Calculus of Variations Introduction to...
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236861 Numerical Geometry of Images
Tutorial 1
Introduction to the
Calculus of Variations
Alex Bronstein
c©2005
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Calculus Calculus of variations
1. Function Functional
f : Rn → R f : F → R, in particular
f(u) =∫Ω
F (x, u(x), u′(x), ...) dx
Example: Example:
f(x, y) = x2 + y2 f (u) =∫Ω‖∇u(x, y)‖2 dxdy
2. Derivative Variationdf(x)dx
= lim∆x→0
f(x + ∆x) − f(x)∆x
δf(u)δu
= limε→0
f(u + ε δu) − f(u)ε
df = ∂∂εf(x + ε ∆x)
∣∣ε=0
δf = ∂∂εf(u + ε δu)
∣∣ε=0
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Calculus Calculus of variations
3. Local (relative) minimum
f(x∗) ≤ f(x) f(u∗) ≤ f(u)
∀x : ‖x − x∗‖ ≤ α ∀u : maxx∈Ω
|u(x) − u∗(x)| ≤ α
4. Necessary condition for local extremumdf(x)dx
= 0δf(u)
δu= 0
5. Sufficient condition for local extremumd2f(x)
dx2≥ 0 More complex theory
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Calculus Calculus of variations
3. Constrained local minimum
minx
f(x) minu(x)
∫Ω
F (x, u(x), u′(x))dx
s.t. g(x) = 0 s.t. G(x, u(x), u′(x)) = 0
4. Lagrangian
(x) = f(x) + λg(x) (u) =∫
Ω
(F + λ(x)G) dx
5. Method of Lagrange multipliers
d(x∗)dx
= 0δ(u∗)
δu= 0
g(x) = 0 G(x, u∗(x), u∗′(x)) = 0
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Examples of functionals
1. Curve length L(y) =
∫ x1
x0
√1 + y′2dx
The length of a non-parametric curve y(x).
2. Curve length L(x, y) =
∫ 1
0
√x′2 + y′2dt
The length of a parametric curve (x(t), y(t)).
3. Surface area A(z) =
∫S
√1 + z2
x + z2ydxdy
The area of a non-parametric surface S = (x, y, z(x, y)).
4. Total variation TV (y) =
∫ x1
x0
|y′|dx
The “oscillation strength” of a non-parametric curve y(x).
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The Euler-Lagrange equation
Let us be given the functional
f(u) =
∫ x1
x0
F (x, u(x), u′(x)) dx
with F ∈ C3 and all admissible u(x) having fixed boundary values
u(x0) = u0 and u(x1) = u1.
An extremum of f(u) satisfies the differential equation
Fu − d
dxFu′ = 0
with the boundary conditions u(x0) = u0 and u(x1) = u1.
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The second Euler-Lagrange equation
A second, less known Euler-Lagrange equation is
d
dx(F − u′Fu′)− Fx = 0
satisfied along u∗(x).
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The E-L equation (independent on x)
If the integrand does not depend on the independent variable x,
f(u) =
∫ x1
x0
F (u(x), u′(x)) dx,
the second E-L equation becomes a first-order differential equation
d
dx(F − u′Fu′) = 0
or
F − u′Fu′ = const.
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The E-L equation (independent on u(x))
If the integrand does not depend on u(x),
f(u) =
∫ x1
x0
F (u(x), u′(x)) dx,
the E-L equation becomes a first-order differential equation
d
dxFu′ = 0
or
Fu′ = const.
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The E-L equation (independent on u′(x))
If the integrand does not depend on u′(x),
f(u) =
∫ x1
x0
F (u(x), u′(x)) dx,
the E-L equation becomes an algebraic equation
Fu(x, u(x)) = 0.
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The E-L equation (high-order functionals)
Given the functional
f(u) =
∫ x1
x0
F(x, u(x), u′(x), ..., u(n)(x)
)dx
with F ∈ Cn+2 and fixed boundary values u(i)(x0) = u(i)0 and
u(i)(x1) = u(i)1 for i = 0, ..., n − 1.
