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Chapter 12 Nonlinear equations with time-variant linear parts
Chapter 12
Nonlinear Equations with Time-Variant Linear Parts 12.1
Equations with general linear parts
Let X be a Banach space. Recall that Ω(R) = {x ∈ X : kxk ≤ R} for a positive R ≤ ∞. Consider in X the initial problem x(k + 1) = A(k)x(k) + F (x(k), k) (k = 0, 1, 2, ....) ,
(1.1)
x(0) = x0 ∈ X
(1.2)
where A(k) are linear operators and F (., k) : Ω(R) → X (k = 0, 1, 2, ...) are given continuous functions. Assume that there are nonnegative constants ν = ν(R) and l = l(R), such that kF (h, k)k ≤ ν khk + l (h ∈ Ω(R); k = 0, 1, 2, ...). Denote by U (t, j) the evolution operator of the linear part v(k + 1) = A(k)v(k) (k = 0, 1, 2, ...) of (1.1). That is, U (t, j) = A(t − 1)A(t − 2) · · · A(j) (t > j ≥ 0); U (t, t) = I. Recall that I is the unit operator. 173
(1.3)
174
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Theorem 12.1.1 Let the conditions (1.3), η0 := sup t≥1
t X
kU (t, k)k < 1/ν,
(1.4)
k=1
and χ0 kx0 k + lη0 < R(1 − νη0 )
(1.5)
hold, where χ0 := sup kU (t, 0)k. t≥0
Then a solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(t)k ≤
χ0 kx0 k + lη0 (t ≥ 0). 1 − νη0
(1.6)
This theorem is proved in the next section. It immediately yields Corollary 12.1.2 Let the conditions kF (h, k)k ≤ ν khk (h ∈ Ω(R); k = 0, 1, 2, ...)
(1.7)
and (1.4) hold. Then the zero solution to equation (1.1) is stable. Moreover, any vector x0 satisfying the inequality χ0 kx0 k < R(1 − νη0 ), implies that estimate (1.6) is true with l = 0. Corollary 12.1.3 Let the conditions kF (h, k)k ≤ l (h ∈ Ω(R); k = 0, 1, 2, ...) and η0 = sup t≥1
t X
kU (t, k)k < ∞
k=1
hold. Then a solution x(t) of problem (1.1), (1.2) is bounded, provided χ0 kx0 k + η0 l < R. Moreover, kx(t)k ≤ χ0 kx0 k + η0 l (t ≥ 0).
(1.8)
12.1. EQUATIONS WITH GENERAL LINEAR PARTS
175
Let there be constants M1 ≥ 1 and c0 ∈ (0, 1), such that kU (t, k)k ≤ M1 ct−k (t ≥ k ≥ 0). 0
(1.9)
Then we have t X
kU (t, k)k ≤ M1
k=1
t X
ct−k = M1 0
t−1 X
k=1
cj0 ≤
j=0
M1 . 1 − c0
Now the previous theorem and the substitution x(t) = (1 − )t y(t) with a small enough > 0, imply Corollary 12.1.4 Let the conditions (1.7), (1.9) and νM1 + c0 < 1 hold. Then the zero solution to equation (1.1) is exponentially stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(t)k ≤
M1 (1 − c0 )kx0 k (t ≥ 0), 1 − c0 − νM1
provided M1 (1 − c0 )kx0 k < R(1 − c0 − νM1 ). Let there be a positive constant a < 1, such that kA(t)k ≤ a (t ≥ 0). Then kU (t, k)k ≤ at−k and thus the previous corollary yields the stability conditions with c0 = a, M1 = 1. Let A(t) be a normal, in particular Hermitian, operator in a Hilbert space H. Then kA(t)kH = rs (A(t)). Here k.kH is the norm in H. In this case Corollary 12.1.4 implies Corollary 12.1.5 Under condition (1.7) with ν < 1, for all t ≥ 0 let A(t) be a normal operator. In addition, let ρ0 := sup rs (A(t)) < 1 − ν. t≥0
176
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Then the zero solution to equation (1.1) is exponentially stable. Moreover, any solution x(t) of (1.1) satisfies the inequality kx(t)kH ≤
(1 − ρ0 )kx0 kH (t ≥ 0), 1 − ν − ρ0
provided (1 − ρ0 )kx0 kH < R. 1 − ν − ρ0 We will say that (1.1) is a quasilinear equation if kF (h, t)k =0 h→0 khk lim
uniformly in t. Corollary 12.1.6 (Stability by linear approximation) Let (1.1) be a quasilinear equation and t X sup kU (t, k)k < ∞. t≥1
k=1
Then the zero solution to (1.1) is stable. Moreover, if condition (1.9) holds and (1.1) is a quasilinear equation, then the zero solution to equation (1.1) is exponentially stable. This result immediately follows from Corollaries 12.1.2 and 12.1.4, if we take R sufficiently small.
12.2
Proof of Theorem 12.1.1
Due to the Variation of Constants Formula x(t) = U (t, 0)x(0) +
t−1 X
U (t, k + 1)F (x(k), k).
k=0
Hence, kx(t)k ≤ kU (t, 0)x(0)k +
t−1 X
kU (t, k + 1)F (x(k), k)k.
k=0
First let condition (1.3) holds with R = ∞. Then kx(t)k ≤ kU (t, 0)x(0)k +
t−1 X k=0
kU (t, k + 1)k(νkx(k)k + l).
