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A two-cell linear finite element best approximation for Poisson

Example

For I=(0,1) and −u′′=1 with zero endpoint values, the exact weak solution is u(x)=12x(1−x). Take the mesh 0,12,1 and the continuous piecewise-linear basis function ϕ(x)={2x,0≤x≤12,2(1−x),12≤x≤1. The Galerkin space is Vh=span⁡{ϕ}. The unique discrete solution is uh=18ϕ, and the energy error is ∥u−uh∥a2=∫01∣(u−uh)′∣2 dx=148. By the Hermitian best-approximation result, uh is the energy-best element of this one-dimensional space.

Facts & Assumptions

Given: Countable Choice and the Axiom of Choice; I=(0,1); a scalar field K∈{R,C}; f=1, the piecewise-linear function ϕ, and the standard Poisson form a(v,w)=∫01v′w′‾ dx (with conjugation the identity over R).

[F1]

For a Hermitian coercive form on a Hilbert space, the conforming Galerkin solution is the best approximant in the induced energy norm (Hermitian Galerkin solutions are best in the energy norm).

[F2]

The lemma gives explicit real Cc∞ approximants for ϕ and p; these are test functions over either scalar field, so both functions belong to the corresponding H01(0,1;K) (The two-cell hat and quadratic functions belong to H¹₀).

[F3]

The weak Dirichlet Poisson theorem applies on an open nonempty domain bounded in one direction when a Poincar'e constant for W01,2 is specified; under Axiom of Choice and Countable Choice it gives a unique weak solution for each bounded conjugate-linear datum (Existence and uniqueness for the weak Dirichlet Poisson problem).

[F4]

The real or complex L2 integral pairing is well defined, symmetric or conjugate symmetric, and satisfies Cauchy--Schwarz (The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz).

[F5]

The classical second derivative of a smooth function is its weak second derivative, with the bilinear test identity (Classical derivatives agree with weak derivatives).

[F6]

The constant function 1 has L2(0,1) norm 1; its integral pairing defines a bounded conjugate-linear functional on H01(0,1) by Cauchy--Schwarz and ∥v∥L2≤∥v∥H1 (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included, The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz, Integer-order Sobolev spaces and their norms, The negative Sobolev space H−1(Ω)).

[A1]

Countable Choice and the Axiom of Choice are retained from the cited weak-Poisson and coercive-form suppliers (The Axiom of Countable Choice (ACω), The Axiom of Choice).

[F7]

H01(0,1) is the closure of Cc∞(0,1) in the W1,2 norm, and test functions have their classical derivatives as weak derivatives (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms).

[F8]

The real second fundamental theorem evaluates the integral of a continuous derivative, continuous real functions on compact intervals are Riemann integrable, and a bounded Riemann-integrable function has the same Lebesgue integral (The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a), Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral). For integrable real or complex functions, restricting the integral to a measurable set corresponds to multiplying by its indicator; this follows componentwise from the positive/negative-part definition (Integral over a measurable subset, Integrable real and complex functions, and their integrals).

[F9]

λ1([0,x])=x for 0≤x≤1 and λ1((0,1))=1; the integral of an indicator is the measure of its set (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included, The nonnegative Lebesgue integral).

Verification

1.1F2A1given

The function ϕ is absolutely continuous, vanishes at 0 and 1, and has derivative 2 on (0,1/2) and −2 on (1/2,1). Its derivative is in L2, so [F2] gives ϕ∈H01(0,1).

1.2F2A1givenalgebra

Put p(x)=12x(1−x). It is absolutely continuous with zero endpoint values, p′=12−x∈L2, and p′′=−1; [F2] gives p∈H01(0,1).

1.3F4F7F8F9F10givenalgebra

Poincar'e constant for this interval. Let v∈H01(0,1;K) and δ>0. By [F7], choose φ∈Cc∞(0,1;K) with ∥v−φ∥H1<δ. For 0<x≤1, apply the real second fundamental theorem to the real and imaginary parts of φ on [0,x]; their derivatives are continuous and hence Riemann integrable by [F8]. The Riemann--Lebesgue equality in [F8] and the definition of the integral over a set give φ(x)=(φ′,1[0,x])L2(0,1), since φ(0)=0. By [F9], ∥1[0,x]∥2=x, so [F4] gives ∣φ(x)∣≤x∥φ′∥2≤∥φ′∥2; this also holds at x=0. Integrating the squared bound over (0,1) and using λ1((0,1))=1 and [F10] yields ∥φ∥2≤∥φ′∥2. Therefore ∥v∥2≤∥φ∥2+δ≤∥φ′∥2+δ≤∥v′∥2+2δ. Since δ is arbitrary, ∥v∥2≤∥v′∥2 for every v∈H01(0,1;K), so the required Poincar'e constant is CP=1.

2.1F3F5F6F7A1step 1.3givenalgebra

For v∈Cc∞(0,1), apply [F5] to p′ with test v‾. Since p′′=−1, it gives a(p,v)=∫01p′v′‾ dx=−∫01p′′v‾ dx=∫01v‾ dx. Both sides are continuous in the H1 norm by [F4]. Since H01 is the W1,2-closure of these tests, the identity extends to all v∈H01(0,1) by [F7]. By [F6], the datum v↦∫01v‾ lies in H−1(0,1). The interval I is open and nonempty, is bounded in direction e=1 between a=0 and b=1, and step 1.3 supplies CP=1; [A1] supplies the required axioms. Thus [F3] applies and makes p the unique weak solution.

2.2step 1.3givenstep 1.1algebra

Direct integration gives a(ϕ,ϕ)=∫01/24 dx+∫1/214 dx=4,∫01ϕ(x) dx=12. For uh=cϕ, the Galerkin equation tested with ϕ is 4c=∫01ϕ=12, hence c=18 and uh=18ϕ.

3.1F1F3F4A1step 2.2algebra∎

On the first half interval, (p−uh)′=14−x; on the second, (p−uh)′=34−x. Therefore ∥p−uh∥a2=∫01/2(14−x)2 dx+∫1/21(34−x)2 dx=196+196=148. By [F1], with the Hilbert and coercivity hypotheses verified in [F3] and the Hermitian property in [F4], this computed Galerkin solution is the energy-best approximation.

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