Differentiating a boundary condition at infinity

Question

A typical boundary condition for an initial boundary value problem is $$ \lim_{x\rightarrow\infty} T(x,t) = T_\infty.$$

For example, this might be the temperature at the end of a very long rod. Under what conditions is it equivalent to instead enforce the boundary condition $$ \lim_{x\rightarrow\infty} \frac{\partial T}{\partial x} = 0$$

or, perhaps worded differently, is it ever the case, given $T\rightarrow T_\infty$ as $x\rightarrow\infty$, that $\partial T/\partial x\ne0$ as $x\rightarrow\infty$? The alternative boundary condition makes sense physically, but does there exist a formal mathematical treatment of this?

Answer 1

As gerw wrote, in general neither condition implies the other. However, in the context of PDE you may have extra information about the function which allows to pass from one condition to the other.

Fact. If $T\to 0$ and $T''$ is bounded, then $T'\to 0$.

I remember the following proof from Littlewood's Miscellany. Draw the graph of $T'$ (in red):

If $T'$ is not small somewhere, then one can fit a triangle (black) under the graph, so that the height of the triangle is not small, and the slope of its legs is not large (bounded by $|T''|$). Therefore, the area of this triangle is not small, contradicting the assumption $T\to 0$. $\Box$

This fact could be applicable to solutions of second-order equations, which give a way to control $T''$.

Answer 2

Take $T(x) = \sin(x^2)/x$. Then, $T(x) \to 0$ as $x \to +\infty$, but $T'(x) = (2\,x^2\,\cos(x^2)+ \sin(x^2))/x^2 \not\to 0$ as $x \to +\infty$.

Also, the reverse implication is not true: take $T(x) = \ln(x)$. Then, $T'(x) \to 0$, but $T(x) \not\to 0$.

Hence, those two conditions are not related.