We assume that the boundaries of Ω are chosen such that the equation
$$ \left\{ \begin{array}{rll} -\varepsilon\text{div}(M\nabla T_{H})&=\tilde f&\qquad\text{in}~{{\varOmega}},\\ T_{H}&=0&\qquad\text{on}~{\varGamma}_{i},\quad i=1,\ldots,n,\\ \varepsilon M\nabla T_{H}\cdot n &=0&\qquad\text{on}~{\varGamma}_{N} \end{array}\right. $$
(5)
has a unique solution TH ∈ H2(Ω) for any \(\tilde f\in L^{2}({{\varOmega }})\). Moreover we assume that \(M\in W^{1,\infty }({{\varOmega }})^{d\times d}\) and that M(x)v ⋅ v ≥ α|v|2 for a.e. x ∈Ω holds. Further, for any \(u\in \mathbb {R}^{n}\) we assume the existence of g ∈ W2,6(Ω) with \(g|_{{\varGamma }_{i}}=u_{i}\) for i = 1, … , n, g vanishing in a neighbourhood of ΓN, and \(\|g\|_{W^{2,6}({{\varOmega }})}\leq c|u|_{\mathbb {R}^{n}}\), with c independent of u. For example \(g={\sum }_{i=1}^{n}u_{i}g_{i}\) can be chosen where the functions gi are chosen as smooth bump functions which are equal to 1 on \(\bar \omega _{i}\), vanish near ΓN and have the property supp(gi) ∩supp(gj) = ∅ for all i, j = 1, … , n. Moreover for \(\tilde T:=T_{H}+g\in H^{2}({{\varOmega }})\), we have \(\tilde T|_{{\varGamma }_{i}}=u_{i}\) for all i = 1, … , n and
$$ \varepsilon {\int}_{{{\varOmega}}} M\nabla T_{H}\cdot \nabla v~\mathrm{d}x={\int}_{{{\varOmega}}} \tilde fv+ \varepsilon \text{div}(M\nabla g)v~\mathrm{d}x $$
for all \(v\in V:= {H^{1}_{0}}({{\varOmega }}\cup {\varGamma }_{N})=\{v\in H^{1}({{\varOmega }})~|~v|_{{\varGamma }_{i}}=0,~i=1,\ldots ,n\}\). In the subsequent developments (5) will be used with \(\tilde f\) replaced by
$$ -\sqrt{\beta+|\nabla (T_{H} +g)|_{M}^{2}}+ \varepsilon\text{div}(M\nabla g)+1. $$
Theorem 1
For ε > 0 sufficiently large (4) has a unique solution
$$ T\in W^{2,6}({{\varOmega}}). $$
Moreover there exists a constant c, independent of \(u\in \mathbb {R}^{n}\), and β ∈ [0, 1] such that \(\|T\|_{W^{2,6}} \le c(1+ |u|)\).
Proof
1. Existence: Let \(T_{H}\in {H^{1}_{0}}({{\varOmega }}\cup {\varGamma }_{N})\) be fixed. Then we set
$$ f(T_{H})(x):=-\sqrt{\beta+|\nabla(T_{H}(x)+g(x))|_{M}^{2}} $$
with β ∈ [0, 1]. Since TH ∈ V, g ∈ W2,6(Ω) and \(M\in W^{1,\infty }({{\varOmega }})^{d\times d}\) it follows that f(TH) ∈ L2(Ω). Now let w ∈ H2(Ω) be the unique solution of
$$ \left\{ \begin{array}{rll} -\varepsilon\text{div}(M\nabla w)&=f(T_{H})+\varepsilon\text{div}(M\nabla g)+1&\qquad\text{in}~{{\varOmega}},\\ w&=0&\qquad\text{on}~{\varGamma},\\ \varepsilon M\nabla w\cdot n &=0&\qquad\text{on}~{\varGamma}_{N} \end{array}\right. $$
(6)
with the estimate
$$ \|w\|_{H^{2}({{\varOmega}})}\leq c(\|f(T_{H})\|_{L^{2}({{\varOmega}})}+|u|+1). $$
