A Model Problem for Nematic-Isotropic Transitions with Highly Disparate Elastic Constants

Abstract

Continuing the program initiated in Golovaty et al. (SIAM J Math Anal 51(1):276–320, 2018), we analyze a model problem based on highly disparate elastic constants that we propose in order to understand corners and cusps that form on the boundary between the nematic and isotropic phases in a liquid crystal. For a bounded planar domain $\Omega$ we investigate the $\varepsilon \rightarrow 0$ asymptotics of the variational problem

$$\begin{aligned} \inf \displaystyle \frac{1}{2}\int _\Omega \left( \frac{1}{\varepsilon } W(u)+\varepsilon |\nabla u|^2 + L_\varepsilon (\mathrm {div}\,u)^2 \right) \,\hbox {d}x \end{aligned}$$

within various parameter regimes for $L_\varepsilon > 0.$ Here $u:\Omega \rightarrow \mathbb {R}^2$ and W is a potential vanishing on the unit circle and at the origin. When $\varepsilon \ll L_\varepsilon \rightarrow 0$, we show that these functionals $\Gamma$-converge to a constant multiple of the perimeter of the phase boundary and the divergence penalty is not felt. However, when $L_\varepsilon \equiv L > 0$, we find that a tangency requirement along the phase boundary for competitors in the conjectured $\Gamma$-limit becomes a mechanism for development of singularities. We establish criticality conditions for this limit and under a non-degeneracy assumption on the potential we prove the compactness of energy bounded sequences in $L^2$. The role played by this tangency condition on the formation of interfacial singularities is investigated through several examples: each of these examples involves analytically rigorous reasoning motivated by numerical experiments. We argue that generically, “wall” singularities between $\mathbb {S}^1$-valued states of the kind analyzed in Golovaty et al. (SIAM J Math Anal 51(1):276–320, 2018) are expected near the defects along the phase boundary.

Appendix

We present here the Proof of Theorem 3.15. See Fig. 4 for a guide to the notation.

Proof

The derivation of (5.18) follows the same general lines as those appearing in the proof of Theorem 3.14. However, a major complicating consideration is that it is no longer possible to assume that the deforming vector field X is normal to all four curves $\Gamma _{ij}$ since they all meet at p. Instead we will have to incorporate tangential components of X along these four curves as well.

To this end, we assume simply that $X\in C_0^1(B(p,R);\mathbb {R}^2)$ and again introduce the map $\Psi$ via (3.56). We assume that each $\Gamma _{ij}$ is smoothly parametrized by arclength through a map $r_{ij}:[0,s_0]\rightarrow \Gamma _{ij}$ for some $s_0>0$ with $r_{ij}(0)=p$. Then we replace (3.55) by

$$\begin{aligned} X(r_{ij}(s))=h^{\tau }_{ij}(s)\tau _{ij}(s)+h^{\nu }_{ij}(s)\nu _{ij}(s)\quad \hbox {for}\;s\in [0,s_0], \end{aligned}$$

(5.1)

where

$$\begin{aligned} h^{\tau }_{ij}:=X(r_{ij}(s))\cdot \tau _{ij}(s)\quad \hbox {and}\quad h^{\nu }_{ij}(s):=X(r_{ij}(s))\cdot \nu _{ij}(s). \end{aligned}$$

As a consequence of the compact support of X, we have that

$$\begin{aligned} h^{\tau }_{ij}(s_0)=h^{\nu }_{ij}(s_0)=0\quad \hbox {for all functions}\;h^{\tau }_{ij}\;\hbox {and}\;h^{\nu }_{ij} \end{aligned}$$

(5.2)

but we stress that none of these functions is assumed to vanish at $s=0$, namely at the location of the junction P.

We now deform each region $\Omega _j$, for $j=0,1,2,3$ by the map $\Psi$ to form four contiguous regions $\Omega _j^t:=\Psi (\Omega _j,t)$ and we deform the four boundary curves $\Gamma _{ij}$ to form four new boundary curves $\Gamma _{ij}^t:=\Psi (\Gamma _{ij},t).$ Of course the junction point P is also carried along by this flow.

