Topological Defects and Metric Anomalies as Sources of Incompatibility for Piecewise Smooth Strain Fields

Abstract

The incompatibility of linearized piecewise smooth strain field, arising out of volumetric and surface densities of topological defects and metric anomalies, is investigated. First, general forms of compatibility equations are derived for a piecewise smooth strain field, defined over a simply connected domain, with either a perfectly bonded or an imperfectly bonded interface. Several special cases are considered and discussed in the context of existing results in the literature. Next, defects, representing dislocations and disclinations, and metric anomalies, representing extra matter, interstitials, thermal, and growth strains, etc., are introduced in a unified framework which allows for incorporation of their bulk and surface densities, as well as for surface densities of defect dipoles. Finally, strain incompatibility relations are derived both on the singular interface, and away from it, with sources in terms of defect and metric anomaly densities. With appropriate choice of constitutive equations, the incompatibility relations can be used to determine the state of internal stress within a body in response to the given prescription of defects and metric anomalies.

Appendices

Appendix A: Proof of Identities in Section 2.5

1.1 A.1 Proof of Identities 2.1

• (a) For $B \in \mathcal{B}(\varOmega )$ and $\boldsymbol{\psi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\nabla B (\boldsymbol{\psi })= - B(\operatorname{div}\boldsymbol{\psi }) =\int _{\varOmega }\langle \nabla b , \boldsymbol{\psi }\rangle dx - \int _{S} \langle [\!\![b ]\!\!]\boldsymbol{n}, \boldsymbol{\psi } \rangle da$.

• (b) For $C \in \mathcal{C}(\varOmega )$ and $\boldsymbol{\psi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, let $\overline{c} \in C^{\infty }(\varOmega )$ be a smooth extension of ${c} \in C^{\infty }(S)$ so as to write $\nabla C (\boldsymbol{\psi })= - C(\operatorname{div} \boldsymbol{\psi }) = -\int _{S} c (\operatorname{div} \boldsymbol{\psi }) da = -\int _{S} (\operatorname{div}(\overline{c} \boldsymbol{\psi }) - \langle \nabla \overline{c}, \boldsymbol{\psi } \rangle ) da$. Subsequently, use $\operatorname{div}(\overline{c} \boldsymbol{\psi }) = \operatorname{div}_{S} (c\boldsymbol{\psi }) + \langle \nabla (\overline{c}\boldsymbol{\psi })\boldsymbol{n}, \boldsymbol{n} \rangle$, $\langle \nabla \overline{c}, \boldsymbol{\psi } \rangle = \langle \nabla _{S} c,\boldsymbol{\psi } \rangle + \langle \nabla \overline{c} , \boldsymbol{n} \rangle \langle \boldsymbol{\psi },\boldsymbol{n} \rangle$ on $S$, and the divergence theorem to get the desired result.

• (c) For $F \in \mathcal{F}(\varOmega )$ and $\boldsymbol{\psi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\nabla F (\boldsymbol{\psi })= - F(\operatorname{div}\boldsymbol{\psi }) = -\int _{S} f {\partial ( \operatorname{div}\boldsymbol{\psi })}/{\partial n} da$. But ${\partial (\operatorname{div}\boldsymbol{\psi })}/{\partial n} = \langle \nabla (\operatorname{div}\boldsymbol{\psi }) , \boldsymbol{n}\rangle =\langle \operatorname{div}_{S} (\nabla \boldsymbol{\psi })^{T},\boldsymbol{n}\rangle + \langle (\nabla ( \nabla \boldsymbol{\psi }) ) \boldsymbol{n}\otimes \boldsymbol{n}, \boldsymbol{n}\rangle$, on one hand, and $\langle \operatorname{div} _{S} ( (\nabla \boldsymbol{\psi })^{T} ),\boldsymbol{n} \rangle =\operatorname{div}_{S} ({\partial \boldsymbol{\psi }}/{\partial n}) - \langle \nabla _{S} \boldsymbol{n}, \nabla \boldsymbol{\psi } \rangle$, on the other. Upon substitution, and using the chain rule for derivatives, we can obtain $\nabla F (\boldsymbol{\psi }) =$