The Euler-Lagrange equation (also known as the Euler-Poisson
equation) is
Fu − d
dxFu′ +
d2
dx2Fu′′ + (−1)n dn
dxnFu(n) = 0.
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The E-L equation (multiple independent variables)
Given the functional
f(u) =
∫Ω
F (x, u(x), ux1(x), ..., uxn(x)) dx
with x ∈ Rn and u(∂Ω) = u∂Ω.
An extremum of f(u) satisfies the differential equation
∂F
∂u− ∂
∂x1
∂F
∂ux1
− ... − ∂
∂xn
∂F
∂uxn
= 0
with the boundary condition u(∂Ω) = u∂Ω.
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The E-L equation (multiple functions)
Given the functional
f(u) =
∫ x1
x0
F (x, u1(x), ..., un(x), u′1(x), ..., u
′n(x)) dx
with F ∈ C3 and all admissible u(x) having fixed boundary values
ui(x0) = u0i and ui(x1) = u1
i .
An extremum of f(u) satisfies the system of differential equations
Fui− d
dxFu′
i= 0
with the boundary conditions u(x0) = u0 and u(x1) = u1.
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Problem 1: Minimum distance on a plane
Prove that the family of curves minimizing the distance
L =
∫ −1
−1
√1 + y′2dx,
in the plane are straight lines.
Solution
The Euler-Lagrange equation
0 =d
dxFy′ − Fy =
d
dx
(y′√
1 + y′2
).
Integration w.r.t. x yields
y′√1 + y′2
= γ = const.
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Problem 1: Minimum distance on a plane (cont.)
Substitute y′ = tan θ:
y′√1 + y′2
=sin θcos θ√
1 + sin2 θcos2 θ
=sin θcos θ√
sin2 + cos2 θcos2 θ
=sin θ
cos θ
√cos2
1= sin θ,
from where sin θ = γ. Substituting again yields
y′ = tan θ =sin θ
cos θ=
± sin θ√1 − sin2 θ
=±γ√1 − γ
= α = const,
from where
y = αx + β,
β = const. The solution describes a line in the plane; the exact values of α, β are
determined from the endpoint values.
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Problem 2: Constrained maximum
Find a curve y(x) with fixed endpoints y(±1) = 0 and perimeter
S =
∫ −1
−1
√1 + y′2dx,
which maximizes the area
A(y) =
∫ −1
−1
ydx.
Solution
Construct the Lagrangian
L(y) =
∫ −1
−1
(y + λ
√1 + y′2
)dx + const,
and denote
F (x, y, y′) = y + λ√
1 + y′2.
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Problem 2: Constrained maximum (cont.)
The Euler-Lagrange equation
0 =d
dxFy′ − Fy =
d
dx
(λy′√1 + y′2
)− 1.
Integration w.r.t. x yields
λy′√1 + y′2
= x − α,
where α = const. Substitute y′ = tan θ:
λy′√
1 + y′2= λ
sin θcos θ√
1 + sin2 θcos2 θ
= λsin θcos θ√
sin2 + cos2 θcos2 θ
= λsin θ
cos θ
√cos2
1= λ sin θ,
from where
sin θ =x − α
λ.
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Problem 2: Constrained maximum (cont.)
Substituting again
y′ = tan θ =sin θ
cos θ=
± sin θ√1 − sin2 θ
=±(x − α)
λ√
1 − (x−α)2
λ2
=±(x − α)√
λ2 − (x − α)2.
Integration w.r.t. x yields (table of integrals or Mathematica)
y =√
λ2 − (x − α)2 + β2,
or
(x − α)2 + (y − β)2 = λ2
where β = const. The latter equation describes a part of a circle with radius λ centered
at (α, β). The exact values of the constants are determined by demanding the endpoint
conditions and the perimeter constraint.
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