The four curves $\Gamma _{ij}^t$ are parametrized by $s\mapsto \Psi (r_{ij}(s),t)$ which we denote by $r_{ij}^t(s)$ though s no longer represents arclength. Indeed one calculates that

$$\begin{aligned} r_{ij}^t(x)\sim r_{ij}(s)+t\big (h^{\tau }_{ij}(s)\tau _{ij}(s)+h^{\nu }_{ij}(s)\nu _{ij}(s)\big ) \end{aligned}$$

(5.3)

from which it follows that

$$\begin{aligned} \left| {r_{ij}^t\,'(s)}\right| \sim 1+t\big (h^{\tau }_{ij}\,'(s)-h^{\nu }_{ij}(s)\kappa _{ij}(s)\big ), \end{aligned}$$

(5.4)

where $\kappa _{ij}(s)$ denotes the curvature of $\Gamma _{ij}$ at $r_{ij}(s)$ (compare with (3.61)) and we have invoked the Frenet relations $\tau _{ij}'=\kappa _{ij}\nu _{ij}$ and $\nu _{ij}'=-\kappa _{ij}\tau _{ij}$. A related calculation goes to show that the unit normal $\nu _{ij}^t$ to $\Gamma _{ij}^t$ is given by

$$\begin{aligned} \nu _{ij}^t\sim \nu _{ij}-t\big (h^{\nu }_{ij}\,'+\kappa _{ij}h^{\tau }_{ij}\big )\tau _{ij}. \end{aligned}$$

(5.5)

Now in the ball B(p, R) the unperturbed critical point is given by

$$\begin{aligned} u(x)=\left\{ \begin{array}{ll} 0&{}\quad \hbox {for}\;x\in \Omega _0,\\ u_1(x)&{}\quad \hbox {for}\;x\in \Omega _1,\\ u_2(x)&{}\quad \hbox {for}\;x\in \Omega _2,\\ u_3(x)&{}\quad \hbox {for}\;x\in \Omega _3\end{array}\right. \end{aligned}$$

and we wish to perturb it into a new function $u^t$ given by

$$\begin{aligned} u^t(x)=\left\{ \begin{matrix} 0&{}\quad \hbox {if}\;x\in \Omega _0^t,\\ u_1^t(x)&{}\quad \hbox {for}\;x\in \Omega _1^t,\\ u_2^t(x)&{}\quad \hbox {for}\;x\in \Omega _2^t,\\ u_3^t(x)&{}\quad \hbox {for}\;x\in \Omega _3^t\end{matrix}\right. . \end{aligned}$$

To carry this out, as in the previous proof, we extend the domain of definition of $u_{j}$ to a neighborhood of $\Omega _{j}$ in such a way that the extension is constant along the normals to the boundary of its original domain of definition. Then we introduce three functions $\phi _1,\phi _2$ and $\phi _3$ such that

$$\begin{aligned} u_j^t(x)\sim u_j(x)+t\phi _j(x)u_j(x)^{\perp }\quad \hbox {for}\;x\in \Omega _{j}^t\;\hbox {and for}\;j=1,2,3 \end{aligned}$$

(5.6)

so as to preserve the required $\mathbb {S}^1$-valued nature of $u_j^t.$

We must also take care to preserve the property $u^t\in H_\mathrm {div}\,$ in the sense of (3.38) and this requires that the following four conditions hold to O(t) along $\Gamma _{01},\Gamma _{12},\Gamma _{23}$ and $\Gamma _{03}$ respectively:

$$\begin{aligned}&u_1^t(r_{01}^t(s))\cdot \nu _{01}^t(s)=0,\qquad u_1^t(r_{12}^t(s))\cdot \nu _{12}^t(s)=u_2^t(r_{12}^t(s))\cdot \nu _{12}^t(s),\nonumber \\&u_2^t(r_{23}^t(s))\cdot \nu _{23}^t(s)=u_3^t(r_{23}^t(s))\cdot \nu _{23}^t(s)\qquad \hbox {and}\nonumber \\&u_3^t(r_{03}^t(s))\cdot \nu _{03}^t(s)=0\; \hbox {for}\; s\in [0,s_0]. \end{aligned}$$

