Variational Methods &
Optimal Control
lecture 11
Matthew Roughan
<matthew.roughan@adelaide.edu.au>
Discipline of Applied Mathematics
School of Mathematical Sciences
University of Adelaide
April 14, 2016
Variational Methods & Optimal Control: lecture 11 – p.1/??
Extension 3: several
independent variables
When there are several independent variables, e.g., (x, y) and the extremal
we wish to find represents, for instance, a surface z(x, y), and f is a
function f (x, y, z(x, y), zx, zy ), then the E-L equation generalizes to give
∂ ∂f
∂f
∂ ∂f
−
=0
−
∂z ∂x ∂zx ∂y ∂zy
Variational Methods & Optimal Control: lecture 11 – p.2/??
Several independent variables
Consider a surface minimization problem. We have a surface in 3D that is
a function of (x, y), e.g. z = z(x, y) then x and y are both independent
variables.
Examples:
minimal area surfaces
soap films and bubbles
for construction
problems of the form, minimize
F{z} =
ZZ
Ω
z2x + z2y dx dy
Variational Methods & Optimal Control: lecture 11 – p.3/??
Notation
region
boundary
z
region = Ω
boundary = δΩ
surface = z(x, y)
y
x
Variational Methods & Optimal Control: lecture 11 – p.4/??
Formalisms
Ω is a simply connected, bounded region of IR2
δΩ is the boundary of Ω
Ω̄ = Ω ∪ δΩ is the closure of Ω
C2 (Ω̄) = {z : Ω̄ → IR | z has 2 continuous derivatives}
C2 (δΩ) = {z0 : δΩ → IR | z0 has 2 continuous derivatives}
ZZ
f (x, y) dx dy is an area integral of f over the region Ω
I
f (x, y) dx is a contour integral around the boundary δΩ.
Ω
δΩ
Variational Methods & Optimal Control: lecture 11 – p.5/??
The problem
Find extremals for the functional
F{z} =
ZZ
Ω
f (x, y, z(x, y), zx, zy ) dx dy
Analogy of fixed end points is a fixed boundary, e.g.
z(x, y) = z0 (x, y) for all (x, y) ∈ δΩ
for some specified function z0 ∈ C2 (δΩ).
Variational Methods & Optimal Control: lecture 11 – p.6/??
Solution
As before we consider perturbations, though in this case they are
perturbations to a surface, with fixed edge, e.g.
ẑ(x, y) = z(x, y) + εη(x, y)
where η(x, y) = 0 for all (x, y) ∈ δΩ.
Taylor’s theorem gives
f (x, y, z + εη, zx + εηx , zy + εηy )
∂f
∂f
∂f
+ ηy
+ O(ε2 )
= f (x, y, z, zx, zy ) + ε η + ηx
∂z
∂zx
∂zy
Variational Methods & Optimal Control: lecture 11 – p.7/??
The First Variation
As before we demand that at an extremal, the First Variation δF(η, z) = 0
for all possible η, and small ε
F{z + εη} − F{z}
δF(η, z) = lim
ε→0
ε
ZZ
∂f
∂f
∂f
=
η + ηx
+ ηy
dx dy
∂z
∂zx
∂zy
Ω
We next need to do the equivalent of integration by parts, but its a bit more
complicated — we need to use Green’s theorem.
Variational Methods & Optimal Control: lecture 11 – p.8/??
Green’s theorem
One form of Green’s theorem states
ZZ Ω
∂φ ∂ψ
+
∂x ∂y
dx dy =
I
δΩ
φ dy −
I
δΩ
ψ dx
for φ, ψ : Ω̄ → IR such that φ, ψ, φx and ψy are continuous.
This converts an area integral over a region into a line integral around the
boundary.
Variational Methods & Optimal Control: lecture 11 – p.9/??
