= Solution
{c}
Subtract the constant <steady state> profile: set $v=T-T_0$. Then
$$
v_t=\kappa v_{xx},\qquad v(0,t)=0,\qquad v_x(L,t)=0.
$$
The last equation follows from <Fourier's law> and zero <heat flux> at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, <separation of variables> gives
$$
\frac{H'}{\kappa H}=\frac{X''}{X}=-\mu,\qquad
-X''=\mu X,\qquad X(0)=0,\quad X'(L)=0.
$$
To see that $\mu>0$ for a nonzero <eigenfunction>, multiply the spatial equation by $X$ and use <integration by parts>:
$$
\mu\int_0^L X^2\,dx=\int_0^L(X')^2\,dx.
$$
Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu=\lambda^2$ with $\lambda>0$. Solving gives $X=A\sin(\lambda x)$, and $X'(L)=0$ requires $\cos(\lambda L)=0$. The <mixed Dirichlet-Neumann modes on an interval> therefore have
$$
\lambda_n=\frac{(2n-1)\pi}{2L},\qquad X_n(x)=\sin(\lambda_nx),\qquad
H_n(t)=e^{-\kappa\lambda_n^2t},\qquad n=1,2,\ldots.
$$
The <eigenvalues> of $-d^2/dx^2$ are $\mu_n=\lambda_n^2$; $\lambda_n$ are their positive square roots. By <superposition>, with coefficients for which the resulting <series> and its required <derivatives> converge at positive times,
$$
\boxed{T(x,t)=T_0+\sum_{n=1}^{\infty}b_n\sin\left(\frac{(2n-1)\pi x}{2L}\right)\exp\left[-\frac{\kappa(2n-1)^2\pi^2t}{4L^2}\right].}
$$
The <eigenfunctions> are <orthogonal> on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess <temperature> $v(x,0)=T_0x/L$ gives
$$
\begin{aligned}
b_n&=\frac2L\int_0^L\frac{T_0x}{L}\sin(\lambda_nx)\,dx\\
&=\frac{2T_0}{L^2}\left[-\frac{x\cos(\lambda_nx)}{\lambda_n}+\frac{\sin(\lambda_nx)}{\lambda_n^2}\right]_0^L\\
&=\frac{2T_0(-1)^{n-1}}{L^2\lambda_n^2}
=\boxed{\frac{8T_0(-1)^{n-1}}{(2n-1)^2\pi^2}}.
\end{aligned}
$$
Thus the requested <temperature> is
$$
\boxed{T(x,t)=T_0+\frac{8T_0}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^2}\sin\left(\frac{(2n-1)\pi x}{2L}\right)e^{-\kappa(2n-1)^2\pi^2t/(4L^2)}.}
$$
At the insulated endpoint $\sin(\lambda_nL)=(-1)^{n-1}$ cancels the alternating coefficient sign, leaving
$$
\boxed{T(L,t)=T_0+\frac{8T_0}{\pi^2}\sum_{n=1}^{\infty}\frac{e^{-\kappa(2n-1)^2\pi^2t/(4L^2)}}{(2n-1)^2}>T_0\quad\text{for every finite }t\geq0.}
$$
Every summand is positive. \b[The endpoint temperature never drops below $T_0$]; it decreases towards $T_0$ as $t\to\infty$ and starts at $T(L,0)=2T_0$.
For completeness, the <Sturm-Liouville eigenfunction expansion> gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the <Weierstrass M-test> also gives <uniform convergence> on $[0,L]\times[0,\infty)$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise <derivatives>, giving a <classical solution>. There is, however, an <initial incompatibility with a Neumann boundary condition> at $(L,0)$: the initial slope is $T_0/L$, while the insulated <Neumann boundary condition> requires zero slope for $t>0$. The <temperature> is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial <series> term by term at $x=L$.
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