(5.7)

We note that the first and last of these conditions implies at $t=0$ that either $u_1\equiv \tau _{01}$ or $\equiv -\tau _{01}$ along $\Gamma _{01}$ and likewise either $u_3\equiv \tau _{03}$ or $\equiv -\tau _{03}$ along $\Gamma _{03}$.

Substituting (5.3) and (5.5) into the four conditions of (5.7), and expanding to O(t), a tedious but straight-forward calculation leads to the following requirements relating the traces of the $\phi _j$ to $h^{\nu }_{ij,s}$:

$$\begin{aligned}&\phi _1\big (r_{01}(s)\big )=h^{\nu }_{01,s}(s), \end{aligned}$$

(5.8)

$$\begin{aligned}&\frac{1}{2}\bigg (\phi _1\big (r_{12}(s)\big )+\phi _2\big (r_{12}(s)\big )\bigg )=h^{\nu }_{12,s}(s),\end{aligned}$$

(5.9)

$$\begin{aligned}&\frac{1}{2}\bigg (\phi _2\big (r_{23}(s)\big )+\phi _3\big (r_{23}(s)\big )\bigg )=h^{\nu }_{23,s}(s),\end{aligned}$$

(5.10)

$$\begin{aligned}&\phi _3\big (r_{03}(s)\big )=h^{\nu }_{03,s}(s), \end{aligned}$$

(5.11)

for $s\in [0,s_0]$, where s in the subscript denotes the derivative with respect to s

With the perturbations of the four curves $\Gamma _{ij}$ and three functions $u_j^t$ defined, we are ready to compute the variation of $E_0$ in a neighborhood of the junction point P. Carrying out the calculation (3.70) in $\Omega _j$ for $j=1,2,3$ and then applying the divergence theorem we find with the aid of (3.39) that

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\sum _{j=1}^3\left( \int _{\Omega ^t_j}(\mathrm {div}\,\,u^t_j)^2\,\hbox {d}x\right) \right| _{t=0}\\&\quad =-\int _{\Gamma _{01}}\bigg ( (\mathrm {div}\,\,u_1)^2\,h^{\nu }_{01}+2(\mathrm {div}\,\, u_1)(u_1\cdot \tau _{01})\phi _1 \bigg )\,\hbox {d}s\\&\qquad +\,\int _{\Gamma _{12}}\bigg ( (\mathrm {div}\,\,u_1)^2\,h^{\nu }_{12}+2(\mathrm {div}\,\, u_1)(u_1\cdot \tau _{12} )\phi _1 \bigg )\,\hbox {d}s\\&\qquad -\,\int _{\Gamma _{12}}\bigg ( (\mathrm {div}\,\,u_2)^2\,h^{\nu }_{12}+2(\mathrm {div}\,\, u_2)(u_2\cdot \tau _{12} )\phi _2 \bigg )\,\hbox {d}s\\&\qquad +\,\int _{\Gamma _{23}}\bigg ( (\mathrm {div}\,\,u_2)^2\,h^{\nu }_{23}+2(\mathrm {div}\,\, u_2)(u_2\cdot \tau _{23} )\phi _2 \bigg )\,\hbox {d}s\\&\qquad -\,\int _{\Gamma _{23}}\bigg ( (\mathrm {div}\,\,u_3)^2\,h^{\nu }_{23}+2(\mathrm {div}\,\, u_3)(u_3\cdot \tau _{23} )\phi _3 \bigg )\,\hbox {d}s\\&\qquad -\,\int _{\Gamma _{03}}\bigg ( (\mathrm {div}\,\,u_3)^2\,h^{\nu }_{03}+2(\mathrm {div}\,\, u_3)(u_3\cdot \tau _{03})\phi _3, \bigg )\,\hbox {d}s. \end{aligned}$$

where we have used the fact that $u_1^\perp \cdot \nu _{01}=u_1\cdot \tau _{01}$, $u_1^\perp \cdot \nu _{12}=u_1\cdot \tau _{12}$, etc.

