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A nonsymmetric coercive elliptic form

Example

Assume the Axiom of Choice and Countable Choice. Let n≥2, let Ω⊆Rn be nonempty, open and bounded in one direction, and let CP be its Poincar'e constant for W01,2. Set b=e1 and define a(u,v):=∫Ω∇u⋅∇v‾ dx+∫Ω∂1u v‾ dx(u,v∈H01(Ω)). This is a bounded sesquilinear form with bound 2 and is coercive with constant 1/(1+CP2). It is not symmetric: choose x0∈Ω with x0,2≠0, a ball Br(x0)⊂Ω, a nonzero real radial bump η supported in that ball, and put u=x2η, v=x1η. Then a(u,v)−a(v,u)‾=−∫Ωx2η2 dx=−x0,2∫Ωη2 dx≠0. Thus Lax--Milgram (The Lax--Milgram theorem) applies to this weak Dirichlet problem for Lu=−Δu+∂1u, while the minimisation characterisation of Symmetric Lax--Milgram is energy minimisation does not apply. This is the companion example of Nonsymmetric Lax--Milgram is not a scalar minimisation principle and of the drift term in A large adverse zero-order term destroys Dirichlet coercivity.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; n≥2; a nonempty open Ω⊆Rn bounded in one direction; b=e1; the form a above; and the stated bump supported in a ball about x0 with x0,2≠0.

[F1]

Since b=e1 has ∥b∥L∞=1, Cauchy--Schwarz gives ∣a(u,v)∣≤∥∇u∥2∥∇v∥2+∥∇u∥2∥v∥2≤2∥u∥H01∥v∥H01; the form is bounded and sesquilinear (The elliptic form is well defined and bounded on H1, Uniformly elliptic divergence-form operators and their sesquilinear forms, Holder's inequality for integrals, including the endpoint cases).

[F2]

Poincar'e gives ∥u∥H012≤(1+CP2)∥∇u∥22, so the principal form has coercivity constant 1/(1+CP2) (Coercivity of the principal Dirichlet form, The Sobolev space H1 is a Hilbert space, Integer-order Sobolev spaces and their norms).

[F3]

For φ∈Cc∞(Ω), its zero extension is smooth and compactly supported in Rn. Choose R>0 so that its support lies in (−R,R)n. For each fixed x′=(x2,…,xn), the function t↦∣φ(t,x′)∣2 has compact support in (−R,R), so the one-dimensional fundamental theorem gives ∫−RR∂1∣φ(t,x′)∣2 dt=0. Fubini on the cube then gives ∫Ω∂1∣φ∣2=0, hence Re⁡∫Ω∂1φ φ‾=0. Also Cc∞(Ω) is dense in H01(Ω), and u↦∫Ω∂1u u‾ is continuous in the H1 norm by Cauchy--Schwarz and [F1] (The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a), Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Test function space d of an open set, Zero-boundary Sobolev space as a norm closure, Holder's inequality for integrals, including the endpoint cases).

[F4]

The coordinate product rule gives ∂1(x2η)=x2∂1η and ∂1(x1η)=η+x1∂1η (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0). For a radial bump about x0, reflection x2↦2x0,2−x2 leaves η2 unchanged and has absolute Jacobian 1; applying The published Riemann change-of-variables theorem already gives the Lebesgue formula for continuous compactly supported integrands on Rn to (x2−x0,2)η2 shows its integral equals its negative, hence is zero.

[F6]

The real or complex Hilbert space H01(Ω) is complete, and Lax--Milgram applies to every bounded coercive sesquilinear form without symmetry; the energy-minimisation conclusion requires symmetry (The Sobolev space H1 is a Hilbert space, The Lax--Milgram theorem, Symmetric Lax--Milgram is energy minimisation, Nonsymmetric Lax--Milgram is not a scalar minimisation principle, Weak Dirichlet solutions for a divergence-form operator).

Proof

Given: The Axiom of Choice and Countable Choice; the stated Ω, b=e1 and form; and the bump η.

1.1F1

Boundedness: by [F1] the form is bounded with ∣a(u,v)∣≤2∥u∥H01∥v∥H01 and is linear in its first argument and conjugate-linear in its second.

1.2F3

The drift has zero real part: for φ∈Cc∞(Ω), [F3] gives 2Re⁡∫Ω∂1φ φ‾=∫Ω∂1∣φ∣2=0. By continuity and density in [F3], this extends to every u∈H01(Ω), so Re⁡a(u,u)=∥∇u∥22.

1.3F3F4F5algebraconstruct

Nonsymmetry: openness and nonemptiness of Ω give x0∈Ω with x0,2≠0 and a ball Br(x0)⊂Ω. Set s=3r/4 and η(x)=σ((s2−∣x−x0∣2)/(s2−r2/4)). By [F5] this is a smooth real radial function, equals 1 on B‾r/2(x0) and vanishes outside Bs(x0); hence its support is contained in B‾s(x0)⊂Br(x0) and ∫η2>0. Thus u=x2η and v=x1η are admissible smooth compactly supported tests. Their principal parts cancel, and [F4] gives a(u,v)−a(v,u)‾=∫Ω(∂1u v−∂1v u) dx=−∫Ωx2η2 dx=−x0,2∫Ωη2 dx≠0.

2.1F1F2F6step 1.2step 1.3

Coercivity and solvability: by [F2] and step 1.2, Re⁡a(u,u)=∥∇u∥22≥(1+CP2)−1∥u∥H012, so the bounded form is coercive. Lax--Milgram gives the weak Dirichlet solution, while the minimisation result does not apply to this nonsymmetric form.

3.1F6step 2.1step 1.3∎

Conclusion: b=e1 gives a concrete bounded, coercive, nonsymmetric form on every such nonempty open Ω in dimension n≥2; the example demonstrates exactly why symmetry is required for the energy-minimisation characterization.

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