Norm convergence of the resolvent for wild perturbations

Norm convergence of the resolvent for wild perturbations Laboratoire de mathématiques Jean Leray, Nantes Mathematik, Universität Trier Analysis and Geometry on Graphs and Manifolds Potsdam, July, 31 August,4 2017

framework (X m, g) complete connected Riemannian manifold. Energy form q(f ) = X df 2 dvol g for f C0 (X ). This form is closable and defines by polarization the operator Laplacian, with local expression (f ) = if dvol g = ρ.dx 1 dx n. is selfadjoint, non negative. 1 i,j m 1 ρ x i (ρg ij xj f )

a typical result of this paper : Ω R m open bounded with some regularity : H0 1(Ω) = {u H1 (R m ), supp(u) Ω}. K a compact set included in Ω. Ω n n Ω \ K metrically (every compact in Ω is in Ω n for n large enough, every compact outside Ω is outside Ω n for n large enough) Ω, Ωn Laplacians with Dirichlet boundary conditions Theorem ([RT]) If K has capacity zero then for any real continuous and bounded function F and any u L 2 (Ω) F ( Ωn )P n u F ( Ω)u in n L 2 (R m ).

O. Post, Spectral analysis on graph-like spaces, Lecture Notes in Mathematics, 2039, Springer, Heidelberg, 2012. H (H ε, ε > 0), separable Hilbert spaces closed non negative quadratic forms (q, H 1 ) in H and (q ε, H 1 ε) in H ε. 0, ε corresponding selfadjoint operators. In H Sobolev type spaces H k with norms f k = ( 0 + 1) k/2 f. (H 1,. 1 ) and (H 1 ε,. 1 ) are complete by assumption.

Theorem ([Post]) Suppose there is δ ε > 0, such that f H 1, u H 1 ε:

Theorem ([Post]) Suppose there is δ ε > 0, such that f H 1, u H 1 ε: 1. < J u, f > < u, Jf > δ ε f 1 u 1 2. f J Jf δ ε f 1 and u JJ u δ ε u 1 3. (J 1 J)f δ ε f 1 and (J 1 J )u δ ε u 1 4. q ε (J 1 f, u) q(f, J 1 u) δ ε f k u 1 (the quadratic forms are δ ε -quasi-unitary equivalent) then 0 and ε satisfy ( ε + 1) 1 J J( 0 + 1) 1 (k 2) 0 0 7δ ε. Moreover, if δ ε 0 we obtain the convergence of functions of the operators in norm, of the spectrum, and of the eigenfunctions, in energy norm. ( ε converge to 0 in the resolvent sense). for us, k=2

(X, g) complete Riemannian manifold of dimension m 2 X ε = X B ε with B ε = j Jε B(x j, ε).*2 ε > 0, (x j ) j Jε such that d(x j, x k ) 2b ε ε (b ε = ε α, 0 < α < 1).

(X, g) complete Riemannian manifold of dimension m 2 X ε = X B ε with B ε = j Jε B(x j, ε).*2 ε > 0, (x j ) j Jε such that d(x j, x k ) 2b ε ε (b ε = ε α, 0 < α < 1). H = L 2 (X, g) H ε = L 2 (X ε, g) H 1 = H 1 (X, g) Hε 1 = H0 1 (X ε, g) J : H H ε J 1 : H 1 H 1 ε Jf = f Xε, J 1 f = χ ε f J : H ε H J 1 : H 1 ε H 1 J u Bε = J 1u Bε = 0 χ ε cut-off function on ε < r < ε +, ε ε + b ε, χ ε (r) = 1/r m 2 1/ε m 2 resp. χ 1/ε +(m 2) 1/ε m 2 ε (r) = log(r/ε) log(ε + /ε).

q ε (J 1 f, u) q(f, J 1u) δ ε f 2 u 1??

q ε (J 1 f, u) q(f, J 1u) δ ε f 2 u 1?? q ε (J 1 f, u) q(f, J 1u) = (d(χ ε f ) df, du) X ε = (χ ε 1)(df, du) + f (dχ ε, du) B ε + u 1 ( B ε + df 2 + B ε + B ε + f 2 dχ ε 2 ) *3

assumption of bounded geometry (X, g) has bounded geometry: there exist, i 0 > 0 and k 0 such that x X, Inj(x) i 0 Ric(x) k 0.g E. Hebey, Nonlinear analysis on manifolds: Sobolev spaces and inequalities, Courant Lecture Notes in Mathematics 5, American Mathematical Society, Providence, 1999. this assure the existence of a uniform harmonic radius uniform control on the metric C 0 (X ) is dense in H2 (X, g) = H 2 Theorem (Dirichlet fading) If b ε = ε α, 0 < α < m 2 m (and for m=2 b ε = log ε α, 0 < α < 1/2) The Laplacian with Dirichlet boundary conditions on X ε converge (in the resolvent sense) to the Laplacian on X.

crushed ice problem also results for the solidifying situation: X ε ε 0 X \ Ω 0 Ω 0 open in X, with regular boundary ; B ε Ω 0.

crushed ice problem also results for the solidifying situation: X ε ε 0 X \ Ω 0 Ω 0 open in X, with regular boundary ; B ε Ω 0. k N, b + ε ε,, α ε > 0 such that *4 Ω αε = {x X, d(x, Ω 0 ) < α ε } B b + ε x X, {j J, x B(x j, b + (ε))} k Theorem If lim ε 0 α ε λ ε = +, the Laplacian ε with Dirichlet boundary conditions on X ε = X B ε converge in the sense of the resolvent, to the Laplacian 0 with Dirichlet boundary conditions on X Ω 0.

