Solvability for Schrödinger equations with discontinuous coefficients

By Guixia Pan, Lin Tang

http://www.sciencedirect.com/science/article/pii/S0022123615004127

We reflect on the Lp solvability for divergence and non-divergence shape Schrödinger equations with discontinuous coefficients. As an software, we provide the worldwide Morrey regularity for divergence and non-divergence shape Schrödinger operators with VMO coefficients in a bounded area.

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By means of the definition of the coefficients and the knowledge, now we have Mu(−x1 , x ) + λu(−x1 , x ) = −divg − f in Rn . accordingly, −u(−x1 , x ) is usually an answer to (5. 7). through the individuality of the answer, we receive u(x) = −u(−x1 , x ). this suggests that, as a functionality on Rn+ , u has 0 hint at the boundary and obviously u satisfies (5. 4). The life of the answer is proved. nonetheless, you can still see that if u ∈ W 1,p (Rn+ ) is an answer to (5. 4), then its bizarre extension with appreciate to x1 is an answer to (5.

19] A. Hinz, H. Kalf, Subsolution estimates and Harnack’s inequality for Schrödinger operators, J. Reine Angew. Math. 404 (1990) 118–131. [20] F. John, L. Nirenberg, On services of bounded suggest oscillation, Comm. natural Appl. Math. four (1961) 415–426. [21] D. Kim, N. Krylov, Elliptic differential equations with coefficients measurable with appreciate to 1 variable and VMO with appreciate to the others, SIAM J. Math. Anal. 39 (2007) 489–506. [22] N. Krylov, Parabolic and elliptic equations with VMO coefficients, Comm. Partial Differential Equations 32 (2007) 453–475.

The functionality m(x, V ) is defined through ⎧ ⎪ ⎨ ρ(x) = 1 1 = sup r : n−2 m(x, V ) r>0 ⎪ r ⎩ ⎫ ⎪ ⎬ V (y)dy ≤ 1 B(x,r) ⎪ ⎭ . (2. 1) evidently, zero < m(x, V ) < ∞ if V = zero. specifically, m(x, V ) = 1 with V = 1 and m(x, V ) ∼ (1 + |x|) with V = |x|2 . Lemma 2. 1. (See [28]. ) If V ∈ Bq for a few q ≥ such that n 2, then there exist l0 > zero and C0 > 1 m(x, V ) 1 −l l /(l +1) ≤ C0 (1 + |x − y|m(x, V )) zero zero (1 + |x − y|m(x, V )) zero ≤ . C0 m(y, V ) particularly, m(x, V ) ∼ m(y, V ) if |x − y| < C/m(x, V ). n Lemma 2.

Four. think that −a0ij Dij u + V u = zero in B(x0 , 2R) and V ∈ Bn/2 N ≥ zero, |∇2 u(x)| ≤ sup B(x0 ,2R) CN 1 . N [1 + Rm(x0 , V )] Rn 2 facts. allow η ∈ C0∞ (B(x0 , 2R)) such that η = 1 on B(x0 , 3R 2 ), |∇η| ≤ C/R and |∇ η| ≤ 2 C/R . be aware that, for x ∈ B(x0 , R), we have now Γ0 (x, y)(−a0ij Dij )(uη)(y)dy u(x)η(x) = Rn Γ0 (x, y)(−V uη − a0ij ∂i u∂j η − a0ij ∂j u∂i η − a0ij u∂ij η)dy = Rn Γ0 (x, y)(−V uη)dy − a0ij = Rn Γ0 (x, y)∂i u∂j ηdy Rn − a0ij Γ0 (x, y)∂j u∂i ηdy − a0ij Rn Γ0 (x, y)u∂ij ηdy, Rn the place Γ0 (x, y) denotes the basic resolution for the operator −a0ij Dij in Rn .

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