= Solution
{c}
Use a temperature-dependent <strain energy density> $W(F,\vartheta)$ with different minimizing <deformation gradients> in the two <crystal> phases. Above the transition temperature its wells describe <austenite>; below it the wells are unions $SO(3)U_i$ of <martensitic variants>. Left rotational orbits reflect <material frame indifference>, while the <crystal>'s discrete material symmetries relate the variants. Minimizing the integral of this <elastic energy>, together with loading and boundary <constraints>, models the change of lattice shape. Compatibility restricts which low-energy gradients can meet continuously; fine compatible mixtures can accommodate macroscopic boundary shapes.
For the jump condition, write the two affine pieces as $Ax+c_+$ and $Bx+c_-$. <Continuity> on the interface means
$$
(A-B)x+(c_+-c_-)=0\qquad(x\cdot N=k).
$$
Taking differences along the interface gives $(A-B)t=0$ for every <tangent vector> $t$ with $t\cdot N=0$. Decompose any vector $v$ as $t+(v\cdot N)N$. Then
$$
(A-B)v=(v\cdot N)(A-B)N.
$$
With $a=(A-B)N$ this proves
$$
\boxed{A-B=a\otimes N.}
$$
Conversely this <rank-one connection> permits <continuity> by choosing $c_+-c_-=-ka$. A coherent <twin> interface is such a compatible planar boundary between different <martensitic variants>, possibly after a rigid <rotation> of one variant. The rank-one condition is therefore the geometric compatibility condition for a simple <twin>.
For the unrotated pair, direct subtraction gives
$$
U_1-U_2=2\delta e_3\otimes e_1,
$$
so $R=I$ is one solution, with reference interface <normal> $e_1$.
Put $d=1+\delta^2$, $c=(1-\delta^2)/d$ and $s=2\delta/d$. Since $c^2+s^2=1$, the stated $\overline R$ is a <proper rotation>. Multiplication gives
$$
\overline R U_2=
\begin{pmatrix}c+\delta s&0&-s\\0&1&0\\s-\delta c&0&c\end{pmatrix}
=\begin{pmatrix}1&0&-s\\0&1&0\\\delta&0&c\end{pmatrix}.
$$
Here $c+\delta s=1$ and $s-\delta c=\delta$. Consequently
$$
U_1-\overline R U_2
=\begin{pmatrix}0&0&s\\0&0&0\\0&0&1-c\end{pmatrix}
=\boxed{\frac{2\delta}{1+\delta^2}(e_1+\delta e_3)\otimes e_3.}
$$
This is the second <rank-one connection>, now with <normal> $e_3$. Because $\delta>0$, the two <rotations> are distinct. The permitted assertion that there are at most two proves completeness.
Finally suppose $QU_1$ is rank-one <connected> to $\overline R U_2$, with $Q\in SO(3)$. Left multiplication of their difference by $Q^T$ preserves <rank> and gives $U_1-Q^T\overline R U_2$. The classification just proved forces $Q^T\overline R$ to be either $I$ or $\overline R$, hence $Q=\overline R$ or $Q=I$. The two requested matrices are therefore
$$
\boxed{U_1\quad\text{and}\quad\overline R U_1.}
$$
Their connections are explicit: the first was calculated above, and $\overline R U_1-\overline R U_2=2\delta(\overline R e_3)\otimes e_1$. If desired the second <matrix> is
$$
\overline R U_1=
\begin{pmatrix}c-\delta s&0&-s\\0&1&0\\s+\delta c&0&c\end{pmatrix}.
$$
\Image[/past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2007-compatible-twins.png]
{title=The two compatible <twin> interfaces have reference normals in different directions}
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