NONLINEAR VARIATIONAL EVOLUTION INEQUALITIES IN HILBERT SPACES

The regular problem for solutions of the nonlinear functional differential equations with a nonlinear hemicontinuous and coercive operator A and a nonlinear term f(·,·): x′(t)+Ax(t)+∂φ(x(t)) f(t,x(t))+h(t) is studied. The existence, uniqueness, and a variation of solutions of the equation are given.


Introduction.
Let H and V be two real separable Hilbert spaces such that V is a dense subspace of H. Let the operator A be given a single-valued operator, which is hemicontinuous and coercive from V to V * .Here V * stands for the dual space of V .Let φ : V → (−∞, +∞] be a lower semicontinuous, proper convex function.Then the subdifferential operator ∂φ : V → V * of φ is defined by ∂φ(x) = x * ∈ V * ; φ(x) ≤ φ(y) + (x * ,x − y), y ∈ V , (1.1) where (•, •) denotes the duality pairing between V * and V .We are interested in the following nonlinear functional differential equation on H: where the nonlinear mapping f is a Lipschitz continuous from R × V into H.Equation (1.2) is caused by the following nonlinear variational inequality problem: ≤ f t, x(t) + h(t), x(t) − z , a.e., 0 < t ≤ T , z ∈ V , x(0) = x 0 . (1.3) If A is a linear continuous symmetric operator from V into V * and satisfies the coercive condition, then equation (1.2), which is called the linear parabolic variational inequality, is extensively studied in Barbu [5,Sec. 4.3.2](also see [4,Sec. 4.3.1]).The existence of solutions for the semilinear equation with similar conditions for nonlinear term f have been dealt with in [1,2,6].Using more general hypotheses for nonlinear term f (•,x), we intend to investigate the existence and the norm estimate of a solution of the above nonlinear equation on L 2 (0,T ; V ) ∩ W 1,2 (0,T ; V * ), which is also applicable to optimal control problem.A typical example was given in the last section.

Perturbation of subdifferential operator.
Let H and V be two real Hilbert spaces.Assume that V is a dense subspace in H and the injection of V into H is continuous.If H is identified with its dual space we may write V ⊂ H ⊂ V * densely and the corresponding injections are continuous.The norm on V (respectively H) will be denoted by For the sake of simplicity, we may consider where • * is the norm of the element of V * .
Remark 2.1.If an operator A 0 is bounded linear from V to V * and generates an analytic semigroup, then it is easily seen that Therefore, in terms of the intermediate theory we can see that where (V , V * ) 1/2,2 denotes the real interpolation space between V and V * .
We note that nonlinear operator A is said to be hemicontinuous on where "w-lim" indicates the weak convergence on V .Let A : V → V * be given a single valued and hemicontinuous from V to V * such that for every u, v ∈ V , where ω 2 ∈ and ω 1 ,ω 3 are some positive constants.Here, we note that if A(0) = 0 we need the following assumption: It is also known that A+ω 2 I is maximal monotone and R(A+ω 2 I) = V * where R(A+ ω 2 I) is the range of A + ω 2 I and I is the identity operator.
First, let us be concerned with the following perturbation of subdifferential operator: To prove the regularity for the nonlinear equation (1.2) without the nonlinear term f (•,x) we apply the method in [5,Sec. 4.3.2].
Proposition 2.1.Let h ∈ L 2 (0,T ; V * ) and x 0 ∈ V satisfying that φ(x 0 ) < ∞.Then (2.7) has a unique solution where C 1 is a constant and Proof.Substituting v(t) = e ω 2 t x(t) we can rewrite (2.7) as follows: (2.10) Then the regular problem for (2.7) is equivalent to that for (2.10).Consider the operator L : (2.12) Since for every real number c, so using Gronwall's inequality, the inequality (2.12) implies that for some positive constant C 1 , that is, (2.15) Hence we have proved (2.9).Let Then there exist sequences {x 0n } ⊂ D(L) and ∞ (0,T ; H) be the solution of (2.7) with initial value x 0n and with h n instead of h.Since ∂φ is monotone, we have 1 2 for every real number c. Therefore, if we choose ω 1 − (c/2) then by integrating over [0,T ] and using Gronwall's inequality it follows that (2.17) and hence, we have that lim n→∞ x n (t) = x(t) exists in H. Furthermore, x satisfies (2.7).Indeed, for all 0 ≤ s < t ≤ T and y ∈ ∂φ(x), multiplying (2.7) by x(t) − x and integrating over [s, t] we have and, therefore, a.e., t ∈ (0,T ), that is, Thus, the proof is complete.
Corollary 2.1.Assume the hypotheses as in Proposition 2.1, in addition, assume that ∂φ satisfies the growth condition as follows: (2.23) Then equation (2.7) has a unique solution which satisfies (2.25) Proof.From (2.7) and (2.23) it follows that Hence, by virtue of (2.15) we have that (2.27) Hence, the mapping h x is compact from L 2 (0,T ; H) to L 2 (0,T ; H).

