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Identification of the Dirichlet Laplacian with the sine diagonal operator

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ , let $e_{n} (x) = \sqrt{\frac{2}{L}} \sin ⁣ (\frac{n π x}{L}), λ_{n} : = (\frac{n π}{L})^{2} (n \geq 1),$ let $(e_{n})_{n \geq 1}$ be the complete orthonormal family of Completeness of the Dirichlet sine system on an interval, and let $A_{D}$ be the diagonal operator of Diagonal adjoint and graph-core lemma associated with $(e_{n})$ and $(λ_{n})$ : $A_{D} x = \sum_{n \geq 1} λ_{n} ⟨ x, e_{n} ⟩ e_{n}$ with domain $D (A_{D}) = {x \in L^{2} (0, L; C) : \sum_{n \geq 1} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty} .$ Let $B$ be the Dirichlet Laplacian $B u : = - D^{2} u, D (B) = {u \in H^{2} (0, L) : u^{*} (0) = u^{*} (L) = 0},$ with $H^{2}$ and the trace $u^{*}$ of One-dimensional Sobolev spaces on intervals and Integration by parts and the endpoint-vanishing space on intervals clause (b). Then:

(a) $D (A_{D}) = D (B)$ and $A_{D} = B$ ; in particular $D (A_{D})$ is a linear subspace of $H^{2} (0, L)$ whose members have vanishing endpoint traces;

(b) $A_{D}$ is self-adjoint (Symmetric, self-adjoint and essentially self-adjoint operators), positive, and its spectrum is $σ (A_{D}) = {λ_{n} : n \geq 1}$ ; each $λ_{n}$ is an eigenvalue with eigenvector $e_{n}$ ; the eigenvalues are simple in the sense that $\ker (A_{D} - λ_{n})$ contains no nonzero vector orthogonal to $e_{n}$ ;

(c) for every $u \in D (A_{D})$ , $⟨ A_{D} u, u ⟩ = ∥ D u ∥_{L^{2} (0, L)}^{2}$ ; consequently $A_{D} \geq λ_{1} = (π / L)^{2}$ in the quadratic-form sense on its domain.

Facts & Assumptions

Given: $L > 0$ ; the complete orthonormal sine family $(e_{n})$ and numbers $λ_{n} = (n π / L)^{2}$ of Completeness of the Dirichlet sine system on an interval; the diagonal operator $A_{D}$ of Diagonal adjoint and graph-core lemma.

[F1]

$(e_{n})_{n \geq 1}$ is an orthonormal family with $⟨ x, e_{n} ⟩$ the Fourier coefficients and $x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}$ in $L^{2}$ for every $x \in L^{2} (0, L)$ (Completeness of the Dirichlet sine system on an interval, Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases). Also $e_{n} \in H^{2} (0, L)$ , $e_{n} (0) = e_{n} (L) = 0$ , and the weak derivatives satisfy $D e_{n} (x) = \frac{n π}{L} \sqrt{\frac{2}{L}} \cos (\frac{n π x}{L})$ , $D^{2} e_{n} = - λ_{n} e_{n}$ , with $⟨ D e_{n}, D e_{m} ⟩ = λ_{n} δ_{n m}$ (Integration by parts and the endpoint-vanishing space on intervals and the product-to-sum computation of Completeness of the Dirichlet sine system on an interval).

[F2]

The diagonal operator lemma Diagonal adjoint and graph-core lemma states: the finite-support diagonal operator is symmetric, $A_{D}$ is self-adjoint with the displayed domain, $A_{D} e_{n} = λ_{n} e_{n}$ , the finite spans are a graph core, and $⟨ A_{D} x, x ⟩ = \sum_{n} λ_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2}$ . The companion Diagonal spectrum identifies $σ (A_{D})$ with the closure of the sequence $(λ_{n})$ .

[F3]

For $u, v \in H^{1} (0, L)$ , $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v}$ (Integration by parts and the endpoint-vanishing space on intervals clause (a)); for $u, v \in H^{2} (0, L)$ the second-order identity holds; and $H_{0}^{1} (0, L) = {u \in H^{1} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ (clause (b)).

[F4]

Bessel's inequality for an orthonormal family: for every $x \in L^{2}$ , $\sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} \leq ∥ x ∥^{2}$ (Parseval equivalences for an orthonormal family).

