Let H and V be two real Hilbert spaces. Assume and V is dense subspace in H and the injection of V into H is continuous. If H is identified with its dual space, then we may write V ⊂ H ⊂ V* densely, and the corresponding injections are continuous. The norm on V (resp. H) will be denoted by || · || (resp. |· |). The duality pairing between the element v
1 of V* and the element v
2 of V is denoted by (v
1
, v
2), which is the ordinary inner product in H if v
1
, v
2 ∈ H. For the sake of simplicity, we may consider
where || · ||* is the norm of the element of V*. If an operator A is bounded linear from V to V* and generates an analytic semigroup, then it is easily seen that
for the time T > 0. Therefore, in terms of the intermediate theory we can see that
where denotes the real interpolation space between V and V*.
We note that a nonlinear operator A is said to be hemicontinuous on V if
for every x, y ∈ V where "w - lim" indicates the weak convergence on V*. Let A : V → V* be given a single-valued, monotone operator and hemicontinuous from V to V* such that
(A1) A(0) = 0, (Au - Av, u - v) ≥ ω
1 ||u - v||2
- ω
2 |u - v|
2,
(A2) ||Au||* ≤ ω
3(||u|| + 1)
for every u, v ∈ V where ω
2 ∈ ℝ and ω
1
, ω
3 are some positive constants.
Here, we note that if 0 ≠ A(0), then we need the following assumption:
for every u ∈ V . It is also known that A is maximal monotone, and R(A) = V* where R(A) denotes the range of A.
Let the controller B is a bounded linear operator from a Banach space L
2(0, T; U) to L
2(0, T; H) where U is a Banach space.
For each t ∈ [0, T], we define x
t
: [ -h, 0] → H as
We will set
Let ℒ and ℬ be the Lebesgue σ-field on [0, ∞) and the Borel σ-field on [- h, 0] respectively. Let k : ℝ+ × ℝ+ × ∏ → H be a nonlinear mapping satisfying the following:
(K1) For any x. ∈ ∏ the mapping k(·, ·, x.) is strongly ℒ × ℬ -measurable;
(K2) There exist positive constants K
0, and K
1 such that
for all (t, s) ∈ ℝ+ × [ -h, 0] and x., y. ∈ ∏.
Let g : ℝ+ × ∏ × H → H be a nonlinear mapping satisfying the following:
(G1) For any x ∈ ∏, y ∈ H the mapping g(·, x., y) is strongly ℒ -measurable;
(G2) There exist positive constants L
0
, L
1, and L
2 such that
for all t ∈ ℝ+, x, , and y, .
Remark 2.1. The above operator g is the semilinear case of the nonlinear part of quasi-linear equations considered by Yong and Pan[13].
For x ∈ L
2(-h, T; V ), T > 0 we set
Here, as in[13], we consider the Borel measurable corrections of x(·).
Lemma 2.1. Let x ∈ L
2(- h, T; V ). Then, the mapping t ↦ x
t
belongs to C([0, T ]; ∏) and
(2.1)
Proof. It is easy to verify the first paragraph and (2.1) is a consequence of the estimate:
Lemma 2.2. Let x ∈ L
2(- h, T; V ), T > 0. Then, G(·, x) ∈ L
2(0, T; H) and
(2.2)
Moreover, if x
1
, x
2 ∈ L
2(- h, T; V ), then
(2.3)
Proof. It follows from (K2) and (2.1) that
and hence, from (G2), (2.1), and the above inequality, it is easily seen that
Similarly, we can prove (2.3).
Let us consider the quasi-autonomous differential equation
(E)
where A satisfies the hypotheses mentioned above. The following result is from Theorem 2.6 of Chapter III in [1].
Proposition 2.1. Let Φ
0 ∈ H and f ∈ L
2(0, T; V* ). Then, there exists a unique solution x of (E) belonging to
and satisfying
(2.4)
(2.5)
where C
1
is a constant.
Acting on both sides of (E) by x(t), we have
As is seen Theorem 2.6 in [1], integrating from 0 to t, we can determine the constant C
1 in Proposition 2.1.
