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Variational Methods &
Optimal Control
lecture 11
Matthew Roughan
<[email protected]>
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:
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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
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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/??
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