derivation of heat equation

Let us consider the heat conduction in a $\varrho$ and specific heat capacity $c$.  Denote by  $u(x,y,z,t)$  the temperature in the point  $(x,y,z)$  at the time $t$.  Let $a$ be a surface in the matter and $v$ the spatial region by it.

When the growth of the temperature of a volume element $dv$ in the time $dt$ is $du$, the element releases the amount

$$-duc\varrho dv=-u_t^{\prime }dtc\varrho dv$$

of heat, which is the heat flux through the surface of $dv$.  Thus if there are no sources and sinks of heat in $v$, the heat flux through the surface $a$ in $dt$ is

$-dt{\displaystyle {\int }_v}c\varrho u_t^{\prime }𝑑v.$(1)

On the other hand, the flux through $da$ in the time $dt$ must be proportional to $a$, to $dt$ and to the derivative of the temperature in the direction of the normal line of the surface element $da$, i.e. the flux is

$$-k\nabla u\cdot d\overrightarrow{a}dt,$$

where $k$ is a positive (because the heat always from higher temperature to lower one).  Consequently, the heat flux through the whole surface $a$ is

$$-dt{\oint }_{a}k\nabla u\cdot d\overrightarrow{a},$$

which is, by the Gauss’s theorem, same as

$-dt{\displaystyle {\int }_v}k\nabla \cdot \nabla udv=-dt{\displaystyle {\int }_v}k{\nabla }^{2}udv.$(2)

Equating the expressions (1) and (2) and dividing by $dt$, one obtains

$${\int }_{v}k{\nabla }^{2}udv={\int }_{v}c\varrho u_t^{\prime }𝑑v.$$

Since this equation is valid for any region $v$ in the matter, we infer that

$$k{\nabla }^{2}u=c\varrho u_t^{\prime }.$$

Denoting  ${\displaystyle \frac{k}{c\varrho }}={\alpha }^2$,  we can write this equation as

${\alpha }^2{\nabla }^{2}u={\displaystyle \frac{\partial u}{\partial t}}.$(3)

This is the differential equation of heat conduction, first derived by Fourier.