Now we appeal to the relations (5.8)–(5.11), along with the conditions $u_2\cdot \tau _{12}=-u_1\cdot \tau _{12}$ and $u_3\cdot \tau _{23}=-u_2\cdot \tau _{23}$ and perform an integration by parts to find

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\sum _{j=1}^3\left( \int _{\Omega ^t_j}(\mathrm {div}\,\,u^t_j)^2\,\hbox {d}x\right) \right| _{t=0}\nonumber \\&\quad =\int _{\Gamma _{01}}\bigg \{ -(\mathrm {div}\,\,u_1)^2+2(\mathrm {div}\,\,u_1)'(u_1\cdot \tau _{01}) \bigg \}\,h^{\nu }_{01}\,\hbox {d}s\nonumber \\&\qquad +\,\int _{\Gamma _{12}}\bigg \{\bigg ((\mathrm {div}\,\,u_1)^2- (\mathrm {div}\,\,u_2)^2\nonumber \\&\qquad -\,4\big [(\mathrm {div}\,\,u_2)'(u_1\cdot \tau _{12})+ (\mathrm {div}\,\,u_2)(u_1\cdot \tau _{12})'\big ]\bigg )\,h^{\nu }_{12}\nonumber \\&\qquad +\,2(\mathrm {div}\,\, u_1-\mathrm {div}\,\,u_2)(u_1\cdot \tau _{12} )\phi _1 \bigg \}\,\hbox {d}s\nonumber \\&\qquad +\,\int _{\Gamma _{23}}\bigg \{\bigg ((\mathrm {div}\,\,u_2)^2- (\mathrm {div}\,\,u_3)^2\nonumber \\&\qquad -\,4\big [(\mathrm {div}\,\,u_3)'(u_2\cdot \tau _{23})+ (\mathrm {div}\,\,u_3)(u_2\cdot \tau _{23})'\big ]\bigg )\,h^{\nu }_{23}\nonumber \\&\qquad +\,2(\mathrm {div}\,\, u_2-\mathrm {div}\,\,u_3)(u_2\cdot \tau _{23} )\phi _2 \bigg \}\,\hbox {d}s\nonumber \\&\qquad +\,\int _{\Gamma _{03}}\bigg \{ -(\mathrm {div}\,\,u_3)^2+2(\mathrm {div}\,\,u_3)'(u_3\cdot \tau _{03}) \bigg \}\,h^{\nu }_{03}\,\hbox {d}s\nonumber \\&\qquad +\,2(\mathrm {div}\,\,u_1(p))(u_1(p)\cdot \tau _{01}(0)) \,h^{\nu }_{01}(0) -4\,\mathrm {div}\,\,u_2(p)(u_1(p)\cdot \tau _{12}(0))\,h^{\nu }_{12}(0)\nonumber \\&\qquad -\,4\,\mathrm {div}\,\,u_3(p)(u_2(p)\cdot \tau _{23}(0))h^{\nu }_{23}(0)\nonumber \\&\qquad +\,2\,(\mathrm {div}\,\,u_3(p))(u_3(p)\cdot \tau _{03}(0))\,h^{\nu }_{03}(0). \end{aligned}$$

(5.12)

We turn now to calculating the variations of the four jump energies. We begin by invoking (5.4) to compute