Green’s theorem in use
Green’s theorem:
ZZ Ω
∂φ ∂ψ
+
∂x ∂y
dx dy =
I
δΩ
φ dy −
I
δΩ
ψ dx
For instance, take
∂f
φ=η
∂zx
∂φ
∂x
∂ψ
∂y
and
∂f
ψ=η
∂zy
∂ ∂f
∂f
+η
= ηx
∂zx
∂x ∂zx
∂ ∂f
∂f
+η
= ηy
∂zy
∂y ∂zy
Variational Methods & Optimal Control: lecture 11 – p.10/??
Green’s theorem in use
Green’s theorem:
ZZ Ω
∂φ ∂ψ
+
∂x ∂y
dx dy =
Z
δΩ
φ dy −
Z
δΩ
ψ dx
So
ZZ Ω
=
ηx
I
∂f
∂ ∂f
∂ ∂f
∂f
+ ηy
+η
+η
∂zx
∂zy
∂x ∂zx
∂y ∂zy
∂f
dy −
η
δΩ ∂zx
I
dx dy
∂f
η
dx
δΩ ∂zy
Notice that η(x, y) = 0 for all (x, y) ∈ δΩ, and so the right hand side
integrals are both zero.
Variational Methods & Optimal Control: lecture 11 – p.11/??
Given the RHS of the equation was zero, we can rearrange to get
ZZ Ω
∂f
∂f
ηx
+ ηy
∂zx
∂zy
ZZ
∂ ∂f
∂ ∂f
dx dy = −
η
+
dx dy
∂x ∂zx ∂y ∂zy
Ω
With the result that the First Variation can be written
ZZ
∂ ∂f
∂ ∂f
∂f
−
dx dy
−
δF(η, z) =
η
∂z ∂x ∂zx ∂y ∂zy
Ω
This step is the analogy of integration by parts in the derivation of the
standard Euler-Lagrange equation.
Variational Methods & Optimal Control: lecture 11 – p.12/??
Euler-Lagrange equation
Given that F(η, z) = 0 for all allowable η, Lemma 2.2.2 (see last page)
can be extended directly to the 2D case, with the result that
∂ ∂f
∂ ∂f
∂f
−
=0
−
∂z ∂x ∂zx ∂y ∂zy
This is also called the Euler-Lagrange equation.
The general case of the Euler-Lagrange equations with 2 independent
variables (and the boundary conditions) produces a Dirichlet boundary
value problem.
these can be very hard to solve.
Variational Methods & Optimal Control: lecture 11 – p.13/??
Simple example
Let Ω be the disk defined by x2 + y2 < 1, and the functional of interest be
ZZ
1 2 1 2
F{z} =
1 + zx + zy dx dy
2
2
Ω
with boundary conditions
z0 (x, y) = 2x2 − 1
for all (x, y) such that x2 + y2 = 1.
Variational Methods & Optimal Control: lecture 11 – p.14/??
Simple example: solution
The Euler-Lagrange equation is
∂ ∂f
∂ ∂f
∂f
−
=0
−
∂z ∂x ∂zx ∂y ∂zy
Note that in this example, f has no explicit dependence on x, y or z, and so
we get
∂2 z ∂2 z
+ 2 =0
2
∂x
∂y
This equation is called Laplace’s equation.
Consider the function z = x2 − y2 . This satisfies Laplace’s equation, and
on the boundary y2 = 1 − x2 , so z = 2x2 − 1, which satisfies our boundary
condition.
Variational Methods & Optimal Control: lecture 11 – p.15/??
Example: vibrating string
Imagine a taut string
flexible
uniform mass
small deflections
Equilibrium solution
the string sits in a straight line
consider small perturbations
Variational Methods & Optimal Control: lecture 11 – p.16/??
Example: vibrating string
Model:
length of string is L
position along the string is x ∈ [0, L]
constant tension τ
points on string move up/down perpendicular to x-axis
displacement at x at time t is w(x,t) ≪ L
no friction or other damping
only force occurs to stretch string
constant density σ along the string’s length
Variational Methods & Optimal Control: lecture 11 – p.17/??
