Direct method in the calculus of variations

The direct method in the calculus of variations proves existence by passing to a limit of a minimizing sequence. A useful abstract version takes a nonempty weakly sequentially closed admissible subset $\mathcal{A}$ of a reflexive Banach space $X$ and a functional $E:X\to \mathbb{R}\cup \{+\infty \}$. Assume $E$ is finite at some admissible point, bounded below, coercive on $\mathcal{A}$, and has weak lower semicontinuity.

Choose $u_n\in \mathcal{A}$ with $E(u_n)\to m={\inf }_{\mathcal{A}}E$. Coercivity bounds $\Vert u_n{\Vert }_X$. Reflexivity supplies a weakly convergent subsequence, and closure puts its limit $u$ in $\mathcal{A}$. Finally,so $u$ attains the infimum. Each hypothesis has a distinct role: compactness must preserve the constraints, and lower semicontinuity must hold for the convergence actually obtained.

For the Dirichlet energy with linear forcing on $H_0^1(\Omega )$ over a bounded domain, the Poincaré inequality controls the forcing and gives coercivity, while the squared gradient norm is weakly lower semicontinuous. The resulting minimizer solves the weak Poisson equation by its first variation.^[ref] Strict convexity gives uniqueness when the admissible class is convex. The oscillating minimizing sequence with no minimizer shows what can fail without suitable lower semicontinuity.

A Lavrentiev gap is a strict change in an infimum when the admissible regularity class is restricted, despite retaining the same endpoint values. For example $I(u)={\int }_0^1(u^3-x{)}^2{u}^{\prime 6}dx$ with $u(0)=0,u(1)=1$ has value zero at the absolutely continuous function $u=x^{1/3}$. Every Lipschitz competitor has a uniform positive lower energy bound. Indeed its ratio to $x^{1/3}$ starts at zero, so let $x_0$ be its first crossing of $u(x)=x^{1/3}/2$. Before that point, $|u^3-x|\ge 7x/8$. Holder inequality givesThus $I(u)\ge (49/4096)(3/5{)}^5/x_0\ge (49/4096)(3/5{)}^5>0$. This proves a gap directly. The example's integrand lacks a uniform superlinear lower bound along its zero set; it illustrates the admissible-class issue rather than by itself furnishing a counterexample under stronger coercivity assumptions.

For $I_{\epsilon }(u)={\int }_0^1[(u^4-x{)}^2+\epsilon |u^{\prime }{|}^{4/3}]dx$ with endpoint values zero and one, the zero-penalty minimizer $x^{1/4}$ has infinite derivative penalty. Use $u=x^{1/4+\delta }$ instead. The mismatch is at most $(32/27){\delta }^2$, while the derivative penalty is $(3\epsilon /(4\delta ))(1/4+\delta {)}^{4/3}$. Balancing ${\delta }^2$ and $\epsilon /\delta$ by $\delta ={\epsilon }^{1/3}$ gives an upper energy bound of order ${\epsilon }^{2/3}$. Nonnegativity then proves convergence of minimum energies to zero. This is a recovery construction; no claim of a matching lower asymptotic constant is needed.

For a scalar fixed-endpoint problem on a bounded interval, suppose the continuous integrand $f(x,y,p)$ is convex in $p$ and satisfies $f(x,y,p)\ge \Theta (|p|)-C$, where $\Theta$ is nonnegative convex and $\Theta (r)/r\to \infty$. If there is one admissible absolutely continuous function of finite energy, a global minimizer exists in $W^{1,1}$.

The lower bound gives uniform integrability of derivatives in a minimizing sequence. Fixed endpoint values then give boundedness and equicontinuity of the functions. Extract uniform convergence of the functions and weak $L^1$ convergence of their derivatives; the limit retains the endpoint values. Convexity in the derivative gives weak lower semicontinuity of convex gradient energies, so the limit attains the infimum. Strict convexity of the full integrand on a convex admissible class additionally gives uniqueness.

Suppose the domain has finite measure and $F\in C^1$ is convex, with $|F(p)|\le C(1+|p{|}^2)$ and $|DF(p)|\le C(1+|p|)$. The supporting inequality $F(p)\ge F(q)+DF(q)\cdot (p-q)$, integrated with $q=Du$, gives $\int F(D{u}_k)\ge \int F(Du)+\int DF(Du)\cdot (D{u}_k-Du)$. The last term tends to zero by weak convergence of the gradients and the fixed $L^2$ test field $DF(Du)$. This proves weak lower semicontinuity without requiring pointwise convergence of gradients.

