Sensitivity Analysis for a Class of ${{H_{0}^{1}}}$-Elliptic Variational Inequalities of the Second Kind

Abstract

We study the stability of solutions to ${H_{0}^{1}}$-elliptic variational inequalities of the second kind that contain a non-differentiable Nemytskii operator. The local Lipschitz continuity of the solution map with respect to perturbations of the right-hand side and perturbations of the coefficient of the Nemytskii operator is proved for a large class of problems, and Hadamard directional differentiability results are obtained under comparatively mild structural assumptions. It is further shown that the directional derivatives of the solution map are typically characterized by elliptic variational inequalities in weighted Sobolev spaces whose bilinear forms contain surface integrals.

Appendix: A One-Dimensional Counterexample

In what follows, we demonstrate by means of a simple one-dimensional counterexample that the approach that has been employed by Sokolowski in [28,29,30] for the study of frictional contact problems indeed fails in case of the problem (P) due to a lack of polyhedricity. The variational inequality that we consider takes the following form

$$ \begin{array}{cc} w \in {H_{0}^{1}}(-1,1), \quad \int_{-1}^{1} w^{\prime}(v^{\prime}- w^{\prime}) \mathrm{d}x+ \int_{-1}^{1} |v| \mathrm{d}x - \int_{-1}^{1}|w|& \mathrm{d}x \geq \left\langle f, v- w \right\rangle \\ &\forall v\in {H_{0}^{1}}(-1,1) . \end{array} $$

(Q)

We use a prime here to denote a weak derivative. Note that the VI (Q) is indeed a special instance of (P) with Ω = (− 1, 1), $a(v_{1}, v_{2}) = (v_{1}^{\prime }, v_{2}^{\prime })_{L^{2}}$, c ≡ 1 and j(x) = |x|. Throughout this appendix, the right-hand side f is assumed to be an element of the space H^− 1(− 1, 1). Recall that the approach in [28,29,30] is based on the idea to transform the VI of the second kind at hand into a VI of the first kind by dualization and to subsequently study the differentiability of the solution operator to the dual problem with the classical results of Mignot (cf. [15, 22]). To dualize the problem (Q), we note that, according to [8, Lemma 3.2] (cf. also with the proof of Theorem 2.4 d)), for every f ∈ H^− 1(− 1, 1) with associated solution w = w(f) (see Theorem 2.4 for the unique solvability of (Q)), there exists a unique slack variable q ∈ L^∞(− 1, 1) such that

$$-w^{\prime\prime} + q = f \in H^{-1}(-1,1), \ \ q = \text{sgn}(w) \text{ a.e.\ in } \ \{w \neq 0\}, \ \ |q| \leq 1 \text{ a.e.\ in } \ (-1,1). $$

Consider now the solution operator T to the one-dimensional Poisson problem, i.e., the operator $T : H^{-1}(-1,1) \to {H_{0}^{1}}(-1,1)$ satisfying

$$\begin{array}{@{}rcl@{}} \int_{-1}^{1} T(p)^{\prime} v^{\prime} \mathrm{d}x = \left\langle p, v \right\rangle\quad \forall v \in {H_{0}^{1}}(-1,1) \quad \forall p \in H^{-1}(-1,1), \end{array} $$

and define

$$\begin{array}{@{}rcl@{}} {\Lambda} := \{p \in L^{\infty}(-1,1) : -1 \leq p \leq 1 \text{ a.e.\ in \ } (-1,1)\} \subset H^{-1}(-1,1). \end{array} $$

Then the properties of q yield

$$\begin{array}{@{}rcl@{}} \left\langle q, T(f - q) \right\rangle = \left\langle q, w \right\rangle = \int_{-1}^{1} |w| \mathrm{d}x \geq \int_{-1}^{1} w p \mathrm{d}x = \left\langle p, w \right\rangle = \left\langle p, T(f - q) \right\rangle \end{array} $$

for all p ∈ Λ. If we define b(p₁, p₂) := 〈p₂,T(p₁)〉, then the last inequality can be rewritten as

$$ \begin{array}{cc} q \in {\Lambda} , \quad b (q , p - q) \geq b(f, p - q)\quad \forall p \in {\Lambda} . \end{array} $$

