Conceptual Question About the Nature of PDE Boundary Conditions In this $1D$ boundary value explanation, it is stated that "Taking the example of a straight line, whose slope at the boundary points is decidedly not $0$ [the animation shows a mostly-unlabeled graph of temperature as a function of time, for the sake of concreteness we'll call it the line segment $T = .8t$ from $t = 0$ to $t = L$], as soon as the clock starts, those boundary values will shift infinitesimally, such that the slope there suddenly becomes $0$, and remains that way through the remainder of the evolution. In other word, finding a function satisfying the heat equation itself is not enough; it must also the property that it's flat at each of those endpoints for all times greater than $0$."
The idea seems to be that the slope of the line at the boundaries is not equal to $0$ at time $t = 0$, but becomes $0$ immediately after time begins progressing, as a consequence of the Laplacian seeing only one neighboring point at each boundary point instead of two.
However, when an example is worked, it seems like the PDE solution satisfying the boundary condition is engineered precisely AT $t = 0$, the one value of $t$ where the boundary condition DOESN'T need to be satisfied.
Is this "flattening of the boundaries at the first nonzero instant" feature incidental to the first linked animation, or is it a general feature of boundary conditions? Is it hiding somewhere in the worked example of the heat equation?
 A: The infinitesimal boundary shift is caused by the additional assumption that is the boundary condition, namely that we require
$$ \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) = 0 $$
for all $t>0.$ As the video explains this is an additional constraint that comes from the model (corresponding to no heat flow on the rod ends), and something we need to specify to obtain the "correct" solution when we solve the equation.
What is happening with the example where $u(x,0) = x$ is an incompatibility between the initial and boundary conditions; if the partial derivatives of $u$ were to be continuous up to the boundary, then by the above boundary conditions we would expect that
$$ \lim_{x \to 0} \frac{\partial u}{\partial x}(x,0) = \frac{\partial u}{\partial x}(0,0) = \lim_{t \to 0} \frac{\partial u}{\partial x}(0,t) = 0 $$
and similarly as $x \to L.$ However our initial condition $u(x,0)=x$ evidently does not satisfy this, so we get this jump discontinuity in $\frac{\partial u}{\partial x}$ at the corner point - this is the instantaneous smoothing condition. In general this will always occur if the initial condition is not compatible with the boundary conditions in the above sense.
How exactly this jump occurs when solving the equation is what you will see in the next part - you will write the solution as an infinite series, where it will not be a-priori obvious if this converges (i.e. if the infinite sum evaluates to a reasonable limit), or if you can differentiate it. It will turn out this subtlety with the discontinuous derivative is caused by delicate issues relating to differentiating this infinite sum of functions - however from the above discussion you see that this is inevitable due to the incompatibility.
As for the worked example, the family of solutions
$$ u_n(x,t) = \cos\left(\frac{n\pi}L x\right) \exp\left(-\frac{\alpha n^2\pi^2}{L^2}t\right) $$
can be derived from a general technique known as separation of variables, where we seek solutions of the form $u(x,t) = X(x)T(t).$ We then require that $u$ solves the heat equation subject to the boundary condition, which in this case reduces to requiring $X'(0)=X'(L)=0.$ Hence any solution of the above form must necessarily satisfy the boundary condition for all $t \geq 0.$
It turns out that the above $u_n$ are the only solutions that can be written as a product $X(x)T(t),$ and this is also why we can obtain series solutions using these, but the full story requires some more involved theory.
