Constrained Variational Problems and Variational Inequalities — Examples

The function $u_0$ is smooth on $[0,1]$, with classical derivative $u_0^{\mathrm{\prime }}(x)=\sqrt{2}\pi \cos (\pi x)$ by [F2]. For each test function $\psi \in C_c^{\mathrm{\infty }}(0,1)$, integration of $(u_0\psi {)}^{\mathrm{\prime }}$ over an interior interval containing its support gives ${\int }_0^1{u}_0{\psi }^{\mathrm{\prime }}=-{\int }_0^1{u}_0^{\mathrm{\prime }}\psi$ [F3], so $u_0^{\mathrm{\prime }}$ is its weak derivative; both $u_0$ and $u_0^{\mathrm{\prime }}$ are bounded, hence $u_0\in H^1(0,1)$. Take $\chi$ from [F4] and put $M=\mathrm{\parallel }{\chi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(\mathbb{R})}<\mathrm{\infty }$. For integers $m\ge 5$, define ${\xi }_m(x)=(1-\chi (mx))(1-\chi (m(1-x)))$. Then ${\xi }_m\in C_c^{\mathrm{\infty }}(0,1)$, ${\xi }_m=1$ on $[2\mathrm{/}m,1-2\mathrm{/}m]$, $\mathrm{\parallel }{\xi }_m^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2Mm$, and both $1-{\xi }_m$ and ${\xi }_m^{\mathrm{\prime }}$ are supported in $B_m=(0,2\mathrm{/}m)\cup (1-2\mathrm{/}m,1)$. On $B_m$, $\mathrm{\mid }{u}_0(x)\mathrm{\mid }\le 2\sqrt{2}\pi \mathrm{/}m$, while $\mathrm{\mid }{u}_0^{\mathrm{\prime }}(x)\mathrm{\mid }\le \sqrt{2}\pi$ everywhere; since $\mathrm{\mid }{B}_m\mathrm{\mid }\le 4\mathrm{/}m$, these bounds give $\mathrm{\parallel }(1-{\xi }_m)u_0{\mathrm{\parallel }}_{L^2}\to 0$ and $\mathrm{\parallel }(1-{\xi }_m)u_0^{\mathrm{\prime }}-{\xi }_m^{\mathrm{\prime }}{u}_0{\mathrm{\parallel }}_{L^2}\to 0$. Hence ${\xi }_m{u}_0\to u_0$ in $H^1(0,1)$, and the closure definition of $H_0^1$ gives $u_0\in H_0^1(0,1)$. Finally $u_0(0)=u_0(1)=0$ because $\sin 0=\sin \pi =0$ [F5].

Normalisation and energy: by the power-reduction identities and the vanishing of ${\int }_0^1\cos (2\pi x)dx$ [F5], ${\int }_0^1{\sin }^2(\pi x)dx=\frac{1}{2}{\int }_0^1(1-\cos (2\pi x))dx=\frac{1}{2}$ and ${\int }_0^1{\cos }^2(\pi x)dx=\frac{1}{2}$; hence $\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{L^2}^2=2\cdot \frac{1}{2}=1$, so $u_0\in S$, and $E(u_0)={\int }_0^1{u}_0^{\mathrm{\prime }2}=2{\pi }^2{\int }_0^1{\cos }^2(\pi x)dx={\pi }^2$.

Minimality: for every $v\in S$ the sharp inequality [F1] gives $1=\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le {\pi }^{-1}\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}$, that is $E(v)=\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2\ge {\pi }^2$; since $u_0\in S$ with $E(u_0)={\pi }^2$ by step 2.1, the infimum over $S$ is the minimum ${\lambda }_1={\pi }^2$, attained at $u_0$.