(Q’)

This is the desired dualization of (Q) (cf. with the VI (4.124) in [30, Section 4.5]). Note that the bilinear form $b : H^{-1}(-1,1) \times H^{-1}(-1,1) \to \mathbb {R}$ in (Q’) satisfies

$$\begin{array}{c} |b(p_{1}, p_{2})| \leq \|T\| \|p_{1}\|_{H^{-1}}\|p_{2}\|_{H^{-1}}, \\ b(p_{1}, p_{2}) = \left\langle p_{2}, T(p_{1})\right\rangle = \int_{-1}^{1} T(p_{2})^{\prime} T(p_{1})^{\prime} \mathrm{d}x = \left\langle p_{1}, T(p_{2})\right\rangle = b(p_{2}, p_{1}), \\ \|p\|_{H^{-1}} = \sup\limits_{v \in {H_{0}^{1}}(-1,1), \|v\|_{H^{1}} \leq 1} (T(p)^{\prime}, v^{\prime})_{L^{2}} \leq \|T(p)^{\prime}\|_{L^{2}} = \sqrt{\left\langle p, T(p) \right\rangle } = \sqrt{b(p,p)} \end{array} $$

for all p,p₁,p₂ ∈ H^− 1(− 1, 1). Thus, b is continuous, symmetric and H^− 1-elliptic. Further, it is easy to see that Λ is a closed, non-empty, convex subset of H^− 1(− 1, 1). This shows that (Q’) is indeed an elliptic VI of the first kind in H^− 1(− 1, 1). Recall that, to be able to apply the classical differentiability results in [15, 22] to (Q’), we have to check the condition of polyhedricity, i.e., we have to check if the radial cone

$$\begin{array}{@{}rcl@{}} T_{\text{rad}}(q, {\Lambda} ) := \left\{z \in H^{-1}(-1,1) : \exists \varepsilon > 0 \text{ s.t. } \ q + \varepsilon z \in {\Lambda} \right\} \subset L^{\infty}(-1,1) \end{array} $$

and the critical cone

$$\begin{array}{@{}rcl@{}} T_{\text{crit}}(q, {\Lambda} ):= \left\{ z \in \text{cl}_{H^{-1}}\left( T_{\textup{rad}}(q, {\Lambda} ) \right) : b(q - f, z) = 0 \right\} \end{array} $$

in q to Λ satisfy (see, e.g., [15])

$$\begin{array}{@{}rcl@{}} T_{\textup{crit}}(q,{\Lambda} )= \text{cl}_{H^{-1}} \left( T_{\textup{crit}}(q, {\Lambda} ) \cap T_{\textup{rad}}(q, {\Lambda} ) \right)\text{.} \end{array} $$

(A.1)

We claim that the above is in general not true. To see this, we consider the right-hand side $\tilde f(x) := \pi ^{2} \sin (\pi x) +\text {sgn}(x) \in L^{\infty }(-1,1)$. Note that the solution $\tilde w$ and the multiplier $\tilde q$ associated with $\tilde f$ are given by $\tilde w (x) = \sin (\pi x)$ and $\tilde q (x) = \text {sgn}(x)$. This follows immediately from $- \tilde w^{\prime \prime } + \tilde q = \tilde f$, $\tilde q = \text {sgn}(\tilde w)$ and the estimate

$$\begin{array}{@{}rcl@{}} \left\langle \tilde f, v- \tilde w \right\rangle &=& \left\langle - \tilde w^{\prime\prime} + \tilde q, v- \tilde w \right\rangle = \int_{-1}^{1} \tilde w^{\prime}(v^{\prime} - \tilde w^{\prime}) \mathrm{d}x + \int_{-1}^{1} \tilde q v - |\tilde w| \mathrm{d}x \\ &&\qquad\qquad\quad \leq \int_{-1}^{1} \tilde w^{\prime}(v^{\prime} - \tilde w^{\prime}) \mathrm{d}x + \int_{-1}^{1} |v| - |\tilde w| \mathrm{d}x \quad \forall v \in {H_{0}^{1}}(-1,1). \end{array} $$

