Newton's law of cooling says the temperature of an object satisfies
$$
\frac{dT}{dt} = -k(T(t) – T_0),\tag{1}
$$
where $T_0$ is the surrounding temperature. See these HTML notes for example.
Now if $u(x,t)$ denotes the temperature of a laterally insulated bar at a point $x$ and time $t$, then heat equation says
$$
\frac{\partial u}{\partial t} = \alpha\, \frac{\partial^2 u}{\partial x^2}
$$
If the left end $x = 0$ is exposed to an environment at temperature $T_0$, all mathematical physics books say that Newton's law of cooling is
$$
-c \frac{\partial u}{\partial x}(0,t) = -k( u(0,t) – T_0), \tag{2}
$$
where $c$ is a constant. See Problem 5 in this Google book preview of Boundary Value Problems: and Partial Differential Equations by David Powers. See also page 131 of the same book. (Three pages before the problem 5, where Newton's law of cooling first appears.)
However, according to Newton's law of cooling (1), we get
$$
\frac{\partial u}{\partial t}(0,t) = -k( u(0,t) – T_0),
$$
By the heat equation, $\frac{\partial u}{\partial t} = \alpha\, \frac{\partial^2 u}{\partial x^2}$, we conclude that
$$
\alpha \frac{\partial^2 u}{\partial x^2}(0,t) = -k( u(0,t) – T_0).\tag{3}
$$
This boundary condition is not the same as (2). Why is (3) not correct?
Best Answer
Newton's law of cooling actually comes from the more general equation for heat $Q$ transferred between a system (temeperature $T$) and it's surroundings(temperature $T_0$): $$\frac{dQ}{dt} = -hA(T-T_0)$$ where $A$ is the area through which heat transfer occurs (see, for example, here). For an ordinary macroscopic object, where $dQ = mc\ dT$, we get the conventional Newton's law of cooling in terms of temperature: $$mc\frac{dT}{dt} = -hA(T-T_0)$$ For the case of the conducting bar however, from Fourier's law: $$\frac{1}{A}\frac{dQ}{dt} = -k\frac{\partial u}{\partial x}$$ The boundary condition is therefore: $$-k\frac{\partial u(0, t)}{\partial x} = -h(u(0, t) - T_0)$$