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The Direct Method and Euler--Lagrange Equations — Examples

1 · Prerequisites

2 · Summary

These companions compute the direct method and the Euler--Lagrange equation on explicit energies and delimit each hypothesis by a witness. On the one hand two worked computations: an affine function on a bounded C1 domain is the unique minimiser of the Dirichlet energy among all H1 functions with the same boundary trace, and on a compact interval the energy ∫ab(12u′(x)2+V(u(x)))dx with fixed endpoint values has its local minimisers satisfying −u′′+V′(u)=0, while free endpoints add the natural conditions fu′=0 at each free endpoint. The fixed-trace and free-trace versions of the L2-forced Dirichlet energy are then contrasted: the same energy yields the Dirichlet problem in the first case and the Neumann condition ∂νu=0 in the second. Constant shifts obey I(u+c)=I(u)−c∫Ωf, so the free energy is invariant under constants exactly when ∫Ωf=0, and is never coercive on the whole H1 space. Under the stated connected extension-domain hypotheses, compatible zero-mean forcing gives a unique mean-zero weak Neumann solution. Nonzero mean forbids a free local minimiser and cannot be repaired by normalising the solution; on disconnected domains compatibility and normalisation are required on each component.

On the other hand the hypotheses are shown to be load-bearing: coercivity alone does not attain its infimum when weak lower semicontinuity fails at the only candidate (ℓ2 with I(0)=1, I(u)=∥u∥2 otherwise); a minimising sequence for a weakly lower semicontinuous convex functional need not converge strongly to the minimiser (the standard basis of ℓ2 in the unit ball); a norm-closed set that is not convex need not be weakly closed (the unit sphere of an infinite-dimensional Hilbert space is weakly dense in the unit ball); strict convexity is genuinely needed for uniqueness of a minimiser (I(x,y)=x2 on R2 minimises along a line); a nonconvex gradient integrand can lose weak lower semicontinuity altogether (the sawtooth sequence uk(x)=1kw(kx) makes ∫01(u′2−1)2 vanish on the sequence but equal 1 at the weak limit 0); and stationarity in the Euler--Lagrange equation does not imply a minimum, since J(u)=−12∫Ω∣Du∣2 has u=0 as its only stationary point on H01(Ω) yet is unbounded below.

The conventions are those of the main page: bounded C1 domains in Euclidean space, 1<p<∞, weak limits taken sequentially, and the Euler--Lagrange equation written −div⁡fξ+fs=0. The Dirichlet and free-trace comparison and the nonconvex-gradient counterexample declare the choice principles used by their suppliers. The Hilbert-space counterexamples assume AC for the counting-measure L2 reflexivity and duality interfaces; the unit-sphere example also invokes the HB-dependent weak-closure theorem. The weighted-energy example checks its explicit weak limit directly, without a subsequence extraction. The scalar convexity witness I(x,y)=x2 uses no choice principle, while the one-dimensional variation examples explicitly assume Countable Choice for their measure and fundamental-lemma interfaces.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Non-strict convexity allows many minimisers

Statement refuted

Counterexample. On the real Banach space R2 consider I(x,y)=x2. Then I is convex (Convex and strictly convex functionals on a convex subset of a real vector space) and arg min⁡R2I={(0,y):y∈R}, an affine line of minimisers: I≥0 and I(0,y)=0 for every y, while for x≠0, I(x,y)>0. The functional is not strictly convex, so the strictness hypothesis in Strict convexity gives uniqueness of a minimiser is genuinely needed: convexity alone does not force uniqueness of a minimiser.

Facts & Assumptions

Given: The functional I:R2→R, I(x,y)=x2, on the real Banach space R2.

[F1]

A function I on a convex set is convex when I(λu+(1−λ)v)≤λI(u)+(1−λ)I(v) for all u,v and λ∈[0,1], and strictly convex when the inequality is strict for u≠v and λ∈(0,1) with finite values (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

A proper, strictly convex functional has at most one minimiser on a convex set; in particular at most one finite minimiser (Strict convexity gives uniqueness of a minimiser).

Counterexample

technique · direct verification
1.1F1algebra

I is convex. For u=(x1,y1), v=(x2,y2) and λ∈[0,1] one computes I(λu+(1−λ)v)=(λx1+(1−λ)x2)2 and λI(u)+(1−λ)I(v)=λx12+(1−λ)x22, whose difference is λ(1−λ)(x1−x2)2≥0; hence I is convex by [F1].

1.2algebra

The minimisers. Since I(x,y)=x2≥0 for every (x,y), and x2=0 exactly when x=0, the infimum of I on R2 is 0 and the set of minimisers is exactly the affine line {(0,y):y∈R}.

1.3F1algebra

I is not strictly convex. Take the distinct points u=(0,0) and v=(0,1), which satisfy I(u)=I(v)=0<+∞, and λ=12. Then I(12u+12v)=I(0,12)=0=12I(u)+12I(v), so the strict inequality required by [F1] fails; hence I is not strictly convex.

2.1F2step 1.2step 1.3∎

Uniqueness genuinely needs strictness. The line {(0,y):y∈R} consists of pairwise distinct minimisers of the convex functional I by step 1.2, so convexity alone does not force uniqueness; by [F2] the uniqueness conclusion requires the strict convexity hypothesis, which fails for this functional by step 1.3. This is exactly the sharpness recorded in the statement refuted.

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A coercive functional need not attain without weak lower semicontinuity

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let H=ℓ2(N,R) (Square-summable families on an arbitrary index set and the space ℓ2(I)), let ek be the family that is 1 at k and 0 elsewhere, and define I:H→R by I(0):=1,I(u):=∥u∥2  (u≠0). Then I is proper and coercive (Proper, coercive and weakly lower semicontinuous extended-real functionals) and inf⁡HI=0, but no point of H minimises I: the value 0 is not attained, because only u=0 gives ∥u∥2=0 and I(0)=1. The functional is not weakly sequentially lower semicontinuous at 0: for j≥1, ej/j⇀0 while I(ej/j)=1/j2→0<1=I(0). Hence the weak lower semicontinuity hypothesis in The direct method in a reflexive Banach space cannot be replaced by coercivity alone, even in a reflexive space.