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \int _{\Gamma _{01}^t}1\,\hbox {d}s+\int _{\Gamma _{03}^t}1\,\hbox {d}s\right) \right| _{t=0}\\&\quad = \left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \int _0^{s_0}\left| {r_{01,s}^t(s)}\right| \,\hbox {d}s+\int _0^{s_0}\left| {r_{03,s}^t}\right| \,\hbox {d}s\right) \right| _{t=0}\\&\quad =\frac{{\hbox {d}}}{\hbox {d}t}\left( \int _0^{s_0}1+t\big (h^{\tau }_{01,s}-h^{\nu }_{01}(s)\kappa _{01}(s)\big )\,\hbox {d}s+\int _0^{s_0}1\right. \\&\qquad \left. \left. +\,t\big (h^{\tau }_{03,s}-h^{\nu }_{03}(s)\kappa _{03}(s)\big )\,\hbox {d}s\right) \right| _{t=0}\\&\quad =\int _0^{s_0}\big (h^{\tau }_{01,s}-h^{\nu }_{01}(s)\kappa _{01}(s)\big )\,\hbox {d}s+\int _0^{s_0}\big (h^{\tau }_{03,s}-h^{\nu }_{03}(s)\kappa _{03}(s)\big )\,\hbox {d}s \end{aligned}$$

Thus,

$$\begin{aligned}&\frac{K(0)}{2} \left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \mathcal {H}^1(\Gamma _{01}^t)+\mathcal {H}^1(\Gamma _{03}^t)\right) \right| _{t=0}\nonumber \\&\quad =-\frac{K(0)}{2} \bigg (\int _{\Gamma _{01}}h^{\nu }_{01}\kappa _{01}\,\hbox {d}s+\int _{\Gamma _{03}}h^{\nu }_{03}\kappa _{03}\,\hbox {d}s+h^{\tau }_{01}(0)+h^{\tau }_{03}(0)\bigg ). \qquad \quad \end{aligned}$$

(5.13)

To compute the variation in the jump energies over $\Gamma _{12}^t$ and $\Gamma _{23}^t$ requires an expansion to O(t) of the quantities $u^t\cdot \nu _{12}^t$ and $u^t\cdot \nu _{23}^t.$ Substituting the expression for $r_{12}^t$ from (5.3) into the formula for $u_1^t$ from (5.6) and Taylor expanding in t we find with the use of (5.5) that along $\Gamma _{12}^t$ we have

$$\begin{aligned}&u^t\cdot \nu _{12}^t\sim \bigg (u_1(r_{12}^t(s))+tu_1^\perp (r_{12}(s))\phi _1(r_{12}(s))\bigg )\cdot \nu _{12}^t\nonumber \\&\quad \sim \bigg (u_1\big (r_{12}+t\left[ h^{\tau }_{12}\tau _{12} +h^{\nu }_{12}\nu _{12}\right] \big )+t\phi _1(r_{12})u_1^\perp (r_{12})\bigg )\nonumber \\&\qquad \cdot \bigg (\nu _{12}-t\big (h^{\nu }_{12,s}+\kappa _{12}h^{\tau }_{12}\big )\tau _{12}\bigg )\nonumber \\&\quad \sim u_1\cdot \nu _{12}+t\bigg [ \left( \phi _1-h^{\nu }_{12,s}-\kappa _{12}h^{\tau }_{12}\right) (u_1\cdot \tau _{12})+h^{\tau }_{12}(u_1'\cdot \nu _{12})\bigg ],\qquad \quad \end{aligned}$$

(5.14)

where $u_1$ and $\phi _1$ in the expression above are evaluated at $x=r_{12}(s)$ and $u_1'=\frac{{\hbox {d}}}{\hbox {d}s}u_1(r_{12}(s)).$ In the last line we have also used that our extension of $u_1$ was constant along $\nu _{12}$ to eliminate the term $\nabla u_1\cdot \nu _{12}$ that would other have been present upon Taylor expanding.

