 Solution

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.