12.3. SLOWLY VARYING EQUATIONS IN BANACH SPACES
177
Thus sup kx(t)k ≤ χ0 kx(0)k + η0 (l + ν sup kx(t)k). t≥0
t≥0
Now condition (1.4) yields inequality (1.6). Now let R < ∞. According to the Urysohn theorem let us define the function F (x, k) , kxk ≤ R, e f (x, k) = 0 , kxk > R. Since kfe(x, k)k ≤ ν kxk (k = 0, 1, ...; x ∈ X), ∞
then the sequence {e x(k)}=0 defined by x e(0) x e(k + 1)
= x(0) and = A(k)e x(k) + fe(e x(k), k), k = 0, 1, ...
satisfies the inequality sup ke x(k)k ≤ k=0,1,...
χ0 kx0 k + lη0
according to the above arguments and condition (1.5). But F and fe coincide on Ω(R). So x(k) = x e(k) for k = 0, 1, 2, .... Therefore, (1.6) is satisfied, concluding the proof. Q. E. D.
12.3
Slowly varying equations in a Banach space
Let us suppose that there is a scalar sequence q(t) (t = 0, ±1, ±2, ...), such that q(k) = q(−k) ≥ 0 (k = 1, 2, ...) and q(0) = 0, and kA(k) − A(j)k ≤ q(k − j) (j, k = 0, 1, 2, ... ).
(3.1)
kF (h, k)k ≤ ν khk (h ∈ Ω(R); k = 0, 1, 2, ...)
(3.2)
Assume that and β0 :=
sup k,s=0,1,...
k
A (s) < ∞.
178
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Theorem 12.3.1 Under conditions (3.1) and (3.2), let S0 :=
∞ X
(q(k) + ν)
sup s=0,1,2,...
k=0
k
A (s) < 1.
Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)k ≤
β0 kx0 k (k = 1, 2, ...), 1 − S0
(3.3)
provided β0 kx0 k < R(1 − S0 ).
(3.4)
This theorem is proved in the next section. Furthermore, let there be a positive constant q˜, such that kA(k + 1) − A(k)k ≤ q˜ (k = 0, 1, 2, ...).
(3.5)
Then condition (3.1) holds with q(k) = |k|˜ q (k = 0, ±1, ±2, ...). Now the previous theorem implies Corollary 12.3.2 Under conditions (3.2) and (3.5), let S˜ :=
∞ X
(k q˜ + ν)
sup s=0,1,2,...
k=0
k
A (s) < 1.
Then the zero solution to equation (1.1) is stable. Moreover, if an initial vector x0 , satisfies the inequality ˜ β0 kx0 k < R(1 − S), then the corresponding solution x(t) of problem (1.1), (1.2) is subject to the estimate β0 kx0 k kx(k)k ≤ (k = 1, 2, ...). 1 − S˜
12.4
Proof of Theorem 12.3.1
Rewrite equation (1.1) as x(k + 1) − A(s) x(k) = (A(k) − A(s)) x(k) + F (x(k), k) with a fixed integer s. The Variation of Constants Formula yields x(m + 1) = Am+1 (s)x(0) +
m X j=0
Am−j (s)[(A(j) − A(s)) x(j) + F (x(j), j)].
12.4. PROOF OF THEOREM 12.3.1
179
Take s = m: x(m + 1) = Am+1 (m)x(0) +
m X
Am−j (m)[(A(j) − A(m)) x(j) + F (x(j), j)].
j=0
There are two cases to consider: R = ∞ and R < ∞. First, assume that (1.3) is valid with R = ∞, then by (3.1) and (3.2) kx(m + 1)k ≤ β0 kx(0)k +
m X
m−j
A (m) [q(m − j) kx(j)k + ν kx(j)k] ≤ j=0
β0 kx(0)k +
m X
m−j
A (m) (q(m − j) + ν) kx(j)k j=0
≤ β0 kx(0)k + max kx(k)k k=0,...,m
≤ β0 kx(0)k + max kx(k)k k=1,...,m
m X
k
A (m) (q(k) + ν) k=0
∞ X
(q(k) + ν)
sup s=0,1,2,...
k=0
!
k
A (s) .
Thus, max
k=0,...,m+1
kx(k)k ≤ β0 kx(0)k + S0
max
k=0,...,m+1
kx(k)k .
Hence, by (1.4) we arrive at the inequality sup
kx(k)k ≤
k=0,...,m+1
β0 kx(0)k . 1 − S0
But the right-hand part of this inequality does not depend on m. So we get the required inequality (3.3). Now let R < ∞. Define the function F (x, k) , kxk ≤ R fe(x, k) = 0 , kxk > R . Since
e
f (x, k) ≤ ν kxk (k = 0, 1, ...; x ∈ X ), ∞
then the sequence {e x(k)}=0 defined by x e(0) x e(k + 1)
= x(0) and = A(k)e x(k) + fe(e x(k), k), k = 0, 1, ...
satisfies the inequality sup ke x(k)k ≤ k=0,1,...
β0 kx(0)k
180
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
according to the above arguments and condition (3.4). But F and fe coincide on Ω(R). So x(k) = x e(k) for k = 0, 1, 2, .... Therefore, (3.3) is satisfied. This concludes the proof. Q. E. D.
12.5
Slowly varying equations in a Hilbert space
Let X = H be a separable Hilbert space with a norm k.kH . In the present section we make the results of the previous section sharper under the condition A(s) − A∗ (s) ∈ C2 (H) (s = 0, 1, 2, ...). (5.1) Recall that gI (A) is defined in Section 4.5 and √ g(AI ) ≤ 2N2 (A − A∗ ). Assume that ρ0 :=
sup
rs (A(m)) < 1
(5.2)
gI (A(m)) < ∞.