  $$\begin{aligned} &- \int _{S} \biggl(\operatorname{div}_{S} \biggl(f \frac{\partial \boldsymbol{\psi }}{\partial n} \biggr)- \biggl\langle \nabla _{S} f, \frac{\partial \boldsymbol{\psi }}{\partial n} \biggr\rangle - \operatorname{div}_{S}\bigl(f( \nabla _{S} \boldsymbol{n}) \boldsymbol{\psi }\bigr)+ \bigl\langle \operatorname{div}_{S} (f\nabla _{S} \boldsymbol{n}), \boldsymbol{\psi } \bigr\rangle \\ &\quad{}+ \bigl\langle \bigl(\nabla (\nabla \boldsymbol{ \psi })\bigr) \boldsymbol{n}\otimes \boldsymbol{n}, \boldsymbol{n}\bigr\rangle \biggr) da, \end{aligned}$$

  which immediately yields the result.

• (d) For $H \in \mathcal{H}(\varOmega )$ and $\boldsymbol{\psi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, we have $\nabla H ( \boldsymbol{\psi }) = - H(\operatorname{div}\boldsymbol{\psi }) = -\int _{L} h (\operatorname{div}\boldsymbol{\psi }) dl = -\int _{L} (h \langle \nabla \boldsymbol{\psi },(\boldsymbol{I}-\boldsymbol{t} \otimes \boldsymbol{t})\rangle + \langle h \boldsymbol{t}, {\partial \boldsymbol{\psi }}/{\partial t}\rangle ) dl$, leading to the desired identity.

1.2 A.2 Proof of Identities 2.2

• (a) For $\boldsymbol{B} \in \mathcal{B}(\varOmega ,\mathbb{R}^{3})$ and ${\psi } \in \mathcal{D}(\varOmega )$, $\operatorname{Div}\boldsymbol{B} ({\psi })=-\boldsymbol{B}(\nabla {\psi }) = -\int _{\varOmega }\langle \boldsymbol{b},\nabla {\psi } \rangle dv$, which on using the divergence theorem yields the result.

• (b) For $\boldsymbol{C} \in \mathcal{C}(\varOmega ,\mathbb{R}^{3})$ and ${\psi } \in \mathcal{D}(\varOmega )$, $\operatorname{Div}\boldsymbol{C} ({\psi })=-\boldsymbol{C}(\nabla {\psi })=- \int _{S} \langle \boldsymbol{c}, \nabla {\psi } \rangle da = -\int _{S} \operatorname{div}_{S} (\boldsymbol{c} {\psi }) da + \int _{S} ( \operatorname{div}_{S} \boldsymbol{c} ){\psi } da - \int _{S} \langle \boldsymbol{c}, \boldsymbol{n} \rangle ({\partial {\psi}}/{\partial {n}}) da$. The desired identity follows upon using the divergence theorem.

• (c) For $\boldsymbol{F} \in \mathcal{F}(\varOmega ,\mathbb{R}^{3})$ and ${\psi } \in \mathcal{D}(\varOmega )$, $\operatorname{Div}\boldsymbol{F}( {\psi }) = -\boldsymbol{F}(\nabla {\psi }) = -\int _{S} \langle \boldsymbol{f},\nabla (\nabla {\psi }) \boldsymbol{n} \rangle da$. Using $\nabla (\nabla {\psi }) \boldsymbol{n} = (\boldsymbol{I}- \boldsymbol{n}\otimes \boldsymbol{n})(\nabla (\nabla {\psi }) \boldsymbol{n}) +(\boldsymbol{n}\otimes \boldsymbol{n})(\nabla ( \nabla {\psi }) \boldsymbol{n})$ and $(\boldsymbol{I}-\boldsymbol{n} \otimes \boldsymbol{n})(\nabla (\nabla {\psi }) \boldsymbol{n}) = \nabla _{S} ({\partial \varPsi }/{\partial n}) - \nabla _{S} \boldsymbol{n} \nabla {\psi }$ we get

  $$ \operatorname{Div}\boldsymbol{F}({\psi }) = - \int _{S} \biggl\langle \boldsymbol{f}, \biggl(\nabla _{S} \biggl(\frac{\partial \psi }{\partial n} \biggr) - \nabla _{S} \boldsymbol{n}\nabla {\psi } \biggr) \biggr\rangle da - \int _{S} \langle \boldsymbol{f},\boldsymbol{n} \rangle \bigl\langle \nabla (\nabla {\psi }), \boldsymbol{n} \otimes \boldsymbol{n} \bigr\rangle da, $$

  which after some manipulation produces the required identity.

