Heat conduction equation

2 Heat conduction equation

Consider a long thin bar, or wire, of constant cross-section and of homogeneous material oriented along the $x$ -axis (see Figure 4).

Figure 4

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Imagine that the bar is thermally insulated laterally and is sufficiently thin that heat flows (by conduction) only in the $x$ -direction. Then the temperature $u$ at any point in the bar depends only on the $x$ -coordinate of the point and the time $t$ . By applying the principle of conservation of energy it can be shown that $u (x, t)$ satisfies the PDE

$\frac{\partial u}{\partial t} = k \frac{\partial^{2} u}{\partial x^{2}} \begin{matrix} 0 \leq x \leq L \\ t > 0 \end{matrix}$ (3)

where $k$ is a positive constant. In fact $k$ , sometimes called the thermal diffusivity of the bar, is given by

$k = \frac{κ}{s ρ}$

where $κ =$ thermal conductivity of the material of the bar

$s =$ specific heat capacity of the material of the bar

$ρ =$ density of the material of the bar.

The PDE (3) is called the one-dimensional heat conduction equation (or, in other contexts where it arises, the diffusion equation ).

Task!

What is the obvious difference between the wave equation (1) and the heat conduction equation (3)?

Both equations involve second derivatives in the space variable $x$ but whereas the wave equation has a second derivative in the time variable $t$ the heat conduction equation has only a first derivative in $t$ . This means that the solutions of (3) are quite different in form from those of (1) and we shall study them separately later. (3) is first order in $t$ means that only one initial condition at $t = 0$ is needed, together with two boundary conditions, to obtain a unique solution. The usual initial condition specifies the initial temperature distribution in the bar

$u (x, 0) = f (x)$

where $f (x)$ is known. Various types of boundary conditions at $x = 0$ and $x = L$ are possible. For example:

$u (0, t) = T_{1}$ and $u (L, T) = T_{2}$ (ends of the bar are at constant temperatures $T_{1}$ and $T_{2}$ ).
$\frac{\partial u}{\partial x} (0, t) = \frac{\partial u}{\partial x} (L, t) = 0$ which are insulation conditions since they tell us that there is no heat flow through the ends of the bar.

As you would expect, there are two-dimensional and three-dimensional forms of the heat conduction equation. The two dimensional version of $(3)$ is

$\frac{\partial u}{\partial t} = k (\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}})$ (4)

where $u (x, y, t)$ is the temperature in a flat plate. The plate is assumed to be thin and insulated on its top and bottom surface so that no heat flow occurs other than in the $O x y$ plane. Boundary conditions and an initial condition are needed to give unique solutions of (4). For example if the plate is rectangular as in Figure 5:

Figure 5

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typical boundary conditions might be

$u (x, 0) = T_{1} 0 \leq x \leq a$ (bottom side at fixed temperature)

$\frac{\partial u}{\partial x} (a, y) = 0 0 \leq y \leq b$ (right-hand side insulated)

$u (x, b) = T_{2} 0 \leq x \leq a$ (top side at fixed temperature)

$u (0, y) = 0 0 \leq y \leq b$ (left hand side at zero fixed temperature).

An initial condition would have the form $u (x, y, 0) = f (x, y)$ , where $f$ is a given function.

2 Heat conduction equation

Task!

Answer