(5.3)
m=0,1,...
and v0 :=
sup m=0,1,2,...
Due to Corollary 4.5.4 kAm (t)kH ≤
m X
m!v0k ρm−k 0 (m, t = 0, 1, 2, ...). 3/2 (m − k)!(k!) k=0
Put ˜ := M
(5.4)
m X
m!v0k ρm−k 0 . 3/2 m=0,1,2,... (m − k)!(k!) k=0 max
Now the Theorem 12.3.1 implies Theorem 12.5.1 Let inequalities (3.1) and (3.2) be fulfilled with k.k = k.kH -the norm in H. In addition, let the conditions (5.1)-(5.3) and sH (F, A) :=
∞ X k X
ρk−j v j Ckj 0√ 0 (q(k) + ν) < 1 j! k=0 j=0
hold. Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)kH ≤
˜ kx(0)k M H (t ≥ 0), 1 − sH (F, A)
provided ˜ kx0 k < R(1 − sH (F, A)). M H
12.5. SLOWLY VARYING EQUATIONS IN A HILBERT SPACE
181
Recall that Ckj are the binomial coefficients. Let us turn now to condition (3.5). First we will prove the inequality ∞ X
(k q˜ + ν)
sup m=0,1,2,...
k=0
where θ˜ :=
∞ X
√
k=0
and θ1 :=
∞ X
v0k (k + 1) k!(1 − ρ0 )k+2
√
k=0
k
A (m) ≤ q˜θ˜ + νθ1 , H
v0k . k!(1 − ρ0 )k+1
Indeed the inequality ∞ X k=1
k
sup m=0,1,2,...
k
A (m) ≤ θ˜ H
is proved in Section 10.3. So we need only to check that ∞ X k=0
sup m=0,1,2,...
k
A (m) ≤ θ1 . H
To this end note that from (5.4) it follows that ∞ X
kAm (s)kH ≤
m=0
∞ X ∞ X
m!v0k ρm−k 0 = 3/2 (m − k)!(k!) m=0 k=0
∞ X
∞ v0k X m!ρm−k 0 (s ≥ 0). 3/2 (m − k)! (k!) m=0 k=0
But
∞ ∞ X m!xm−k dk X m x = = k (m − k)! dx m=0 m=0
dk k! (1 − x)−1 = (x ∈ (0, 1)). k dx (1 − x)k+1 Thus
∞ X m!ρm−k (A) 0 = (m − k)! m=0
=
k! . (1 − ρ0 )k+1
This relation proves inequality (5.5). Now Corollary 12.3.2 yields
(5.5)
182
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Corollary 12.5.2 Let inequalities (3.2) and (3.5) be fulfilled with k.k = k.kH -the Hilbert norm. In addition, let the conditions (5.1)-(5.3) and sH1 (F, A) := q˜θ˜ + νθ1 < 1 hold. Then the zero solution to equation (1.1) is stable. Moreover, any initial vector x0 , satisfying the condition ˜ kx0 k M H < R, 1 − sH1 (F, A) implies that the corresponding solution x(t) of (1.1) subjects the estimate kx(t)kH ≤
12.6
˜ kx0 k M H (t = 0, 1, ...). 1 − sH1 (F, A)
The finite dimensional case
In the present section X = Cn with the Euclidean norm k.kC n and A(m) (m = 0, 1, 2, ...) are n × n-matrices. Recall that g (A) is defined in Section 3.2. It is supposed that the conditions (5.2) and vn :=
sup
g (A(m)) < ∞
(6.1)
m=0,1,2,...
hold. As it was proved in Section 10.4, β0 = supk,s=0,1,... Ak (s) C n ≤ Mn , where Mn := 1 +
n−1 X
k (ψk + k)k ρψ 0
k=1
with ψk = max{0, −k
vnk (k!)3/2
ln (eρ0 ) }. ln (ρ0 )
Due to Corollary 3.2.4 kAm (t)kC n ≤
n−1 X k=0
m!vnk ρm−k 0 (t, m = 0, 1, ...). (m − k)!(k!)3/2
(6.2)
Now Theorem 12.3.1 implies Theorem 12.6.1 Let X = Cn and inequalities (3.1) and (3.2) be fulfilled. In addition, let the conditions (5.2), (6.1) and sC n (F, A) :=
∞ n−1 X X
ρk−j v j Ckj 0√ n (q(k) + ν) < 1 j! k=0 j=0
12.7. EQUATIONS IN ORDERED SPACES
183
hold. Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)kC n ≤
Mn kx0 kC n (t ≥ 0) 1 − sC n (F, A)
provided Mn kx0 kC n < R. 1 − sC n (F, A) Now let us turn to condition (3.5). According to (6.2) and (5.5) ∞ X k=0
(k q˜ + ν)
sup m=0,1,2,...
