Section 11.2 Physical Interpretation of the Divergence
We next want to derive a physical interpretation of \(\divergence\vect f\text{.}\) Let \(\vect f\) be the velocity of a fluid or a gas in a region \(D\text{.}\) Fix a point \(\vect a=(a_1,a_2)\) in \(D\text{,}\) and denote by \(B_\delta\) a disc centred at \(\vect a\) with radius \(\delta\text{.}\) Then
\begin{equation*}
\int_{\partial B_\delta}\vect f\cdot\vect n\,ds
\end{equation*}
is the net flow out of \(B_\delta\) (if it is negative then the net flow is into \(B_\delta\)). By the Divergence Theorem
\begin{equation*}
\frac{1}{\area(B_\delta)}
\int_{B_\delta}\divergence\vect f\,d\vect x
=\frac{1}{\area(B_\delta)}
\int_{\partial B_\delta}\vect f\cdot\vect n\,ds
\end{equation*}
models the net loss or increase of fluid (or gas) from
\(B_\delta\) per area. If we shrink
\(B_\delta\) to
\(\vect a\) we get a density for the loss or increase of fluid (or gas) at the point
\(\vect a\text{.}\) It indicates how much fluid (or gas) disappears or appears at that point. It turns out that (see
Lemma 11.6 below)
\begin{equation*}
\lim_{\delta\to 0}\frac{1}{\area(B_\delta)}
\int_{B_\delta}\divergence\vect f\,d\vect x
=\divergence\vect f(\vect a)
\end{equation*}
Hence \(\divergence\vect f\) is often called the source strength of the vector field. If we have a gas, then by compressing it is possible that less flows out than in, or if we expand it more flows out than in. Hence if the sign of \(\divergence\vect f\) is positive there is a net flow out, if it is negative there is a net flow in. For that reason \(\divergence\vect f\) is also sometimes called the compressibility of a vector field. In any case we talk about a source if \(\divergence\vect f>0\text{,}\) and about a sink if \(\divergence\vect f<0\text{.}\)
To make the above precise we prove a lemma that in some sense is a generalisation of the proof of the fundamental theorem of calculus to higher dimensions. It also works for \(N\geq 2\text{,}\) we only need to replace area by volume.
Lemma 11.6.
Suppose that \(D\subset\mathbb R^2\) is an open set, and that \(g\colon D\to\mathbb R\) is continuous at \(\vect a\in D\text{.}\) Moreover, let \(B_\delta:=B_\delta(\vect a)\text{.}\) Then
\begin{equation}
\lim_{\delta\to 0}\frac{1}{\area(B_\delta)}
\int_{B_\delta} g(\vect x)\,d\vect x
=g(\vect a)\text{.}\tag{11.2}
\end{equation}
Proof.
As \(\int_{B_\delta}g(\vect a)\,d\vect x=\area(B_\delta)g(\vect a)\) we have that
\begin{equation*}
\frac{1}{\area(B_\delta)}\int_{B_\delta} g(\vect x)\,d\vect x
-g(\vect a)
=\frac{1}{\area(B_\delta)}\int_{B_\delta} g(\vect x)
-g(\vect a)\,d\vect x\text{.}
\end{equation*}
Now fix
\(\varepsilon>0\) arbitrary. As
\(g\) is continuous at
\(\vect a\) there exists
\(\delta_0>0\) such that
\(|g(\vect x)-g(\vect a)|<\varepsilon\) whenever
\(\|\vect x-\vect a\|<\delta_0\) (see
Proposition 3.24). Hence for all
\(\delta<\delta_0\) we have
\begin{align*}
\Bigl|\frac{1}{\area(B_\delta)}\amp\int_{B_\delta}g(\vect x)\,d\vect x
-g(\vect a)\Bigr|\\
\amp\leq\frac{1}{\area(B_\delta)}\int_{B_\delta}|g(\vect x)-g(\vect a)| \,d\vect x\\
\amp\leq\frac{1}{\area(B_\delta)}\int_{B_\delta} \varepsilon \,d\vect x\\
\amp=\frac{\area(B_\delta)}{\area(B_\delta)}\varepsilon\\
\amp=\varepsilon
\end{align*}
As
\(\varepsilon>0\) was arbitrary
(11.2) follows.
