A short proof of the existence of the solution to elliptic boundary problem

There are several methods for proving the existence of the solution to the elliptic boundary problem Lu = f in D, u | S = 0 , ( ∗ ). Here L is an elliptic operator of second order, f is a given function, and uniqueness of the solution to problem (*) is assumed. The known methods for proving the existence of the solution to (*) include variational methods, integral equation methods, method of upper and lower solutions. In this paper a method based on functional analysis is proposed. This method is conceptually simple and technically is easy. It requires some known a priori estimates and a continuation in a parameter method.


Consider the boundary problem
where D ⊂ R 3 is a bounded domain with a C 2 −smooth boundary S, L is an elliptic operator, Lu = −∂ i (a ij (x)∂ j u) + q(x)u. (1.3) Here and below ∂ i = ∂ ∂x i , over the repeated indices summation is understood, 1 ≤ i, j ≤ 3, a ij (x) = a ji (x), Im a ij (x) = 0, where c 0 , c 1 > 0 are constants independent of x and |ξ| 2 = 3 j=1 |ξ j | 2 .We assume that q(x) is a real-valued bounded function and |∇a ij (x)| ≤ c.One may easily consider by our method the case of complex-valued q, see Remark 2.2 in Section 2. By c > 0 various estimation constants are denoted.In this paper the Hilbert space H := H 0 := L 2 (D), the Sobolev space H 1 0 , the closure of C ∞ 0 (D) in the norm of the Sobolev space H 1 = H 1 (D), and the Sobolev space H 2 0 := H 2 (D) ∩ H 1 0 are used.We assume for simplicity that problem (1.1)- (1.2) has no more than one solution.This, for example, is the case if where c 2 > 0 is a constant, and D(L) = H 2 0 .The norm in the Sobolev space H ℓ is denoted by the symbol || • || ℓ .For example, . (1.6) By |∂u| 2 the sum of the squares of the derivatives of the first order is denoted, and |∂ 2 u| 2 is understood similarly.
The goal of this paper is to suggest a method for a proof of the existence of the solution to problem (1.1) -(1.2), based on functional analysis.This method is simple, short, and does not require too much of a background knowledge from the reader.
The background material, that is used in our proof, includes the notions of closed linear unbounded operators and symmetric operators (see [4]) and second basic elliptic inequality (see [1], [2], [5], [6]): and the definition and basic properties of the mollification operator, see, for example, [1].
Let us outline the ideas of our proof.Let R(L) denote the range of L.
2) has a solution.Uniqueness of the solution follows trivially from the assumption (1.5).
Let us summarize our result.This result is known (see, for example, [2], [5], [6]), but we give a short and essentially self-contained proof of it.
Theorem 1.1.Assume that S is C 2 -smooth, inequalities (1.4), (1.5) hold, and q is a real-valued bounded function.Then problem (1.1) -(1.2) has a solution in H 2 0 for any f ∈ H 0 , and this solution is unique.The operator L is an isomorphism of H 2 0 (D) onto Remark 1.1.We are not trying to formulate the result in its maximal generality.For example, one may consider by the same method elliptic operators which are non-selfadjoint.In Section 2, Remark 2.2 addresses this question.
In Section 2 proofs are given.

Proofs
It follows from (1.5) that Therefore, if Lu = 0 then u = 0.This proves the uniqueness of the solution.
To prove the existence of the solution it is sufficient to prove that the range of L is closed and its orthogonal complement in H 0 is just the zero element.Indeed, one has where R(L) ⊥ denotes the orthogonal complement in H = H 0 and the over-line denotes the closure.Therefore, if (2.6) Therefore, formula (2.6) implies Lu = f .This argument proves that R(L) is a closed subspace of H 0 and the operator L is closed on D(L).
Let us now prove that R(L) ⊥ = {0}.Assume the contrary.Then there is an element Let us derive from (2.7) that h = 0. To do this, first assume that L = L 0 := −∆, where ∆ is the Dirichlet Laplacian, and prove that L 0 is an isomorphism of H 2 0 (D) onto H 0 = L 2 (D).This will prove Theorem 1.1 for L = L 0 .Then we use continuation in a parameter method and prove that the same is true for L, which will prove Theorem 1.1.Take an arbitrary point x ∈ D, choose ǫ > 0 so that the distance d(x, S) from x to S is larger than ǫ, and set u = w ǫ (|x − y|), where w ǫ (|x|) is a mollification kernel (see, for example, [1], p.Let us now prove that R(L) = H 0 for the operator (1.3).This is proved by a continuation in a parameter.Define We prove that R(L s ) = H 0 for all s ∈ [0, 1] and the map L s : H 2 0 → H 0 is an isomorphism.For s = 0 this is proved above.
Consider equation (1.1) with L = L s and apply the operator This equation is in the space H 2 0 .The norm of the operator sL −1 0 (L − L 0 ) in H 2 0 is less than one if s is sufficiently small.Indeed, inequality similar to (1.7) holds for L s for any s ∈ [0, 1] with the same constant c 3 , because this constant depends only on the bounds on the coefficients of L s and these bounds are independent of s ∈ [0, 1].Thus, (2.13) Therefore, because ||(L − L 0 )u|| 0 ≤ c||u|| 2 , where c > 0 is a constant not depending on s, c depends only on the bounds on the coefficients of L. Consequently, if sc ′ 3 < 1, that is, if s < (c ′ 3 ) −1 , then equation (2.12) is uniquely solvable in H 2 0 for any f ∈ H 0 , and . One has L s 0 +s ′ = L s 0 as s ′ = 0 and L s 0 +s ′ = L as s ′ = 1.Applying the same argument and using the fact that ||L −1 s 0 || H 0 →H 2 0 (D) does not depend on s 0 , one gets (2.15) Therefore, for s ′ < (c ′ 3 ) −1 , one has Then R(L s ) = H 0 for s < s 0 + s 1 and L s : H 2 0 → H 0 is an isomorphism.Consequently, repeating the above argument finitely many times one reaches the operator L and gets both conclusions: R(L) = H 0 and L is an isomorphism of H 2 0 onto H 0 .Theorem 1.1 is proved.✷ Remark 2.1.The method of continuation in a parameter goes back to [8], see also [6].
Remark 2.2.Consider the operator L 1 = L + L ′ , where L ′ is an arbitrary first order differential operator and L is the same as in Section 2. The operator L 1 is not necessarily symmetric.Problem (1.1) -(1.2) is equivalent to the operator equation where is a compact operator in H 0 .This follows from the Sobolev embedding theorem ([1], [2]).Therefore, the Fredholm alternative holds for equation (2.17).So, if the homogeneous version of the equation (2.17) has only the trivial solution (zero solution) then equation (2.17) is solvable for any f , and its solution u ∈ H 2 0 .