Weak eigenvalue equation: the weak identity of [F1] for $\varphi =\sin (\pi x)$ scales by $\sqrt{2}$ to ${\int }_0^1{u}_0^{\mathrm{\prime }}{h}^{\mathrm{\prime }}={\pi }^2{\int }_0^1{u}_{0}h$ for every $h\in H_0^1(0,1)$, and by [F2] $u_0^{\mathrm{\prime }\mathrm{\prime }}=-{\pi }^2{u}_0$ classically with $u_0(0)=u_0(1)=0$; thus $-u_0^{\mathrm{\prime }\mathrm{\prime }}={\pi }^2{u}_0$ holds in the weak sense, with ${\lambda }_1={\pi }^2$ as the eigenvalue.

Steps 1.1-4.1 exhibit the minimiser, the minimum and the eigenvalue equation explicitly, so the constrained minimisation of [F6] on $(0,1)$ has $u_0(x)=\sqrt{2}\sin (\pi x)$ as a minimiser with ${\lambda }_1={\pi }^2$, in agreement with the general statement; the only choice principle consumed by this computation is Countable Choice through the sharp interval inequality [F1].

The polynomial $u_0$ is smooth on $[0,1]$ with $u_0(0)=u_0(1)=0$; its class on $(0,1)$ is absolutely continuous with weak derivative $u_0^{\mathrm{\prime }}=6A(1-2x)$ by [F2], and its endpoint trace vanishes, so $u_0\in \ker T=H_0^1(0,1)$ by [F1]. Moreover ${\int }_0^1{u}_{0}dx=6A{\int }_0^1(x-x^2)dx=6A(1\mathrm{/}2-1\mathrm{/}3)=A$ by [F3], so $u_0$ is admissible.

For every $h\in H_0^1(0,1)$ the derivative formulae are $DJ(u_0)h=2{\int }_0^1{u}_0^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ and $DG(u)h={\int }_0^{1}h$ by [F5]; in particular $DG(u_0)$ is a nonzero bounded functional, because $DG(u_0)u_0=A\mathrm{\ne }0$ by step 1.1, so the constraint is regular.

We compute $2{\int }_0^1{u}_0^{\mathrm{\prime }}{h}^{\mathrm{\prime }}=24A{\int }_0^{1}h$ for every $h\in H_0^1(0,1)$: by [F1] the absolutely continuous representative $h^{\ast }$ vanishes at both endpoints and $(h^{\ast }{)}^{\mathrm{\prime }}=Dh$, so [F4] applied to $F=1-2x$ and $H=h^{\ast }$ gives ${\int }_0^1(1-2x)Dh=-{\int }_0^1(-2)h=2{\int }_0^{1}h$; multiplying by $6A$ gives ${\int }_0^1{u}_0^{\mathrm{\prime }}{h}^{\mathrm{\prime }}=12A{\int }_0^{1}h$, hence the displayed identity. Thus $DJ(u_0)=24ADG(u_0)$.

Let $v\in H_0^1(0,1)$ satisfy the constraint ${\int }_0^{1}vdx=A$; then $h:=v-u_0\in H_0^1(0,1)$ by step 1.1, and [F5] gives $J(v)=J(u_0)+2{\int }_0^1{u}_0^{\mathrm{\prime }}{h}^{\mathrm{\prime }}+{\int }_0^1(h^{\mathrm{\prime }}{)}^2=J(u_0)+24A{\int }_0^{1}h+\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2=J(u_0)+\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2$ by step 3.1, because ${\int }_0^{1}h=0$; hence $J(v)\ge J(u_0)$, with equality exactly when $\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2=0$.

If $\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2=0$, then $h$ has weak derivative $0$, so its absolutely continuous representative is constant by [F7] and [F1]; that constant is $h^{\ast }(0)=0$ because $h\in H_0^1(0,1)$ has vanishing trace, so $h=0$ and $v=u_0$. Therefore $u_0$ is the unique admissible minimiser, in particular a local minimiser, and the multiplier rule [F6] applies with the regular constraint $G$; comparing its conclusion $DJ(u_0)=\lambda DG(u_0)$ with the identity of step 3.1 and the fact that $DG(u_0)\mathrm{\ne }0$ gives the unique multiplier $\lambda =24A$. This proves the example.