Thus we can define the operator \(G\colon V \rightarrow H^{2}({{\varOmega }}) \subset V \), G: TH↦w which satisfies the inequality
$$ \|G(T_{H})\|_{H^{2}({{\varOmega}})}\leq c(M,\varepsilon)(\|T_{H}\|_{V}+|u|+1), $$
(7)
with c(M, ε) independent of β ∈ [0, 1] and TH. In the following we shall utilize Schaefer’s fixed point theorem in order to prove that G has a fixed point. At first we prove that \(G\colon V\rightarrow V\) is continuous and compact. Let {TH, k}k ⊂ V be a convergent sequence with limit TH in V. We set wk := G(TH, k) and have
$$ \sup_{k}\|w_{k}\|_{H^{2}({{\varOmega}})}<\infty, $$
according to (7). The compact embedding of H2(Ω) ∩ V in V implies the existence of a subsequence {wk} and of a w ∈ V with \(w_{k}\rightarrow w\) in V. By taking the limit in the weak formulation of (6) we see that G(TH) = w. Thus \(G\colon V\rightarrow V\) is continuous. A similar argument shows that \(G\colon V \rightarrow V\) is compact. In order to apply Schaefer fixed point theorem we have to further show that the set
$$ \{T\in V ~|~T=\lambda G(T)~\text{for some}~0<\lambda\leq1\} $$
is bounded in V. Let TH ∈ V be such that TH = λG(TH) for some 0 < λ ≤ 1. Then we have
$$ -\varepsilon\text{div}(M\nabla T_{H})=\lambda(f(T_{H})+\varepsilon \text{div}(M\nabla g)+1)\quad\text{a.e. in }{{\varOmega}}. $$
Multiplying this equation with TH and integrating over Ω, we obtain by Young’s inequality and the fact that 0 < λ ≤ 1:
$$ \begin{array}{@{}rcl@{}} \varepsilon\alpha\|\nabla T_{H}\|_{L^{2}({{\varOmega}})}^{2}&\leq&\varepsilon{\int}_{{{\varOmega}}} M\nabla T_{H}\cdot\nabla T_{H}~\mathrm{d}x\\ & =&\lambda{\int}_{{{\varOmega}}}\left( -\sqrt{\beta+|\nabla(T_{H}+g)|_{M}^{2}}+1+\varepsilon \text{div}(M\nabla g)\right)T_{H}~\mathrm{d}x\\ &\leq& \|M\|_{\infty}^{2}\|\nabla T_{H}\|_{L^{2}({{\varOmega}})}^{2}+\frac 32\|T_{H}\|_{L^{2}({{\varOmega}})}^{2} + \frac{\varepsilon^{2}}{2}\|\text{div}(M\nabla g)\|_{L^{2}({{\varOmega}})}^{2}\\ && +\|M\|_{\infty}^{2}\|\nabla g\|_{L^{2}({{\varOmega}})}^{2}+\frac{|{{\varOmega}}|}{2}(\beta+1)\\ &\leq& c(M)\left( \|\nabla T_{H}\|_{L^{2}({{\varOmega}})}^{2}+\varepsilon^{2}|u|^{2}+1+\beta\right), \end{array} $$
with c(M) independent of λ and ε. Thus if ε is sufficiently large, we have \(\|T_{H}\|_{V}\leq \tilde c(M,\varepsilon )(1+|u|)\), for some constant \(\tilde c(M,\varepsilon )\) independent of λ ∈ (0, 1] and β ∈ [0, 1]. Then Schaefer’s fixed point theorem can be applied to G and yields the existence of an element TH ∈ V with G(TH) = TH which is a solution of (6). Setting T = TH + g we have obtained a solution to (4), for which by (7) we have \(|T|_{H^{2}({{\varOmega }})} \le C(M,\varepsilon )\), with C(M, ε) independent of β ∈ [0, 1].