Facts & Assumptions

Given: The Axiom of Choice; the real Hilbert space H=ℓ2(N,R) (Square-summable families on an arbitrary index set and the space ℓ2(I)), its coordinate vectors ek, namely the family that is 1 at k and 0 elsewhere, and the functional I:H→R with I(0):=1 and I(u):=∥u∥2 for u≠0. The counting-measure dictionary ℓp is the Lp space of counting measure identifies H with real L2(#), which is reflexive (and thus Banach) by Reflexivity of Lp for one less p less infinity under Countable Choice, supplied here by AC; the series pairing is its inner product.

[F1]

Proper, coercive and weakly sequentially lower semicontinuous functionals are defined as in Proper, coercive and weakly lower semicontinuous extended-real functionals; a functional is proper when its effective domain is nonempty, equivalently when its infimum is less than +∞ (it may be −∞).

[F2]

In the direct method, weak sequential lower semicontinuity is a hypothesis alongside coercivity; The direct method in a reflexive Banach space states all of its hypotheses explicitly, so a coercive proper functional on a reflexive space need not attain when that hypothesis fails.

[F3]

Each coordinate vector satisfies ek∈H and ∥ek∥=1, and ek⇀0: by the duality of ℓp and ℓq every bounded linear functional on ℓ2 is Λ(a)=∑kakbk for a unique b∈ℓ2 (Counting measure specializes the representation theorem to ℓp and ℓq), so Λ(ek)=bk, and bk→0 because a square-summable family has small tails, that is, for every ε>0 there is a finite F with ∑k∉F∣bk∣2<ε (Square-summable families on an arbitrary index set and the space ℓ2(I)), whence ∣bk∣2<ε for every k beyond all elements of F; weak convergence means convergence against every bounded linear functional (Weak convergence of nets and sequences).

Counterexample

technique · direct computation of the values along the sequence $e_j/j$, $j\ge1$
1.1F1algebra

I is proper. The effective domain of I is all of H, which is nonempty, and I≥0 with I(e1/j)=1/j2→0 as j→∞, so inf⁡HI=0<+∞; by [F1] the functional is proper.

1.2F1algebra

I is coercive. For ∥u∥≥1 one has I(u)=∥u∥2≥∥u∥ (and the value I(0)=1 does not affect large norms). Given M∈R, put R:=max⁡{1,M+1}; then ∥u∥≥R implies I(u)≥∥u∥≥M+1>M, so every sublevel set is bounded and I is coercive by [F1].

1.3F1algebra

None of the values 0 is attained. If I(u)=0 then u≠0 and ∥u∥2=0, hence u=0, a contradiction; and I(0)=1≠0. Since inf⁡HI=0, the infimum is not attained.

1.4F3algebra

Failure of weak lower semicontinuity at 0. Let uj:=ej/j for j≥1. Then ∥uj∥=1/j→0 by [F3], and for every bounded linear functional Λ on H one has Λ(uj)=Λ(ej)/j→0 by [F3]; hence uj⇀0. On the other hand uj≠0 gives I(uj)=∥uj∥2=1/j2→0<1=I(0). Hence I(0)>lim inf⁡jI(uj), and I is not weakly sequentially lower semicontinuous at 0.

2.1F2step 1.3step 1.4∎

Conclusion. The functional is proper and coercive on the reflexive space H, yet attains no minimum and fails weak lower semicontinuity; hence the weak lower semicontinuity hypothesis of [F2] cannot be dropped, and coercivity alone does not give attainment even in a reflexive space.

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A minimising sequence need not converge strongly

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let H=ℓ2(N,R), let B={u∈H:∥u∥≤1} be its closed unit ball and let I(u)=∑j≥01j+1uj2(u∈H). Then I is nonnegative, convex and continuous, I(0)=0, and uk:=ek is a minimising sequence for inf⁡BI=0 with I(ek)=1/(k+1)→0; moreover ek⇀0∈B and 0 is the unique minimiser of I on B, but ∥ek−0∥=1 for every k, so no subsequence of the minimising sequence converges strongly to the minimiser. The compactness recovered in the direct method is therefore genuinely weak compactness (The direct method in a reflexive Banach space).

Facts & Assumptions

Given: The Axiom of Choice; the real Hilbert space H=ℓ2(N,R) (Square-summable families on an arbitrary index set and the space ℓ2(I)), its coordinate vectors ek, namely the family that is 1 at k and 0 elsewhere, its closed unit ball B={u∈H:∥u∥≤1}, and the functional I(u)=∑j≥0uj2/(j+1). The counting-measure dictionary ℓp is the Lp space of counting measure identifies H with real L2(#), which is reflexive (and thus Banach) by Reflexivity of Lp for one less p less infinity under Countable Choice, supplied here by AC; the series pairing is its inner product.

[F1]

The series defining I converges absolutely for every u∈H because 0≤uj2/(j+1)≤uj2 and ∑juj2=∥u∥2<∞; moreover ∣I(u+v)−I(u)∣≤2∥u∥∥v∥+∥v∥2, so I is continuous, and I is convex and nonnegative (Square-summable families on an arbitrary index set and the space ℓ2(I)).

[F2]

Each coordinate vector satisfies ek∈H, ∥ek∥=1 and ek⇀0: by the duality of ℓp and ℓq every bounded linear functional on ℓ2 is Λ(a)=∑kakbk for a unique b∈ℓ2 (Counting measure specializes the representation theorem to ℓp and ℓq), so Λ(ek)=bk, and bk→0 because a square-summable family has small tails, that is, for every ε>0 there is a finite F with ∑k∉F∣bk∣2<ε (Square-summable families on an arbitrary index set and the space ℓ2(I)), whence ∣bk∣2<ε for every k beyond all elements of F; weak convergence means convergence against every bounded linear functional (Weak convergence of nets and sequences).