By [F1] choose a maximal, equivalently complete, orthonormal set $B\subseteq H$.

The set $B$ is infinite: were $B$ finite, its linear span would be finite dimensional and closed by [F2], and completeness would force it to equal the closed linear span of $B$, namely $H$; then $H$ would admit an ordered basis of finite length, contrary to the hypothesis.

Let $\mathcal{T}$ be the nonempty set of all finite tuples of distinct elements of $B$, including the empty tuple. Every tuple has an extension by one new element because its range is finite and $B$ is infinite. The Axiom of Choice in [A1] selects one such extension for each tuple in $\mathcal{T}$. Starting with the empty tuple, iterate this fixed extension function recursively over $\mathbb{N}$; the successive appended elements give distinct $e_j\in B$. This proves [F3] locally. By [F4], $(e_j)$ is orthonormal, so $\mathrm{\parallel }{e}_j\mathrm{\parallel }=1$ and $e_j\in S$ for every $j$.

We claim $e_j\rightharpoonup 0$. Fix $y\in H$; by Bessel's inequality [F5] the series ${\sum }_j\mathrm{\mid }⟨e_j,y⟩{\mathrm{\mid }}^2$ has finite sum, so its terms tend to $0$, that is $⟨e_j,y⟩\to 0$. Given a bounded linear functional $f$, write $f(x)=⟨x,y_f⟩$ by [F6]; then $f(e_j)=⟨e_j,y_f⟩\to 0$, which by the definition of weak convergence [F7] is exactly $e_j\rightharpoonup 0$.

The set $S$ is norm closed, because $S$ is the preimage of the closed singleton $\{1\}$ under the continuous norm [F8], and it is norm bounded because $\mathrm{\parallel }u\mathrm{\parallel }=1$ for every $u\in S$. Since $e_j\in S$ for every $j$ by step 3.1 while $0\mathrm{\notin }S$ because $\mathrm{\parallel }0\mathrm{\parallel }=0\mathrm{\ne }1$, and $e_j\rightharpoonup 0$ by step 4.1, the sphere $S$ is not weakly sequentially closed; the refuted claim is therefore false.

By [F7] the principal form is bounded and coercive. Put $s:=\sqrt{1-2\epsilon }$, so $0<s<1$ and $t=1-s\in (0,1)$; from $s^2=1-2\epsilon$ one gets $\epsilon =\frac{1}{2}(1-s^2)=\frac{1}{2}(1-s)(1+s)=\frac{1}{2}t(2-t)=t-\frac{1}{2}{t}^2$, hence $\psi (\pm t)=\epsilon -\frac{1}{2}{t}^2=t-t^2=t(1-t)$ and ${\psi }^{\mathrm{\prime }}(\pm t)=\mp t$.

Define $u(x)=\epsilon -\frac{1}{2}{x}^2$ for $\mathrm{\mid }x\mathrm{\mid }\le t$ and $u(x)=t(1-\mathrm{\mid }x\mathrm{\mid })$ for $t\le \mathrm{\mid }x\mathrm{\mid }\le 1$. At $x=\pm t$ the two formulas agree by step 1.1, and the one-sided derivatives agree as well because the inner derivative is ${\psi }^{\mathrm{\prime }}(x)=-x$ with ${\psi }^{\mathrm{\prime }}(\pm t)=\mp t$ and the outer derivative is $\mp t$; hence $u$ is $C^1$ on $[-1,1]$ by [F5], with $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \max \{t,1\}=1$ and $u(\pm 1)=0$. Therefore $u\in H^1(-1,1)$ with weak derivative $u^{\mathrm{\prime }}$ and $Tu=(0,0)$, so $u\in \ker T=H_0^1(-1,1)$ by [F2]. Finally $u-\psi =0$ on $[-t,t]$, while for $t\le \mathrm{\mid }x\mathrm{\mid }\le 1$ one has $u(x)-\psi (x)=t(1-\mathrm{\mid }x\mathrm{\mid })-\epsilon +\frac{1}{2}{x}^2=\frac{1}{2}(\mathrm{\mid }x\mathrm{\mid }-t{)}^2$ by step 1.1; hence $u\ge \psi$ on $(-1,1)$ with equality exactly on $[-t,t]$, so $u\in K$ and $\{u=\psi \}=[-t,t]$.