Moreover, since ∇T ∈ H1(Ω)d and thus ∇T ∈ L6(Ω)d, and since also g ∈ W2,6(Ω) we have that
$$ -\sqrt{\beta+|\nabla T|_{M}^{2}}+1+\varepsilon\text{div}(M\nabla g)\in L^{6}({{\varOmega}}), $$
and thus
$$ \|T\|_{W^{2,6}({{\varOmega}})} \le \tilde C(M,\varepsilon)(1+|u|) \text { with } \tilde C(M,\varepsilon) \text{ independent of } \beta\in [0, 1]. $$
2. Uniqueness: Let Ti ∈ W2,6(Ω), i = 1, 2 be two solutions of (4) and define δT = T1 − T2. Then δT satisfies the equation
$$ \left\{ \begin{array}{rll} -\varepsilon\text{div}(M\nabla \delta T)+\sqrt{\beta+|\nabla T_{1}|_{M}^{2}} - \sqrt{\beta+|\nabla T_{2}|_{M}^{2}}&=0&\qquad\text{in}~{{\varOmega}},\\ \delta T&=0&\qquad\text{on}~{\varGamma},\\ \varepsilon\nabla \delta T\cdot n&=0&\qquad\text{on}~{\varGamma}_{N}. \end{array}\right. $$
(8)
Let us define for \((x,v) \in {{\varOmega }} \times \mathbb {R}^{d}\) the function
$$ B(x,v):= \left\{\begin{array}{ll} \frac{M(x)v}{\sqrt{\beta+|v|_{M(x)}^{2}}},&\quad v\neq 0,\\ 0,&\quad v=0. \end{array}\right. $$
It is easy to see, that
$$ B(x,\bar v)\cdot(v-\bar v)\leq \sqrt{\beta+|v|_{M}^{2}}-\sqrt{\beta+|\bar v|_{M}^{2}} $$
holds. Indeed, in case \(\beta +|\bar v|_{M}^{2} =0\) the inequality is correct by the definition of B. Otherwise we have
$$ \begin{array}{@{}rcl@{}} \frac{M\bar v\cdot(v-\bar v)}{\sqrt{\beta+|\bar v|_{M}^{2}}} &=& \frac{M\bar v\cdot v+\beta}{\sqrt{\beta+|\bar v|_{M}^{2}}}-\frac{\beta+|\bar v|_{M}^{2}}{\sqrt{\beta+|\bar v|_{M}^{2}}}\\ &\leq& \frac{\sqrt{\beta+|v|_{M}^{2}}\sqrt{\beta+|\bar v|_{M}^{2}}}{\sqrt{\beta+|\bar v|_{M}^{2}}}-\sqrt{\beta+|\bar v|_{M}^{2}} = \sqrt{\beta+|v|_{M}^{2}}-\sqrt{\beta+|\bar v|_{M}^{2}}. \end{array} $$
Here we have used that \(\left (\begin {array}{cc} M & 0 \\ 0 & 1 \end {array} \right )\) defines a scalar product for the vectors \((v,\sqrt {\beta })\). Alternatively we can note that B(x, v) is an element of the subdifferential of the convex function \(v\to \sqrt {\beta + |v|_{M}^{2}}\). Thus we have
$$ B(x,\nabla T_{2})\cdot\nabla\delta T\leq \sqrt{\beta+|\nabla T_{1}|_{M}^{2}}-\sqrt{\beta+|\nabla T_{2}|_{M}^{2}}. $$
Consequently,
$$ -\varepsilon\text{div}(M\nabla \delta T)+B(x,\nabla T_{2})\cdot \nabla\delta T\leq 0, $$
where \(B(x,\nabla T_{2})\in L^{\infty }({{\varOmega }})^{d}\), since T2 is an element of W2,6(Ω). Then the maximum principle implies that δT ≤ 0 in Ω, see [25, Theorem 3.27]. Exchanging the roles of T1 and T2 in the above argument leads to δT ≥ 0 in Ω, and consequently to δT = 0, which implies the desired uniqueness. □
This proof is inspired from [8, Section 9.2, Theorem 5]. Henceforth it will be assumed that ε is large enough so that the solution to (4) according to Theorem 1 exists.
Theorem 2
We have
$$ T_{\beta}\to T_{0}\quad \text{ in }~ H^{2}({{\varOmega}}), $$
where Tβ denotes the solution to (4) as a function of β.