Similarly, we calculate that along $\Gamma _{23}$ we have

$$\begin{aligned} u^t\cdot \nu _{23}^t\sim u_2\cdot \nu _{23}+t\bigg [ \left( \phi _2-h^{\nu }_{23,s}-\kappa _{23}h^{\tau }_{23}\right) (u_2\cdot \tau _{23})+h^{\tau }_{23}(u_2'\cdot \nu _{23})\bigg ].\nonumber \\ \end{aligned}$$

(5.15)

From (5.14) and (5.15), along with (5.4) we can compute that

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \int _{\Gamma _{12}}K\big (u^t\cdot \nu _{12}^t\big )\,\hbox {d}s+\int _{\Gamma _{23}}K\big (u^t\cdot \nu _{23}^t)\,\hbox {d}s\right) \right| _{t=0}\\&\quad =\frac{{\hbox {d}}}{\hbox {d}t}\left( \int _0^{s_0}K\big (u^t(r_{12}^t(s))\cdot \nu _{12}^t(s)\big )\left| {r_{12,s}^t}\right| \,\hbox {d}s\right. \\&\qquad \left. \left. +\, \int _0^{s_0}K\big (u^t(r_{23}^t(s))\cdot \nu _{23}^t(s)\big )\left| {r_{23,s}^t}\right| \,\hbox {d}s\right) \right| _{t=0}\\&\quad = \int _{\Gamma _{12}} K(u_1\cdot \nu _{12})\big (h^{\tau }_{12,s}-h^{\nu }_{12}\kappa _{12}\big )\,\hbox {d}s\\&\qquad +\,\int _{\Gamma _{12}} K'(u_1\cdot \nu _{12})\bigg (\big (\phi _1-h^{\nu }_{12,s}-h^{\tau }_{12}\kappa _{12}\big )(u_1\cdot \tau _{12})+ h^{\tau }_{12}(u_1'\cdot \nu _{12})\bigg )\,\hbox {d}s\\&\qquad +\,\int _{\Gamma _{23}} K(u_2\cdot \nu _{23})\big (h^{\tau }_{23,s}-h^{\nu }_{23}\kappa _{23}\big )\,\hbox {d}s\\&\qquad +\,\int _{\Gamma _{23}} K'(u_2\cdot \nu _{23})\bigg (\big (\phi _2-h^{\nu }_{23,s}-h^{\tau }_{23}\kappa _{23}\big )(u_2\cdot \tau _{23})+ h^{\tau }_{23}(u_2'\cdot \nu _{23})\bigg )\,\hbox {d}s. \end{aligned}$$

Now, since

$$\begin{aligned} \frac{{\hbox {d}}}{\hbox {d}s}\big [K\big (u_1\cdot \nu _{12})\big ]=(u_1'\cdot \nu _{12})-\kappa _{12}(u_1\cdot \tau _{12}) \end{aligned}$$

and

$$\begin{aligned} \frac{{\hbox {d}}}{\hbox {d}s}\big [K\big (u_2\cdot \nu _{23})\big ]=(u_2'\cdot \nu _{23})-\kappa _{23}(u_2\cdot \tau _{23}), \end{aligned}$$

we have that

$$\begin{aligned}&K(u_1\cdot \nu _{12})h^{\tau }_{12,s}+K'(u_1\cdot \nu _{12})\big ( (u_1'\cdot \nu _{12})-\kappa _{12}(u_1\cdot \tau _{12}) \big )h^{\tau }_{12}\\&\quad =\frac{{\hbox {d}}}{\hbox {d}s}\big [K\big (u_1\cdot \nu _{12})h^{\tau }_{12}\big ] \end{aligned}$$

and

$$\begin{aligned}&K(u_2\cdot \nu _{23})h^{\tau }_{23,s}+K'(u_2\cdot \nu _{23})\big ( (u_2'\cdot \nu _{23})-\kappa _{23}(u_2\cdot \tau _{23}) \big )h^{\tau }_{23}\nonumber \\&\quad =\frac{{\hbox {d}}}{\hbox {d}s}\big [K\big (u_2\cdot \nu _{23})h^{\tau }_{23}\big ]. \end{aligned}$$