• (d) For $\boldsymbol{H} \in \mathcal{H}(\varOmega ,\mathbb{R}^{3})$ and ${\psi } \in \mathcal{D}(\varOmega )$, we have $\operatorname{Div} \boldsymbol{H} ({\psi }) = -\boldsymbol{H}(\nabla {\psi }) = -\int _{L} \langle \boldsymbol{h},\nabla {\psi } \rangle dl = - \int _{L} \langle \boldsymbol{h},(\boldsymbol{I}-\boldsymbol{t} \otimes \boldsymbol{t}) \nabla {\psi } \rangle dl -\int _{L} \langle \boldsymbol{h},({\partial {\psi }}/{\partial t})\boldsymbol{t} \rangle dl$. The final identity is immediate.

1.3 A.3 Proof of Identities 2.3

• (a) For $\boldsymbol{B} \in \mathcal{B}(\varOmega ,\mathbb{R}^{3})$ and $\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\operatorname{Curl}\boldsymbol{B} (\boldsymbol{\phi })= \boldsymbol{B} (\operatorname{curl}\boldsymbol{\phi })= \int _{\varOmega } \langle \boldsymbol{b},\operatorname{curl}\boldsymbol{\phi } \rangle dv = \int _{\varOmega } (\operatorname{div}(\boldsymbol{\phi }\times \boldsymbol{b})+\langle \operatorname{curl}\boldsymbol{b}, \boldsymbol{\phi } \rangle ) dv$. The result follows after using the divergence theorem.

• (b) For $\boldsymbol{C} \in \mathcal{C}(\varOmega ,\mathbb{R}^{3})$ and $\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, we have $\operatorname{Curl}\boldsymbol{C} (\boldsymbol{\phi })= \boldsymbol{C} (\operatorname{curl}\boldsymbol{\phi })= \int _{S} \langle \boldsymbol{c},\operatorname{curl}\boldsymbol{\phi } \rangle da = \int _{S} \langle \boldsymbol{c},\operatorname{curl}_{S} \boldsymbol{\phi }-({\partial \boldsymbol{\phi }}/{\partial n}) \times \boldsymbol{n} \rangle da$. Recall the identity $\operatorname{div}_{S} (\boldsymbol{u}\times \boldsymbol{v})=\langle \operatorname{curl}_{S} \boldsymbol{u},\boldsymbol{v}\rangle -\langle \boldsymbol{u},\operatorname{curl}_{S} \boldsymbol{v} \rangle$, for $\boldsymbol{u},\boldsymbol{v} \in C^{\infty }(S,\mathbb{R}^{3})$, to get

  $$ \operatorname{Curl}\boldsymbol{C} (\boldsymbol{\phi })= \int _{S} \operatorname{div}_{S} (\boldsymbol{ \phi }\times \boldsymbol{c}) da + \int _{S} \langle \boldsymbol{\phi },\operatorname{curl}_{S} \boldsymbol{c}\rangle da- \int _{S} \biggl\langle \boldsymbol{c} , \frac{ \partial \boldsymbol{\phi }}{\partial n} \times \boldsymbol{n} \biggr\rangle da, $$

  which immediately lead to the pertinent identity.

