## Derivation of the Fundamental Equation of Heat Conduction

The Fundamental Heat Flow equation is
$c \rho \frac{\partial T}{\partial t} - k \nabla^2 T=0$
, where
$c$
is thew specific heat capacity
$\rho$
is the density of the conducting material
$k$
is the thermal conductivity of the material.
If the flow of heat is represented by a vector
$\mathbf{u}$
then the rate of heat flow through a surface is equal to
$\int \int_S \mathbf{u} \cdot \mathbf{n} dS$

By the Divergence Theorem
$\int \int_S \mathbf{u} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \cdot \mathbf{u}dV$

The basic law of heat conduction is that
$\mathbf{u}= - k \mathbf{\nabla} T$

Hence
$\int \int_S - k ( \mathbf{\nabla} T) \cdot \mathbf{n} dS = -k \int \int \int_V \mathbf{\nabla} \cdot (\mathbf{\nabla} T) dV$

The rate at which heat is being gained per unit mass from the material is
$c \frac{\partial T}{\partial t}$
and the rate at which heat is bring gained by the region
$V$
is
$\int \int \int_V \frac{\partial T}{\partial t} dV$

Hence
$k \int \int \int_V \mathbf{\nabla} \cdot (\mathbf{\nabla} T) dV = \int \int \int_V \frac{\partial T}{\partial t} dV$

Then
$\int \int \int_V (k \mathbf{\nabla} \cdot (\mathbf{\nabla} T) - \frac{\partial T}{\partial t} )dV=0$

Finally
$( k \mathbf{\nabla} \cdot (\mathbf{\nabla} T) - \frac{\partial T}{\partial t} )=0$ 