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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 $t=0$.

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You can think of this problem in two parts. The first part takes place while the rod is connected to the heat baths, i.e. $t < 0$. During this part, the boundary conditions on the rod are \begin{gather} \Theta(x = 0,t < 0) = \Theta_0\\ \Theta(x = L,t < 0) = \Theta_L. \end{gather} As a result of these boundary conditions, the temperature in the cylinder will evolve according to the heat equation until the temperature in the cylinder is described by \begin{equation} \Theta(x,t < 0)=\Theta_0+\frac{\Theta_L-\Theta_0}{L}x \end{equation}

The second part of this problem takes place at $t = 0$ where the ends of the rod are removed from the baths and insulated so that there is no heat transfer out of the ends. The boundary conditions, as stated in your question are \begin{align} \frac{\partial\Theta(x,t)}{\partial x}\Bigg{\rvert}_{x=0}= 0\\ \frac{\partial\Theta(x,t)}{\partial x}\Bigg{\rvert}_{x=L}= 0. \end{align} The rod is no longer at steady state because the boundary conditions have changed.

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 $t > 0$)?

These boundary conditions only state that there is no temperature gradient at the ends of the rods ($x = 0$ and $x = L$) but does not say that heat cannot be transferred within the rod ($0<x<L$). As you can see from your initial condition at $t = 0$, the temperature in the rod is not uniform and in the absence of any sources of heat, the heat will diffuse until it is the same temperature everywhere.

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  • $\begingroup$ Thanks for your answer, I think we are getting somewhere, what I am asking here is at the point $x=0$ at $t=0$, the BC states that no heat enters (from the outside) or leaves (from inside the rod). But why no heat transfer away or towards the point $x=0$ at $t=0$? It is almost as if the BC only takes into account heat transfer in or out of the rod, and ignores heat transfer through it. $\endgroup$ – BLAZE Feb 6 at 2:23
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    $\begingroup$ @BLAZE Boundary conditions are named so because they are only valid exactly at the boundary. The BC does not say that there is no temperature gradient within the rod, only that none of the heat leaves from the ends of the rod. In this case $\partial \Theta/\partial x \neq 0$ for $0<x<L$ (at least until equilibrium) and therefore heat does move towards/away from the ends. So you are right, the BC only tells you what is happening at the boundaries and one needs to use the heat transfer equation to determine what is happening between the boundaries. $\endgroup$ – Ragnar Feb 6 at 3:14
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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 $t=0$.

There is indeed heat flow along the rod for $t>0$, flowing from hot to cold.

But there's no heat flowing in or out at $x=0$ or $x=L$. And because heat flow is driven by a temperature gradient, acc. Fourier:

$$\dot{q}=-k\nabla \Theta$$

If $\dot{q}=0$, and with $k\neq 0$, then by definition also $\nabla \Theta=0$, hence the zero temperature gradient boundary conditions for $x=0,L$.

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