= Solution
{c}
Write $V=H_0^1(\Omega;\mathbb R^3)$. For any $v\in V$ and any real $t$, $u+tv$ is admissible. Expanding the functional gives
$$
J(u+tv)-J(u)=t\bigl(a(u,v)-\ell(v)\bigr)+\frac{t^2}2a(v,v).
$$
Since the left side is nonnegative for positive and negative $t$, its <derivative> at zero vanishes. Thus the <first variation> gives the <weak formulation>
$$
\boxed{u\in V,\qquad a(u,v)=\ell(v)\quad(v\in V).}
$$
To identify its differential equation, define the <stress tensor>
$$
\sigma(u)=2\mu\varepsilon(u)+\lambda(\operatorname{div}u)I.
$$
This is a <symmetric matrix> field. The <Frobenius inner product> of a <symmetric matrix> with a <skew-symmetric matrix> is zero, so $\sigma(u):\nabla v=\sigma(u):\varepsilon(v)$. Since $\operatorname{tr}\varepsilon(v)=\operatorname{div}v$, we obtain
$$
a(u,v)=\int_\Omega\sigma(u):\nabla v.
$$
For smooth compactly supported <test functions>, <integration by parts> in each row gives
$$
\int_\Omega\sigma(u):\nabla v=-\langle\operatorname{div}\sigma(u),v\rangle.
$$
Hence $-\operatorname{div}\sigma(u)=f$ as a <distributional identity>, that is,
$$
-2\mu\operatorname{div}\varepsilon(u)-\lambda\nabla(\operatorname{div}u)=f.
$$
Membership in $V$ supplies the zero <Sobolev trace>. This identifies precisely the required <weak solution> of the <linear elasticity> problem, with row-wise <divergence>.
For uniqueness, use the exact form of the <Korn inequality> supplied in the paper. Put $E=\varepsilon(v)$ and $d=\operatorname{tr}E=\operatorname{div}v$. The <deviatoric strain> is $D=E-\tfrac13dI$. Because $\operatorname{tr}D=0$ and $I:I=3$, <orthogonality> for the <Frobenius inner product> gives
$$
|E|_F^2=|D|_F^2+\frac13d^2,\qquad |D|_F^2\leq|E|_F^2.
$$
If $C_K$ is the supplied <Korn inequality> constant and $K$ the permitted <Poincaré inequality> constant, then
$$
\|v\|_1^2\leq(1+K)\|\nabla v\|_0^2
\leq(1+K)C_K\|D\|_0^2
\leq(1+K)C_K\|\varepsilon(v)\|_0^2.
$$
Consequently,
$$
\boxed{a(v,v)\geq\alpha\|v\|_1^2,\qquad\alpha=\frac{2\mu}{(1+K)C_K}>0.}
$$
If $u_1,u_2$ are two <weak solutions>, their difference $d_u\in V$ satisfies $a(d_u,v)=0$ for every $v\in V$. Choosing $v=d_u$ and using this <coercive bilinear form> estimate forces $d_u=0$. The minimizer therefore gives the unique <weak solution>.
For later use, $|\varepsilon(v)|_F\leq|\nabla v|_F$ and $|\operatorname{div}v|\leq\sqrt3|\nabla v|_F$ also give the <bounded bilinear form> and <linear functional> estimates
$$
|a(p,v)|\leq(2\mu+3\lambda)\|p\|_1\|v\|_1,\qquad
|\ell(v)|\leq\|f\|_0\|v\|_1.
$$
Together with the <coercive bilinear form> estimate and the <Lax-Milgram theorem>, these would establish existence independently of the assumed minimizer.
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