Example: vibrating string
w(x,t)
0
x
L
end points are fixed so w(0,t) = w(L,t) = 0
velocity of particle is wt = ∂w
∂t
kinetic energy of string
σ
T=
2
Z
0
L
wt2 dx
Variational Methods & Optimal Control: lecture 11 – p.18/??
Example: vibrating string
w(x,t)
0
x
L
slope of string ∂w
∂x
potential energy of the string depends on how much it is stretch
from its original length L
length at time t is given by
J(t) =
Z
0
Lq
1 + w2x dx
Variational Methods & Optimal Control: lecture 11 – p.19/??
Example: vibrating string
potential is V = τ(J − L), so
V (t) = τ
Z
0
Lq
1 + w2x − 1 dx
we assumed that w is small, so we can approximate
q
1 2
2
1 + wx ≃ 1 + wx
2
so we use
τ
V (t) =
2
Z
0
L
w2x dx
Variational Methods & Optimal Control: lecture 11 – p.20/??
Example: vibrating string
The system is conservative so we apply the “principle of least action”
(Hamilton’s principle), which says the shape will be an extremum with
respect to
F{w} =
Z
t2
t1
1
(T −V ) dt =
2
Z
t2 Z L
t1
0
σwt2 − τw2x dx dt
The Euler-Lagrange equation is
∂ ∂f
∂ ∂f
∂f
−
=0
−
∂w ∂x ∂wx ∂t ∂wt
which gives
∂
∂
τwx = σwt
∂x
∂t
Variational Methods & Optimal Control: lecture 11 – p.21/??
Example: vibrating string
∂
∂
τwx = σwt
∂x
∂t
or
∂2 w σ ∂2 w
=
2
∂x
τ ∂t 2
which is the classic wave equation, which you have no doubt seen solved
in other contexts.
Variational Methods & Optimal Control: lecture 11 – p.22/??
Example: Plateau’s problem
We want to find the surface with minimal area stretched between a
boundary.
this is what a soap film does
architecture influenced by
minimal surfaces
architect Frei Otto
Munich Olympic Stadium
Variational Methods & Optimal Control: lecture 11 – p.23/??
Surface area minimization
The functional of interest is the surfaceZarea
F{z} =
Ω
dS
As before, we can’t compute this integral, so we must convert it to a convenient form:
11111111111111111111
00000000000000000000
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
B
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
A
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
A = (0, dy, zy dy)
B = (dx, 0, zx dx)
A × B = (zx dx dy, zy dx dy, −dx dy)
area = |A xB|
z
x
dy
y
dx
q
dS = |A × B| = (zx dx dy)2 + (zy dx dy)2 + (−dx dy)2
q
= dx dy 1 + z2x + z2y
Variational Methods & Optimal Control: lecture 11 – p.24/??
Surface area minimization
So we may rewrite the functional as
F{z} =
ZZ q
Ω
1 + z2x + z2y dx dy
The Euler-Lagrange equation is
∂f
∂ ∂f
∂ ∂f
−
=0
−
∂z ∂x ∂zx ∂y ∂zy
Which in this context is




zy
∂ 
∂ 
zx

=0
q
q
−
−
∂x
∂y
1 + z2 + z2
1 + z2 + z2
x
y
x
y
Variational Methods & Optimal Control: lecture 11 – p.25/??
Surface area minimization
Continuing the derivation


zx
∂ 
 =
q
∂x
1 + z2 + z2
x
y


y
1 + z2x + z2y
−
zx (zx zxx + zy zyx )
(1 + z2x + z2y )3/2
=
zxx (1 + z2x + z2y ) − zx (zx zxx + zy zyx )
(1 + z2x + z2y )3/2
=
zxx (1 + z2y ) − zx zy zyx
(1 + z2x + z2y )3/2
zy
∂ 
 =
q
∂y
1 + z2 + z2
x
zxx
q
zyy (1 + z2x ) − zx zy zyx
(1 + z2x + z2y )3/2
Variational Methods & Optimal Control: lecture 11 – p.26/??
