Multidimensional Heat Equation ~ Math Crumbs

Before moving onto the method of separation of variables, let us enhance our one-dimensional heat equation to three dimensions by first introducing the gradient operator:

\nabla = \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}

With that in mind, let us enhance the heat energy to encompass a three-dimensional, arbitrary subregion

R

∭_{R} c ρ u d V

In three dimensions, the heat flux becomes a vector instead of a scalar 'direction' (

φ > 0

, etc.). Because of this, our heat flux becomes

\int \int_{} ◯ φ \cdot \hat{n} d S

where

\hat{n}

is the unit normal vector to the region's surface.
It clearly follows that our multidimensional heat equation then turns into

\frac{d}{d t} ∭_{R} c ρ u d V = - \int \int_{} ◯ φ \cdot \hat{n} d S + ∭_{R} Q d V

But this is pretty hard to work with since the closed surface integral is expressed in terms of the scalar product of two vectors. However, there is this really cool theorem called the divergence theorem which, simply put, states that

∭_{R} \nabla \cdot \vec{A} d V = \int \int_{} ◯ \vec{A} \cdot \hat{n} d S

We can use it to rewrite our heat equation above like this:

\frac{d}{d t} ∭_{R} c ρ u d V = - ∭_{R} \nabla \cdot φ d V + ∭_{R} Q d V

After manipulating this equation a bit, we end up with

c ρ \frac{\partial u}{\partial t} = - \nabla \cdot φ + Q

This is analogous to the one-dimensional heat equation we derived in the last entry! All we need now is the three-dimensional equivalent of Fourier's law of heat conduction.

φ = - K_{0} \nabla u

That was pretty straightforward. Now, substituting, treating

c

ρ

and

K_{0}

as constants and letting

Q = 0

yields our desired, three-dimensional heat equation

\frac{\partial u}{\partial t} = k \nabla^{2} u

where

\nabla^{2}

is often called the Laplacian.

I had to get sidetracked there a little bit, but stay tuned for the next entry, where we will be discussing the method of separation of variables!