Norm convergence of the resolvent for wild perturbations
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1 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
2 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
3 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 )
4
5 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 ).
6 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 ). P n (u) = χ Ω Ωn.u.
7 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 ). P n (u) = χ Ω Ωn.u. *1this assure convergence of the discret spectrum. examples : small holes.
8 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 ). P n (u) = χ Ω Ωn.u. *1this assure convergence of the discret spectrum. examples : small holes. question: can we have convergence in norm which works also for unbounded domains?
9 O. Post, Spectral analysis on graph-like spaces, Lecture Notes in Mathematics, 2039, Springer, Heidelberg, 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.
10 O. Post, Spectral analysis on graph-like spaces, Lecture Notes in Mathematics, 2039, Springer, Heidelberg, 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. J : H H ε J 1 : H 1 Hε 1 J : H ε H J 1 : Hε 1 H 1. bounded operators of transplantation.
11 Theorem ([Post]) Suppose there is δ ε > 0, such that f H 1, u H 1 ε:
12 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)
13 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.
14 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
15 (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).
16 (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)
17 (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).
18 (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). To prove : q ε (J 1 f, u) q(f, J 1u) δ ε f 2 u 1
19 q ε (J 1 f, u) q(f, J 1u) δ ε f 2 u 1??
20 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
21 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, 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
22 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, 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.
23 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.
24 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
25 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.
26 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. λ ε = λ 1 ( N D (B R m(0, b+ ε ) B R m(0, ε)) Cεm b ε +(m 2). [RT]
27
28
29 (X, g) Riemannian complete, dimension m 2 and with bounded geometry
30 (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
31 Let C = j J [0, 1] S m 1.
32 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
33 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 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
34 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) = 0. ε 0
35 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) = 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 )
36 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) = 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 )
37 gluing condition
38 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 +
39 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}
40 Thank you!
41 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.
42 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) δ ε.
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