Difference-differential Equations with Fredholm Operator in the Main ...

More documents

Recommendations

Info

The main part of operator ˜L has the order q 0 , the main part of operator Lks ˜ - the order q 1 . Because q 1 < q 0 then the boundary conditions on projections P u, (I − P )u may be different in the general case. If q 1 = 0 then boundary conditions may be given only on the projection (I − P )u. Let ω = Ω 1 ∩ Ω 2 ⊂ Ω. We shall find the solution u : ω ⊂ R r → E 1 of equation (1), whose projections P u, (I − P )u are subjected to the following boundary conditions: τ 1 (I − P )u = α(x), x ∈ ∂Ω 1 , (11) τ 2 P u = β(x), x ∈ ∂Ω 2 . (12) Here τ 1 , τ 2 are linear difference-differential operators in the space of abstract functions, which commute with the projection operator P and with the operators B, Γ. So the functions α(x), β(x) must satisfy conditions P α(x) = 0, (I −P )β(x) = 0. Let the conditions of theorem 1 be satisfied. We shall search for the solution u of equation (1), which satisfies boundary conditions (11), (12). Based on theorem 1 the solution has the form u(x) = Γv(x) + (C(x), Φ), (13) where v satisfies (8) and the condition Qv = 0. The components of the vector C can be find from system (10). Lemma 2. If v, C satisfy the boundary conditions τ 1 v = Bα(x), x ∈ ∂Ω 1 , (14) τ 2 (C, Φ) = β(x), x ∈ ∂Ω 2 , (15) then solution (13) satisfies boundary conditions (11), (12). Proof. Since P u = (C, Φ), then condition (12) holds if and only if condition (15) holds. Since Qv = 0, P Γ = ΓQ, then (I − P )u = Γv. By applying the operator B to the both sides of this equality, we obtain v = B(I − P )u. ¿From here τ 1 v = τ 1 B(I − P )u = Bα(x), and so B[τ 1 (I −P )u−α(x)] = 0, i.e., τ 1 (I −P )u−α(x) ∈ ∑ c i φ i . So if v satisfies condition (14) then v satisfies (11). Besides, it is not difficult to understand if v satisfies (11) then v satisfies (14). Lemma 3. If Ker ˜L = {0} then any solution vof equation (8) with condition (14) lies in the subspace E 2∞−k . Proof. Since QB = BP, P α(x) = 0, QA i Γ = A i ΓQ, then the projection Qv satisfies homogeneous problem ˜LQv = 0. As Ker( ˜L) = {0}, then Qv = 0 and v ∈ E 2∞−k . Suppose 4. Operators ˜L, Lks ˜ have the bounded inverses. Then based on lemma 2, lemma 3 and theorems 1 we obtain the following results Theorem 3. Let conditions 1, 2, 3, 4 be satisfied. Then problem (1) with conditions (11), (12) has the unique solution u(x) = Γv(x) + (C(x), Φ), where v satisfies problem (8), (14), vector C- equations (10) with condition (15). 6
5 Examples Theorem 3 can be used for the statement and the investigation nonclassical boundary value problems for partial differential and difference equations with degeneration. 1) Suppose that the coefficients a i (x, y), b i (x, y) of elliptic equation (∆ def = ∂2 + ∂2 ): ∂x 2 ∂y 2 (△+a 1 (x, y) ∂ ∂x +a 2(x, y) ∂ ∂y +a 3(x, y))Bu+( ∂ ∂x +b 1(x, y) ∂ ∂y +b 2(x, y))Au = f(x, y) (16) are defined in bounded domain Ω ⊂ R 2 and belong to the space C l−2,α (Ω), l ≥ 2, α ∈ (0, 1), f : Ω → E 2 belongs to the space C l−2,α (Ω). Fredholm operator B has a complete A -Jordan set, A is a compact operator. Let the domain ω ⊆ Ω, ∂ω = l 1 ∪ ∂ 1 , where l 1 is the part of the straight line x = x 0 (fig.1). y ✻ ✬ ✟ ✟ Ω ✟ ✟ ∂ ✡ 1 ☎ l 1 ❤❤❤❤❤ ✌ ✫ x 0 ✩ ✪ ✲ x Fig. 1. Example of a domain We shall consider the ”weakened” Dirichlet problem for equation (23) in the domain ω, i.e. the problem of finding function u, satisfying equation (23) in ω, and on the boundary ∂ω the conditions u| x=x0 = φ 0 (y), (17) (I − P )u| ∂1 = φ 1 (x, y), (18) where φ 0 , φ 1 ∈ C l,α , P φ 1 = 0, (I − P )Φ 0 (y) = Φ 1 (x 0 , y), φ 0 , φ 1 ∈ D(B). Note that weakening the classical statement of the Dirichlet problem is that there is no condition on projection P u along ∂ 1 of the boundary ∂ω. It is connected with that P u will be defined from the system of the first order. Based on theorem 1 the solution has form (4). If condition 4 is satisfied and l ≥ 2max 1≤i≤l p i then based on theorem 3 problem (16)-(18) has the solution in the class C l,α (¯ω). Suppose that the homogeneous problem ˜Lu ≡ (△ + a 1 (x, y) ∂ ∂x + a 2(x, y) ∂ ∂y + a 3(x, y))u + ( ∂ ∂x + b 1(x, y) ∂ ∂y + b 2(x, y))AΓu = 0, u| ∂ω = 0 7
Page 1 and 2: Difference-differential Equations w
Page 3 and 4: Definition 1. If u ∈ D(A), P u
Page 5: So it is proved Theorem 1. Suppose
Page 9 and 10: ✻ ✬ ✗ ✖ ✫ x 0 1 ω ∂
Page 11 and 12: [6] N.A.Sidorov and E.B. Blagodatsk

Difference-differential Equations with Fredholm Operator in the Main ...

Create successful ePaper yourself

Delete template?

Save as template?