• (c) For $\boldsymbol{F} \in \mathcal{F}(\varOmega ,\mathbb{R}^{3})$ and $\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\operatorname{Curl}\boldsymbol{F} (\boldsymbol{\phi }) = \boldsymbol{F}(\operatorname{curl}\boldsymbol{\phi } ) = \int _{S} \langle \boldsymbol{f},{\partial (\operatorname{curl} \boldsymbol{\phi })}/{\partial n} \rangle da$. Use the skew part of the identity $\nabla _{S} ({\partial \boldsymbol{\phi }}/{\partial n})= \nabla (\nabla \boldsymbol{\phi })\boldsymbol{n}-(\nabla (\nabla \boldsymbol{\phi })\boldsymbol{n}\otimes \boldsymbol{n})\otimes \boldsymbol{n} +\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n}$ to obtain $\operatorname{curl}_{S}({\partial \boldsymbol{\phi }}/{\partial n})={\partial (\operatorname{curl}\boldsymbol{\phi })}/{\partial n}+( \nabla (\nabla \boldsymbol{\phi })\boldsymbol{n}\otimes \boldsymbol{n})\times \boldsymbol{n} + ax(\nabla \boldsymbol{\phi } \nabla _{S} \boldsymbol{n}-(\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n})^{T})$. Furthermore, we note that

  $$\begin{aligned} \int _{S} \biggl\langle \boldsymbol{f}, \operatorname{curl}_{S} \biggl(\frac{\partial \boldsymbol{\phi }}{\partial n} \biggr) \biggr\rangle da =& \int _{S} \biggl\langle -\kappa (\boldsymbol{f}\times \boldsymbol{n} ) +\operatorname{curl}_{S} (\boldsymbol{f} ), \frac{ \partial \boldsymbol{\phi }}{\partial n} \biggr\rangle da \\ &{}+ \int _{\partial S - \partial \varOmega } \biggl\langle ( \boldsymbol{f}\times \boldsymbol{ \nu } ), \frac{\partial \boldsymbol{\phi }}{\partial n} \biggr\rangle dl, \\ \int _{S} \bigl\langle \boldsymbol{f}, \bigl(\nabla (\nabla \boldsymbol{\phi })\boldsymbol{n}\otimes \boldsymbol{n} \bigr)\times \boldsymbol{n} \bigr\rangle da =&- \int _{S} \bigl\langle \boldsymbol{f}\times \boldsymbol{n}, \bigl(\nabla (\nabla \boldsymbol{\phi }) \boldsymbol{n}\otimes \boldsymbol{n} \bigr) \bigr\rangle da, \end{aligned}$$

  and $\langle \boldsymbol{f} ,ax(\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n}-(\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n})^{T}) \rangle = \langle \tilde{\boldsymbol{f}},\nabla \boldsymbol{\phi } \nabla _{S} \boldsymbol{n} \rangle = -\langle (\nabla _{S} \boldsymbol{n} \times \boldsymbol{f})^{T},\nabla _{S} \boldsymbol{\phi } \rangle = \langle \operatorname{div}_{S} (\nabla _{S} \boldsymbol{n} \times \boldsymbol{f})^{T},\boldsymbol{\phi } \rangle - \operatorname{div}_{S} ((\nabla _{S} \boldsymbol{n} \times \boldsymbol{f}) \boldsymbol{\phi })$, where $\tilde{\boldsymbol{f}}$ is the skew symmetric tensor whose axial vector is $\boldsymbol{f}$. Consequently, $\int _{S} \langle \boldsymbol{f} ,ax(\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n}-(\nabla \boldsymbol{\phi }\nabla _{S} \boldsymbol{n})^{T}) \rangle da =$

  $$ \int _{S} \bigl\langle \operatorname{div}_{S} ( \nabla _{S} \boldsymbol{n} \times \boldsymbol{f} )^{T}, \boldsymbol{\phi } \bigr\rangle da - \int _{\partial S - \partial \varOmega } \bigl\langle ( \nabla _{S} \boldsymbol{n} \times \boldsymbol{f})\boldsymbol{\phi }, \boldsymbol{\nu } \bigr\rangle dl + \int _{S} \kappa \bigl\langle ( \nabla _{S} \boldsymbol{n}\times \boldsymbol{f})\boldsymbol{\phi }, \boldsymbol{n} \bigr\rangle da. $$

  The desired identity follows upon combining the above results.