Note that the definition of Λ and the identity $\tilde q (x) = \text {sgn}(x)$ yield

$$\begin{array}{@{}rcl@{}} T_{\textup{rad}}(\tilde q, {\Lambda} ) = \{ z \in L^{\infty}(-1,1) : z \geq 0 \text{ a.e.\ in } \ (-1,0), \ z \leq 0 \text{ a.e.\ in }\ (0,1)\}. \end{array} $$

(A.2)

Further, the definition of b and the identity $\tilde w = T(\tilde f - \tilde q)$ imply

$$\begin{array}{@{}rcl@{}} b(\tilde q - \tilde f, z) = \left\langle z, T(\tilde q) - T (\tilde f) \right\rangle = -\left\langle z, \tilde w\right\rangle \quad \forall z \in H^{-1}(-1,1). \end{array} $$

If we combine the above, then we obtain that every $z \in T_{\textup {crit}}(\tilde q,{\Lambda } ) \cap T_{\textup {rad}}(\tilde q, {\Lambda } )$ satisfies

$$\begin{array}{@{}rcl@{}} b(\tilde q - \tilde f, z) = -\left\langle z, \tilde w\right\rangle = -\int_{-1}^{1} z(x)\sin(\pi x) \mathrm{d}x = \int_{-1}^{1} |z(x)\sin(\pi x)| \mathrm{d}x = 0. \end{array} $$

This shows that $\text {cl}_{H^{-1}} (T_{\textup {crit}}(\tilde q, {\Lambda } ) \cap T_{\textup {rad}}(\tilde q, {\Lambda } ) ) = T_{\textup {crit}}(\tilde q, {\Lambda } ) \cap T_{\textup {rad}}(\tilde q, {\Lambda } ) = \{0\}$. On the other hand, it follows from (A.2) that the functions

are contained in $T_{\textup {rad}}(\tilde q, {\Lambda } )$ for all α₁,α₂ ∈ [0,∞) and all t ∈ (0, 1), and these functions satisfy $z_{t, \alpha _{1}, \alpha _{2}} \to (\alpha _{1} - \alpha _{2}) \delta _{0}$ in H^− 1(− 1, 1), where δ₀ is the Dirac delta at the origin. Thus, $\mathbb {R} \delta _{0} \subset \text {cl}_{H^{-1}}\left (T_{\textup {rad}}(\tilde q, {\Lambda } ) \right )$. Moreover, we clearly have $b(\tilde q - \tilde f, z) = -\left \langle z, \tilde w \right \rangle = 0$ for all $z \in \mathbb {R} \delta _{0}$. Consequently, $\mathbb {R} \delta _{0} \subset T_{\textup {crit}}(\tilde q, {\Lambda } )$, and we arrive at

$$\begin{array}{@{}rcl@{}} \mathbb{R} \delta_{0} \subset T_{\textup{crit}}(\tilde q, {\Lambda} ), \qquad \text{cl}_{H^{-1}} \left( T_{\textup{crit}}(\tilde q, {\Lambda} ) \cap T_{\textup{rad}}(\tilde q, {\Lambda} ) \right) = \{0\}. \end{array} $$

The above shows that the polyhedricity condition (A.1) is violated in $\tilde q$, that the results of Mignot in [15, 22] cannot be employed and that the approach of [28,29,30] is indeed not applicable. Note that the set Λ is trivially polyhedric as a subset of the Dirichlet space L²(− 1, 1), see, e.g., [31]. Our example shows that this is not the case when Λ is considered as a subset of the space H^− 1(− 1, 1). Moreover, it should be noted that, in the above, we have $\mathcal{M} = \{x \in (-1,1) : \tilde w(x) = 0, \tilde w^{\prime }(x) \neq 0\} = \{0\}$. This shows that the surface integrals in (5.1) appear exactly on those parts of the domain Ω that are responsible for the loss of polyhedricity in the dual setting.