Surface area minimization
Add the two terms above to get the E-L equation
zxx (1 + z2y ) − 2zx zy zyx + zyy (1 + z2x )
=0
2C =
3/2
2
2
(1 + zx + zy )
where we call C the mean curvature (which is 0 on the extremals).
We multiply both sides of the E-L equation by the denominator to get
zxx (1 + z2y ) − 2zx zy zyx + zyy (1 + z2x ) = 0
This is a hard PDE in general.
Variational Methods & Optimal Control: lecture 11 – p.27/??
Approximate solutions
If the surfaces are almost planes (e.g. if z is small), then we can take
squared derivate terms like z2x , z2y and zx zy to be zero. In this case the
general equation
zxx (1 + z2y ) − 2zx zy zyx + zyy (1 + z2x ) = 0
simplifies to give us
zxx + zyy = 0
the Laplace equation again. We know from the previous example that this
is equivalent to approximating
q
1 2 1 2
2
2
f (x, y, z, zx, zy ) = 1 + zx + zy ≃ 1 + zx + zy
2
2
Variational Methods & Optimal Control: lecture 11 – p.28/??
Example
Design a surfaces of minimum surface
area over a stadium with small
curved walls, of shape z = ε cos π2 by , located at x = ±a, and with no end
walls at y = ±b.
y
11111111
00000000
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
b
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
1111111111111111111
0000000000000000000
0000000000000000000
1111111111111111111
0
1
0000000000000000000
1111111111111111111
0
1
ε
0000000000000000000
1111111111111111111
0
1
0000000000000000000
1111111111111111111
0
1
0000000000000000000
1111111111111111111
0
1
1111111111
0000000000
11111111
00000000
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
00000000
11111111
x
a
b
Variational Methods & Optimal Control: lecture 11 – p.29/??
Example
Use the approximation, so we wish to solve
zxx + zyy = 0
z(±a, y) = ε cos
z(x, ±b) = 0
π y 2b
Assume a solution with separation of variables, e.g. z(x, y) = X(x)Y (y),
then the DE implies that
cosh
cos
z∝
(λy) ×
(λy)
sinh
sin
π
to match the boundary conditions, and choose
Choose cos with λ = 2b
cosh because we expect the solution to be even.
Variational Methods & Optimal Control: lecture 11 – p.30/??
Example: solution
So the solution is
z(x, y) = A cos
πy 2b
cosh
πx 2b
Determine A using the end-points, e.g.
ε cos
πy 2b
= A cos
πy So
A = ε/ cosh
and
z(x, y) = ε cos
πy 2b
2b
cosh
πa 2b
πa cosh
2b
πx 2b
/ cosh
πa 2b
Variational Methods & Optimal Control: lecture 11 – p.31/??
Example: solution
0.2
0.1
0
1
0
1
0
−1 −1
Variational Methods & Optimal Control: lecture 11 – p.32/??
Example: solution
In fact, once we realize it will have a cosine cross-section, we know that
the “area” of the curve for any given x will be proportional to the height,
so we are in fact solving a problem that looks a lot like that of the
catenary. So we should be surprised to see that the result has the same
cosh function.
Variational Methods & Optimal Control: lecture 11 – p.33/??
But this is hard...
Solving the PDE form of the EL equations can be very hard. What can we
do to make it easier? Surely computers can help?
Variational Methods & Optimal Control: lecture 11 – p.34/??
Plateau’s laws
A little bit extra:
Soap films are made of entire smooth surfaces
The average curvature of a portion of a soap film is always constant
on any point on the same piece of soap film
Soap films always meet in threes, and they do so at an angle of
cos−1 (−1/2) = 120 degrees forming an edge called a Plateau
Border.
Plateau Borders meet in fours at an angle of cos−1 (−1/3) ≃ 109.47
degrees (the tetrahedral angle) to form a vertex.
Variational Methods & Optimal Control: lecture 11 – p.35/??