• (d) For $\boldsymbol{H} \in \mathcal{H}(\varOmega ,\mathbb{R}^{3})$ and $\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\operatorname{Curl}\boldsymbol{H} (\boldsymbol{\phi })= \boldsymbol{H}(\operatorname{curl}\boldsymbol{\phi })= \int _{L} \langle \boldsymbol{h} , \operatorname{curl}\boldsymbol{\phi }) \rangle dl= \int _{L} \langle \boldsymbol{h} , \operatorname{curl}_{t} \boldsymbol{\phi } \rangle dl -\int _{L} \langle \boldsymbol{h} , ( {\partial \boldsymbol{\phi }}/{\partial t} \times \boldsymbol{t}) \rangle dl$. The required result is imminent.

Appendix B: A Lemma for Theorem 2.2

A distribution $T\in \mathcal{D}'(\varOmega )$ is said to be of order $m$ if, for any compact set $K \subset \varOmega$, there exists a finite $M \in \mathbb{R}$ such that, for any smooth function $\phi$ supported in $K$, $|T(\phi )| \leq M \varSigma _{|\alpha | \leq m} |\text{sup} ( \partial ^{\alpha }\phi )|$, where $\partial ^{\alpha }$ denotes the $\alpha$ order derivative of $\phi$. In particular, $T$ is of order 0 if $|T(\phi )| \leq M |\text{sup} (\phi )|$.

Lemma B.1

For a$\boldsymbol{T} \in \mathcal{D}'(\varOmega ,\mathbb{R}^{3})$, which satisfies$\operatorname{Div}\boldsymbol{T}=0$, there exists$\boldsymbol{u} \in C^{\infty } (\varOmega , \mathbb{R}^{3})$and$\boldsymbol{S} \in \mathcal{D}'(\varOmega ,\mathbb{R}^{3})$such that

$$ \boldsymbol{T_{u}} - \boldsymbol{T} = \operatorname{Curl} \boldsymbol{S}, $$

(114)

where$\boldsymbol{T_{u}} \in \mathcal{D}'(\varOmega ,\mathbb{R}^{3})$is given by$\boldsymbol{T_{u}} (\boldsymbol{\phi })=\int _{\varOmega } \langle \boldsymbol{u}, \boldsymbol{\phi } \rangle dv$for all$\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$.

Proof

Consider a map $H^{y} : [0,1]\times \mathbb{R}^{3} \to \mathbb{R}^{3}$ given by $H^{y} (t,x)=x + t \psi (x) y$, where $\psi$ is a smooth scalar field over $\mathbb{R}^{3}$ such that $\psi (x) = 0$ for ${x} \notin \varOmega$ but $0<\psi \leq 1$, $|\nabla \psi | \leq 1$ whenever $x \in \varOmega$, and ${y} \in \mathbb{R}^{3}$ is such that $|{y}| <1$. It can be shown that, for any $t \in [0,1]$, $H^{y} : [0,1] \times \varOmega \to \varOmega$. For $\boldsymbol{\phi } \in \mathcal{D}( \varOmega ,\mathbb{R}^{3})$, we introduce

$$ \boldsymbol{S}^{y} (\boldsymbol{\phi }) = \int _{0}^{1} \bigl\langle \boldsymbol{T} , \bigl(\boldsymbol{\phi } \bigl(H^{y}(t,{x})\bigr) \times y\bigr)\psi (x) \bigr\rangle dt. $$

(115)

To check that $\boldsymbol{S}^{y} \in \mathcal{D}'(\varOmega ,\mathbb{R} ^{3})$ it is sufficient to note that ${S}^{y}_{i}$ defines a linear functional on $\mathcal{D}(\varOmega )$ and that a sequence of smooth functions $\phi _{m}$ converging to 0 implies the convergence of $(\boldsymbol{\phi } (H^{y}(t,x)) \times y)_{i}\psi (x)$, and consequently of ${S}^{y}_{i}(\phi _{m})$, to 0. Moreover, for $\boldsymbol{\phi } \in \mathcal{D}(\varOmega ,\mathbb{R}^{3})$, $\operatorname{Curl}\boldsymbol{S}^{y} (\boldsymbol{\phi }) = \boldsymbol{S}^{y} (\operatorname{curl}\boldsymbol{\phi })=$

