Caculus of Variations

Variational Methods and Optimal Control (VMOC): 2012
Tutorial 5: Wednesday 10th October

1. Capillaries: Imagine a slender, open-end cylinder dipped into a large water bath. It
is well known that the cylinder acts as a capillary tube, and the water will rise up the
tube, and moreover that the shape of the surface of the water inside the tube will have
a curved shape, as shown in the figure.
z
zh
z0
z=0
r
R
For convenience, we assume the water level outside the cylinder is at height z = 0, and
that the cylinders center of rotation is the z-axis. We will consider it in cylindrical coordinates (r, , z), where the cylinder will have radius R. Given the radial symmetry
of the problem, we will describe the height of water in the cylinder by z(r), and we
denote z0 = z(0) and zh = z(R), but note that these are not fixed boundary conditions
(the end-points are free).
At equilibrium, the potential energy of this system will be minimized. The potential
energy is made up of the following components:
The gravitational potential:
G{z} = 2g
r
0
s ds dr = 2g
z2
dr,
2
where is the difference between the density of the liquid and air, and g is the
gravitational constant.
The surface energy in the interfaces between the liquid and solid (the cylinder
walls), the gas and solid, and the air and liquid, given by
S{z} = A(SL ) + LG A(LG )
Z R
= 2Rzh + 2LG
r 1 + z 2 r dr,
0
where
The constant parameters SG , SL and LG are the respective parameters determining the strength of affiliation or attraction between these three components. We can think of a parameter as the tension on the surface with units
corresponding to force along a line of unit length. We take = SL SG .
SG , SL and LG are the surfaces between the respective phases.
A() denotes the surface area. A() represents the surface area deformation
from the case without the cylinder, for instance, the undeformed surface
R R LG
would have area given by a circular disk (radius R, area R2 = 2 0 r dr),
and the undeformed surface LS would correspond to the liquid in the cylinder at the same height as the water bath.
so the integrals above are trying to minimize the energy resulting from tension
in the surfaces due to their deformation from the case without the cylinder. The
first integral is the energy in the gas-liquid surface, and the second is the energy
resulting in the liquid-solid surface minus the energy from the solid-gas interface
it replaces.
Use calculus of variations to determine the height and shape of the water surface inside
the cylinder.
2. Inequality constraints and broken extremals: solve the isoperimetric problem inside a square region, i.e., what is the shape that contains the largest area without exceeding a given perimeter L, given that the shape must be entirely contained in a square
with sides 2W in length.
Note that the problem is uninteresting for W > L/2 because a circle of radius
R = L/2 satisfies the isoperimetric constraint, and fits inside the square, and by
previous work this is clearly the maximal area region (though there are actually multiple possible circles that might fit). Likewise, 8W < L is uninteresting, because we
cannot meet the perimeter constraint without having a concave shape, so the obvious
solution is to contain the entire area of the square, but have the perimeter dip into the
shape along a line enclosing zero area. So we consider the case
L/8 < W < L/2.
We will simplify the problem in a few ways. Firstly, the reflective symmetries of the
problem suggest that we could consider 1/8 of the square, rather than the whole square
(see the figure, which shows 1/8 of the square with sides 2W .).
y
W
111111111111111
000000000000000
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
000000000000000
111111111111111
So the inequality constraints on the problem become:

y(x) W
y(x) x
where the left end point may move along the y-axis (between the constraints), and the
right end-point is free to move along the boundary x = y. We will define the end-point
to be (x1 , y1 ).
The area and perimeter of the region are easily measured by
Z x1
Z x1 p
A{y} = 8
y x dx. and L{y} = 8
1 + y 2 dx.
0
Ignoring the factor of 8 in each term, and including the isoperimetric constraint into
the problem via a Lagrange multiplier, we obtain an objective function
Z x1
p
J{y} =
y x + 1 + y 2 dx.
0
Find the shape that maximizes the area without exceeding the perimeter constraint.
3 Terminal costs and optimal control: We originally posed Newtons aerodynamic

nose-cone problem for a nose cone pointed upwards (with flow downwards). We could
equally have posed it with the flow from right to left as in the figure below, where the
shape is described by r(u), u being the horizontal axis.
In this case, a similar simplification of the problem reduces the function of interest to
Z L
rr 3
1
1
2
F {r} = r(L) +
dw
2
2
2
0 1+r
Questions:
Show that the above function results from a simple transformation of the previous
problem.
Use natural end-point conditions for a problem with a terminal cost to determine
an equation to find r(L).

Caculus of Variations

Uploaded by

Copyright:

Available Formats

Caculus of Variations

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Caculus of Variations

Uploaded by

Copyright:

Available Formats

Variational Methods and Optimal Control (VMOC): 2012

Tutorial 5: Wednesday 10th October

Variational Methods and Optimal Control (VMOC): 2012

Variational Methods and Optimal Control (VMOC): 2012

So the inequality constraints on the problem become:

Variational Methods and Optimal Control (VMOC): 2012

3 Terminal costs and optimal control: We originally posed Newtons aerodynamic

You might also like