Using these last two identities in (5.16) and integrating by parts implies that

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \int _{\Gamma _{12}}K\big (u^t\cdot \nu _{12}^t\big )\,\hbox {d}s+\int _{\Gamma _{23}}K\big (u^t\cdot \nu _{23,s}^t)\right) \right| _{t=0}\nonumber \\&\quad = -\int _{\Gamma _{12}} K(u_1\cdot \nu _{12})h^{\nu }_{12}\kappa _{12}\,\hbox {d}s+ \int _{\Gamma _{12}} K'(u_1\cdot \nu _{12})(\phi _1-h^{\nu }_{12,s})(u_1\cdot \tau _{12})\,\hbox {d}s\nonumber \\&\qquad -\,\int _{\Gamma _{23}} K(u_2\cdot \nu _{23})h^{\nu }_{23}\kappa _{23}\,\hbox {d}s+ \int _{\Gamma _{23}} K'(u_2\cdot \nu _{23})(\phi _2-h^{\nu }_{23,s})(u_2\cdot \tau _{23})\,\hbox {d}s\nonumber \\&\qquad -\,K\big (u_1(p)\cdot \nu _{12}(0))h^{\tau }_{12}(0)-K\big (u_2(p)\cdot \nu _{23}(0))h^{\tau }_{23}(0). \end{aligned}$$

(5.16)

Then invoking the criticality condition (3.40) from Theorem 3.11 and integrating by parts we can rewrite this identity as

$$\begin{aligned}&\left. \frac{{\hbox {d}}}{\hbox {d}t}\left( \int _{\Gamma _{12}}K\big (u^t\cdot \nu _{12}^t\big )\,\hbox {d}s+\int _{\Gamma _{23}}K\big (u^t\cdot \nu _{23}^t)\,\hbox {d}s\right) \right| _{t=0}\nonumber \\&\quad = -\int _{\Gamma _{12}} K(u_1\cdot \nu _{12})h^{\nu }_{12}\kappa _{12}\,\hbox {d}s\nonumber \\&\qquad +\, L\int _{\Gamma _{12}} \big ( \mathrm {div}\,\,u_2-\mathrm {div}\,\,u_1 \big )(\phi _1-h^{\nu }_{12,s})(u_1\cdot \tau _{12})\,\hbox {d}s\nonumber \\&\qquad -\,\int _{\Gamma _{23}} K(u_2\cdot \nu _{23})h^{\nu }_{23}\kappa _{23}\,\hbox {d}s\nonumber \\&\qquad +\, L\int _{\Gamma _{23}} \big ( \mathrm {div}\,\,u_3-\mathrm {div}\,\,u_2 \big )(\phi _2-h^{\nu }_{23,s})(u_2\cdot \tau _{23})\,\hbox {d}s\nonumber \\&\qquad -\,K\big (u_1(p)\cdot \nu _{12}(0))h^{\tau }_{12}(0)-K\big (u_2(p)\cdot \nu _{23}(0))h^{\tau }_{23}(0)\nonumber \\&\quad = \int _{\Gamma _{12}}\bigg \{ L \big ( \mathrm {div}\,\,u_2-\mathrm {div}\,\,u_1 \big )'(u_1\cdot \tau _{12})+L\big ( \mathrm {div}\,\,u_2-\mathrm {div}\,\,u_1 \big )(u_1\cdot \tau _{12})'\nonumber \\&\qquad -\,K(u_1\cdot \nu _{12})\kappa _{12}\bigg \}h^{\nu }_{12}\,\hbox {d}s +L\int _{\Gamma _{12}} \big ( \mathrm {div}\,\,u_2-\mathrm {div}\,\,u_1 \big )(u_1\cdot \tau _{12})\phi _1\,\hbox {d}s\nonumber \\&\qquad +\,\int _{\Gamma _{23}}\bigg \{L \big ( \mathrm {div}\,\,u_3-\mathrm {div}\,\,u_2 \big )'(u_2\cdot \tau _{23})+L\big ( \mathrm {div}\,\,u_3-\mathrm {div}\,\,u_2 \big )(u_2\cdot \tau _{23})'\nonumber \\&\qquad -\,K(u_2\cdot \nu _{23})\kappa _{23}\bigg \}h^{\nu }_{23}\,\hbox {d}s + L\int _{\Gamma _{23}} \big ( \mathrm {div}\,\,u_3-\mathrm {div}\,\,u_2 \big )(u_2\cdot \tau _{23})\phi _2\,\hbox {d}s\nonumber \\&\qquad -\,K\big (u_1(p)\cdot \nu _{12}(0))h^{\tau }_{12}(0)+L\big ( \mathrm {div}\,\,u_2(p) \nonumber \\&\qquad -\,\mathrm {div}\,\,u_1(p) \big )(u_1(p)\cdot \tau _{12}(0))h^{\nu }_{12}(0)\nonumber \\&\qquad -\,K\big (u_2(p)\cdot \nu _{23}(0))h^{\tau }_{23}(0)+L\big ( \mathrm {div}\,\,u_3(p) \nonumber \\&\qquad -\,\mathrm {div}\,\,u_2(p) \big )(u_2(p)\cdot \tau _{23}(0))h^{\nu }_{23}(0). \end{aligned}$$

