Calculus of Variations

Summer Term 2015

Lecture 5

Universität des Saarlandes

6. Mai 2015

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 1 / 25

 Purpose of Lesson

Purpose of Lesson:

To finish discussion about catenary of fixed length.

Consider possible pathologic cases, discuss rigid extremals and

give interpretation of the Lagrange multiplier λ

To solve the more general case of Dido’s problem with general

shape and parametrically described perimeter.

To discuss why does the Lagrange multiplier approach work.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 2 / 25

 Isoperimetric problems (continued) Catenary of fixed length (continued)

Example 5.1 (cf. Example 4.2)

From Example 4.2 we know that a solution of the catenary
problem with length constraint has the form
x 
y = Ccosh − λ,
C
and y satisfy the additional conditions

Z1 q  
0 2
1
y (−1) = y (1) = 2, L= 1 + (y ) dx = 2C sinh .
C
−1

Using Maple we calculate y for the natural catenary (without

length constraint), as well as for L = 2.05 L = 2.9 and L = 5. See
Worksheet 1 for the detailed calculation.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 3 / 25

 Isoperimetric problems (continued) Catenary of fixed length (continued)

All catenaries are valid, but one is natural

The red curve shows the natural catenary (without length

constraint), and the green, yellow and blue curves show other
catenaries with different lengths.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 4 / 25

 Isoperimetric problems (continued) Pathologies

Pathologies
Note that in both cases (”simple Dido’s problem” and ”catenary of fixed
length”)

the approach only works for certain ranges of L.

If L is too small, there is no physically possible solution

e.g., if wire length L < x1 − x0

e.g., if oxhide length L < x1 − x0

If L is too large in comparison to y1 = y (x1 ), the solution may have

our wire dragging on the ground.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 5 / 25

 Isoperimetric problems (continued) Rigid Extremals

A particular problem to watch for are rigid extremals

Rigid extremals are extremals that cannot be perturbed, and stiil

satisfy the constraint.

Example 5.2
For example
Z1 q √
G[y ] = 1 + (y 0 )2 dx = 2
0

with the boundary constraints y (0) = 0 and y (1) = 1.

The only possible y to satisfy this constraint is y (x) = x, so we

cannot perturb around this curve to find conditions for viable
extremals.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 6 / 25

 Isoperimetric problems (continued) Rigid Extremals

Rigid extremals cases have some similarities to maximization of a

function, where the constraints specify a single point:

Example 5.3
Maximize f (x, y ) = x + y , under the constraint that x 2 + y 2 = 0.

In Example 5.2, the constraint, and the end-point leave only one
choice of function, y (x) = x.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 7 / 25

 Isoperimetric problems (continued) Interpretation of λ

Interpretation of λ:
Consider again to finding extremals for

H[y ] = J[y ] + λG[y ], (5.1)

where we include G to meet an isoperimetric constraint G[y ] = L.

One way to think about λ is to think of (5.3) as trying to minimize

J[y ] and G[y ] − L.
1 λ is a tradeoff between J and G.
2 If λ is big, we give alot of weight to G.
3 If λ is small, then we give most weight to J.

So, λ might be thought of as how hard we have to ”pull” towards

the constraint in order to make it.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 8 / 25

 Isoperimetric problems (continued) Interpretation of λ

Interpretation of λ (cont.)

For example,

in the catenary problem, the size of λ is the amount we have to

shift the cosh function up or down to get the right length.

when λ = 0 we get the natural catenary,

i.e., in this case, we didn’t need to change anything to get the right
shape, so the constraint had no affect.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 9 / 25

 Isoperimetric problems (continued) Interpretation of λ

Interpretation of λ (cont.)
Write the problem (including the constant) as minimize
Z
H[y ] = F + λ(G − k )dx,

L
for the constant k = R , then
1dx
∂H
= λ,
∂k

we can also think of λ as the rate of change of the value of the

optimum with respect to k .