$$ \int _{0}^{1} \biggl\langle {T}_{i}, (\frac{\partial }{\partial t} \biggl( \phi _{i}\bigl(H^{y} (t,x)\bigr)+ \phi _{j} \bigl(H^{y} (t,x)\bigr) y_{j} t \frac{\partial \psi }{\partial x_{i}} \biggr) - \frac{\partial }{\partial x_{i}} \bigl( \phi _{j} \bigl(H^{y} (t,x) y_{j} \psi \bigr) \bigr) \biggr\rangle dt, $$

which, on using $\operatorname{Div}\boldsymbol{T}=0$ and $H^{y} (0,x)=x$, yields

$$ \operatorname{Curl}\boldsymbol{S}^{y} (\boldsymbol{\phi }) = \biggl\langle {T} _{i}, \biggl(\phi _{i}\bigl({x}+\psi ({x}) y \bigr)+ \phi _{j}\bigl({x}+\psi ({x}) y\bigr) y _{j} \frac{\partial \psi }{\partial x_{i}} \biggr) \biggr\rangle - \boldsymbol{T}(\boldsymbol{\phi }). $$

(116)

Let $\rho \in C^{\infty }( \mathbb{R}^{3} )$ be a smooth function supported over a ball of unit radius, centred at the origin, such that it depends only on $|\boldsymbol{x}|$ and satisfies $\int _{\mathbb{R} ^{3}} \rho ({x}) dv =1$. Given $\epsilon >0$, the function $\rho _{\epsilon }= \epsilon ^{-3} \rho (x/{\epsilon })$ is supported in a ball of radius $\epsilon$ such that $\int _{\mathbb{R}^{3}} \rho _{\epsilon }({x}) dv =1$. For $\boldsymbol{S} \in \mathcal{D}'( \varOmega ,\mathbb{R}^{3})$, defined as $\boldsymbol{S}=\int _{B(0,\epsilon )} \boldsymbol{S}^{y} \rho _{\epsilon }(y) dv_{y}$, where $B(0,\epsilon )$ is a ball of radius $\epsilon$ centred at the origin, $\operatorname{Curl}\boldsymbol{S}(\boldsymbol{\phi })= \int _{B(0,\epsilon )} \operatorname{Curl}\boldsymbol{S}^{y} ( \boldsymbol{\phi })\rho _{\epsilon }(y) dv_{y} =$

$$ \int _{B(0,\epsilon )} \biggl( \biggl\langle {T}_{i}, \biggl(\phi _{i}\bigl({x}+\psi ( {x}) y\bigr)+ \phi _{j} \bigl({x}+\psi ({x}) y\bigr) y_{j} \frac{\partial \psi }{ \partial x_{i}}\biggr)\biggr\rangle \rho _{\epsilon }(y) \biggr) dv_{y} - \boldsymbol{T}( \boldsymbol{\phi }). $$

We can henceforth write $\operatorname{Curl}\boldsymbol{S}= \boldsymbol{T}_{1}-\boldsymbol{T}$, where $\boldsymbol{T}_{1} ( \boldsymbol{\phi })= \boldsymbol{T} (\boldsymbol{\phi }^{\epsilon })$,

$$ \phi ^{\epsilon }_{i}(x) = \int \rho _{\epsilon } \biggl(\frac{z-x}{ \psi (x)} \biggr) \frac{\phi _{i}(z)}{\psi (x)} dv_{z} + \int \biggl(\rho _{\epsilon } \biggl(\frac{z-x}{\psi (x)} \biggr) \frac{z_{j} - x_{j}}{ \psi (x)}\frac{\partial \psi }{\partial x_{i}} \biggr) \frac{\phi _{j}(z)}{ \psi (x)} dv_{z}, $$

and $z = x+\psi (x) y$. Since $\rho _{\epsilon }$ is smooth, its derivatives remain bounded and the supremum norm of $\boldsymbol{\phi }^{\epsilon }$ and all the partial derivatives of $\boldsymbol{\phi }^{\epsilon }$ are controlled by the supremum norm of $|\boldsymbol{\phi }|$. Therefore, there exist a $\boldsymbol{u} \in C^{\infty } (\varOmega , \mathbb{R}^{3})$ such that $\boldsymbol{T} _{1}=\boldsymbol{T_{u}}$ leading us to our assertion. □