A 1-D slab, $0 ≤ x ≤ L$, is initially at a temperature of $F(x)$. For times t > 0, both of the boundary surfaces are perfectly insulated. Obtain an expression for the temperature $T(x,t)$ in the slab. Clearly show the steady-state temperature in your expression. Independently derive the steady-state temperature from simple conservation of energy and show that this agrees with your above expression.
Heat Diffusion Equation: $\frac{\partial^2 T}{\partial x^2}= \frac{1}{\alpha} \frac{\partial T}{\partial t}$
Boundary Conditions: $\frac{\partial T}{\partial x} = 0 $ at $x=0,L$
Initial Condition: $T(x,0)=F(x)$
Let $T(x,t) = X(x) \Gamma(t)$.
From the heat diffusion equation, $\begin{align*} \frac{1}{X} \frac{\partial ^2 X}{\partial x^2} = \frac{1}{\partial \Gamma} \frac{\partial \Gamma}{\partial t} = - \lambda^2 \end{align*} $
$X(x) = C_1 \cos \lambda x + C_2 \sin \lambda x $
$\frac{\partial X}{\partial x} = - \lambda C_1 \sin \lambda x + \lambda C_2 \cos \lambda x $
$\frac{\partial X}{\partial x} =C_2 \lambda=0 $ at $x = 0 $, then $C_2 =0$
$\frac{\partial X}{\partial x} = -C_1 \lambda \sin \lambda L =0 $ at $ x = L$, then $\lambda_n = \frac{n\pi}{L}, n=0,1,2,...$.
$X_n(x) = \cos \lambda_n x$
From $\frac{1}{\alpha \Gamma_n} \frac{ \partial \Gamma_n}{\partial t} = - \lambda_n^2$, $$ \Gamma_n(t) = \exp(-\alpha \lambda_n^2 t) $$ $$T(x,t) = \sum_{n=0}^\infty C_n X_n(x) \Gamma_n(t) = C_0 + \sum_{n=0}^\infty X_n(x) \Gamma_n(t).$$ From $T(x,0)=F(x)$, $$ C_n = \frac{\int_0^L \cos \lambda_n x F(x) dx }{\int_0^L \cos^2 \lambda_n x dx}$$ $$ C_0 = \frac{\int_0^L F(x) dx }{ L }$$
Therfore, $\lim_{t\to \infty} T(x,t) = C_0 = \frac{1}{L} \int_0^L F(x) dx$ is a steady temperature.
Since $T(x,0)=F(x)$ and the slab is insulated, steady temperature is a mean of $F(x)$.