when λ = 0, the functional H has a stationary point

e.g., in the catenary problem this is a local minimum
corresponding to the natural catenary.
c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 10 / 25
 Isoperimetric problems (continued) Dido’s problem - traditional

Consider now the more general case of Dido’s problem:

a general shape,

Ω  

∂Ω  

without a coast,

so that the perimeter must be parametrically described.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 11 / 25
 Isoperimetric problems (continued) Dido’s problem - traditional

Dido’s problem is usual posed as follows:

Problem 5-1 (Dido’s problem -traditional)

To find the curve of length L which encloses the largest possible area,
i.e., maximize ZZ
Area = 1dxdy
Ω

subject to the constraint I

1ds = L
∂Ω

Of course Problem 5-1 is not yet in a convinient form.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 12 / 25

 Isoperimetric problems (continued) Green’s Theorem

Green’s Theorem converts an integral over the area Ω to a contour

integral around the boundary ∂Ω.

Green’s Theorem
ZZ   I
∂φ ∂ϕ
+ dxdy = φdy − ϕdx
∂x ∂y
Ω ∂Ω

for φ, ϕ : Ω → R such that φ, ϕ, φx and ϕy are continuous.

This converts an area integral over a region into a line integral around
the boundary.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 13 / 25

 Isoperimetric problems (continued) Geometric representation of area

The area of a region is given by

ZZ
Area = 1dxdy .
Ω

x y
In Green’s theorem choose φ = and ϕ = , so that we get
2 2
ZZ I
1
Area = 1dxdy = xdy − ydx
Ω 2 ∂Ω

Previous approach to Dido, was to use y = y (x), but in more

general case where the boundary must be closed, we can’t define
y as a function of x (or visa versa).

So, we write the boundary curve parametrically as (x(t), y (t)).

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 14 / 25

 Isoperimetric problems (continued) Dido’s Problem

If the boundary ∂Ω is represented parametrically by (x(t), y (t))

then
ZZ
Area = 1dxdy
Ω
I
1
= xdy − ydx
2
∂Ω
I
1
= (x ẏ − y ẋ) dt
2
∂Ω

So, now the problem is written in terms of

one independent variable = t

two dependent variables = (x, y ).

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 15 / 25

 Isoperimetric problems (continued) Isoperimetric constraint

Previously we wrote the isoperimetric constraint as

Z Zx 1 q
G[y ] = 1ds = 1 + (y 0 )2 dx = L
x0

Now we must also modify the constraint using

s
 2  2
ds dx dy
= +
dt dt dt

to get I I q
G[y ] = 1ds = ẋ 2 + ẏ 2 dt = L

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 16 / 25

 Isoperimetric problems (continued) Dido’s problem: Lagrange multiplier

Hence, we look for extremals of

I  
1
q
H[x, y ] = 2 2
(x ẏ − y ẋ) + λ ẋ + ẏ dt
2

p
So, H(t, x, y , ẋ, ẏ ) = 12 (x ẏ − y ẋ) + λ ẋ 2 + ẏ 2 , and there are two
dependent variables, with derivatives