(5.17)

Now we can combine (5.12), (5.13) and (5.17), and through a use of the criticality conditions (3.53) and (3.54) of Theorem 3.14 we find that all integrals over the four curves $\Gamma _{ij}$ drop, leaving only

$$\begin{aligned}&\frac{{\hbox {d}}}{\hbox {d}t}E_0(u^t)|_{t=0}\\&\quad = -\frac{K(0)}{2}\big (h^{\tau }_{01}(0)+h^{\tau }_{03}(0)\big )-K\big (u_1(p)\cdot \nu _{12}(0)\big )h^{\tau }_{12}-K\big (u_2(p)\cdot \nu _{23}(0)\big )h^{\tau }_{23}\\&\quad \quad +\,L\bigg \{ \mathrm {div}\,\,u_1(p)(u_1(p)\cdot \tau _{01}(0)) h^{\nu }_{01}(0) + \mathrm {div}\,\,u_3(p)(u_3(p)\cdot \tau _{03}(0)) h^{\nu }_{03}(0) \bigg \} \\&\qquad -\,L\bigg \{\big (\mathrm {div}\,\,u_1(p)+\mathrm {div}\,\,u_2(p)\big )\big (u_1(p)\cdot \tau _{12}(0)\big )h^{\nu }_{12}+ \big (\mathrm {div}\,\,u_2(p)\\&\qquad +\,\mathrm {div}\,\,u_3(p)\big )\big (u_2(p)\cdot \tau _{23}(0)\big )h^{\nu }_{23}\bigg \} \end{aligned}$$

Recall now that $h^{\tau }_{01}(0)=X(p)\cdot \tau _{01}(0)$, $h^{\nu }_{01}(0)=X(p)\cdot \nu _{01}(0)$, etc. Thus, the arbitrary value of the vector X(p), implies that a vanishing first variation $\frac{{\hbox {d}}}{\hbox {d}t}_{|_t=0} E_0(u^t)=0$ leads to the necessary condition at a junction P of the form

$$\begin{aligned}&\frac{K(0)}{2}\big (\tau _{01}+\tau _{03}\big )+K\big (u_1\cdot \nu _{12}\big )\tau _{12}+K\big (u_2\cdot \nu _{23}\big )\tau _{23}\nonumber \\&\quad =L\bigg \{ \mathrm {div}\,\,u_1(u_1\cdot \tau _{01}) \nu _{01} + \mathrm {div}\,\,u_3(u_3\cdot \tau _{03}) \nu _{03} \bigg \} \nonumber \\&\qquad -\,L\bigg \{\big (\mathrm {div}\,\,u_1+\mathrm {div}\,\,u_2\big )\big (u_1\cdot \tau _{12}\big )\nu _{12}+ \big (\mathrm {div}\,\,u_2+\mathrm {div}\,\,u_3\big )\big (u_2\cdot \tau _{23}\big )\nu _{23} \bigg \},\nonumber \\ \end{aligned}$$

(5.18)

where all quantities above are evaluated at the junction P. $\quad \square$