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?

  • 1
    $\begingroup$ Unlike your equation (2), your equation (3) is not a boundary condition. $\endgroup$
    – Amey Joshi
    Apr 26, 2015 at 16:47
  • 1
    $\begingroup$ @AmeyJoshi Why? By definition, a "boundary condition" is a condition on the boundary required of the function. By this definition, (3) is certainly a "boundary condition". $\endgroup$
    – Curiosity
    Apr 26, 2015 at 17:42
  • 2
    $\begingroup$ Part of the problem here is your source for (1). It's not wrong per se, but it uses a very simplified heat transfer model. In that case, $k$ is not conductivity used in (2), it's more like the heat transfer coefficient, $c$. It's treating the object as lumped (constant temperature) rather than resolving heat transfer within it, as the subsequent equations you've used do. That's also why they've used the notation for ODEs ($d/dt$) rather than PDEs ($\partial/\partial t$). $\endgroup$ Apr 28, 2015 at 5:56
  • 1
    $\begingroup$ @user3823992, you should elaborate this a little in an answer; it will be more to the point than the existing answer, I think. $\endgroup$
    – Ilja
    Apr 23, 2016 at 19:26

1 Answer 1


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: $$-k\frac{\partial u(0, t)}{\partial x} = -h(u(0, t) - T_0)$$

  • 1
    $\begingroup$ But, this still does not answer "why" --- why can't you use the formula $dQ = mc dT$ for the case of the bar to conclude equation (3)? That is, assume you did not know about Fourier's law. What would prevent you from using the formula $dQ = mc dT$ to deduce the formula $\partial u/\partial t (0,t) = -k(u(0,t) - T_0)$? Would you not think this formula was valid if you were not told to use Fourier's law instead? $\endgroup$
    – Curiosity
    Apr 28, 2015 at 12:30
  • $\begingroup$ At the boundary, there is an incoming heat (by conduction) $Q_1$ and heat transfer to the surroundings $Q_2$ per unit time. On considering a segment of length $dx$ at the boundary, we see that the net heat entering the region is in fact $dQ = Q_1 - Q_2 = Q(dx) - Q(0)$. It is this that contributes to the change in temperature - not the heat loss to the surroundings alone. This change is given by $\frac{\partial Q}{\partial x}dx$ while $dQ = mc\dot{T} = \rho Ac dx \dot{T}$. Thus, we get $\dot{T} = \frac{1}{\rho Ac} \frac{\partial Q}{\partial x}$. Thus is equivalent to the heat equation... $\endgroup$
    – AV23
    Apr 28, 2015 at 12:56
  • $\begingroup$ ...with the missing piece of what $Q$ actually is being given by Fourier's law. $\endgroup$
    – AV23
    Apr 28, 2015 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.