where θ0n :=
k
A (m) n ≤ q˜θ0n + νθ1n , C
n−1 X
√
k=0
and θ1n :=
n−1 X k=0
√
vnk (k + 1) k!(1 − ρ0 )k+2
vnk . k!(1 − ρ0 )k+1
Now Corollary 12.3.2 and a small perturbation imply Corollary 12.6.2 Let X = Cn and inequalities (3.2), and (3.5) be fulfilled with the Euclidean norm. In addition, let the conditions (5.2), (6.1) and sC n (F, A) := q˜θ˜0n + νθ1n < 1 hold. Then the zero solution to equation (1.1) is exponentially stable. Moreover, any initial vector x0 , satisfying the condition Mn kx0 kC n < R (1 − sC n (F, A)), belongs the region of attraction, and the corresponding solution x(t) of equation (1.1) is subject to the inequality kx(t)kC n ≤
12.7
Mn kx0 kC n (t ≥ 0). 1 − sC n (F, A)
Equations in ordered spaces
Let X be a Banach lattice. Consider in X the equation x(t + 1) = Φ(x(t), t) (t ≥ 0)
(7.1)
184
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
where Φ(., t) : Ω(R) → X (0 < R ≤ ∞) is a continuous mapping. Assume that Φ has a majorant on Ω(R) . That is, there is a positive constant linear operator B, such that |Φ(h, t)| ≤ B|h| (h ∈ Ω(R); t ≥ 0). (7.2) Theorem 12.7.1 Let the conditions (7.2) and rs (B) < 1
(7.3)
hold. Then the zero solution to equation (7.1) is exponentially stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality |x(t)| ≤ B t |x0 | (t ≥ 0),
(7.4)
provided sup kB t kkx0 k < R. t≥0
The proof is obvious. For example, let X = l∞ (R), and ∞ Φ(h, t) = (fj (h))∞ j=1 (h = (hk )j=1 ∈ Ω(R)),
where fj : Ω(R) → R are continuous functions and |fj (h)| ≤
∞ X
mjk |hk |
k=1
provided khk ≤ R. So in this case B = (mjk )nj,k=1 . Let us suppose that sup j
∞ X
mjk < 1.
k=1
Then rs (B) ≤ kBk < 1. So the considered equation is exponentially stable. Besides sup kB t k = 1. t≥0
12.8
Perturbations of nonlinear equations
Consider in a Banach space X the equation x(t + 1) = Q(x(t), t)x(t) + F (x(t), t) (t ≥ 0),
(8.1)
12.8. PERTURBATIONS OF NONLINEAR EQUATIONS
185
where Q(z, t) are linear operators in X continuously dependent on z ∈ Ω(R) and F (., t) : Ω(R) → X are continuous mappings for each t = 0, 1, .... Denote by ω∞ (R) the set of sequences with values in Ω(R): ω∞ (R) := {h(t) ∈ Ω(R); t = 0, 1, ...}. Let Uh (t, s) := Q(h(t − 1), t − 1) · · · Q(h(s), s) (t ≥ s ≥ 0) with a h ∈ ω∞ (R). That is, Uh is the evolution operator of the linear equation y(t + 1) = Q(h(t), t)y(t) (t ≥ 0).
(8.2)
Let us assume that there are a positive continuous function φ(t, s) = φ(R, t, s), and nonnegative constants ν = ν(R) and l = l(R) independent of h, such that kUh (t, j)k ≤ φ(t, j) (h ∈ ω∞ (R)) and kF (w, j)k ≤ νkwk + l (w ∈ Ω(R); t, j ≥ 0).
(8.3)
Additionally, suppose that t X
ηφ := sup t=1,2,...
φ(t, k) < 1/ν
(8.4)
k=1
and put $ :=
sup
φ(t, 0).
t=0,1,2,...
Theorem 12.8.1 Let conditions (8.3) and (8.4) hold. If, in addition, an initial vector x(0) satisfies the inequality $kx(0)k + lηφ < R(1 − νηφ ),
(8.5)
then the corresponding solution x(t) of (8.1) is subject to the estimate kx(t)k ≤ Proof:
kx(0)k$ + lηφ (t = 1, 2, ...). 1 − νηφ
(8.6)
First let R = ∞. Consider the equation y(t + 1) = Q(h(t), t)y(t) + F (y(t), t) (t ≥ 0).
(8.7)
By the Variation of Constants Formula this equation is equivalent to the following one: t−1 X y(t) = Uh (t, 0)y(0) + Uh (t, k + 1)F (y(k), k) (t ≥ 0). k=0
186
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Consequently, ky(t)k ≤ kUh (t, 0)y(0)k +
t−1 X
kUh (t, k + 1)kkF (y(k), k)k (t ≥ 0).
k=0
Hence, ky(t)k ≤ φ(t, 0)ky(0)k +
t−1 X
φ(t, k + 1)[νky(k)k + l].
k=0
We thus get sup ky(t)k ≤ $ky(0)k + ηφ (ν sup ky(t)k + l). t=0,1,...
t=0,1,...
Thanks to (8.4) $ky(0)k + ηφ l . 1 − νηφ t Take h(t) = x(t)-the solution of (8.1). Then (8.1) and (8.7) coincide, and therefore (8.6) is valid. The case R < ∞ can be considered exactly as in the proof of Theorem 12.1.1. Q. E. D. sup ky(t)k ≤
Clearly the preceding theorem gives us stability and boundedness conditions. Suppose that kQ(z, t)k ≤ a < 1 (z ∈ Ω(R); t = 1, 2, ...).
