= Solution
Taking traces of the <constitutive equation> gives
$$
\lambda\operatorname{tr}e_\lambda=\frac13\left(\operatorname{tr}S_\lambda-2\mu\operatorname{tr}e_\lambda\right).
$$
The assumed finite limits of <stress> and <strain> make the right side bounded. Therefore $\operatorname{tr}e_\lambda=O(\lambda^{-1})$, and the limiting <strain> satisfies $\operatorname{tr}e=0$. The permitted interchange of <derivatives> and limits gives $\operatorname{tr}e=\operatorname{div}u$. Define
$$
q=\lim_{\lambda\to\infty}\lambda\operatorname{tr}e_\lambda=\tfrac13\operatorname{tr}S,\qquad p=-q.
$$
Then the limit <stress> is $S=-pI+2\mu e$. Passing to the limit in <force> balance and using $2\operatorname{Div}e=\Delta u+\nabla\operatorname{div}u$ gives
$$
\boxed{-\mu\Delta u+\nabla p=f,\qquad\operatorname{div}u=0\quad\text{in }\Omega,\qquad u=0\text{ on }\partial\Omega.}
$$
The <pressure> acts as the <Lagrange multiplier> enforcing the <volume> constraint. The <differential equations> determine it up to an additive constant, while a specified absolute limit <stress> can fix that constant.
Nonlinear <elastic incompressibility> is $\det Dy=1$. Expanding the <determinant> for $Dy=I+Du$ gives $\det(I+Du)=1+\operatorname{div}u+O(|Du|^2)$, so its first-order form is precisely $\operatorname{div}u=0$. The volumetric energy term $\lambda(\operatorname{div}u)^2/2$ becomes increasingly expensive as $\lambda$ grows, while the fixed positive $\mu$ leaves <shear strain> possible. Thus \b[the infinite-bulk-stiffness limit is the <incompressible limit of isotropic linear elasticity>], with the same equilibrium operator as the steady <Stokes equations>.
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