Let \(\Omega\) be a fixed volume and denote the boundary of this volume by \(\partial \Omega\). The rate at which \(u\) is changing in time throughout \(\Omega\) needs to be balanced by the rate at which \(u\) leaves the volume plus any sources of \(u\). Mathematically, this means that \begin{gather*} \pd{ }{t} \iiint_{\Omega} u dV = -\iint_{\partial \Omega} \bq \cdot n dA + \iiint_\Omega f dV. \end{gather*}

This is a global balance law in the sense that it holds for all volumes \(\Omega\). The troubles here are two fold: (1) there are many integrals, and (2) there are really two variables (\(u\) and \(q\) since \(f=f(u,x,t)\)) so the equation is not closed. In order to mitigate that fact we apply the divergence theorem to get \begin{gather*} \pd{ }{t} \iiint_{\Omega} u dV = -\iiint_{\Omega} \nabla \cdot \bq dV + \iiint_\Omega f dV. \end{gather*}

Gathering all of the terms on the right of <<Unresolved xref, reference "eqn_global_balance2"; check spelling or use "provisional" attribute>>, interchanging the integral and the derivative on the left (since the volume is not changing in time), and rewriting gives \begin{gather*} \iiint_\Omega \left( \pd{u}{t} + \nabla \cdot \bq \right) dV = \iiint_\Omega f dV \end{gather*}

If we presume that this equation holds for all volumes \(\Omega\) then the integrands must be equal and we get the local balance law \begin{gather*} \pd{u}{t} + \nabla \cdot \bq = f. \end{gather*}

If equation <<Unresolved xref, reference "eqn_local_balance"; check spelling or use "provisional" attribute>> it is often assumed that the system is free of external sources. In this case we set \(f\) to zero and obtain the source-free balance law \begin{gather*} \pd{u}{t} + \nabla \cdot \bq = 0. \end{gather*}

It is this form of balance law where many of the most interesting and important partial differential equations come from. In particular consider the following two cases: mass balance and energy balance.

In mass balance we take \(u\) to either be the density of a substance (e.g. in the case of liquids) or the concentration of a substance in a mixture (e.g. in the case of gasses). If \(C\) is the mass concentration of a substance in a gas then the flux of that substance is given via Fick's Law as \begin{gather*} \bq = -k \nabla C. \end{gather*}

Combining <<Unresolved xref, reference "eqn_fick"; check spelling or use "provisional" attribute>> with <<Unresolved xref, reference "eqn_local_source_free"; check spelling or use "provisional" attribute>> (and assuming that \(k\) is independent of space, time, and concentration) gives \begin{gather*} \pd{C}{t} = k \nabla \cdot \nabla C. \end{gather*}

In the presenence of external sources of mass, <<Unresolved xref, reference "eqn_fick2_simp"; check spelling or use "provisional" attribute>> is \begin{gather*} \pd{C}{t} = k \nabla \cdot \nabla C + f(x). \end{gather*}

What does <<Unresolved xref, reference "eqn_fick3"; check spelling or use "provisional" attribute>> equation look like in terms of spatial derivatives on the right-hand side? \begin{align*} \pd{C}{t} \amp = \underline{\hspace{2in}} \text{ (1 Spatial Dimension) }\\ \pd{C}{t} \amp = \underline{\hspace{2in}} \text{ (2 Spatial Dimensions) }\\ \pd{C}{t} \amp = \underline{\hspace{2in}} \text{ (3 Spatial Dimensions) } \end{align*}

The energy balance equation is essentially the same as the mass balance equation. If \(u\) is temperature then the flux of temperature is given by Fourier's Law for heat conduction \begin{gather*} q = -k\nabla T. \end{gather*}

Making the same simplifications as in the mass balance equation we arrive at \begin{gather*} \pd{T}{t} = k \nabla \cdot \nabla T. \end{gather*}

In the presence of external sources of heat, <<Unresolved xref, reference "eqn_fourier2"; check spelling or use "provisional" attribute>> becomes \begin{gather*} \pd{T}{t} = k \nabla \cdot \nabla T + f(x). \end{gather*}

What does <<Unresolved xref, reference "eqn_fourier3"; check spelling or use "provisional" attribute>> equation look like in terms of spatial derivatives on the right-hand side? \begin{align*} \pd{T}{t} \amp = \underline{\hspace{2in}} \text{ (1 Spatial Dimension) }\\ \pd{T}{t} \amp = \underline{\hspace{2in}} \text{ (2 Spatial Dimensions) }\\ \pd{T}{t} \amp = \underline{\hspace{2in}} \text{ (3 Spatial Dimensions) } \end{align*}

Equations <<Unresolved xref, reference "eqn_fick3"; check spelling or use "provisional" attribute>> and <<Unresolved xref, reference "eqn_fourier3"; check spelling or use "provisional" attribute>> are the same partial differential equation for two very important physical phenomenon; mass and heat transfer. In the case where time is allowed to run to infinity and no external sources of mass or energy are included these equations reach a steady state solution (no longer changing in time) and we arrive at Laplace's Equation \begin{gather*} \nabla \cdot \nabla u = 0. \end{gather*}

Laplace's equation is actually a statement of minimal energy as well as steady state heat or temperature. We can see this since entropy always drives systems from high energy to low energy, and if we have reached a steady state then we must have also reached a surface of minimal energy.

Equation <<Unresolved xref, reference "eqn_laplace"; check spelling or use "provisional" attribute>> is sometimes denoted as \(\nabla \cdot \nabla u = \nabla^2 u = \Delta u\), and in terms of the partial derivatives it is written as \begin{align*} 0 \amp = \underline{\hspace{2in}} \text{ (1 Spatial Dimension) }\\ 0 \amp = \underline{\hspace{2in}} \text{ (2 Spatial Dimensions) }\\ 0 \amp = \underline{\hspace{2in}} \text{ (3 Spatial Dimensions) } \end{align*}

If there is a time-independent external source the the right-hand side of <<Unresolved xref, reference "eqn_laplace"; check spelling or use "provisional" attribute>> will be non-zero and we arrive at Poisson's equation: \begin{gather*} \nabla \cdot \nabla u = -f(x). \end{gather*}

Note that the negative on the right-hand side comes from the fact that \(\pd{u}{t} = k \nabla \cdot \nabla u + f(x)\) and \(\pd{u}{t} \to 0\). Technically we are taking absorbing the constant \(k\) into \(f\) (that is “\(f\)” is really “\(f/k\)”). Also note that in many instances the value of \(k\) is not contant and cannot therefore be pulled out of the derivative without a use of the product rule.

We will start our exploration of numerical PDEs with Laplace's and Poisson's equations. We will then layer on the temporal derivatives to explore mass and heat transport. Finally, we will explore wave phenomena as well as advection-diffusion transport models.