(i) A slab of thickness $L$ is initially at a uniform temperature $T_{0}$. The face at $x=0$ is perfectly insulated. At time $t=0$, a constant heat flux $q_{s}$ is imposed on the face at $x=L$. Show that the temperature response is
$$
\frac{T-T_{0}}{q_{s} L / k}=\frac{3 x^{2}-L^{2}}{6 L^{2}}+\frac{\alpha t}{L^{2}}-\frac{2}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} e^{-\left(\alpha n^{2} \pi^{2} / L^{2}\right) t} \cos \frac{n \pi x}{L}
$$
(ii) If a negative heat flux equal to $-q_{s}$ is subsequently imposed at time $t=t_{1}$, show that the temperature response for $t>t_{1}$ is
$$
\frac{T-T_{0}}{q_{s} L / k}=\frac{\alpha t_{1}}{L^{2}}-\frac{2}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} e^{-\left(\alpha n^{2} \pi^{2} / L^{2}\right) t}\left[1-e^{\left(\alpha n^{2} \pi^{2} / L^{2}\right) r_{1}}\right] \cos \frac{n \pi x}{L}
$$
(Hint: Use the principle of superposition for a linear differential equation.)