(8.8)
Then kUh (t, j)k ≤ at−j (h ∈ ω∞ (R)). In this case
t X
ηφ ≤ sup t=0,1,...
at−k =
k=0
1 1−a
and $ = 1. Now Theorem 12.8.1 implies Corollary 12.8.2 Let the conditions (8.8) and (8.3) with l = 0 hold. In addition, let ν < 1. 1−a Then the zero solution of equation (8.1) is stable. If, in addition, an initial vector x(0) satisfies the inequality (1 − a)kx(0)k < R(1 − a − ν), then the corresponding solution x(t) of equation (8.1) is subject to the estimate kx(t)k ≤
1−a kx(0)k < R (t ≥ 0). 1−a−ν
Nonlinear Equations with Time-Variant Linear Parts 12.1
Equations with general linear parts
Let X be a Banach space. Recall that Ω(R) = {x ∈ X : kxk ≤ R} for a positive R ≤ ∞. Consider in X the initial problem x(k + 1) = A(k)x(k) + F (x(k), k) (k = 0, 1, 2, ....) ,
(1.1)
x(0) = x0 ∈ X
(1.2)
where A(k) are linear operators and F (., k) : Ω(R) → X (k = 0, 1, 2, ...) are given continuous functions. Assume that there are nonnegative constants ν = ν(R) and l = l(R), such that kF (h, k)k ≤ ν khk + l (h ∈ Ω(R); k = 0, 1, 2, ...). Denote by U (t, j) the evolution operator of the linear part v(k + 1) = A(k)v(k) (k = 0, 1, 2, ...) of (1.1). That is, U (t, j) = A(t − 1)A(t − 2) · · · A(j) (t > j ≥ 0); U (t, t) = I. Recall that I is the unit operator. 173
(1.3)
174
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Theorem 12.1.1 Let the conditions (1.3), η0 := sup t≥1
t X
kU (t, k)k < 1/ν,
(1.4)
k=1
and χ0 kx0 k + lη0 < R(1 − νη0 )
(1.5)
hold, where χ0 := sup kU (t, 0)k. t≥0
Then a solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(t)k ≤
χ0 kx0 k + lη0 (t ≥ 0). 1 − νη0
(1.6)
This theorem is proved in the next section. It immediately yields Corollary 12.1.2 Let the conditions kF (h, k)k ≤ ν khk (h ∈ Ω(R); k = 0, 1, 2, ...)
(1.7)
and (1.4) hold. Then the zero solution to equation (1.1) is stable. Moreover, any vector x0 satisfying the inequality χ0 kx0 k < R(1 − νη0 ), implies that estimate (1.6) is true with l = 0. Corollary 12.1.3 Let the conditions kF (h, k)k ≤ l (h ∈ Ω(R); k = 0, 1, 2, ...) and η0 = sup t≥1
t X
kU (t, k)k < ∞
k=1
hold. Then a solution x(t) of problem (1.1), (1.2) is bounded, provided χ0 kx0 k + η0 l < R. Moreover, kx(t)k ≤ χ0 kx0 k + η0 l (t ≥ 0).
(1.8)
12.1. EQUATIONS WITH GENERAL LINEAR PARTS
175
Let there be constants M1 ≥ 1 and c0 ∈ (0, 1), such that kU (t, k)k ≤ M1 ct−k (t ≥ k ≥ 0). 0
(1.9)
Then we have t X
kU (t, k)k ≤ M1
k=1
t X
ct−k = M1 0
t−1 X
k=1
cj0 ≤
j=0
M1 . 1 − c0
Now the previous theorem and the substitution x(t) = (1 − )t y(t) with a small enough > 0, imply Corollary 12.1.4 Let the conditions (1.7), (1.9) and νM1 + c0 < 1 hold. Then the zero solution to equation (1.1) is exponentially stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(t)k ≤
M1 (1 − c0 )kx0 k (t ≥ 0), 1 − c0 − νM1
provided M1 (1 − c0 )kx0 k < R(1 − c0 − νM1 ). Let there be a positive constant a < 1, such that kA(t)k ≤ a (t ≥ 0). Then kU (t, k)k ≤ at−k and thus the previous corollary yields the stability conditions with c0 = a, M1 = 1. Let A(t) be a normal, in particular Hermitian, operator in a Hilbert space H. Then kA(t)kH = rs (A(t)). Here k.kH is the norm in H. In this case Corollary 12.1.4 implies Corollary 12.1.5 Under condition (1.7) with ν < 1, for all t ≥ 0 let A(t) be a normal operator. In addition, let ρ0 := sup rs (A(t)) < 1 − ν. t≥0
176
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Then the zero solution to equation (1.1) is exponentially stable. Moreover, any solution x(t) of (1.1) satisfies the inequality kx(t)kH ≤
(1 − ρ0 )kx0 kH (t ≥ 0), 1 − ν − ρ0
provided (1 − ρ0 )kx0 kH < R. 1 − ν − ρ0 We will say that (1.1) is a quasilinear equation if kF (h, t)k =0 h→0 khk lim
uniformly in t. Corollary 12.1.6 (Stability by linear approximation) Let (1.1) be a quasilinear equation and t X sup kU (t, k)k < ∞. t≥1
k=1
Then the zero solution to (1.1) is stable. Moreover, if condition (1.9) holds and (1.1) is a quasilinear equation, then the zero solution to equation (1.1) is exponentially stable. This result immediately follows from Corollaries 12.1.2 and 12.1.4, if we take R sufficiently small.
12.2
Proof of Theorem 12.1.1
Due to the Variation of Constants Formula x(t) = U (t, 0)x(0) +
t−1 X
U (t, k + 1)F (x(k), k).
k=0
Hence, kx(t)k ≤ kU (t, 0)x(0)k +
t−1 X
kU (t, k + 1)F (x(k), k)k.
k=0
First let condition (1.3) holds with R = ∞. Then kx(t)k ≤ kU (t, 0)x(0)k +
t−1 X k=0
kU (t, k + 1)k(νkx(k)k + l).