The derivative $u^{\mathrm{\prime }}$ equals $t$ on $(-1,-t)$, $-x$ on $(-t,t)$ and $-t$ on $(t,1)$; it is continuous and piecewise affine, hence Lipschitz and absolutely continuous on $[-1,1]$, with $u^{\mathrm{\prime }\mathrm{\prime }}=0$ a.e. on $(-1,-t)\cup (t,1)$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-1$ a.e. on $(-t,t)$. Let $v\in K$ and let $q:=v-u$ be represented by its absolutely continuous representative vanishing at $\pm 1$, which exists by step 2.1 and [F2]. Applying integration by parts [F4] to $F=q$ and $G=u^{\mathrm{\prime }}$ gives ${\int }_{-1}^1{q}^{\mathrm{\prime }}{u}^{\mathrm{\prime }}+{\int }_{-1}^{1}q{u}^{\mathrm{\prime }\mathrm{\prime }}=q(1)u^{\mathrm{\prime }}(1)-q(-1)u^{\mathrm{\prime }}(-1)=0$, that is ${\int }_{-1}^1{u}^{\mathrm{\prime }}{q}^{\mathrm{\prime }}={\int }_{-t}^{t}qdx$.

Since $v\in K$ one has $v\ge \psi$ a.e. on $(-1,1)$ [F1], and $u=\psi$ on $[-t,t]$ by step 2.1, so the representative $q=v-u$ of step 3.1 satisfies $q\ge 0$ a.e. on $(-t,t)$ and ${\int }_{-t}^{t}qdx={\int }_{-t}^t(v-\psi )dx\ge 0$.

For every $v\in K$, [F1] expands $J(v)-J(u)=a(u,q)+\frac{1}{2}a(q,q)={\int }_{-1}^1{u}^{\mathrm{\prime }}{q}^{\mathrm{\prime }}+\frac{1}{2}{\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2$ with $q=v-u$; by steps 3.1 and 4.1 this equals $\frac{1}{2}{\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\int }_{-t}^t(v-\psi )dx\ge 0$. Hence $u$ minimises $J$ on $K$.

If $v\in K$ satisfies $J(v)=J(u)$, then both nonnegative terms in step 5.1 vanish, so ${\int }_{-1}^1\mathrm{\mid }{q}^{\mathrm{\prime }}{\mathrm{\mid }}^2=0$ and $q^{\mathrm{\prime }}=0$ a.e.; the absolutely continuous representative of $q$ is then constant by [F6], and its endpoint values $Tq=(0,0)$ (it lies in $H_0^1(-1,1)$) force that constant to be $0$. Hence $q=0$ a.e. and $v=u$: the minimiser is unique.

By [F1] the obstacle problem has exactly one minimiser on $K$ and it is the unique solution of the variational inequality; steps 4.1 and 5.1 identify this minimiser with the explicit $u$, so $u$ is the obstacle solution. Step 2.1 gives the contact set $\{u=\psi \}=[-t,t]$ and the noncontact set $\{t<\mathrm{\mid }x\mathrm{\mid }<1\}$, and step 1.1 gives the matching slopes $u^{\mathrm{\prime }}(\pm t)=\mp t={\psi }^{\mathrm{\prime }}(\pm t)$ at the free boundary. The Axiom of Choice enters through the obstacle setting and the trace lemma [F1, F2], and it supplies the Countable and Dependent Choice consumed by the integration by parts [F3, F4]; no further choice principle is used.