[F3]

In the direct method the compactness recovered is weak sequential compactness: a norm-bounded sequence in a reflexive Banach space has a weakly convergent subsequence (A bounded sequence in a reflexive Banach space has a weakly convergent subsequence), and the limit need not be a strong limit (The direct method in a reflexive Banach space, Proper, coercive and weakly lower semicontinuous extended-real functionals).

Counterexample

technique · direct computation along the standard basis
1.1F1algebra

I is nonnegative, convex and continuous. Nonnegativity and convexity are immediate from the formula. For continuity, expanding the squares gives I(u+v)−I(u)=∑j(2ujvj+vj2)/(j+1), and by Cauchy-Schwarz, ∑j∣2ujvj∣/(j+1)≤2(∑juj2)1/2(∑jvj2)1/2=2∥u∥∥v∥ while ∑jvj2/(j+1)≤∥v∥2; hence ∣I(u+v)−I(u)∣≤2∥u∥∥v∥+∥v∥2→0 as v→0.

1.2F1F2algebra

The minimising sequence. The point 0∈B has I(0)=0, and I≥0, so inf⁡BI=0. By [F2] the vectors ek lie in B and satisfy I(ek)=1/(k+1)→0, so (ek) is a minimising sequence for inf⁡BI.

2.1step 1.1algebra

The unique minimiser. If u∈B has I(u)=0, then uj2/(j+1)=0 for every j, hence uj=0 for every j and u=0. So 0 is the unique minimiser of I on B.

3.1F2step 2.1

No strong convergence of the minimising sequence. By [F2] one has ek⇀0 with ∥ek∥=1 for every k, so ∥ek−0∥=1 does not tend to 0; a fortiori no subsequence of (ek) converges in norm to 0, the unique minimiser of step 2.1.

4.1F2F3step 1.2step 3.1∎

Conclusion. The bounded minimising sequence (ek) converges weakly to 0 by [F2], but no subsequence converges in norm to the unique minimiser by step 3.1. Thus the weak subsequence conclusion described in [F3], under that theorem's stated principles, cannot be upgraded to strong convergence of minimising sequences. No additional extraction is needed for this explicit witness.

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A norm-closed nonconvex set need not be weakly closed

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let H=ℓ2(N,R) (Hilbert space, Square-summable families on an arbitrary index set and the space ℓ2(I)) and let S={u∈H:∥u∥=1} be its unit sphere. Then S is norm closed and norm bounded, but it is not weakly sequentially closed: the coordinate vectors ej, the families that are 1 at j and 0 elsewhere, lie in S and satisfy ej⇀0∉S. Indeed the weak closure of S is the closed unit ball (Weak closure of the unit sphere is the closed unit ball). So the convexity hypothesis in A norm-closed convex set is weakly sequentially closed cannot be dropped, and a norm-closed admissible set need not be weakly closed for The direct method in a reflexive Banach space.

Facts & Assumptions

Given: The Axiom of Choice; the real Hilbert space H=ℓ2(N,R) (Hilbert space, Square-summable families on an arbitrary index set and the space ℓ2(I)), its coordinate vectors ej, the families that are 1 at j and 0 elsewhere, and its unit sphere S={u∈H:∥u∥=1}. The counting-measure dictionary ℓp is the Lp space of counting measure identifies H with real L2(#), which is reflexive (and thus Banach) by Reflexivity of Lp for one less p less infinity under Countable Choice, supplied here by AC; the series pairing is its inner product.

[F1]

The unit sphere of an infinite-dimensional Hilbert space is norm closed and norm bounded; its weak closure is the closed unit ball (Weak closure of the unit sphere is the closed unit ball).

[F2]

Each coordinate vector satisfies ej∈H, ∥ej∥=1 and ej⇀0: by the duality of ℓp and ℓq every bounded linear functional on ℓ2 is Λ(a)=∑jajbj for a unique b∈ℓ2 (Counting measure specializes the representation theorem to ℓp and ℓq), so Λ(ej)=bj, and bj→0 because a square-summable family has small tails, that is, for every ε>0 there is a finite F with ∑j∉F∣bj∣2<ε (Square-summable families on an arbitrary index set and the space ℓ2(I)), whence ∣bj∣2<ε for every j beyond all elements of F; weak convergence means convergence against every bounded linear functional (Weak convergence of nets and sequences). Hence the sequence (ej) lies in S and converges weakly to the origin, which is not in S.

[F3]

A norm-closed convex set is weakly closed under the Axiom of Choice (A norm-closed convex set is weakly sequentially closed); the direct method's admissibility requirement is weak sequential closedness (The direct method in a reflexive Banach space).

Counterexample

technique · direct verification with an orthonormal sequence
1.1F1algebra

S is norm closed and norm bounded. The map u↦∥u∥ is continuous for the norm topology and {1} is closed in R, so S, its preimage, is norm closed; and ∥u∥=1 for u∈S, so S is norm bounded.

1.2F2algebra

S is not weakly sequentially closed. By [F2] the orthonormal sequence (ej) lies in S, satisfies ej⇀0, and 0∉S; hence a sequence in S converges weakly to a point outside S, so S is not weakly sequentially closed.

1.3F1algebra

The failure is not an artefact of the sequence. By [F1] the full weak closure of S is the closed unit ball, which strictly contains S; so S is not weakly closed. This is a property of the set, independent of which witness sequence is used.

2.1F3step 1.2step 1.3∎

Conclusion. The set S is norm closed (even bounded) but not weakly sequentially closed, so the convexity hypothesis in [F3] cannot be dropped; consequently a norm-closed nonconvex admissible set need not qualify for the direct method on the strength of norm closedness alone.

