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The adjoint solution operator solves the adjoint form problem

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, $μ \geq β$ , and let $a^{*}$ be the adjoint form of The formal adjoint and the adjoint weak Dirichlet problem. Define $K_{μ}^{*} f$ , for $f \in L^{2} (Ω)$ , to be the unique $v \in H_{0}^{1} (Ω)$ with $a_{μ}^{*} (v, w) = (f, w)_{L^{2}}$ for every $w \in H_{0}^{1} (Ω)$ (existence and uniqueness by The Lax--Milgram theorem). Then $K_{μ}^{*}$ is well defined and linear on $L^{2} (Ω)$ , and it is the Hilbert-space adjoint of the shifted solution operator $K_{μ}$ of The shifted elliptic solution operator: $(K_{μ} f, g)_{L^{2}} = (f, K_{μ}^{*} g)_{L^{2}} for all f, g \in L^{2} (Ω) .$ If $Ω$ is bounded and the Axiom of Choice is also assumed, then $K_{μ}^{*}$ is compact on $L^{2} (Ω)$ (The shifted solution operator is compact on $L^{2}$ is proved by the same bounded-map/Rellich composition for $a^{*}$ ). Moreover, for $v \in L^{2} (Ω)$ one has $v \in \ker (I - μ K_{μ}^{*})$ if and only if $v \in H_{0}^{1} (Ω)$ and $a^{*} (v, w) = 0$ for every $w \in H_{0}^{1} (Ω)$ .

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ ; the divergence-form operator $L$ , its form $a$ , the adjoint form $a^{*}$ , a fixed $μ \geq β$ , and the operators $K_{μ}, K_{μ}^{*}$ on $L^{2} (Ω)$ .

[F1]

The adjoint form $a^{*}$ and its shift: $a^{*} (v, w) = \overline{a (w, v)}$ , $a_{μ}^{*} = a^{*} + μ (\cdot, \cdot)_{L^{2}}$ , and $Re a_{μ}^{*} (u, u) = Re a_{μ} (u, u) \geq \frac{θ}{2} ∥ u ∥_{H_{0}^{1}}^{2}$ , so $a_{μ}^{*}$ is bounded and coercive on $H_{0}^{1} (Ω)$ with the same constants as $a_{μ}$ ; the datum $w \mapsto (f, w)_{L^{2}}$ is a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ (The formal adjoint and the adjoint weak Dirichlet problem, The shifted elliptic solution operator, The space $L^{p} (μ)$ as the quotient by null functions, Zero-boundary Sobolev space as a norm closure).

[F2]

Lax--Milgram: a bounded coercive sesquilinear form on a Hilbert space and a bounded conjugate-linear functional have a unique solution, and the solution map is linear with norm at most $1 / α$ (The Lax--Milgram theorem, A bounded linear operator between normed spaces, Integer-order Sobolev spaces and their norms).

[F3]

Hilbert-space adjoints: $S^{*}$ is the operator with $(S f, g) = (f, S^{*} g)$ for all $f, g$ , and it is unique (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F4]

The $L^{2}$ realization of $K_{μ}$ is compact when $Ω$ is bounded: a bounded linear map $L^{2} (Ω) \to H_{0}^{1} (Ω)$ followed by the compact Rellich inclusion $H_{0}^{1} (Ω) ↪ L^{2} (Ω)$ is compact (The shifted solution operator is compact on $L^{2}$ , Compositions with a compact operator are compact, Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets, Compact linear operator, The Axiom of Choice).

Proof

technique · direct

1.1F1F2given

Well-definedness and linearity. By [F1] and [F2] applied to $a_{μ}^{*}$ , for each $f \in L^{2} (Ω)$ there is a unique $v \in H_{0}^{1} (Ω)$ with $a_{μ}^{*} (v, w) = (f, w)_{L^{2}}$ for all $w \in H_{0}^{1} (Ω)$ ; uniqueness makes $K_{μ}^{*}$ independent of any choice, and linearity of $f \mapsto K_{μ}^{*} f$ follows from uniqueness exactly as for $K_{μ}$ , since $a_{μ}^{*}$ and the datum functional are linear in that slot.

2.1F1F3step 1.1givenalgebra

Adjoint identity. For $f, g \in L^{2} (Ω)$ put $v : = K_{μ}^{*} g$ , so that $a_{μ}^{*} (v, w) = (g, w)_{L^{2}}$ for every $w \in H_{0}^{1} (Ω)$ . Testing at $w = K_{μ} f$ and conjugating, and using $a_{μ}^{*} (v, u) = \overline{a_{μ} (u, v)}$ , $(K_{μ} f, g)_{L^{2}} = \overline{(g, K_{μ} f)_{L^{2}}} = \overline{a_{μ}^{*} (v, K_{μ} f)} = a_{μ} (K_{μ} f, v) = (f, v)_{L^{2}} = (f, K_{μ}^{*} g)_{L^{2}},$ where the penultimate identity is the defining equation of $K_{μ}$ . Hence $K_{μ}^{*}$ is the Hilbert-space adjoint of $K_{μ}$ by [F3].

2.2F4step 1.1given

Compactness on bounded $Ω$ . If $Ω$ is bounded, [F1] and [F2] make $K_{μ}^{*} : L^{2} (Ω) \to H_{0}^{1} (Ω)$ bounded linear, and composing with the compact Rellich inclusion $H_{0}^{1} (Ω) ↪ L^{2} (Ω)$ expresses $K_{μ}^{*}$ on $L^{2} (Ω)$ as a bounded map followed by a compact one, hence compact by [F4]; the Axiom of Choice is inherited through the Rellich supplier.

3.1F1F2step 1.1givenalgebra∎

Kernel at $μ$ . For $v \in L^{2} (Ω)$ one has $v \in \ker (I - μ K_{μ}^{*})$ if and only if $v = μ K_{μ}^{*} v$ , and since $ran K_{μ}^{*} \subseteq H_{0}^{1} (Ω)$ this forces $v \in H_{0}^{1} (Ω)$ and, by the defining equation of $K_{μ}^{*}$ with datum $μ v$ , $a_{μ}^{*} (v, w) = μ (v, w)_{L^{2}} for every w \in H_{0}^{1} (Ω),$ which is exactly $a^{*} (v, w) = 0$ for every $w$ . Conversely, if $v \in H_{0}^{1} (Ω)$ satisfies $a^{*} (v, w) = 0$ for all $w$ , then $a_{μ}^{*} (v, w) = μ (v, w)_{L^{2}}$ for all $w$ , so uniqueness in [F2] gives $K_{μ}^{*} (μ v) = v$ , that is $μ K_{μ}^{*} v = v$ .

Depends on

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Sources

John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page notes) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)