(X, g) Riemannian complete, dimension m 2 and with bounded geometry

(X, g) Riemannian complete, dimension m 2 and with bounded geometry ε > 0, (x ± j ) j J such that d(x ± j, x ± k ) 2b ε ε We define X ε = X B ε with B ε = s=±,j J B(x s j, ε) a set of handles of length l ε > 0 : C ε = j J [0, l ε ] εs m 1. j J ε, we glue [0, l ε ] εs m 1 between x j and x + j (almost isometric):*5

Let C = j J [0, 1] S m 1.

Let C = j J [0, 1] S m 1. Φ : L 2 (C ε ) L 2 (C) h ε m 1 l ε h isometry

Let C = j J [0, 1] S m 1. Φ : L 2 (C ε ) L 2 (C) h ε m 1 l ε h isometry quadratic form on study: q ε (u, h) = du 2 + X ε j J C ( 1 s h j 2 + 1 ε 2 θh j 2) with domain: { D(q ε ) = (u, h) H 1 (X ε ) H 1 (C); j J ε h j (0, θ) = ε m 1 l ε u(x j, εθ), h j (1, θ) = } ε m 1 l ε u(x + j, εθ) l 2 ε

fading condition Theorem If b ε = ε β (β < 1/2), lim ε 0 l ε = 0 and lim ε 0 ε 1 2β l ε = 0, then lim ( ε + 1) 1 J J( + 1) 1 0 0 = 0. ε 0

fading condition Theorem If b ε = ε β (β < 1/2), lim ε 0 l ε = 0 and lim ε 0 ε 1 2β l ε = 0, then lim ( ε + 1) 1 J J( + 1) 1 0 0 = 0. ε 0 H = L 2 (X ), H ε = L 2 (X ε ) j J L 2 (C), H 1 = H 1 (X ), H 1 ε = D(q ε )

fading condition Theorem If b ε = ε β (β < 1/2), lim ε 0 l ε = 0 and lim ε 0 ε 1 2β l ε = 0, then lim ( ε + 1) 1 J J( + 1) 1 0 0 = 0. ε 0 H = L 2 (X ), H ε = L 2 (X ε ) j J L 2 (C), H 1 = H 1 (X ), H 1 ε = D(q ε ) J : H H ε J 1 : H 1 Hε 1 f (1 Xε.f, 0) f (1 Xε.f, Φ ε (f ))(harmonic on C ε ) J = J : H ε H J 1 : H 1 ε H 1 u 1 Xε.u u ũ(harmonic on B ε )

gluing condition

gluing condition Ω ± regular isometric (d(ω +, Ω ) > 0), let: Ω α = Ω α Ω + α, Ω = Ω Ω +. α 0 > 0 φ : Ω α 0 Ω + α 0 isometry such that φ(ω ) = Ω +

gluing condition Ω ± regular isometric (d(ω +, Ω ) > 0), let: Ω α = Ω α Ω + α, Ω = Ω Ω +. α 0 > 0 φ : Ω α 0 Ω + α 0 isometry such that φ(ω ) = Ω + Theorem If B ± ε solidify in Ω ± and if lim α ε + b+m ε (l ε + ε) ε 0 α ε ε m 1 + εl ε b +2 ε = 0 then the Laplacian on X ε C ε converges, in the resolvent sense, to the Laplacian on functions on X which coincide on Ω + and Ω : D( 0 ) = {f H 2, (f f φ) Ω = 0}

Thank you!

complements Φ ε (f ) = (h j ) j J satisfies 2 s (h j ) + ( lε ε )2 S m 1(h j ) = 0. Gluing condition applies for l ε = ε if there exists β, m 2 m 1 < β < 1 and γ, β < γ < βm (m 2) such that Ω ± α(ε) B± ε as a + k-regular cover, for α(ε) = ε γ and ε + = ε β.

recall transplantations: J : H H ε J 1 : H 1 Hε 1 J : H ε H J 1 : Hε 1 H 1. assumption of quasi-unitary equivalence: f H 1, u H 1 ε: 1. < J u, f > < u, Jf > δ ε f 1 u 1 2. f J Jf δ ε f 1 and u JJ u δ ε u 1 3. (J 1 J)f δ ε f 1 and (J 1 J )u δ ε u 1 4. q ε (J 1 f, u) q(f, J 1 u) δ ε f 2 u 1 conclusion: ( ε + 1) 1 J J( 0 + 1) 1 0 0 4δ ε.