Then y = e^γtx and by definition,

||xk − x ||γ → 0.■