Abstract: In this research, we study nonlinear pseudo parabolic inequalities with initial-boundary conditions. We show that if the nonlinear operators satisfy in some conditions then the pseudo parabolic Inequalities has a unique solution.
INTRODUCTION
Let Vbe a Branch space and K⊂V be a nonempty, closed convex subset of V. The main problem is finding uεK such that for all vεK:
| (1) |
where, A:V → V′ and f∈V′.
This type f inequalities of elliptic and parabolic variational problem for monotone operators were investigated (Evans, 1998). Ptashnyk (2002) studied pseudo parabolic variational inequalities for nonlinear operators and proved some existence and uniqueness theorems for their solutions. Showalter and Ting (1970) have investigated elliptic and parabolic inequalities for pseudo monotone operators (Ptashnyk, 2004). These inequalities appear in the study of the free boundary problems (Cufner and Fuchik, 1998; Showalter, 1997).
The present research studied initial-boundary value problem for pseudo parabolic inequalities of type (1,1) where the operator A is a nonlinear and pseudo-monotone operator.
PROBLEM STATEMENTS
Let Ω⊂
| (2) |
| (3) |
| (4) |
where,
We assume that functions bij(x) = bji(x) (i, j = 1, 2,....,n-1),
bnn (x), ai (x, ξ) are continuous for any
| (i) |
The quadratic from:
is positively defined, for all xεΩ and its rank is equal to (n-1) for all xεΓ0:
(ii) |
(iii) |
We start with following definitions:
Definition 1: The function A:V → V′ is coercive if:
Definition 2: The operator A:V→V′ is pseudo-monotone if un → u and lim sup (Aun-u)≤0 imply (Au, u-v)≤ lim inf (Aun, u-v) for all u εV.
Definition 3: The operator A:D (A)⊂V→V′ is strongly monotone if there is a c>0 for which:
Now, we require that the operator M be pseudo-monotone and coercive and also the linear operator L be monotone and bounded.
Let KL be a set of functions which satisfy the following conditions.
| • |
| • | u(x) vanishes on some neighborhood of Γ1 |
In that case, we prove following theorems:
Theorems 1: Operator L, defined as a mapping from KL into space L2 (Ω), is symmetric and positively defined.
Proof: Let u (x) and v(x) ε KL. Then:
We prove that:
To prove the equation above, it’s enough to show that:
And the equality
The existence of limit above explicitly follows from the definition of set KL.
Consider there exists an element
Hence, the integral:
is divergent and we obtain a contradiction to condition u (x)εKL.
Therefore, we proved that (L(u), v) = (u, L(v)). As L (x, ξ) is positive, then the operator L is positive. It is not hard to prove that operator L is positively defined (Hakobyan and Shakhbaghyan, 1995; Lotfikar and Hakobyan, 2009).
Let’s denote by the same L the Friedreich extension of operator L, which will be self-adjoint and also define a new scalar product on linear manifold KL by the formula:
| (5) |
And we denote by HL the closure of manifold KL by the new norm (derived from scalar product (5)). So, the functions of HL will have first generalized derivatives by Sobolev and will vanish on boundary Γ1.
Theorem 2: The operator L:HL →L2(Ω) is bounded and monotone.
Proof: The proof is very easy because from (5) we will have:
(Lu, v) = [u, v]≤2y||HL||v||HL |
And easily from theorem (1) we can show that (Lotfikar and Hakobyan, 2009):
(Lu-Lv, u-v)≥0 |
Theorem 3: The operator M from KL is bounded.
Proof: Let u ε KL, v ε KL:
We can show (similarly as in theorem (1) that:
hence:
From the condition (ii) it follows that:
Therefore ||Mu||HL≤c||u||HL||v||HL.
Theorem 4: The operator M from KL is strongly monotone.
Proof: Suppose u, v ε KL. Then:
From the equation:
It follows that:
Therefore, the operator M is strongly monotone.
Lemma 1: If A:V →V' is strongly monotone and hemi-continues then A is pseudo-monotone and coercive.
WEAK AND STRONG PROBLEMS
We start this section which following definitions:
Definition 4: The family of operators {G (s):s≥0} is said to be a linear semigroup over the branch space V if F (s): V →V is a linear, continuous operator for all S≥0 and:
G (0) = I, G (s+t) = G (s) + G(t), S, t≥0 G (.)x ε C ([0, ∞), V), x ε V |
Definition 5: Let {G (s): s≥0} be a linear semigroup over the branch space V and let:
Then the operator B: D (B) →V for which:
is said to be a generator of the semigroup G (s).
We consider operators represent able in the form:
| (6) |
where, d = ∂/∂t and the operators L and M satisfy in the conditions of upper section.
Also, we can show that following conditions hold:
| • | The operator (-d) is a generator for linear semigroup G (s) over the space V = L2 (Ω), with a definition domain D (d). |
Defining the operator dh: V →V by the formula:
From definition we get:
| (7) |
| • | D (A) = D (d) |
| • | For any φ ε KL⊂V we have: |
| (8) |
| • | For any φ, ψ ε KL |
| (9) |
Remark: The operators satisfying (8) and (9) are monotone.
Besides, the requirement:
And continuity of the operator L imply that:
Thus under (8) and (9) letting h → 0 yields:
| (10) |
| (11) |
Here, from (6), we state a variational problem similar to (1):
| (12) |
From classical theory we know this problem have a solution.
Suppose the all of the conditions are fulfilled and u is a solution of the problem (12), then the conditions (10) and (11) imply that for all v ε KL ∩ D (d)
| (13) |
From (13) it is evident that a solution of (12) is also a solution of a problem:
| (14) |
The variational problem (12) and (14) we call correspondingly strong and weak, and accordingly we have strong and weak solutions.
Theorem 5: Letthe all of the conditions are hold then for any f ε V' = L2(Ω) the weak problem has a unique solution in KL and this solution will be unique solution for strong problem (Petrosyan and Hakobyan, 2008; Petrosyan, 2008).
CONCLUSION
In this research we studied initial-boundary value problem for degenerate nonlinear pseudo-parabolic inequality as (Au, v-u)≥(f, v-u). We assumed that A = Ld + M where d = ∂/∂t and the operators L and M satisfy in some conditions, so we changed our system with pseudo parabolic variational inequality. Finally, we know (Petrosyan and Hakobyan, 2008; Petrosyan, 2008) our system have a unique solution.