\[c \rho \frac{\partial T}{\partial t} - k \nabla^2 T=0\]
, 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
\[ ( k \mathbf{\nabla} \cdot (\mathbf{\nabla} T) - \frac{\partial T}{\partial t} )=0\]