= Solution
{c}
Choose the positive weights
$$
\boxed{a=\rho_1c_1,\qquad b=\rho_2c_2.}
$$
These are the <volumetric heat capacities>. The <heat equations> and <integration by parts> give
$$
\begin{aligned}
I'(t)
&=2k_1\int_{-L}^0T_1T_{1xx}\,dx+2k_2\int_0^LT_2T_{2xx}\,dx\\
&=2k_1[T_1T_{1x}]_{-L}^{0}+2k_2[T_2T_{2x}]_0^L
-2k_1\int_{-L}^0T_{1x}^2\,dx-2k_2\int_0^LT_{2x}^2\,dx.
\end{aligned}
$$
The exterior <boundary terms> vanish because the <temperatures> there are zero. At the <thermal interface>, the remaining term is
$$
2T(0,t)\,[k_1T_{1x}(0_-,t)-k_2T_{2x}(0_+,t)]=0
$$
by continuity of <temperature> and <heat flux>. Hence the weighted <energy estimate> is
$$
\boxed{I'(t)=-2k_1\int_{-L}^0T_{1x}^2\,dx-2k_2\int_0^LT_{2x}^2\,dx\leq0.}
$$
Both weights are positive, so $I\geq0$. 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>.
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