12.3. SLOWLY VARYING EQUATIONS IN BANACH SPACES
177
Thus sup kx(t)k ≤ χ0 kx(0)k + η0 (l + ν sup kx(t)k). t≥0
t≥0
Now condition (1.4) yields inequality (1.6). Now let R < ∞. According to the Urysohn theorem let us define the function F (x, k) , kxk ≤ R, e f (x, k) = 0 , kxk > R. Since kfe(x, k)k ≤ ν kxk (k = 0, 1, ...; x ∈ X), ∞
then the sequence {e x(k)}=0 defined by x e(0) x e(k + 1)
= x(0) and = A(k)e x(k) + fe(e x(k), k), k = 0, 1, ...
satisfies the inequality sup ke x(k)k ≤ k=0,1,...
χ0 kx0 k + lη0
according to the above arguments and condition (1.5). But F and fe coincide on Ω(R). So x(k) = x e(k) for k = 0, 1, 2, .... Therefore, (1.6) is satisfied, concluding the proof. Q. E. D.
12.3
Slowly varying equations in a Banach space
Let us suppose that there is a scalar sequence q(t) (t = 0, ±1, ±2, ...), such that q(k) = q(−k) ≥ 0 (k = 1, 2, ...) and q(0) = 0, and kA(k) − A(j)k ≤ q(k − j) (j, k = 0, 1, 2, ... ).
(3.1)
kF (h, k)k ≤ ν khk (h ∈ Ω(R); k = 0, 1, 2, ...)
(3.2)
Assume that and β0 :=
sup k,s=0,1,...
k
A (s) < ∞.
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CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Theorem 12.3.1 Under conditions (3.1) and (3.2), let S0 :=
∞ X
(q(k) + ν)
sup s=0,1,2,...
k=0
k
A (s) < 1.
Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)k ≤
β0 kx0 k (k = 1, 2, ...), 1 − S0
(3.3)
provided β0 kx0 k < R(1 − S0 ).
(3.4)
This theorem is proved in the next section. Furthermore, let there be a positive constant q˜, such that kA(k + 1) − A(k)k ≤ q˜ (k = 0, 1, 2, ...).
(3.5)
Then condition (3.1) holds with q(k) = |k|˜ q (k = 0, ±1, ±2, ...). Now the previous theorem implies Corollary 12.3.2 Under conditions (3.2) and (3.5), let S˜ :=
∞ X
(k q˜ + ν)
sup s=0,1,2,...
k=0
k
A (s) < 1.
Then the zero solution to equation (1.1) is stable. Moreover, if an initial vector x0 , satisfies the inequality ˜ β0 kx0 k < R(1 − S), then the corresponding solution x(t) of problem (1.1), (1.2) is subject to the estimate β0 kx0 k kx(k)k ≤ (k = 1, 2, ...). 1 − S˜
12.4
Proof of Theorem 12.3.1
Rewrite equation (1.1) as x(k + 1) − A(s) x(k) = (A(k) − A(s)) x(k) + F (x(k), k) with a fixed integer s. The Variation of Constants Formula yields x(m + 1) = Am+1 (s)x(0) +
m X j=0
Am−j (s)[(A(j) − A(s)) x(j) + F (x(j), j)].
12.4. PROOF OF THEOREM 12.3.1
179
Take s = m: x(m + 1) = Am+1 (m)x(0) +
m X
Am−j (m)[(A(j) − A(m)) x(j) + F (x(j), j)].
j=0
There are two cases to consider: R = ∞ and R < ∞. First, assume that (1.3) is valid with R = ∞, then by (3.1) and (3.2) kx(m + 1)k ≤ β0 kx(0)k +
m X
m−j
A (m) [q(m − j) kx(j)k + ν kx(j)k] ≤ j=0
β0 kx(0)k +
m X
m−j
A (m) (q(m − j) + ν) kx(j)k j=0
≤ β0 kx(0)k + max kx(k)k k=0,...,m
≤ β0 kx(0)k + max kx(k)k k=1,...,m
m X
k
A (m) (q(k) + ν) k=0
∞ X
(q(k) + ν)
sup s=0,1,2,...
k=0
!
k
A (s) .
Thus, max
k=0,...,m+1
kx(k)k ≤ β0 kx(0)k + S0
max
k=0,...,m+1
kx(k)k .
Hence, by (1.4) we arrive at the inequality sup
kx(k)k ≤
k=0,...,m+1
β0 kx(0)k . 1 − S0
But the right-hand part of this inequality does not depend on m. So we get the required inequality (3.3). Now let R < ∞. Define the function F (x, k) , kxk ≤ R fe(x, k) = 0 , kxk > R . Since
e
f (x, k) ≤ ν kxk (k = 0, 1, ...; x ∈ X ), ∞
then the sequence {e x(k)}=0 defined by x e(0) x e(k + 1)
= x(0) and = A(k)e x(k) + fe(e x(k), k), k = 0, 1, ...
satisfies the inequality sup ke x(k)k ≤ k=0,1,...
β0 kx(0)k
180
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
according to the above arguments and condition (3.4). But F and fe coincide on Ω(R). So x(k) = x e(k) for k = 0, 1, 2, .... Therefore, (3.3) is satisfied. This concludes the proof. Q. E. D.
12.5
Slowly varying equations in a Hilbert space
Let X = H be a separable Hilbert space with a norm k.kH . In the present section we make the results of the previous section sharper under the condition A(s) − A∗ (s) ∈ C2 (H) (s = 0, 1, 2, ...). (5.1) Recall that gI (A) is defined in Section 4.5 and √ g(AI ) ≤ 2N2 (A − A∗ ). Assume that ρ0 :=
sup
rs (A(m)) < 1
(5.2)
gI (A(m)) < ∞.
(5.3)
m=0,1,...
and v0 :=
sup m=0,1,2,...