Proof

1.1F1F2F4givenalgebra

[Series of derivatives] Let $u \in D (A_{D})$ and put $c_{n} : = ⟨ u, e_{n} ⟩$ , so that $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} < \infty$ by [F2]. By [F4], $\sum_{n} ∣ c_{n} ∣^{2} \leq ∥ u ∥^{2}$ , so Cauchy-Schwarz gives $\sum_{n} λ_{n} ∣ c_{n} ∣^{2} \leq (\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2})^{1 / 2} (\sum_{n} ∣ c_{n} ∣^{2})^{1 / 2} < \infty,$ and hence the series $w : = \sum_{n} c_{n} D e_{n}$ converges in $L^{2}$ with $∥ w ∥^{2} = \sum_{n} λ_{n} ∣ c_{n} ∣^{2}$ by [F1].

2.1step 1.1F1F2F3F4algebra

[ $u \in D (B)$ and $A_{D} u = - D^{2} u$ ] We first show $D u = w$ for $u \in D (A_{D})$ . Let $φ \in C_{c}^{\infty} (0, L)$ and put $u_{N} : = \sum_{n \leq N} c_{n} e_{n}$ . Since $u_{N} \to u$ in $L^{2}$ , $\int u_{N} \overline{φ^{'}} \to \int u \overline{φ^{'}}$ . For each $n$ , [F3] clause (a) applied to $e_{n}$ and $φ$ (both in $H^{1} (0, L)$ , and $φ$ compactly supported in the interior) gives $\int e_{n} \overline{φ^{'}} = e_{n}^{*} (L) \overline{φ (L)} - e_{n}^{*} (0) \overline{φ (0)} - \int e_{n}^{'} \overline{φ} = - \int e_{n}^{'} \overline{φ}$ , because $φ$ vanishes near the endpoints and $e_{n} (0) = e_{n} (L) = 0$ . Summing over $n \leq N$ and passing to the limit (justified by absolute convergence: $\sum_{n} ∣ c_{n} ∣ ∣ ⟨ e_{n}^{'}, φ ⟩ ∣ \leq (\sum_{n} λ_{n} ∣ c_{n} ∣^{2})^{1 / 2} (\sum_{n} λ_{n}^{- 1} ∣ ⟨ e_{n}, φ^{'} ⟩ ∣^{2})^{1 / 2} \leq (\sum_{n} λ_{n} ∣ c_{n} ∣^{2})^{1 / 2} λ_{1}^{- 1 / 2} ∥ φ^{'} ∥$ , using $[F 4]$ for $φ^{'}$ and $λ_{n} \geq λ_{1}$ ) gives $\int u \overline{φ^{'}} = - \int w \overline{φ}$ for every test $φ$ ; hence $D u = w \in L^{2}$ and $u \in H^{1} (0, L)$ .

3.1step 1.1step 2.1F1F2F3

[Endpoint traces and second derivative] With $u \in D (A_{D})$ , $D u = w$ , and the partial sums $u_{N}$ of step 2.1, the trace bound of Endpoint trace bound and decay of H1 functions clause (a) gives $∣ u_{N}^{*} (x) - u^{*} (x) ∣ \leq L^{- 1 / 2} ∥ u_{N} - u ∥ + L^{1 / 2} ∥ D u_{N} - D u ∥ (x \in [0, L]),$ and the right side tends to $0$ because $u_{N} \to u$ and $D u_{N} \to D u$ in $L^{2}$ by [F1] and step 1.1; hence $u_{N}^{*} \to u^{*}$ uniformly and, since $u_{N}^{*} (0) = u_{N}^{*} (L) = 0$ , also $u^{*} (0) = u^{*} (L) = 0$ . To identify the weak derivative of $w$ , put $w_{N} : = \sum_{n \leq N} c_{n} D e_{n}$ and $v_{N} : = \sum_{n \leq N} c_{n} D^{2} e_{n} = - \sum_{n \leq N} λ_{n} c_{n} e_{n}$ . By step 1.1, $w_{N} \to w$ in $L^{2}$ ; by [F2] and $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} < \infty$ , $v_{N} \to - A_{D} u$ in $L^{2}$ . For each $φ \in C_{c}^{\infty} (0, L)$ , integration by parts [F3] applied to $D e_{n}$ and $φ$ gives $\int_{0}^{L} w_{N} \overline{φ^{'}} = - \int_{0}^{L} v_{N} \overline{φ}$ . Passing to the two $L^{2}$ limits proves $D w = - A_{D} u$ . Hence $D u = w \in H^{1} (0, L)$ , $D^{2} u = - A_{D} u$ , and $u \in D (B)$ with $B u = - D^{2} u = A_{D} u$ .

