= Solution
{c}
For <separation of variables>, put $U=X(x)H(t)$ and set $X''/X=H'/(\kappa H)=-\lambda$. The <boundary conditions> are $X(0)=0$, $X'(L)=0$. Multiplying $-X''=\lambda X$ by $X$ and using <integration by parts> gives
$$
\lambda\int_0^L X^2\,dx=\int_0^L(X')^2\,dx.
$$
A nonzero solution therefore has $\lambda>0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda=\mu^2$, the <eigenfunctions> and their temporal factors are
$$
X_n(x)=\sin(\mu_nx),\qquad
\mu_n=\frac{(n+1/2)\pi}{L},\qquad
H_n(t)=e^{-\kappa\mu_n^2t},\qquad n=0,1,\ldots.
$$
These <sine> functions are <orthogonal> on $[0,L]$, with squared <norm> $L/2$. Expanding the <initial condition> and multiplying by $\sin(\mu_mx)$ gives, using the permitted interchange of the <sum> and <integral>,
$$
\begin{aligned}
b_n
&=\frac2L\int_0^L\left(-\frac{T_0x}{L}\right)\sin(\mu_nx)\,dx\\
&=-\frac{2T_0}{L^2}
\left[-\frac{x\cos(\mu_nx)}{\mu_n}
+\frac{\sin(\mu_nx)}{\mu_n^2}\right]_0^L\\
&=-\frac{2T_0(-1)^n}{L^2\mu_n^2}
=-\frac{8T_0(-1)^n}{\pi^2(2n+1)^2}.
\end{aligned}
$$
Here $\cos(\mu_nL)=0$ and $\sin(\mu_nL)=(-1)^n$. Thus the <Fourier series> solution is
$$
\boxed{U(x,t)=-\frac{8T_0}{\pi^2}\sum_{n=0}^\infty
\frac{(-1)^n}{(2n+1)^2}
\sin\left(\frac{(2n+1)\pi x}{2L}\right)
\exp\left(-\frac{\kappa(2n+1)^2\pi^2t}{4L^2}\right),\qquad T=T_0+U.}
$$
The <exponential decay> permits differentiation of the <series> for $t>0$, and each term satisfies both <boundary conditions>. At $t=0$ it is the mixed-end <Fourier expansion> of $-T_0x/L$.
The gained <heat> is the increase in <internal energy>. Since the paper specifies no <cross-sectional area>, first give the answer per unit <cross-sectional area>; multiply by $A$ for a rod of <cross-sectional area> $A$. The initial <temperature> has <integral> $T_0L/2$, whereas $\int_0^L\sin(\mu_nx)\,dx=1/\mu_n$. At $t_*=L^2/\kappa$ this gives
$$
\boxed{\frac{\Delta Q}{A}
=\rho cT_0L\left[
\frac12-\frac{16}{\pi^3}\sum_{n=0}^\infty
\frac{(-1)^n}{(2n+1)^3}
\exp\left(-\frac{(2n+1)^2\pi^2}{4}\right)
\right].}
$$
Numerically $\Delta Q/A\simeq0.45623855\,\rho cT_0L$. This is positive and less than the eventual gain $\rho cT_0L/2$, as expected from relaxation towards the uniform <steady state>.
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