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Stationarity of the Euler-Lagrange equation does not imply a minimum

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let Ω⊆Rn be a bounded C1 domain and let J(u)=−12∫Ω∣Du∣2 dx(u∈H01(Ω)). Then u=0 is the only stationary point: its weak Euler-Lagrange (stationarity) equation reads −∫ΩDu⋅Dφ dx=0 for every φ∈H01(Ω), which forces Du=0 and then u=0 almost everywhere by the Poincare inequality (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction). But J is not bounded below: for any fixed nonzero φ∈H01(Ω) one has J(tφ)=−t22∥Dφ∥22→−∞ as ∣t∣→∞. So the Euler-Lagrange equation is a necessary condition only; the functional is concave, not convex, and Stationarity is sufficient for a global minimum of a convex differentiable functional does not apply.

Facts & Assumptions

Given: The Axiom of Choice; a bounded C1 domain Ω⊆Rn, the functional J(u)=−12∫Ω∣Du∣2 dx on H01(Ω), and the Lagrangian f(x,s,ξ)=−12∣ξ∣2.

[F1]

A stationary point of J is a point u∈H01(Ω) whose first variation vanishes in every direction φ∈H01(Ω). The quadratic expansion below computes this variation directly for every n≥1. For n≥2, its vanishing is also the fixed-zero-trace weak Euler-Lagrange formula of The weak Euler-Lagrange equation for integral functionals with fixed trace, since fs=0, fξ=−ξ and the p=2 differentiation growth bounds hold.

[F2]

On the full linear space H01(Ω), at a local minimiser the first variation vanishes (The first variation vanishes at an interior minimiser); the converse requires convexity and stationarity in the sense δI(u;v−u)≥0 for all competitors, by Stationarity is sufficient for a global minimum of a convex differentiable functional.

[F3]

Poincare's inequality controls the L2 norm by the Dirichlet energy on H01(Ω) (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction); in particular ∫Ω∣Du∣2=0 forces u=0 almost everywhere.

[F4]

The Lagrangian f(x,s,ξ)=−12∣ξ∣2 is concave, not convex, in ξ (Convex and strictly convex functionals on a convex subset of a real vector space).

Counterexample

technique · direct computation of the stationary equation and along the line $t\varphi$
1.1F1algebra

The stationary equation. For f(x,s,ξ)=−12∣ξ∣2 one has fξ(x,s,ξ)=−ξ and fs=0, so the exact expansion J(u+εφ)−J(u)=−ε∫Du⋅Dφ−12ε2∫∣Dφ∣2 gives the bounded first variation δJ(u;φ)=−∫Du⋅Dφ by Holder (Holder's inequality for integrals, including the endpoint cases). Thus a point u∈H01(Ω) satisfies the weak Euler-Lagrange equation of [F1] exactly when ∫ΩDu⋅Dφ dx=0 for every φ∈H01(Ω).

1.2F3algebra

0 is not a minimiser, not even locally. Fix any nonzero φ∈H01(Ω), which exists: choose a ball BR(a)⊆Ω and a nonzero smooth bump supported inside it (Compactly supported scaled Euclidean bumps). Poincare [F3] gives ∥Dφ∥2>0, and consider J(tφ)=−t22∫Ω∣Dφ∣2 dx. As ∣t∣→∞ this tends to −∞, so J is not bounded below on H01(Ω); and for every t≠0 one has J(tφ)<0=J(0), and ∥tφ∥H1=∣t∣∥φ∥H1→0 as t→0, so 0 is not a local minimiser either.

2.1F3step 1.1

The only stationary point. If u is stationary, step 1.1 applies with the admissible test function φ=u∈H01(Ω), giving ∫Ω∣Du∣2=0; by [F3] this forces Du=0 and hence u=0 almost everywhere. Conversely u=0 satisfies the equation because D0=0. So 0 is the only stationary point of J.

3.1F2F4step 2.1step 1.2∎

Conclusion. The only stationary point of J fails to be a minimiser by step 1.2, so the Euler-Lagrange equation is a necessary condition only; the failure is consistent with [F2], since J is concave in the gradient by [F4] and the convex stationarity-sufficiency theorem therefore does not apply.

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The one-dimensional Euler-Lagrange equation for an energy with a potential

Example

Example. Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let V∈C1(R), let a<b be real numbers and let I(u)=∫ab(12u′(x)2+V(u(x)))dx. On the admissible class C2([a,b]) with fixed endpoint values u(a)=u0, u(b)=u1, a local minimiser in the C2 norm satisfies the boundary value problem u′′=V′(u)  in (a,b),u(a)=u0,u(b)=u1, the classical Euler-Lagrange equation of the Lagrangian f(x,s,ξ)=12ξ2+V(s) (The classical Euler-Lagrange equation under regularity). In the special case V=0 this is the one-dimensional Laplace equation u′′=0 with the affine solution u(x)=u0+u1−u0b−a(x−a); for V(s)=12ω2s2 it is the equation of an inverted harmonic oscillator u′′=ω2u.

Facts & Assumptions

Given: Countable Choice; a function V∈C1(R), a compact interval [a,b]⊂R with a<b, the functional I(u)=∫ab(12u′(x)2+V(u(x)))dx on the admissible class of u∈C2([a,b]) with fixed endpoint values u(a)=u0, u(b)=u1, and the Lagrangian f(x,s,ξ)=12ξ2+V(s).

[F1]

The conclusion has the shape of the classical Euler-Lagrange equation of The classical Euler-Lagrange equation under regularity for the Lagrangian f(x,s,ξ)=12ξ2+V(s), whose partials are fξ=ξ and fs=V′(s). That corollary also assumes f∈C2 and the growth hypotheses of the weak Euler-Lagrange theorem (The weak Euler-Lagrange equation for integral functionals with fixed trace), which need not hold for a general V∈C1; the verification below therefore computes the equation directly from the minimality of u.