4.1step 3.1F1F3F4algebra

[Reverse inclusion] Let $u \in D (B)$ ; then $u \in H^{2} (0, L)$ and $D^{2} u \in L^{2}$ , and $u^{*} (0) = u^{*} (L) = 0$ . Put $c_{n} : = ⟨ u, e_{n} ⟩$ . For each $n$ , apply the second-order identity [F3] to $u$ and $e_{n}$ : all boundary terms vanish because $u$ and $e_{n}$ vanish at $0$ and $L$ , so $⟨ - D^{2} u, e_{n} ⟩ = \int_{0}^{L} (- D^{2} u) \overline{e_{n}} = - \int_{0}^{L} u \overline{D^{2} e_{n}} = λ_{n} ⟨ u, e_{n} ⟩ = λ_{n} c_{n},$ using $D^{2} e_{n} = - λ_{n} e_{n}$ from [F1]. By Bessel [F4] applied to $- D^{2} u \in L^{2}$ , $\sum_{n} λ_{n}^{2} ∣ c_{n} ∣^{2} = \sum_{n} ∣ ⟨ - D^{2} u, e_{n} ⟩ ∣^{2} \leq ∥ D^{2} u ∥^{2} < \infty,$ so $u \in D (A_{D})$ . Moreover $A_{D} u = \sum_{n} λ_{n} c_{n} e_{n} = \sum_{n} ⟨ - D^{2} u, e_{n} ⟩ e_{n} = - D^{2} u$ by the expansion [F1]. Hence $D (B) \subseteq D (A_{D})$ and $A_{D} u = B u$ .

5.1step 3.1step 4.1

[Conclusion of (a)] Steps 3.1 and 4.1 give $D (A_{D}) = D (B)$ and $A_{D} = B$ .

6.1step 5.1F2

[Self-adjointness, spectrum, positivity] By [F2], $A_{D}$ is self-adjoint with spectrum the closure of ${λ_{n}}$ ; since $λ_{n} = (n π / L)^{2} \to \infty$ and the map $n \mapsto λ_{n}$ is injective, the set ${λ_{n} : n \geq 1}$ has no finite accumulation point and is closed in $R$ , so $σ (A_{D}) = {λ_{n} : n \geq 1}$ , and [F2] gives $A_{D} e_{n} = λ_{n} e_{n}$ . If $x = \sum_{m} c_{m} e_{m} \in D (A_{D})$ lies in $\ker (A_{D} - λ_{n})$ , then $0 = (A_{D} - λ_{n}) x = \sum_{m} (λ_{m} - λ_{n}) c_{m} e_{m}$ , and uniqueness of coefficients in the orthonormal expansion [F1] gives $(λ_{m} - λ_{n}) c_{m} = 0$ for every $m$ , so $c_{m} = 0$ unless $m = n$ and $\ker (A_{D} - λ_{n}) = C e_{n}$ ; this is the simplicity assertion of (b). Finally, for $u \in D (A_{D})$ and $u \neq 0$ , [F2] gives $⟨ A_{D} u, u ⟩ = \sum_{n} λ_{n} ∣ c_{n} ∣^{2} \geq λ_{1} \sum_{n} ∣ c_{n} ∣^{2} = λ_{1} ∥ u ∥^{2} > 0$ , since $λ_{1} > 0$ ; so $A_{D}$ is positive and $A_{D} \geq λ_{1}$ in the quadratic-form sense.

7.1step 5.1F3algebra∎

[Form identity] Let $u \in D (A_{D}) = D (B)$ . Since $u^{*} (0) = u^{*} (L) = 0$ and $A_{D} u = - D^{2} u$ , the first-order identity [F3] applied to the pair $(D u, u)$ gives $⟨ A_{D} u, u ⟩ = \int_{0}^{L} (- D^{2} u) \overline{u} = \int_{0}^{L} ∣ D u ∣^{2} - [D u^{*} \overline{u^{*}}]_{0}^{L} = ∥ D u ∥_{L^{2}}^{2}$ , because the boundary terms contain the factor $u^{*} (0) = u^{*} (L) = 0$ . This is (c).

Identification of the Dirichlet Laplacian with the sine diagonal operator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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