Proof
By Theorem 1 the family \(\{T_{H}^{\beta }\}_{\beta \in [0, 1]}\) is bounded in H2(Ω) ∩ V and hence there exists a subsequence, denoted in the same manner, and \(\hat T_{H} \in H^{2}({{\varOmega }}) \cap V\) such that \(T_{H}^{\beta } \rightharpoonup \hat T_{H}\) in H2(Ω) and \(T_{H}^{\beta } \to \hat T_{H}\) in V. Thus we can pass to the limit in
$$ {\int}_{{{\varOmega}}}\varepsilon M\nabla T_{H}^{\beta}\cdot\nabla \varphi+\sqrt{\beta+|\nabla (T_{H}^{\beta}+g)|_{M}^{2}}\varphi~\mathrm{d}x = {\int}_{{{\varOmega}}}(1+\varepsilon\text{div}(M\nabla g))\varphi~\mathrm{d}x~\text{ for all } \varphi \in V $$
to obtain that
$$ {\int}_{{{\varOmega}}}\varepsilon M \nabla \hat T_{H}\cdot\nabla \varphi +|\nabla (\hat T_{H}+g)|_{M}\varphi~\mathrm{d}x = {\int}_{{{\varOmega}}}(1+\varepsilon\text{div}(M\nabla g)) \varphi~\mathrm{d}x~\text{ for all } \varphi \in V. $$
Moreover, by the trace theorem \(\hat T_{H}=0\) on Γi for i = 1, … , n. Now we set \(T_{0}=\hat T_{H}+g\). By uniqueness, asserted in Theorem 1 we have \(\hat T_{H} = T_{H}\), where TH is the homogenous solution for β = 0 from Theorem 1, and thus the whole family \(T_{\beta }=T_{H}^{\beta }+g\) converges to T0 in V. Moreover we have
$$ \begin{array}{@{}rcl@{}} {\int}_{{{\varOmega}}}\varepsilon^{2} \left|\text{div}(M\nabla (T_{\beta} - T_{0}))\right|^{2}~\mathrm{d}x &=& {\int}_{{{\varOmega}}}\varepsilon^{2} \left|\text{div}(M\nabla (T_{H}^{\beta} - T_{H}))\right|^{2}~\mathrm{d}x\\ &=& {\int}_{{{\varOmega}}}\left( \sqrt{\beta+|\nabla T_{\beta}|_{M}^{2}}-|\nabla T_{0}|_{M}\right)^{2}~\mathrm{d}x \to 0 \end{array} $$
for β → 0+. Since \(T_{H}^{\beta }|_{{\varGamma }_{i}},T_{H}|_{{\varGamma }_{i}}=0\) and \(\big ({\int \limits }_{{{\varOmega }}}|\text {div}(M\nabla \cdot )|^{2}~\mathrm {d}x\big )^{1/2}\) defines an equivalent norm to the H2(Ω)-norm on H2(Ω) ∩ V, the claim follows. □
We close this section by asserting the existence of a solution to (3).
Theorem 3
There exists an optimal solution to problem (3).
Proof
Due to boundedness of Uad and boundedness from below of J, there exists a minimizing sequence {un} in Uad, satisfying
$$ \lim_{n\to\infty} J(u_{n})= \inf_{u\in U_{ad}}J(u). $$
(9)
Compactness and closedness of Uad imply the existence of a subsequence \(\{u_{n_{k}}\}\) with a limit \(\bar u \in U_{ad}\). Let \(T(u_{n_{k}})\) denote the solutions to (4) with \(u=u_{n_{k}}\). By Theorem 1 the sequence \(T(u_{n_{k}})\) is bounded in W2,6(Ω), and hence there exists a subsequence, denoted in the same manner, which converges weakly in W2,6(Ω) and, by Rellich’s compact embedding theorem, strongly in \(C^{1}(\bar {{\varOmega }})\) to some \(\bar T \in W^{2,6}({{\varOmega }})\). We can now pass to the limit in (4) (with \(T=T(u_{n_{k}})\) and \(u=u_{n_{k}}\)) to obtain that the pair \((\bar u, \bar T)\) satisfies (4). By (9) we have that \(\lim _{n\to \infty } J(u_{n_{k}})= J(\bar u) = \min \limits _{u\in U_{ad}}J(u)\). This concludes the proof. □