= Solution
For <antiplane shear>, the <displacement field> is $\mathbf u=(0,0,w(x,y,t))$, independent of $z$. Its <divergence> is zero. Consequently the only possibly nonzero components of the <stress tensor> are the <shear stresses>
$$
\tau_{xz}=\tau_{zx}=\mu w_x,\qquad
\tau_{yz}=\tau_{zy}=\mu w_y.
$$
The $x$ and $y$ components of the <Navier-Cauchy equation> vanish identically. Its $z$ component gives $\rho w_{tt}=\partial_x(\mu w_x)+\partial_y(\mu w_y)$. Each material has constant <shear modulus> and <mass density>, so the required <wave equation> and <wave speeds> are
$$
\boxed{w_{j,tt}=c_j^2(w_{j,xx}+w_{j,yy}),\qquad
c_j=\sqrt{\frac{\mu_j}{\rho_j}},\quad j=1,2.}
$$
At a <perfectly bonded elastic interface>, both the <displacement field> and the <traction> are continuous. Using the same normal $\mathbf e_y$ on both sides, the only nonzero <traction> component is $\tau_{zy}$. Thus the interface <boundary conditions> are
$$
\boxed{w_1(x,h,t)=w_2(x,h,t),\qquad
\mu_1w_{1,y}(x,h,t)=\mu_2w_{2,y}(x,h,t).}
$$
<Continuity> of “stress” here means <continuity> of <traction>; it does not require equality of the full <stress tensors>. The rigid plate supplies the remaining <boundary condition> $w_1(x,0,t)=0$.
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