\[c \rho \frac{\partial T}{\partial t} - k \nabla^2 T=0\]

, 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

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