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:<\/p>\n
\\[ \\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 \\; , \\]<\/p>\n
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 $$\\partial {\\mathcal S}$$.<\/p>\n
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:
\n\\[ \\vec {\\mathcal A} \\doteq [t,-t,-1] \\; ,\\]
\n\\[ \\vec {\\mathcal B} \\doteq [t,t,-1] \\; ,\\]
\n\\[ \\vec {\\mathcal C} \\doteq [t,t,1] \\; ,\\]
\nand
\n\\[ \\vec {\\mathcal B} \\doteq [t,-t,1] \\; .\\]
\nNote 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.<\/p>\n
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<\/p>\n
\\[\\vec {\\mathcal R}(u,v) \\doteq \\left[ t, (2u-1)t, 2v -1 \\right] \\; \\; u,v \\in[0,1] \\; .\\]<\/p>\n
In this parameterization, the outward normal to a differential patch is<\/p>\n
\\[ 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 \\; , \\]<\/p>\n
from which we immediately get the total area as<\/p>\n
\\[ Area(\\vec {\\mathcal S}) = \\int_{(u,v)} |d \\vec {\\mathcal S} | = \\int_0^1 du \\int_0^1 dv \\, 4 t = 4t \\]<\/p>\n
as expected.<\/p>\n
The form of the vector field is<\/p>\n
\\[ \\vec F(\\vec r) \\doteq [x y^2, y z^2 z x^2] \\]<\/p>\n
where $$\\vec r \\doteq [x,y,z] $$.<\/p>\n
Since the form of the surface and the vector field are fairly simple, the flux through the surface due to $$\\vec F$$<\/p>\n
\\[ \\Phi[\\vec F,\\vec {\\mathcal S}] = \\int_{\\vec {\\mathcal S}} \\vec F(\\vec {\\mathcal R}(u,v)) \\cdot d \\vec {\\mathcal S} \\]<\/p>\n
can be explicitly computed as
\n\\[ \\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} \\; .\\]<\/p>\n
The time derivative is then easily obtained as<\/p>\n
\\[ \\frac{d \\Phi[\\vec F,\\vec {\\mathcal S}]}{d t} = \\frac{16t^3}{3} \\; .\\]<\/p>\n
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.<\/p>\n
The divergence of the field is given by<\/p>\n
\\[ div(\\vec F) = x^2 + y^2 + z^2 \\; , \\]<\/p>\n
which, in the $$u-v$$ parametrization becomes<\/p>\n
\\[ div(\\vec F) = t^2 + (2 u – 1)^2 t^2 + (2 v- 1)^2 \\; .\\]<\/p>\n
The time derivative of the field is<\/p>\n
\\[\\frac{\\partial \\vec F}{\\partial t} = 0 \\]<\/p>\n
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.<\/p>\n
The surface integral in the flux transport equation, formally given by<\/p>\n
\\[ 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] \\]<\/p>\n
in the $$u-v$$ parametrization, evaluates to<\/p>\n
\\[ I_0 = \\frac{ 16 t^3}{3} + \\frac{4 t }{3} \\; .\\]<\/p>\n
The velocity of any point on the surface is obtained by taking a time derivative with respect to $$\\vec {\\mathcal R}$$<\/p>\n
\\[ \\vec v = \\frac{\\partial \\mathcal{R} (u,v)}{\\partial t} \\doteq [1, 2 u – 1, 0] \\; .\\]<\/p>\n
The integrand of the line integral is<\/p>\n
\\[ \\vec F \\times \\vec v \\doteq [ – z x^2 (2 u – 1), z x^2 , x y^2 (2 u – 1) – y z^2 ] \\; , \\]<\/p>\n
which, in the $$u-v$$ parametrization, becomes<\/p>\n
\\[ \\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] \\; .\\]<\/p>\n
There are four distinct lines or legs making up the boundary of the surface. These are given by<\/p>\n
\\[ \\vec {\\mathcal L}_1(u) = \\vec {\\mathcal A} + u(\\vec {\\mathcal B} – \\vec {\\mathcal A}) \\; ,\\]
\n\\[ \\vec {\\mathcal L}_2(v) = \\vec {\\mathcal B} + v(\\vec {\\mathcal C} – \\vec {\\mathcal B}) \\; ,\\]
\n\\[ \\vec {\\mathcal L}_3(u) = \\vec {\\mathcal C} + u(\\vec {\\mathcal D} – \\vec {\\mathcal C}) \\; ,\\]<\/p>\n
and<\/p>\n
\\[ \\vec {\\mathcal L}_4(v) = \\vec {\\mathcal D} + v(\\vec {\\mathcal A} – \\vec {\\mathcal D}) \\]<\/p>\n
and there is a corresponding integral for each.<\/p>\n
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
\n\\[ I_1 =-2 t^3 \\int_0^1 du = -2 t^3 \\; , \\]
\n\\[ I_2 = 2 t^3 \\int_0^1 dv – 2t \\int_0^1 dv (2v-1)^2 = \\frac{ 2(3 t^3 – t)}{3} \\; , \\]
\n\\[ I_3 = -2t^3 \\int_0^1 du = -2 t^3 \\; , \\]
\nand<\/p>\n
\\[ I_4 = 2 t^3 \\int_0^1 dv – 2t\\int_0^1 dv (2v-1)^2 = 2 t^3 – \\frac{2}{3} t \\; ,\\]
\nrespectively.<\/p>\n
The total line integral around the boundary is then given as the sum of these terms<\/p>\n
\\[ I_c = – \\frac{4}{3} t \\; .\\]<\/p>\n
Adding this result to the result from surface integral gives<\/p>\n
\\[I_{tot} = \\frac{16 t^3}{3} \\; , \\]<\/p>\n
which is the same result as before.<\/p>\n
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.<\/p>\n","protected":false},"excerpt":{"rendered":"