[F2]

One-variable mean value theorem and Fermat's interior-extremum theorem: a differentiable function on an interval with an interior local extremum has vanishing derivative there, and the mean value theorem identifies (V(u+εφ)−V(u))/ε with V′(u+θεφ)φ for some θ∈(0,1) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a), Fermat's interior extremum theorem: if f has a local extremum at a point c interior to its domain and is differentiable at c, then f′(c)=0).

[F3]

Integration by parts on a compactly supported test function: ∫abu′φ′ dx=−∫abu′′φ dx for φ∈Cc∞((a,b)) and u∈C2([a,b]). This is Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives with F=u′ and G=φ; its continuous integrands have equal Riemann and Lebesgue integrals by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F4]

Fundamental lemma: a continuous function on an open set orthogonal to every compactly supported smooth test function vanishes identically (The fundamental lemma of the calculus of variations).

[F5]

Verification

technique · direct, by testing the minimality of $u$ against compactly supported variations
1.1F1F2givenalgebra

The first variation. Let φ∈Cc∞((a,b)). Since φ vanishes near the endpoints, u+εφ belongs to the admissible class for every ε, and for ∣ε∣ small it is close to u in C2([a,b]), so Φ(ε):=I(u+εφ) has a local minimum at ε=0. Computing the difference quotient, ε−1(Φ(ε)−Φ(0))=∫ab(u′φ′+ε2φ′2+ε−1(V(u+εφ)−V(u)))dx, and by the mean value theorem [F2] the last term equals φ(x)V′(u(x)+θxεφ(x)) with θx∈(0,1); as ε→0 this converges to φV′(u) uniformly on [a,b], because V′ is uniformly continuous by Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous on a compact interval containing the values u(x)+θxεφ(x) and ∣u+θεφ−u∣≤∣ε∣∥φ∥∞. Hence Φ is differentiable at 0 with Φ′(0)=∫ab(u′φ′+V′(u)φ) dx. This is the direct computation announced in [F1].

2.1F2step 1.1

Fermat's theorem. The point 0 is an interior local minimum of the differentiable function Φ, so [F2] gives Φ′(0)=0, that is ∫ab(u′φ′+V′(u)φ) dx=0 for every φ∈Cc∞((a,b)).

3.1F3F4step 2.1

The differential equation. For φ∈Cc∞((a,b)) integration by parts [F3] gives ∫abu′φ′ dx=−∫abu′′φ dx, so the identity of step 2.1 reads ∫ab(−u′′+V′(u))φ dx=0 for every such φ. The function −u′′+V′(u) is continuous on (a,b), being a sum of continuous functions, so the fundamental lemma [F4] gives −u′′+V′(u)=0 on (a,b), that is u′′=V′(u).

4.1F5step 3.1givenalgebra∎

Endpoint conditions and the two instances. The admissible class fixes u(a)=u0 and u(b)=u1, so u solves the boundary value problem of the statement. For V=0 the equation is u′′=0, and [F5] makes u affine, with values determined by the endpoints: u(x)=u0+u1−u0b−a(x−a). For V(s)=12ω2s2 one has V′(s)=ω2s, so the equation reads u′′=ω2u, the equation of an inverted harmonic oscillator.

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A nonconvex gradient energy can lose weak lower semicontinuity

Statement refuted

Counterexample. Assume the Axiom of Choice, the ultrafilter lemma, DC and HB (The Axiom of Choice, The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, The real dominated-extension principle as an additional hypothesis over ZF). On Ω=(0,1) let w:R→R be the 1-periodic function with w(t)=t on [0,12] and w(t)=1−t on [12,1], and put uk(x)=1kw(kx) for integers k≥1. Then uk→0 uniformly, uk′∈{±1} almost everywhere, and uk⇀0 in H1(0,1). For the nonnegative integrand f(ξ)=(ξ2−1)2, which is not convex since f(0)=1>12f(−1)+12f(1)=0, The integral is interpreted in [0,+∞] on H1(0,1); it need not be finite for every H1 function. I(u)=∫01(u′(x)2−1)2dx satisfies I(uk)=0 for every k, while I(0)=∫011 dx=1. Hence I(0)>lim⁡kI(uk)=0 and I is not weakly sequentially lower semicontinuous. Also I is not convex: I(u1)=I(−u1)=0<I((u1−u1)/2)=1, so A convex norm-lower-semicontinuous functional is weakly lower semicontinuous does not apply. Nevertheless I attains its minimum 0 at every uk. This example does not refute the existence conclusion of The direct method for convex integral functionals with convexity removed: it also lies outside that theorem's n≥2 and p=2 upper-growth hypotheses.

Facts & Assumptions

Given: The Axiom of Choice (for ACL), the ultrafilter lemma, DC and HB; the interval Ω=(0,1); the 1-periodic continuous function w with w(t)=t on [0,12] and w(t)=1−t on [12,1]; the functions uk(x)=1kw(kx); and the integrand f(ξ)=(ξ2−1)2 with I(u)=∫01(u′(x)2−1)2dx. The weak compactness step uses the ultrafilter lemma, DC and HB (The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, The real dominated-extension principle as an additional hypothesis over ZF).

[F1]

The function w is continuous, 1-periodic, satisfies ∣w∣≤12 and w(0)=w(1)=0, and is piecewise linear with ∣w′∣=1 almost everywhere. Hence uk(x)=1kw(kx) is absolutely continuous on [0,1] with a.e. derivative uk′(x)=w′(kx)∈{±1}; by the ACL characterisation uk∈W1,2(0,1) with weak derivative uk′ (The ACL characterisation of W1,p, Integer-order Sobolev spaces and their norms).

[F2]

For φ∈L2(0,1) the maps u↦∫01uφ and u↦∫01u′φ are bounded linear functionals on W1,2(0,1) with operator norm at most ∥φ∥L2, by Cauchy-Schwarz against the two components of the Sobolev norm (Integer-order Sobolev spaces and their norms).