For a bounded domain, bounded symmetric uniformly elliptic $A$, bounded real $q\le 0$, $f\in L^2$ and a boundary lift $\psi \in H^1$, minimizing $J$ on $\psi +H_0^1$ solves $\mathrm{div}(A\nabla u)+qu=f$. The nonnegative quadratic terms are weakly lower semicontinuous, and Poincaré inequality makes the gradient norm coercive on the zero-boundary part. The first variation gives the weak equation with the forcing sign displayed above. If $q$ is positive, coercivity may fail at a Dirichlet eigenvalue; orthogonality to the homogeneous kernel is then needed for solvability.

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.

Take $a=3\pi /2$, $L=10$ and $u(x)=a\exp (-|x{|}^2/L^2)$ on ${\mathbb{R}}^3$. Set $f=-\Delta u+u+\sin u$. Both functions are smooth and belong to L2 space, and $u\in H^1$. The strictly increasing reaction $h(s)=s+\sin s$ has $0\le h(u)\le h(a)=a-1$, whileConsequently $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert u{\Vert }_{\infty }$. This $u$ is a minimizer of the screened sine-Gordon energy: it solves the Euler-Lagrange equation and that energy is convex. Thus even the minimizing and smooth hypotheses do not justify the stronger source-size bound. The appropriate general estimate is maximum bound for a monotone reaction term.

A minimizing sequence satisfies $E(u_n)\to \inf E$. Coercivity bounds an appropriate sublevel set, Compactness supplies a convergent subsequence, and sequential lower semicontinuity turns its limit into a global minimizer.

A minimizing sequence can converge uniformly while its derivatives keep oscillating between distinct energy wells. For the functional $I(u)={\int }_0^1[u^4+(u^{\prime 4}-16{)}^4]$, with zero endpoint values and Lipschitz admissible functions, the triangular waves displayed above have slopes $\pm 2$ almost everywhere and height $1/N$. Direct integration gives $I(u_N)=1/(5N^4)$. Nonnegativity therefore makes the infimum zero. Attainment would require both $u=0$ and $|u^{\prime }|=2$ almost everywhere, an impossibility. The uniform limit is zero but its energy is $16^4$; thus the energy fails to be lower semicontinuous for this convergence. The example explains why control of function values alone cannot replace suitable convexity in the derivative variable in the direct method in the calculus of variations.

Boundary data $g$ satisfy the bounded slope condition with constant $K$ if, at every boundary point $x_0$, there are affine functions ${\ell }_{x_0}^{-}$ and ${\ell }_{x_0}^{+}$ such thatand $|\nabla {\ell }_{x_0}^{\pm }|\le K$. Both the touching condition and the common slope bound matter; arbitrary affine upper/lower bounds would be much weaker.

An affine datum $g(x)=a\cdot x+b$ satisfies the condition with $K=|a|$, using that same affine function for both supports. For a suitable convex variational problem on a convex domain, these supports serve as barriers through the comparison principle for convex variational integrals. Translation comparisons then produce a global Lipschitz continuity bound. This consequence requires the structural and boundary hypotheses of the variational problem, rather than following from the boundary inequalities alone.

One precise comparison principle for convex variational integrals concerns minimizers of $E[u]={\int }_{\Omega }F(\nabla u)dx$ on a bounded Lipschitz domain. Let $1<p<\infty$, take differentiable strictly convex $F$ with $|DF(\xi )|\le C(1+|\xi {|}^{p-1})$, and let $u,v\in W^{1,p}$ be weak stationary solutions whose boundary traces satisfy $u\le v$. Then $u\le v$ almost everywhere in the domain.

Use $w=(u-v{)}^{+}$, the positive part of the difference. Its zero trace makes it an admissible test in the two weak Euler-Lagrange equations. Subtracting givesStrict convexity makes the integrand positive unless the two gradients agree. Hence $\nabla w=0$ almost everywhere, and the zero-trace Poincaré inequality forces $w=0$. The growth bound makes these test pairings integrable.

This proves why ordered affine solutions can bound a minimizer in the bounded slope condition argument. Convexity without further strictness or uniqueness is insufficient to order every pair of minimizers: if $F=0$, every admissible function minimizes the energy and two equal-boundary competitors can cross.