A cylindrical rod of length $L$ is insulated over its curved surface. The end of the rod at $x = 0$ is in contact with a heat bath at temperature $\Theta_0$ and the end of the rod at $x = L$ is in contact with a heat bath at temperature $\Theta_L$. After some time, a steady state is reached. The steady-state (time-independent) solution of the heat equation is $$\Theta(x)=\Theta_0+\frac{\Theta_L-\Theta_0}{L}x$$ The heat equation that describes the temperature profile of the rod is $$\frac{1}{D}\frac{\partial\Theta}{\partial t}=\frac{\partial^2 \Theta}{\partial x^2}$$ Where $D$ is a constant and $\Theta(x,t)$ is the temperature at position $x$ and time $t$.
At time $t = 0$, the rod is disconnected from the heat baths. Assuming that no heat $Q$ subsequently leaves or enters the rod, write down the boundary/initial conditions: $$(a) \quad\text{at}\qquad x = 0,$$ $$(b) \quad\text{at}\qquad x = L,$$ $$\qquad\qquad\qquad\qquad(c) \quad\text{at}\qquad t = 0 \quad \text{for}\qquad 0 \le x \le L$$ (Hint: recall Fourier’s law of heat flow $\frac{1}{A}\frac{\partial Q}{\partial t}=−k\frac{\partial \Theta}{\partial x}$ , where $k$ is the conductivity.)
The answer given to parts $(a)$, $(b)$ & $(c)$ (respectively) are
The boundary condition at $x = 0$ is that no heat is flowing in or out of the end of the rod. This implies that the temperature gradient at $x = 0$ is zero: $$\frac{\partial\Theta(x,t)}{\partial x}\Bigg{\rvert}_{x=0}= 0$$
The boundary condition at $x = L$ is that no heat is flowing in or out of the end of the rod. This implies that the temperature gradient at $x = L$ is zero: $$\frac{\partial\Theta(x,t)}{\partial x}\Bigg{\rvert}_{x=L}= 0$$
The initial condition at $t = 0$ for $0 \le x \le L$ is that the initial temperature distribution is the steady-state temperature distribution: $$\Theta(x)=\Theta_0+\frac{\Theta_L-\Theta_0}{L}x$$
I'm struggling to find physical intuition for these boundary/initial conditions. I have learnt from reading the comment below this question that steady state in this context means there is as much heat flowing out of the $\Theta_L$ heat bath as there is flowing into the $\Theta_0$ heat bath, and I acknowledge that this is definitely not the same as thermal equilibrium.
However, if the temperature gradient at $t=0$ is zero at $x=0,L$ after the rod is disconnected how can there be any transfer of heat whatsoever (even for $t \gt 0$)?
Put in another way, I know that the there will be no heat leaving either end of the rod (as it is insulated), and there will be no heat entering either end of the rod (as the heat baths are no longer present). But there must be a transfer of heat from $x=0$ and/or $x=L$ along the rod (in the direction towards the rods centre). If this were not the case how would the temperature profile ever evolve?
And it does evolve as the final answer for $\Theta(x,t)$ (working omitted) is
$$\Theta(x,t)=\frac{\Theta_0+\Theta_L}{2}+ \frac{2 \left(\Theta_L-\Theta_0\right)}{\pi^2} \sum_{n=1}^{\infty}\frac{\left[(-1)^n-1\right]}{n^2}\cos\left(\frac{n \pi x}{L}\right)\exp\left(-\frac{n^2 \pi^2 D}{L^2}t \right)$$
So put simply I don't understand physically why at $t=0$ $$\frac{\partial\Theta(x,t)}{\partial x}\Bigg{\rvert}_{x=0/L}= 0$$ as by my logic there must be heat flow along the rod (not out or in the rod) even at $t=0$.