One of the more confusing things when I was learning vector calculus \& classical field theory was the derivation of the flux transport theorem. The theorem relates the change in flux $$\Phi$$ due to a vector field $$\vec F$$ through a moving surface $${\mathcal S}$$ to certain integrals involving the time rate of change and the divergence of the field over the surface along with a contour integral around the surface’s boundary. Using the notation of ‘Introduction to Vector Analysis, 4th ed.’ by Davis and Snider, the flux transport theorem is given by:
\[ \frac{d \Phi}{d t} = \int_{\mathcal S} d \vec {\mathcal S} \cdot \left[ \frac{\partial \vec F}{\partial t} + (\nabla \cdot \vec F) \vec v \right] + \int_{\partial {\mathcal S}} d \vec \ell \cdot \vec F \times \vec v \; , \]
where $$d \vec {\mathcal S}$$ is the outward normal to a differential portion of surface area, $$\vec v$$ is the velocity at each point on the surface (recall it is moving), and $$d \vec \ell$$ is a differential line element along the boundary of the surface $$\partial {\mathcal S}$$.
Davis and Snider have some nice homework problems but I wanted an example where the surface changed size as well as moved in space. The example I concocted is one with a rectangular shaped surface whose four vertices are given by:
\[ \vec {\mathcal A} \doteq [t,-t,-1] \; ,\]
\[ \vec {\mathcal B} \doteq [t,t,-1] \; ,\]
\[ \vec {\mathcal C} \doteq [t,t,1] \; ,\]
and
\[ \vec {\mathcal B} \doteq [t,-t,1] \; .\]
Note that I represent ($$\doteq$$) the coordinates of the points by row arrays simply for typographical convenience, although whether they are rows or columns doesn’t matter.
The surface, which moves uniformly in the x-direction, is canted 45 degrees with respect to the y-z plane and has a time varying area of $$4t$$. The surface is parametrized by two parameters $$u$$ and $$v$$ ranging from $$0$$ to $$1$$ such that any point on the surface (including the boundary) is given by
\[\vec {\mathcal R}(u,v) \doteq \left[ t, (2u-1)t, 2v -1 \right] \; \; u,v \in[0,1] \; .\]
In this parameterization, the outward normal to a differential patch is
\[ d \vec {\mathcal S} = \frac{\partial \vec {\mathcal R}}{\partial u} \times \frac{\partial \vec {\mathcal R}}{\partial v} du dv \doteq [4t,0,0] du dv \; , \]
from which we immediately get the total area as
\[ Area(\vec {\mathcal S}) = \int_{(u,v)} |d \vec {\mathcal S} | = \int_0^1 du \int_0^1 dv \, 4 t = 4t \]
as expected.
The form of the vector field is
\[ \vec F(\vec r) \doteq [x y^2, y z^2 z x^2] \]
where $$\vec r \doteq [x,y,z] $$.
Since the form of the surface and the vector field are fairly simple, the flux through the surface due to $$\vec F$$
\[ \Phi[\vec F,\vec {\mathcal S}] = \int_{\vec {\mathcal S}} \vec F(\vec {\mathcal R}(u,v)) \cdot d \vec {\mathcal S} \]
can be explicitly computed as
\[ \Phi[\vec F,\vec {\mathcal S}] = \int_0^1 du \int_0^1 dv \, 4 t^4 (2 u – 1)^2 = \frac{4t^4}{3} \; .\]
The time derivative is then easily obtained as
\[ \frac{d \Phi[\vec F,\vec {\mathcal S}]}{d t} = \frac{16t^3}{3} \; .\]
To use the flux transport theorem, we need to compute $$div(\vec F)$$ and $$\frac{\partial \vec F}{\partial t}$$ and then evaluate these terms across the expanse of the surface $$\mathcal{S}$$. Likewise we also need to compute $$\vec F \times \vec v$$, where $$\vec v$$ is the velocity of the surface and then evaluate this terms along the its boundary.
The divergence of the field is given by
\[ div(\vec F) = x^2 + y^2 + z^2 \; , \]
which, in the $$u-v$$ parametrization becomes
\[ div(\vec F) = t^2 + (2 u – 1)^2 t^2 + (2 v- 1)^2 \; .\]
The time derivative of the field is
\[\frac{\partial \vec F}{\partial t} = 0 \]
since the vector field $$\vec F$$ has no explicit time dependence. All of the time rate of change is due to the surface moving within the field.
The surface integral in the flux transport equation, formally given by
\[ I_0 = \int_{\vec {\mathcal S}} d \vec {\mathcal S} \cdot \left[ div\left(\vec F(\vec {\mathcal R}(u,v))\right) \vec v + \frac{\partial \vec F}{\partial t}\left(\vec {\mathcal R}(u,v)\right) \right] \]
in the $$u-v$$ parametrization, evaluates to
\[ I_0 = \frac{ 16 t^3}{3} + \frac{4 t }{3} \; .\]
The velocity of any point on the surface is obtained by taking a time derivative with respect to $$\vec {\mathcal R}$$
\[ \vec v = \frac{\partial \mathcal{R} (u,v)}{\partial t} \doteq [1, 2 u – 1, 0] \; .\]
The integrand of the line integral is
\[ \vec F \times \vec v \doteq [ – z x^2 (2 u – 1), z x^2 , x y^2 (2 u – 1) – y z^2 ] \; , \]
which, in the $$u-v$$ parametrization, becomes
\[ \vec F \times \vec v \doteq [-(2 v – 1)(2 u – 1) t^2, (2 v – 1) t^2, (2 u – 1)^3 t^3 – (2 u – 1)(2 v – 1 )^2 t] \; .\]
There are four distinct lines or legs making up the boundary of the surface. These are given by
\[ \vec {\mathcal L}_1(u) = \vec {\mathcal A} + u(\vec {\mathcal B} – \vec {\mathcal A}) \; ,\]
\[ \vec {\mathcal L}_2(v) = \vec {\mathcal B} + v(\vec {\mathcal C} – \vec {\mathcal B}) \; ,\]
\[ \vec {\mathcal L}_3(u) = \vec {\mathcal C} + u(\vec {\mathcal D} – \vec {\mathcal C}) \; ,\]
and
\[ \vec {\mathcal L}_4(v) = \vec {\mathcal D} + v(\vec {\mathcal A} – \vec {\mathcal D}) \]
and there is a corresponding integral for each.
The evaluation of each is a bit tedious (particularly in taking care to make sure that $$u$$ or $$v$$ take on the appropriate value for the leg being traversed) but straightforward leading to
\[ I_1 =-2 t^3 \int_0^1 du = -2 t^3 \; , \]
\[ I_2 = 2 t^3 \int_0^1 dv – 2t \int_0^1 dv (2v-1)^2 = \frac{ 2(3 t^3 – t)}{3} \; , \]
\[ I_3 = -2t^3 \int_0^1 du = -2 t^3 \; , \]
and
\[ I_4 = 2 t^3 \int_0^1 dv – 2t\int_0^1 dv (2v-1)^2 = 2 t^3 – \frac{2}{3} t \; ,\]
respectively.
The total line integral around the boundary is then given as the sum of these terms
\[ I_c = – \frac{4}{3} t \; .\]
Adding this result to the result from surface integral gives
\[I_{tot} = \frac{16 t^3}{3} \; , \]
which is the same result as before.
While this example is a bit contrived, it does offer a simple combination of both movement and growth that seems absent within the literature. Furthermore, when worked in detail, each piece drives home the content of the flux transport theorem.