[F3]

W1,2(0,1) is a reflexive Banach space, so every norm-bounded sequence in it has a weakly convergent subsequence under the ultrafilter lemma, DC and HB (W^{1,p}(Omega) is reflexive for 1<p<infinity, Reflexivity is equivalent to weak subsequential compactness of bounded sequences).

[F4]

If y∈L2(0,1;R) has ∫yφ=0 for all real φ∈L2, take φ=y to obtain ∥y∥22=0, so y=0 as an L2 class. This directly identifies the subsequential Sobolev limit; uniqueness of weak probability-measure limits is not used.

[F5]

The weak lower semicontinuity lemma assumes convexity of the functional (A convex norm-lower-semicontinuous functional is weakly lower semicontinuous). The convex integral existence theorem also assumes n≥2 and a p-growth upper bound (The direct method for convex integral functionals); the present interval and quartic integrand fail these hypotheses for p=2. Nonconvexity is checked by the midpoint inequality (Convex and strictly convex functionals on a convex subset of a real vector space), and existence here is decided by the explicit values of I.

Counterexample

technique · direct computation along the oscillating sawtooth sequence
1.1F1algebra

The sequence and its bounds. By [F1] the functions uk lie in W1,2(0,1) with ∣uk∣≤12k and ∣uk′∣=1 almost everywhere; hence ∥uk∥L2≤12k→0 and ∥uk′∥L2=1 for every k, so (uk) converges to 0 in L2 and is norm bounded in W1,2(0,1).

1.2F1algebra

The values of I. Since ∣uk′∣=1 almost everywhere, I(uk)=∫01(uk′2−1)2=0 for every k; and I(0)=∫01(0−1)2dx=1. Hence lim inf⁡kI(uk)=0<1=I(0).

2.1F2F3F4step 1.1

uk⇀0 in W1,2(0,1). Suppose not; then there are a bounded linear functional F on W1,2(0,1) and ε>0 with ∣F(uk)∣≥ε for infinitely many k. Along that subsequence, which stays norm bounded, [F3] provides a further subsequence with ukl⇀y for some y∈W1,2(0,1). For every φ∈L2(0,1) the bounded functional of [F2] gives ∫01uklφ→∫01yφ, while ∫01uklφ→0 because ukl→0 in L2 by step 1.1; passing to the limit, ∫01yφ=0 for all φ∈L2, so y=0 by taking φ=y in [F4]. But then F(ukl)→F(0)=0, contradicting ∣F(ukl)∣≥ε. Hence uk⇀0 in W1,2(0,1).

3.1F5step 1.2step 2.1algebra∎

Conclusion and scope. Steps 1.2 and 2.1 give I(0)=1>0=lim inf⁡kI(uk) along a sequence converging weakly to 0, so I is not weakly sequentially lower semicontinuous. Since I(−u1)=I(u1)=0 but I(0)=1, I is not convex. Yet I≥0 and I(u1)=0, so its minimum is attained. Thus the example shows loss of weak lower semicontinuity for a nonconvex gradient energy, without asserting necessity of convexity for existence; the cited convex integral theorem also has dimensional and growth hypotheses absent here.

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The natural Neumann condition from a free endpoint in one dimension

Example

Example. Assume Countable Choice (The Axiom of Countable Choice (ACω)) and let a<b. On C2([a,b]) consider I(u)=∫abf(x,u(x),u′(x)) dx with f∈C2, no boundary conditions, and let u be a free-endpoint local minimiser in the C2 norm. Retaining the boundary term produced by integration by parts gives, in addition to the Euler-Lagrange equation fs−ddxfξ=0 (The classical Euler-Lagrange equation under regularity), the two natural boundary conditions fξ(a,u(a),u′(a))=0,fξ(b,u(b),u′(b))=0, which are precisely the one-dimensional case of The natural boundary condition for free boundary variations. For f(x,s,ξ)=12ξ2−g(x)s they read u′′=−g on (a,b) with u′(a)=u′(b)=0.

Facts & Assumptions

Given: Countable Choice and a<b; a function f∈C2([a,b]×R×R), the functional I(u)=∫abf(x,u(x),u′(x))dx on C2([a,b]) with no boundary conditions, and a free-endpoint local minimiser u: I(u)≤I(v) for all v∈C2([a,b]) with ∥v−u∥C2 small.

[F1]

The endpoint conditions are the one-dimensional analogue of The natural boundary condition for free boundary variations, whose stated domain and Sobolev hypotheses do not cover this example. They will be proved directly below; the outward signs are −1 at a and +1 at b.

[F2]

The interior equation has the form in The classical Euler-Lagrange equation under regularity, but is derived directly in step 2.1 because no global Sobolev growth bound is imposed here.

[F3]

Continuous partials make the integrand totally differentiable (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative), and the chain rule (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)) computes its derivative along (u+tϕ,u′+tϕ′) as fsϕ+fξϕ′. First variation: for every ϕ∈C∞([a,b]) the function Φ(ε):=I(u+εϕ) has an interior local minimum at ε=0 and, by the mean value theorem applied to the C2 integrand, Φ′(0)=∫ab(fs(x,u,u′)ϕ+fξ(x,u,u′)ϕ′)dx=0 (Fermat's interior extremum theorem: if f has a local extremum at a point c interior to its domain and is differentiable at c, then f′(c)=0, The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[F4]

Integration by parts and the fundamental lemma: for u∈C2, g∈C1 one has ∫ab(gϕ′+g′ϕ)dx=[gϕ]ab for every ϕ∈C∞([a,b]) by Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives; the continuous integrands have equal Riemann and Lebesgue integrals by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral. A continuous function orthogonal to all compactly supported test functions vanishes (The fundamental lemma of the calculus of variations).