Nonlinear integrodifferential equation.
Let f : [0,T ] × V → H be a nonlinear mapping satisfying the following variational evolution inequality: for a positive constant L. Furthermore, there exists a constant C 2 such that Thus, by virtue of Proposition 2.1 we know that the problem has a unique solution x y ∈ L 2 (0,T ; V )∩C [0,T ]; H , where x y is the solution of (3.5).
Let us choose a constant c > 0 such that and let us fix T 0 > 0 so that 2cω 1 − c 2 −1 e 2ω 2 T 0 L < 1.
(3.7) Let x i ,i = 1, 2, be the solution of (3.5) corresponding to y i .Then, by the monotonicity of ∂φ, it follows that and hence, using the assumption (2.5), we have that for every c > 0 and by integrating (3.9) over (0,T 0 ) we have and by Gronwall's inequality, (3.12) Thus, from (3.1) it follows that and hence for some positive constant C 2 .Since condition (3.7) is independent of the initial values, the solution of (1.2) can be extended to the interval [0,nT 0 ] for natural number n, i.e., for the initial value x(nT 0 ) in the interval nT 0 ,(n+1)T 0 , as analogous estimate (3.19) holds for the solution in [0,(n+1)T 0 ].Furthermore, similar to (2.12) and (2.15) in Section 2, the estimate (3.3) is easily obtained.Now we prove the last result.If (x 0 ,h) ∈ V ×L 2 (0,T ; V * ) then x belongs to L 2 (0,T ; V ).Let (x 0i ,h i ) ∈ V × L 2 (0,T ; V * ) and x i be the solution of (1.2) with (x 0i ,h i ) in place of (x 0 ,u) for i = 1, 2. Multiplying (1.2) by x 1 (t) − x 2 (t), we have where T 1 < T and as seen in the first part of the proof, it follows that (3.23) Putting we have ), and let x n and x be the solutions of (1.2) with (x 0n ,h n ) and (x 0 ,h), respectively.Then, by virtue of (3.25) and (3.20), we see that Therefore the same argument shows that x n → x in Repeating this process, we conclude that If ∂φ satisfies the growth condition (2.23) as is seen in Corollary 2.1, we can obtain the following result.Furthermore, there exists a constant C 2 such that (3.28) is continuous.

Example.
Let Ω be a region in an n-dimensional Euclidean space R n with boundary ∂Ω and closure Ω.For an integer m ≥ 0, C m (Ω) is the set of all m-times continuously differentiable functions in Ω, and C m 0 (Ω) is its subspace consisting of functions with compact supports in Ω.If m ≥ 0 is an integer and 1 ≤ p ≤ ∞, W m,p (Ω) is the set of all functions f whose derivative D α f up to degree m in the distribution sense belong to L p (Ω).As usual, the norm of W m,p (Ω) is given by where A α (u, ξ) are real functions defined on Ω × R N and satisfy the following conditions: (1) A α are measurable in u and continuous in ξ.There exists k ∈ L 2 (Ω) and a positive constant C such that where ξ = (ξ α ; |α| ≤ m).