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Lax--Milgram solvability for coercive divergence-form equations

Statement

Assume the Axiom of Choice, inherited through the Poincaré supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, nonempty and bounded in one direction, let $L$ and $a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with ellipticity constant $θ$ , coefficient bounds $M_{a}, M_{b}, M_{c}$ (with the componentwise drift bounds $∣ b^{i} ∣ \leq M_{b}$ ), and let $C_{P}$ be the Poincar'e constant of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction for $p = 2$ . Assume the explicit smallness condition $θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0.$ Then for every $F \in H^{- 1} (Ω)$ there is a unique weak solution $u \in H_{0}^{1} (Ω)$ of $L u = F$ (Weak Dirichlet solutions for a divergence-form operator), and with $α_{0} : = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c}$ it satisfies $∥ u ∥_{H_{0}^{1}} \leq \frac{1 + C_{P}^{2}}{α_{0}} ∥ F ∥_{H^{- 1}} .$ When $b \equiv 0$ , taking $M_{b} = 0$ , the condition reduces to $M_{c} < θ / C_{P}^{2}$ , the sign/smallness condition of the plan; in the model case $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ , taking $θ = 1$ and $M_{b} = M_{c} = 0$ , it gives $α_{0} = 1$ and gives existence and uniqueness for the zero-boundary weak Poisson problem.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open, nonempty $Ω \subseteq R^{n}$ bounded in one direction; divergence-form coefficients with ellipticity constant $θ > 0$ and bounds $M_{a}, M_{b}, M_{c}$ , where $∣ b^{i} ∣ \leq M_{b}$ componentwise; the Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ at $p = 2$ ; the smallness assumption $α_{0} : = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0$ ; and the form $a$ on $H_{0}^{1} (Ω)$ .

[F1]

Pointwise ellipticity and coefficient bounds: $Re (a^{i j} D_{j} u \overline{D_{i} u}) \geq θ ∣ D u ∣^{2}$ a.e. and $∣ b^{i} ∣ \leq M_{b}$ , $∣ c ∣ \leq M_{c}$ a.e. (Uniformly elliptic divergence-form operators and their sesquilinear forms, The essential supremum of a measurable function with respect to a measure, The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[F2]

The form $a$ is bounded on $H_{0}^{1} (Ω)$ (The elliptic form is well defined and bounded on $H^{1}$ ) and $H_{0}^{1} (Ω)$ is a Hilbert space with $∥ u ∥_{H_{0}^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2}$ (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

Poincar'e: $∥ u ∥_{L^{2}} \leq C_{P} ∥ D u ∥_{L^{2}}$ for $u \in H_{0}^{1} (Ω)$ , hence $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F4]

Lax--Milgram and the a priori estimate: a bounded coercive form on a Hilbert space with a bounded conjugate-linear datum has a unique solution; any solution satisfies $α ∥ u ∥_{H_{0}^{1}} \leq ∥ F ∥$ for a coercivity constant $α$ (The Lax--Milgram theorem, Testing a coercive weak solution with itself gives the energy bound, The negative Sobolev space $H^{- 1} (Ω)$ , Weak Dirichlet solutions for a divergence-form operator).

[F5]

For $u \in H^{1} (Ω)$ , $\sum_{i = 1}^{n} ∥ D_{i} u ∥_{L^{2}} \leq \sqrt{n} ∥ D u ∥_{L^{2}}$ by Cauchy--Schwarz in the finite coordinate index (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

1.1F1F2F3F5algebra

Coercivity: for $u \in H_{0}^{1} (Ω)$ , pointwise ellipticity and the coefficient bounds give $Re a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} - \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}} - M_{c} ∥ u ∥_{L^{2}}^{2},$ since $∣ b^{i} ∣ \leq M_{b}$ and [F5] bound the coordinate sum. Poincar'e gives $∥ u ∥_{L^{2}} \leq C_{P} ∥ D u ∥_{L^{2}}$ , so the last two terms are at least $- \sqrt{n} C_{P} M_{b} ∥ D u ∥_{L^{2}}^{2}$ and $- C_{P}^{2} M_{c} ∥ D u ∥_{L^{2}}^{2}$ ; hence $Re a (u, u) \geq α_{0} ∥ D u ∥_{L^{2}}^{2}$ with $α_{0} = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0$ . Since $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ , this gives $Re a (u, u) \geq \frac{α_{0}}{1 + C_{P}^{2}} ∥ u ∥_{H_{0}^{1}}^{2}$ : the form is coercive on $H_{0}^{1} (Ω)$ with constant $α_{0} / (1 + C_{P}^{2})$ , and it is bounded by [F2].

2.1F2F4step 1.1

Solvability: applying Lax--Milgram [F4] to the Hilbert space $H_{0}^{1} (Ω)$ , the bounded coercive form $a$ and the datum $F \in H^{- 1} (Ω)$ gives a unique $u \in H_{0}^{1} (Ω)$ with $a (u, v) = F (v)$ for every $v \in H_{0}^{1} (Ω)$ : a unique weak solution of $L u = F$ .

3.1F4step 1.1algebra∎

Estimate: the a priori estimate of [F4] with $α = α_{0} / (1 + C_{P}^{2})$ gives $∥ u ∥_{H_{0}^{1}} \leq \frac{1 + C_{P}^{2}}{α_{0}} ∥ F ∥_{H^{- 1}}$ . When $b \equiv 0$ , taking $M_{b} = 0$ , the condition is $M_{c} < θ / C_{P}^{2}$ , and in the model case $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ , taking $θ = 1$ and $M_{b} = M_{c} = 0$ , one has $α_{0} = 1$ , recovering the zero-boundary Poisson theorem.

Lax--Milgram solvability for coercive divergence-form equations

Statement

Facts & Assumptions

Proof

Depends on

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Sources