Due to Corollary 4.5.4 kAm (t)kH ≤
m X
m!v0k ρm−k 0 (m, t = 0, 1, 2, ...). 3/2 (m − k)!(k!) k=0
Put ˜ := M
(5.4)
m X
m!v0k ρm−k 0 . 3/2 m=0,1,2,... (m − k)!(k!) k=0 max
Now the Theorem 12.3.1 implies Theorem 12.5.1 Let inequalities (3.1) and (3.2) be fulfilled with k.k = k.kH -the norm in H. In addition, let the conditions (5.1)-(5.3) and sH (F, A) :=
∞ X k X
ρk−j v j Ckj 0√ 0 (q(k) + ν) < 1 j! k=0 j=0
hold. Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)kH ≤
˜ kx(0)k M H (t ≥ 0), 1 − sH (F, A)
provided ˜ kx0 k < R(1 − sH (F, A)). M H
12.5. SLOWLY VARYING EQUATIONS IN A HILBERT SPACE
181
Recall that Ckj are the binomial coefficients. Let us turn now to condition (3.5). First we will prove the inequality ∞ X
(k q˜ + ν)
sup m=0,1,2,...
k=0
where θ˜ :=
∞ X
√
k=0
and θ1 :=
∞ X
v0k (k + 1) k!(1 − ρ0 )k+2
√
k=0
k
A (m) ≤ q˜θ˜ + νθ1 , H
v0k . k!(1 − ρ0 )k+1
Indeed the inequality ∞ X k=1
k
sup m=0,1,2,...
k
A (m) ≤ θ˜ H
is proved in Section 10.3. So we need only to check that ∞ X k=0
sup m=0,1,2,...
k
A (m) ≤ θ1 . H
To this end note that from (5.4) it follows that ∞ X
kAm (s)kH ≤
m=0
∞ X ∞ X
m!v0k ρm−k 0 = 3/2 (m − k)!(k!) m=0 k=0
∞ X
∞ v0k X m!ρm−k 0 (s ≥ 0). 3/2 (m − k)! (k!) m=0 k=0
But
∞ ∞ X m!xm−k dk X m x = = k (m − k)! dx m=0 m=0
dk k! (1 − x)−1 = (x ∈ (0, 1)). k dx (1 − x)k+1 Thus
∞ X m!ρm−k (A) 0 = (m − k)! m=0
=
k! . (1 − ρ0 )k+1
This relation proves inequality (5.5). Now Corollary 12.3.2 yields
(5.5)
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CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Corollary 12.5.2 Let inequalities (3.2) and (3.5) be fulfilled with k.k = k.kH -the Hilbert norm. In addition, let the conditions (5.1)-(5.3) and sH1 (F, A) := q˜θ˜ + νθ1 < 1 hold. Then the zero solution to equation (1.1) is stable. Moreover, any initial vector x0 , satisfying the condition ˜ kx0 k M H < R, 1 − sH1 (F, A) implies that the corresponding solution x(t) of (1.1) subjects the estimate kx(t)kH ≤
12.6
˜ kx0 k M H (t = 0, 1, ...). 1 − sH1 (F, A)
The finite dimensional case
In the present section X = Cn with the Euclidean norm k.kC n and A(m) (m = 0, 1, 2, ...) are n × n-matrices. Recall that g (A) is defined in Section 3.2. It is supposed that the conditions (5.2) and vn :=
sup
g (A(m)) < ∞
(6.1)
m=0,1,2,...
hold. As it was proved in Section 10.4, β0 = supk,s=0,1,... Ak (s) C n ≤ Mn , where Mn := 1 +
n−1 X
k (ψk + k)k ρψ 0
k=1
with ψk = max{0, −k
vnk (k!)3/2
ln (eρ0 ) }. ln (ρ0 )
Due to Corollary 3.2.4 kAm (t)kC n ≤
n−1 X k=0
m!vnk ρm−k 0 (t, m = 0, 1, ...). (m − k)!(k!)3/2
(6.2)
Now Theorem 12.3.1 implies Theorem 12.6.1 Let X = Cn and inequalities (3.1) and (3.2) be fulfilled. In addition, let the conditions (5.2), (6.1) and sC n (F, A) :=
∞ n−1 X X
ρk−j v j Ckj 0√ n (q(k) + ν) < 1 j! k=0 j=0
12.7. EQUATIONS IN ORDERED SPACES
183
hold. Then the zero solution to equation (1.1) is stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality kx(k)kC n ≤
Mn kx0 kC n (t ≥ 0) 1 − sC n (F, A)
provided Mn kx0 kC n < R. 1 − sC n (F, A) Now let us turn to condition (3.5). According to (6.2) and (5.5) ∞ X k=0
(k q˜ + ν)
sup m=0,1,2,...