[F5]

Composites of Ck Euclidean maps are Ck, so x↦fξ(x,u(x),u′(x)) is C1 when f∈C2 and u∈C2 (Ck Euclidean maps are closed under componentwise algebra and composition, Ck maps and multi-index derivative notation in Euclidean space).

Verification

technique · direct, testing the first variation with compactly supported and with endpoint-supported variations
1.1F3given

The first-variation identity. Let ϕ∈C∞([a,b]). Since no boundary conditions are imposed, u+εϕ lies in the admissible class for every ε and for ∣ε∣ small it is close to u in the C2 norm, so Φ(ε)=I(u+εϕ) has an interior local minimum at 0; the mean value theorem expresses the integrand difference quotient as fs(x,u+θεϕ,u′+θεϕ′)ϕ+fξ(x,u+θεϕ,u′+θεϕ′)ϕ′ for some 0<θ<1. Heine--Cantor (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous) makes the partials uniformly continuous on a compact set containing these arguments, this quotient therefore converges uniformly to its value at ε=0, and its integral is Φ′(0). Fermat's theorem in [F3] then gives ∫ab(fs(x,u,u′)ϕ+fξ(x,u,u′)ϕ′)dx=0.

2.1F2F4F5step 1.1

The interior equation. Taking ϕ∈Cc∞((a,b)) in step 1.1 and integrating by parts, using that x↦fξ(x,u(x),u′(x)) is C1 by [F5] and has derivative ddxfξ(x,u,u′), gives ∫ab(fs(x,u,u′)−ddxfξ(x,u,u′))ϕ dx=0 for every compactly supported ϕ; the integrand is continuous, so [F4] gives fs−ddxfξ=0 on (a,b), the classical Euler-Lagrange equation.

3.1F4step 1.1step 2.1

The boundary identity. Let now ϕ∈C∞([a,b]) be arbitrary. Writing g:=fξ(x,u,u′) and using the interior equation of step 2.1, step 1.1 becomes 0=∫ab(ddxg ϕ+g ϕ′)dx=[gϕ]ab=g(b)ϕ(b)−g(a)ϕ(a) by [F4], that is fξ(b,u(b),u′(b))ϕ(b)−fξ(a,u(a),u′(a))ϕ(a)=0 for every ϕ∈C∞([a,b]).

4.1F1step 2.1step 3.1∎

Both natural conditions, and the instance. Choosing in step 3.1 ϕ(x)=(b−x)/(b−a) gives fξ(a,u(a),u′(a))=0, and ϕ(x)=(x−a)/(b−a) gives fξ(b,u(b),u′(b))=0; these are the one-dimensional natural boundary conditions, the general form of [F1]. For f(x,s,ξ)=12ξ2−g(x)s one has fξ=ξ and fs=−g(x), so the interior equation of step 2.1 reads u′′=−g on (a,b) and the natural conditions read u′(a)=u′(b)=0.

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The harmonic affine extension minimises the Dirichlet energy

Example

Example. Assume the Axiom of Choice (The Axiom of Choice). Let Ω⊆Rn, n≥2, be a bounded C1 domain and let a(x)=c0+ℓ(x) be affine on Ω‾, so that Δa=0 and Da=ℓ is constant. Then a is the unique minimiser of the Dirichlet energy I(v)=12∫Ω∣Dv∣2 dx on the affine trace class Ka={v∈H1(Ω):Tv=Ta} (The Lp trace operator on a bounded C1 domain): for every v∈Ka, v−a∈H01(Ω) and I(v)=I(a)+12∫Ω∣D(v−a)∣2 dx ≥ I(a), with equality if and only if D(v−a)=0 almost everywhere, and then v=a almost everywhere by the Poincare inequality on H01(Ω) (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

Facts & Assumptions

Given: The Axiom of Choice; a bounded C1 domain Ω⊆Rn, an affine function a(x)=c0+ℓ(x) on Ω‾ (so Da=ℓ is a constant field and Δa=0), the Dirichlet energy I(v)=12∫Ω∣Dv∣2dx, and the affine trace class Ka={v∈H1(Ω):Tv=Ta}.

[F1]

Ka is nonempty, convex and weakly closed, and v−a∈H01(Ω) for every v∈Ka by the kernel description ker⁡T=H01(Ω) (The kernel of the trace is the closure of the test functions, The Lp trace operator on a bounded C1 domain).

[F2]

Poincare's inequality on H01(Ω): if Dη=0 almost everywhere for η∈H01(Ω), then η=0 (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F3]

Under the Axiom of Choice, the ultrafilter lemma, Dependent Choice and Hahn--Banach, the Dirichlet principle also identifies an affine-class minimiser with the unique weak solution (The Dirichlet principle for the Poisson equation). In the zero-forcing case here, a is weakly harmonic with trace Ta, since Da=ℓ is constant and ∫ΩDa⋅Dφ=ℓ⋅∫ΩDφ=0 for every φ∈H01(Ω) (The notation Hk and the reserved zero-boundary symbol).

Verification

technique · direct completion of the square
1.1F1algebra

Orthogonality of the cross term. For v∈Ka one has η:=v−a∈H01(Ω) by [F1], and Da=ℓ is a constant field, so ∫ΩDa⋅Dη dx=ℓ⋅∫ΩDη dx=0: each component of Dη has vanishing integral, because η is the H1-limit of functions in Cc∞(Ω) and for those the integral of each partial derivative vanishes by integration by parts against the smooth constant field (Divergence on a bounded C1 Euclidean domain). Passage to the H1 limit is valid since ∣∫Di(η−ηm)∣≤∣Ω∣1/2∥Di(η−ηm)∥2→0 by Holder (Holder's inequality for integrals, including the endpoint cases).

2.1step 1.1algebra

The energy identity. Expanding the square, ∣Dv∣2=∣Da+Dη∣2=∣Da∣2+2Da⋅Dη+∣Dη∣2, and integrating with step 1.1 gives I(v)=I(a)+12∫Ω∣Dη∣2dx≥I(a).

