Physical Interpretation of the Divergence

Section 11.2 Physical Interpretation of the Divergence

We next want to derive a physical interpretation of

div f .

Let

f

be the velocity of a fluid or a gas in a region

D .

Fix a point

a = (a_{1}, a_{2})

D,

and denote by

B_{δ}

a disc centred at

a

with radius

δ .

Then

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\int_{\partial B_{δ}} f \cdot n d s

is the net flow out of

B_{δ}

(if it is negative then the net flow is into

B_{δ}

). By the Divergence Theorem

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\frac{1}{area (B_{δ})} \int_{B_{δ}} div f d x = \frac{1}{area (B_{δ})} \int_{\partial B_{δ}} f \cdot n d s

models the net loss or increase of fluid (or gas) from

B_{δ}

per area. If we shrink

B_{δ}

a

we get a density for the loss or increase of fluid (or gas) at the point

a .

It indicates how much fluid (or gas) disappears or appears at that point. It turns out that (see Lemma 11.6 below)

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lim_{δ \to 0} \frac{1}{area (B_{δ})} \int_{B_{δ}} div f d x = div f (a)

Hence

div 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

div f

is positive there is a net flow out, if it is negative there is a net flow in. For that reason

div f

is also sometimes called the compressibility of a vector field. In any case we talk about a source if

div f > 0,

and about a sink if

div f < 0 .

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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,

we only need to replace area by volume.

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Lemma 11.6.

Suppose that

D \subset R^{2}

is an open set, and that

g : D \to R

is continuous at

a \in D .

Moreover, let

B_{δ} := B_{δ} (a) .

Then

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\begin{matrix} (11.2) & lim_{δ \to 0} \frac{1}{area (B_{δ})} \int_{B_{δ}} g (x) d x = g (a) . \end{matrix}

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Proof.

\int_{B_{δ}} g (a) d x = area (B_{δ}) g (a)

we have that

\frac{1}{area (B_{δ})} \int_{B_{δ}} g (x) d x - g (a) = \frac{1}{area (B_{δ})} \int_{B_{δ}} g (x) - g (a) d x .

Now fix

ε > 0

arbitrary. As

g

is continuous at

a

there exists

δ_{0} > 0

such that

| g (x) - g (a) | < ε

whenever

‖ x - a ‖ < δ_{0}

(see Proposition 3.24). Hence for all

δ < δ_{0}

we have

\begin{aligned} | \frac{1}{area (B_{δ})} & \int_{B_{δ}} g (x) d x - g (a) | \\ \leq \frac{1}{area (B_{δ})} \int_{B_{δ}} | g (x) - g (a) | d x \\ \leq \frac{1}{area (B_{δ})} \int_{B_{δ}} ε d x \\ = \frac{area (B_{δ})}{area (B_{δ})} ε \\ = ε \end{aligned}

ε > 0

was arbitrary (11.2) follows.

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