where θ0n :=
k
A (m) n ≤ q˜θ0n + νθ1n , C
n−1 X
√
k=0
and θ1n :=
n−1 X k=0
√
vnk (k + 1) k!(1 − ρ0 )k+2
vnk . k!(1 − ρ0 )k+1
Now Corollary 12.3.2 and a small perturbation imply Corollary 12.6.2 Let X = Cn and inequalities (3.2), and (3.5) be fulfilled with the Euclidean norm. In addition, let the conditions (5.2), (6.1) and sC n (F, A) := q˜θ˜0n + νθ1n < 1 hold. Then the zero solution to equation (1.1) is exponentially stable. Moreover, any initial vector x0 , satisfying the condition Mn kx0 kC n < R (1 − sC n (F, A)), belongs the region of attraction, and the corresponding solution x(t) of equation (1.1) is subject to the inequality kx(t)kC n ≤
12.7
Mn kx0 kC n (t ≥ 0). 1 − sC n (F, A)
Equations in ordered spaces
Let X be a Banach lattice. Consider in X the equation x(t + 1) = Φ(x(t), t) (t ≥ 0)
(7.1)
184
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
where Φ(., t) : Ω(R) → X (0 < R ≤ ∞) is a continuous mapping. Assume that Φ has a majorant on Ω(R) . That is, there is a positive constant linear operator B, such that |Φ(h, t)| ≤ B|h| (h ∈ Ω(R); t ≥ 0). (7.2) Theorem 12.7.1 Let the conditions (7.2) and rs (B) < 1
(7.3)
hold. Then the zero solution to equation (7.1) is exponentially stable. Moreover, any solution x(t) of problem (1.1), (1.2) satisfies the inequality |x(t)| ≤ B t |x0 | (t ≥ 0),
(7.4)
provided sup kB t kkx0 k < R. t≥0
The proof is obvious. For example, let X = l∞ (R), and ∞ Φ(h, t) = (fj (h))∞ j=1 (h = (hk )j=1 ∈ Ω(R)),
where fj : Ω(R) → R are continuous functions and |fj (h)| ≤
∞ X
mjk |hk |
k=1
provided khk ≤ R. So in this case B = (mjk )nj,k=1 . Let us suppose that sup j
∞ X
mjk < 1.
k=1
Then rs (B) ≤ kBk < 1. So the considered equation is exponentially stable. Besides sup kB t k = 1. t≥0
12.8
Perturbations of nonlinear equations
Consider in a Banach space X the equation x(t + 1) = Q(x(t), t)x(t) + F (x(t), t) (t ≥ 0),
(8.1)
12.8. PERTURBATIONS OF NONLINEAR EQUATIONS
185
where Q(z, t) are linear operators in X continuously dependent on z ∈ Ω(R) and F (., t) : Ω(R) → X are continuous mappings for each t = 0, 1, .... Denote by ω∞ (R) the set of sequences with values in Ω(R): ω∞ (R) := {h(t) ∈ Ω(R); t = 0, 1, ...}. Let Uh (t, s) := Q(h(t − 1), t − 1) · · · Q(h(s), s) (t ≥ s ≥ 0) with a h ∈ ω∞ (R). That is, Uh is the evolution operator of the linear equation y(t + 1) = Q(h(t), t)y(t) (t ≥ 0).
(8.2)
Let us assume that there are a positive continuous function φ(t, s) = φ(R, t, s), and nonnegative constants ν = ν(R) and l = l(R) independent of h, such that kUh (t, j)k ≤ φ(t, j) (h ∈ ω∞ (R)) and kF (w, j)k ≤ νkwk + l (w ∈ Ω(R); t, j ≥ 0).
(8.3)
Additionally, suppose that t X
ηφ := sup t=1,2,...
φ(t, k) < 1/ν
(8.4)
k=1
and put $ :=
sup
φ(t, 0).
t=0,1,2,...
Theorem 12.8.1 Let conditions (8.3) and (8.4) hold. If, in addition, an initial vector x(0) satisfies the inequality $kx(0)k + lηφ < R(1 − νηφ ),
(8.5)
then the corresponding solution x(t) of (8.1) is subject to the estimate kx(t)k ≤ Proof:
kx(0)k$ + lηφ (t = 1, 2, ...). 1 − νηφ
(8.6)
First let R = ∞. Consider the equation y(t + 1) = Q(h(t), t)y(t) + F (y(t), t) (t ≥ 0).
(8.7)
By the Variation of Constants Formula this equation is equivalent to the following one: t−1 X y(t) = Uh (t, 0)y(0) + Uh (t, k + 1)F (y(k), k) (t ≥ 0). k=0
186
CHAPTER 12. NONLINEAR TIME-VARIANT EQUATIONS
Consequently, ky(t)k ≤ kUh (t, 0)y(0)k +
t−1 X
kUh (t, k + 1)kkF (y(k), k)k (t ≥ 0).
k=0
Hence, ky(t)k ≤ φ(t, 0)ky(0)k +
t−1 X
φ(t, k + 1)[νky(k)k + l].
k=0
We thus get sup ky(t)k ≤ $ky(0)k + ηφ (ν sup ky(t)k + l). t=0,1,...
t=0,1,...
Thanks to (8.4) $ky(0)k + ηφ l . 1 − νηφ t Take h(t) = x(t)-the solution of (8.1). Then (8.1) and (8.7) coincide, and therefore (8.6) is valid. The case R < ∞ can be considered exactly as in the proof of Theorem 12.1.1. Q. E. D. sup ky(t)k ≤
Clearly the preceding theorem gives us stability and boundedness conditions. Suppose that kQ(z, t)k ≤ a < 1 (z ∈ Ω(R); t = 1, 2, ...).
(8.8)
Then kUh (t, j)k ≤ at−j (h ∈ ω∞ (R)). In this case
t X
ηφ ≤ sup t=0,1,...
at−k =
k=0
1 1−a
and $ = 1. Now Theorem 12.8.1 implies Corollary 12.8.2 Let the conditions (8.8) and (8.3) with l = 0 hold. In addition, let ν < 1. 1−a Then the zero solution of equation (8.1) is stable. If, in addition, an initial vector x(0) satisfies the inequality (1 − a)kx(0)k < R(1 − a − ν), then the corresponding solution x(t) of equation (8.1) is subject to the estimate kx(t)k ≤
1−a kx(0)k < R (t ≥ 0). 1−a−ν