3.1F2step 2.1

Equality case. Equality holds exactly when Dη=0 almost everywhere, which by [F2] forces η=0, that is v=a almost everywhere.

4.1F3step 2.1step 3.1∎

a is the minimiser. By steps 2.1 and 3.1 every v∈Ka satisfies I(v)≥I(a) with equality only for v=a; since a∈Ka, it is the unique minimiser of I on Ka. This direct completion-of-the-square argument proves the example's claim. Under the additional choice hypotheses stated in [F3], the general Dirichlet principle also identifies this minimiser with the weak solution; the weak harmonicity of a was checked in [F3].

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Fixed-trace and free-trace variations give different boundary equations

Example

Example. Assume the Axiom of Choice (The Axiom of Choice). Let Ω⊆Rn, n≥2, be a bounded C1 domain, let f∈L2(Ω;R), and put I(u)=12∫Ω∣Du∣2 dx−∫Ωfu dx. (a) Fixed trace: a local minimiser in the H1 norm on the nonempty affine class Kg={v∈H1(Ω):Tv=g}, with g∈H1/2(∂Ω), solves the weak Dirichlet problem −Δu=f, Tu=g (The Dirichlet principle for the Poisson equation, The affine Dirichlet trace class is nonempty, convex and weakly closed). (b) Free trace: if u∈C2(Ω‾) is a local minimiser in the C2 norm on C2(Ω‾), then −Δu=f almost everywhere in Ω and ∂νu=0 on ∂Ω. Choosing the continuous representative f=−Δu makes the interior equation pointwise. This is the boundary condition suggested by The natural boundary condition for free boundary variations, proved directly here since a general L2 forcing need not give a C2 integrand.

The constant-shift identity is I(u+c)=I(u)−c∫Ωf. Thus I is invariant under global constants exactly when ∫Ωf=0, and it is never coercive on all of H1(Ω). If ∫Ωf≠0, no free local minimiser exists. On a connected domain satisfying the extension-domain hypothesis of Weak Neumann solvability on the mean-zero subspace, zero-mean forcing gives a unique mean-zero weak Neumann solution; nonzero mean cannot be repaired merely by normalising the solution. On a disconnected domain compatibility is required on each component and the additive constants are independent on those components.

Facts & Assumptions

Given: The Axiom of Choice; the real domain and data above; local minimality in H1 on Kg in (a), or in C2 on the whole C2 space in (b).

[F1]

The fixed-trace class is a translate of H01(Ω); its admissible directions are exactly that subspace (The affine Dirichlet trace class is nonempty, convex and weakly closed). The weak Euler–Lagrange identity holds for these directions (The weak Euler-Lagrange equation for integral functionals with fixed trace).

[F2]

The Dirichlet principle identifies its energy minimiser with the unique weak Poisson solution (The Dirichlet principle for the Poisson equation).

[F3]

First Green identity holds for u∈C2(Ω‾) and smooth tests under Countable Choice, supplied by AC (First Green identity). A locally integrable function pairing to zero with all compactly supported tests is zero almost everywhere (The fundamental lemma of the calculus of variations). A continuous boundary flux pairing to zero with all ambient smooth tests vanishes on the boundary (The boundary fundamental lemma of the calculus of variations).

[F4]

The Neumann supplier requires a bounded connected extension domain and a bounded forcing functional F with F(1)=0; it gives a unique mean-zero solution and all other solutions differ by constants. It also records the componentwise compatibility needed in the disconnected case (Weak Neumann solvability on the mean-zero subspace).

[F5]

Holder makes v↦∫fv bounded on H1, and bounds all terms in the quadratic expansion below (Holder's inequality for integrals, including the endpoint cases). Fermat's theorem gives a zero derivative at a two-sided interior local minimum (Fermat's interior extremum theorem: if f has a local extremum at a point c interior to its domain and is differentiable at c, then f′(c)=0).

Verification

1.1F1F2F5givenalgebra

Fixed trace. For every φ∈H01(Ω), the curve u+tφ stays in Kg by [F1]. Its energy is exactly I(u)+t∫(Du⋅Dφ−fφ)+12t2∫∣Dφ∣2. Local minimality and [F5] give ∫Du⋅Dφ=∫fφ, the weak Dirichlet equation. Moreover the same expansion at t=1 shows I(u+φ)−I(u)=12∫∣Dφ∣2≥0, so this local minimiser is global and [F2] applies.

1.2F3F5given

Free trace. For every φ∈C∞(Ω‾), the curve u+tφ is admissible and close to u in the C2 norm as t→0. The same quadratic expansion and [F5] give ∫(Du⋅Dφ−fφ)=0. For compactly supported tests, Green identity [F3] then yields ∫(−Δu−f)φ=0, so −Δu=f almost everywhere by the fundamental lemma. Returning to arbitrary smooth tests gives ∫∂Ω(∂νu)φ=0 by Green identity; the continuous field Du and the boundary fundamental lemma force ∂νu=0.

2.1F4F5step 1.2algebra∎

Constants and compatibility. Direct expansion gives I(u+c)=I(u)−c∫f. If ∫f≠0, arbitrarily small constant shifts in the appropriate sign lower the energy, and large shifts make it tend to −∞; if ∫f=0, arbitrarily large shifts leave it fixed. In both cases coercivity on the full space fails. In case (b), testing the first variation with 1 gives ∫f=0. At each boundary point the one-sided C1 graph convention gives a smaller connected subgraph neighbourhood meeting only one component; at interior points use a ball in the component. Thus a component indicator extends locally constantly to Ω‾ and is a C2 admissible direction, giving the componentwise condition. Under the connected extension-domain hypotheses of [F4], the functional F(v)=∫fv is bounded by [F5] and satisfies F(1)=0 precisely for zero-mean forcing, so [F4] supplies the normalised weak solution.

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