Boundary value problem for a fully nonlinear elliptic equation

In this paper, we investigate the well-posedness of a fully nonlinear elliptic problem. By using the acute angle principle for the weakly continuous operator, we obtained the W 2,p -strong solution of the fully nonlinear elliptic problem.


Introduction
In this paper, we study the well-posedness of the following elliptic equation (|∆ | 4 + 1)∆ = ℎ( ) −2| 2 |−| |−| | + ( ), (1) with the boundary condition | Ω = 0, where Ω ⊂ is 2 bounded. The nonlinear elliptic equation has been extensively studied in the past several decades, see [1][2][3][4][5] and the references therein. Among the above literatures, the most mathematical analysis was the wellposedness of the equation, which has been mostly discussed by using some different techniques in the above literatures. Note that Caffarelli [1] obtained a interior 2, -priori estimate for uniformly elliptic equations by using the Aleksandrov-Bakelman-Pucci − ∞ a priori estimate, the Krylov-Safonov Harnack inequality and the Caldeŕn-Zygmund decomposition lemma. Moreover, Winter [2] extended Caffarelli's results on interior 2, -estimates for viscosity solutions and proved 2, -estimates at a flat boundary. In addition, Amendola, Rossi and Vitolo [3] established the weak Harnack inequality and local maximum principle to study continuous viscosity solutions for fully nonlinear elliptic equations ( , , , 2 ) = ( ). Also, Nakagawa researched the maximum principle for -viscosity solutions of nonlinear equations in [4].
It is also worth mentioning that Viaclovsky [5] proved the existence of a fully nonlinear partial differential equation on smooth compact n-dimensional Riemannian manifold with 2 estimate on solutions and the 2, estimate, which come from the work of Evans [6]. Furthermore, Tyagi and Verma [7] researched the existence of nontrivial solutions of a fully nonlinear elliptic equations with gradient nonlinearity. On the other hand, there are also a good amount of interests on the existence of viscosity solutions for elliptic equation, see [8,9]. There are also many other interesting results as for the existence of solutions for fully nonlinear equations, see [10][11][12] for classical results.
But, among the previous studies, most of the researchers are focused on the existence of viscosity solution, mild solutions and the axially symmetric solutions. As far as we know, the existence of a strong solution for problem (1)-(2) has not been considered before. Motivated from the above research result, in this paper, we devote to prove the existence of strong solution for the problem (1)- (2). The method we use is the acute angle principle. This theory was proposed by Ma in [13], in which the author presented a novel inner product operator -compulsorily weakly continuous operators. This theory can solve the weak solutions and strong solutions for many nonlinear problems, see [14][15][16].
The paper is organized as follows. In Section 2, we formulate the theory of the acute angle principle, especially for weakly continuous operator, and also list some basic lemmas. In Section 3. The proof of Theorem 1.1 is given.

Preliminaries
In this part, we introduce the acute angle principle for the compulsorily weakly continuous operator in [13], which can solve the weak solutions and strong solutions for many nonlinear problems.
Let be a linear space, 1 , 2 be the completion of . This Lemma is crucial to the proof of our main result. Next, we recall the following Lemma (see [13]).

Proof of Theorem 1.1
In this section, we prove the Theorem 1.1.
Proof. The prove of Theorem 1.1 will be decomposed into in the following three steps.
By the definition of weak solution, we give the definition of the operator : 2 → 1 * by following: where ∈ 1 , 1 * is the dual space of 1 . It is apparent that the operator was bounded.
This implies that the operator : 2 → 1 * we established in Step 1 meets the acute angle condition. Step 3. Verify the operator is a compulsorily weak continuity operator. Let { } ⊂ 2 , ⇀ 0 in 2 . According to the Definition 2.1, we only need to prove the following for any ∈ 1 . Note that lim Under the assumption ( 2 ), (7) and Lemma 2.3, it is easy to have the following equality lim By (6) and (8), it is easy to derive that where ∆̃ and 2 ̅ lie between ∆ and ∆ 0 , between 2 and 2 0 , respectively. Obviously, combined with (9) and (10), we can derive that lim →∞ ∫ | 2 − 2 0 | 2 d = 0.
which implies that 2 → 2 0 in (Ω) for any 0 < ≤ 2. By the (7), (11) and Lemma 2.3, we prove that the (4) holds true. Then we obtained that the operator is a compulsorily weak continuity operator.