We establish the following result on the solvability of the equation (SE).
Theorem 2.1. Let A and the nonlinear mapping g be given satisfying the assumptions mentioned above. Then, for any (Φ
0
, Φ
1) ∈ H × L
2(- h, 0; V ) and f ∈ L
2(0, T; V*), T > 0, the following nonlinear equation
(2.6)
has a unique solution x belonging to
and satisfying that there exists a constant C
2
such that
(2.7)
Proof. Let y ∈ L
2(0, T; V ). Then, we extend it to the interval (-h, 0) by setting y(s) = Φ
1(s) for s ∈ (-h, 0), and hence, G(·, y(·)) ∈ L
2(0, T; H) from Lemma 2.2. Thus, by virtue of Proposition 2.1, we know that the problem
(2.8)
has a unique solution x
y
∈ L
2(0, T; V ) ∩ W
1,2(0, T; V* ) corresponding to y. Let us fix T
0
> 0 so that
(2.9)
Let x
i
, i = 1, 2, be the solution of (2.8) corresponding to y
i
. Multiplying by x
1(t) - x
2(t), we have that
and hence it follows that
From Lemma 2.2 and integrating over [0,t], it follows
where c is a positive constant satisfying 2ω
1
- c > 0. Here, we used that
for any pair of nonnegative numbers a and b. Thus, from (2.3) it follows that
By using Gronwall's inequality, we get
Taking c = ω
1, it holds that
Hence, we have proved that y ↦ x is a strictly contraction from L
2(0, T
0; V ) to itself if the condition (2.9) is satisfied. It shows that the equation (2.6) has a unique solution in [0, T
0].
From now on, we derive the norm estimates of solution of the equation (2.6). Let y be the solution of
(2.10)
Then,
by multiplying by x(t) - y(t) and using the assumption (A1), we obtain
By integrating over [0, t] and using Gronwall's inequality, we have
and hence, putting
it holds
(2.11)
for some positive constant C
2. Since the condition (2.9) is independent of initial values, the solution of (2.6) can be extended to the internal [0, nT
0] for natural number n, i.e., for the initial value (x(nT
0), x
n
T
0
) in the interval [nT
0, (n + 1)T
0], as analogous estimate (2.11) holds for the solution in [0, (n + 1)T
0].
Theorem 2.2. If (Φ
0, Φ
1) ∈ H × L
2(-h, 0, V )) and f ∈ L
2(0, T; V* ), then x ∈ L
2(-h, T; V ))∩ W
1,2(0, T; V* ), and the mapping
is continuous.
Proof. It is easy to show that if (Φ
0
, Φ
1) ∈ H × L
2(-h, 0; V )) and f ∈ L
2(0, T; V* ) for every T > 0, then x belongs to L
2(- h, T; V )∩W
1,2(0, T; V*). Let
and x
i
be the solution of (2.6) with in place of for i = 1,2.
Then, in view of Proposition 2.1 and Lemma 2.2, we have
(2.12)
If ω
1
- c/ 2 > 0, we can choose a constant c
1
> 0 so that
and
Let T
1
< T be such that
Integrating on (2.12) over [0, T
1] and as is seen in the first part of proof, it follows
Putting that
we have
(2.13)
Suppose that
and let x
n
and x be the solution (2.6) with and respectively.
By virtue of (2.13) with T being replaced by T
1, we see that
This implies that in H × L
2 (-h, 0; V). Hence, the same argument shows that
Repeating this process, we conclude that
Remark 2.2. For x ∈ L
2(0, T; V ), we set
where k belongs to L
2(0, T) and g : [0, T] × V → H be a nonlinear mapping satisfying
for a positive constant L. Let x ∈ L
2(0, T; V ), T > 0. Then, G(·, x) ∈ L
2(0, T; H) and
Moreover, if x
1
, x
2
∈ L
2(0, T; V ), then
Then, with the condition that
in place of the condition (2.9), we can obtain the results of Theorem 2.1.