I don't fully understand why the boundary insulated rod heat problem is mathematically described by the following boundary heat equation on $[0,1]$:
\begin{align*}
u_t &= u_{xx}\\
u_x(0,t) &= u_x(1,t) = 0\\
u(x,0) &= g(x)
\end{align*}
While I vaguely understand this makes sense – we want $0$ heat flow across the boundary, which implies $u_x(0,t) = 0$, it still seems like there's something weird going on. I could wrap my head around something like the requirement for e.g. the left boundary point to have a left sided derivative of $0$, i.e. $\lim_{x \rightarrow 0_{-}} \frac{u(x,0) – u(0,0)}{x} = 0$, but I'm not completely sure it makes sense to require that the right sided derivative is $0$ also. It seems physically plausible that the initial temperature of a rod might not satisfy $u_x(x,0) = 0$, but this can't really be described in this setting.
Here is my question – could it make sense to instead try and treat the physical problem as a heat problem on some open neighborhood of $[0,1]$, and assume that the thermal conductivity $k$ is $1$ on the rod and $0$ outside of it, and the initial temperature values outside of $[0,1]$ are $0$? I.e.:
\begin{align*}
u_t &= k \; u_{xx}\\
k(x) &= 1 \; \forall x \in [0,1]\\
k(x) &= 0 \; \forall x \in [0,1]^C\\
u(x,0) &= g(x)\\
g(x) &= 0 \; \forall x \in [0,1]^C
\end{align*}
Alternatively, if such a discontinuity causes an issue, we can consider some $\delta > 0$ and $k$ could interpolate between $0$ and $1$ on $[-\delta, 0]$ and we would look what the solution look like as $\delta \rightarrow 0$. This could be described as solving the problem on $[-\delta, 1+\delta]$ with boundary conditions
\begin{align*}
u_t &= k \; u_{xx}\\
u(-\delta, t) &= u(1+\delta, t) = 0\\
u(x,0) &= g(x)
\end{align*}
where $k = 1$ on $[0,1]$, $k(-\delta) = k(1+\delta) = 0$, and the values in $[-\delta, 0]$ and $[1, 1+\delta]$ are interpolate between $0$ and $1$.
Would solving a problem like this and then taking $\delta \rightarrow 0$ give the same solutions on $[0,1]$ as using the boundary condition $u_x(0,t) = u_x(1,t) = 0$? How is it related to the solutions that use the boundary condition of $u_x(0,t) = u_x(1,t) = 0$?
Best Answer
This boundary condition can be derived from first principles with relatively few assumptions. First off, the heat equation comes from a general conservation law.
Imagine a short interval of length $\Delta x$ and we want to determine the net rate of change of a quantity $u$ (in units per unit volume). We also assume the interval of material has a fixed cross-sectional area $A$ and the flux of the quantity $u$ per unit area is given by a function $q$. We now determine the conservation law by applying the principle $$\frac{d}{dt}(\mathrm{stuff}) = (\text{stuff in} - \text{stuff out})$$ on this interval. Here we have $$ \begin{aligned} \frac{d}{dt}(\mathrm{stuff}) & = \frac{d}{dt}(A\Delta x~u) = A\Delta x~u_t \\ \text{stuff in} &= Aq(x) \\ \text{stuff out} &= Aq(x+\Delta x) \\ &\implies \\ A\Delta x~u_t &= A(q(x) - q(x+\Delta x)). \end{aligned} $$ Dividing through by $A\Delta x$ and taking limits as $\Delta x\to0$, we obtain $$u_t = -q_x.$$ This conservation law is true for general quantities with fixed control volumes and no sources or sinks, but the "stuff in/out" analysis is how all the more complicated cases are derived.
Notice that even though the system is not closed as written (two unknowns and one equation), we can already apply boundary conditions, e.g., hfixed temperature ends: $u(0) = u_0$, $u(1) = u_1$, or insulated end: $q(0) = q(1) = 0$. One needs to be a bit careful with defining flux boundary conditions, as they are specified generally as $\mathbf{q}\cdot\hat{n}$, where $\hat{n}$ is the outward normal vector to the domain, so signs must be taken into account in $1$-D to determine whether the flux is going in or out of the different boundaries. One could also apply other physical laws to model more complicated boundary interactions, such as Newton's law of cooling, which would state that $q(0) = -hA(u(0,t) - T_0)$ and $q(1) = hA(u(1,t) - T_1)$. Notice the sign change because the outward normal on the left side of the domain is pointing to the left.
To close the system and obtain a single PDE for $u$, we need a closure relation. In this case, we use Fick's law of diffusion, which is an empirical result that relates the flux of a diffusing substance with the gradient of the substance distribution. It is given by $$q = -ku_{x}.$$ Substituting this into our conservation law yields $u_t = ku_{xx}$. Notice that this implies that for spatially varying diffusion coefficients, one should write this equation as $u_t = (k(x)u_x)_x$, not $u_t = k(x)u_{xx}$ asd one might expect. This also allows us to close our boundary conditions that involve the flux. For the case of insulated boundaries, this gives us $q(0) = -ku_{x}(0) = 0\implies u_x(0) = 0$ and similar for the right boundary.
From this, we can see that the Neumann boundary condition is not really a mathematical creation; it is a consequence of applying an empirically determined closure relation to a natural choice of no-flux boundary conditions. The effects on the temperature gradient are effectively coincidental. If we had a different flux that contained, for instance, a velocity advecting the temperature in addition to diffusion, then our closure relation would be something like $q = -ku_x + a$ and we would applying flux-based boundary conditions to that, yielding non-homogeneous boundary conditions for $u$ in the case of no-flux conditions $q=0$.
As for your proposed modifications to the heat equation, it is possible that it gives the same solutions, although I would posit that it would do so in a rather nonphysical way by modifying the physical properties of the material (setting $k=0$) and introducing a singular perturbation and discontinuity to the equation. The previously demonstrated physical motivation for the heat equation is powerful because it allows one to reason and make assumptions on the flux and the density separately, before any closure relation is provided, as opposed to trying to apply all the properties we want directly to $u$ on the already-closed system, which may result in nonphysical phenomena. And if this sounds too non-mathy, know that physical intuition can lead to lots of nontrivial discoveries in systems like these such as, for isntance, entropy conditions in hyperbolic PDEs.