∂H 1 ∂H 1 λẋ
= ẏ =− y+p
∂x 2 ∂ ẋ 2 ẋ 2 + ẏ 2

∂H 1 ∂H 1 λẏ
= − ẋ = x+p
∂y 2 ∂ ẏ 2 ẋ 2 + ẏ 2

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 17 / 25

 Isoperimetric problems (continued) Dido’s problem: EL equations

Leading to the 2 Euler-Lagrange equations

" #
d 1 λẋ 1
− y+p = ẏ
dt 2 2
ẋ + ẏ 2 2
" #
d 1 λẏ 1
x+p = − ẋ
dt 2 2
ẋ + ẏ 2 2

Integrate

1 λẋ 1
− y+p = y +A
2 2
ẋ + ẏ 2 2
1 λẏ 1
x+p =− x −B
2 2
ẋ + ẏ 2 2

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 18 / 25

 Isoperimetric problems (continued) Dido’s problem: solution

After simplification we get

λẋ
p =y +A
ẋ 2 + ẏ 2
λẏ
p = −x − B
ẋ 2 + ẏ 2

Now square the both equations, and add them to get

ẋ 2 + ẏ 2
λ2 = (y + A)2 + (x + B)2
ẋ 2 + ẏ 2

Or, more simply

(y + A)2 + (x + B)2 = λ2 ,

the equation is a circle with center (−B, −A) and radius |λ|.
c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 19 / 25
 Isoperimetric problems (continued) Dido’s problem: End-conditions

Remarks
Note, we can’t set value at end points arbitrarily.

If x(t0 ) = x(t1 ), and y (t0 ) = y (t1 ), then we get a closed curve,

obviously a circle.

These conditions only amount to setting one constant, λ.

On the other hand, if we specify different end-points, we are really

solving a problem such as the simplified problem considered in
Lecture 4.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 20 / 25

 Isoperimetric problems (continued) Why the Lagrange multiplier approach works here?

Why the Lagrange multiplier approach works here?

Consider the approximation of the functional

Zb n  
0
X ∆yi
J[y ] = F (x, y , y )dx ' F xi , yi , ∆x = F (y1 , . . . , yn )
∆xi
a i=1

(b−a)
where ∆x = n , and ∆yi = yi − yi−1 .

The problem of finding an extremal curve now becomes one of

finding stationary points of the function F (y1 , . . . , yn ).

We solve this by looking for

∂F
=0 for all i = 1, 2, . . . , n.
∂yi

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 21 / 25

 Isoperimetric problems (continued) Why the Lagrange multiplier approach works here?

The constraint can be likewise approximted to give

n  
X ∆yi
G[y ] ' G xi , yi , ∆x = G(y1 , . . . , yn ) = L.
∆xi
i=1

Under our usual conditions on J and G, the limit as n → ∞ gives

F (y1 , . . . , yn ) → J[y ]
G(y1 , . . . , yn ) → G[y ]

That is, the functions of the approximation y1 , . . . , yn converge to

the functionals of the curve y (x).

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 22 / 25

 Isoperimetric problems (continued) Why the Lagrange multiplier approach works here?

In the finite dimensional case the constraint is

G(y1 , . . . , yn ) − L = 0

and we use a standard Lagrange multiplier

H(y1 , . . . , yn , λ) = F (y1 , . . . , yn ) + λ [G(y1 , . . . , yn ) − L]

We solve this by looking for

∂H ∂H
= 0, ∀i = 1, 2, . . . , n, and = 0.
∂yi ∂λ

The last equation just gives you back your constraint.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 23 / 25

 Isoperimetric problems (continued) Why the Lagrange multiplier approach works here?

In our formulation of the isoperimetric problem we take

H[y ] = J[y ] + λG[y ]

and we also have

H(y1 , . . . , yn , λ) = F (y1 , . . . , yn ) + λ [G(y1 , . . . , yn ) − L] .

In the limit as n → ∞ we find that

H(y1 , . . . , yn , λ) → H[y ] − λL.

The EL equations for H[y ] − λL and H[y ] are the same, so they
have the same extremals.

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 24 / 25

 Isoperimetric problems (continued) Multiple constraints

Remarks about multiple constraints

We can also handle multiple constraints via multiple Lagrange
multipliers.
Rx1
For instance, if we wish to find extremals of J[y ] = F (x, y , y 0 )dx
x0
with the m constraints
Zx1
Gk [y ] = Gk (x, y , y 0 )dx = Lk
x0

we would look for extremals of

Zx1 Zx1 " m
X
#
0 0 0
H[y ] = H(x, y , y )dx = F (x, y , y ) + λk Gk (x, y , y ) dx
x0 x0 k =1

c Daria Apushkinskaya (UdS) Calculus of variations lecture 5 6. Mai 2015 25 / 25