The Calculus of Variations - Thompson Rivers …The Calculus of Variations All You Need to Know in...

The Calculus of VariationsAll You Need to Know in One Easy Lesson

Richard Taylor

[email protected]

TRU Math Seminar

Putnam 2006: Problem B5

For each continuous function f : [0, 1] → IR let

I(f) =

∫ 1

0

(x2f(x) − xf(x)2

)dx.

Find the maximum value of I(f) over all such functions f .



I(f) =

∫ 1

0

(x2f(x) − xf(x)2

)dx.


0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.4

0.8

x

f(x) = sin(πx)

I(f) = 4π2−π3−164π3 ≈ −0.061



I(f) =

∫ 1

0

(x2f(x) − xf(x)2

)dx.


0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.4

0.8

x

f(x) = sin(πx)

I(f) = 4π2−π3−164π3 ≈ −0.061

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.4

0.8

x

g(x) = x2

I(g) = 130 ≈ 0.033


Could try maximizing I(f) over a family of functions, e.g.

f(x) = cx (c is a parameter).




Then I(f) =

∫ 1

0

(x2(cx) − x(cx)2

)dx




Then I(f) =

∫ 1

0

(x2(cx) − x(cx)2

)dx = 1

4(c − c2).

So, restricted to this family, I(f) ismaximized with c = 1

2 .

0.0 0.2 0.4 0.6 0.8 1.0

0.00

0.02

0.04

0.06

c

I(f)

A Solution:

Claim: The maximum of I(f) is achieved for f(x) = x/2.

A Solution:


Proof. Any f can be written f(x) = x2 + g(x).

A Solution:


Proof. Any f can be written f(x) = x2 + g(x). Then

I(f) =

∫ 1

0

(x2[x2 + g(x)] − x[x2 + g(x)]2

)dx.

A Solution:



I(f) =

∫ 1

0

(x2[x2 + g(x)] − x[x2 + g(x)]2

)dx.

=

∫ 1

0

14x3 dx −

∫ 1

0xg(x)2 dx

A Solution:



I(f) =

∫ 1

0

(x2[x2 + g(x)] − x[x2 + g(x)]2

)dx.

=1

16−

∫ 1

0xg(x)2 dx

︸︷︷︸

≥0

A Solution:



I(f) =

∫ 1

0

(x2[x2 + g(x)] − x[x2 + g(x)]2

)dx.

=1

16−

∫ 1

0xg(x)2 dx

︸︷︷︸

≥0

So the maximum of I(f), achieved with g(x) = 0, is 116 .

Some Multivariable Calculus

That approach works only if you “guess” the right family offunctions (i.e., containing the optimal f(x)) . . . how to tell?



A more reliable method uses ideas from multivariable calculus:

Definition. Given a function f : IRn→ IR, the directional derivative at x,

in the direction of a unit vector u, is

Duf(x) = limh→0

f(x + hu) − f(x)

h= d

dhf(x + hu)∣∣∣h=0



A more reliable method uses ideas from multivariable calculus:

Definition. Given a function f : IRn→ IR, the directional derivative at x,

in the direction of a unit vector u, is

Duf(x) = limh→0

f(x + hu) − f(x)

h= d

dhf(x + hu)∣∣∣h=0

Duf(x) gives the rate of change of f(x) as we move in thedirection u at unit speed. (e.g. rate of change of temperaturealong a given direction).


For f : IR2→ IR, the graph of y = f(x) is a surface, and Duf(x)

is the slope of this surface along the direction u:

xy

z = f(x, y)

x




xy

z = f(x, y)

xu




xy

z = f(x, y)

xu

slope = Duf(x)


The directional derivative helps find extreme values of f(x):


The directional derivative helps find extreme values of f(x):Theorem. If f : IR → IRn has a local extremum at x, then Duf(x) = 0 forevery direction u.

xy

z = f(x, y)

Some Infinite-Dimensional Calculus

Problem: Find f to maximize I(f) =

∫ 1

0

(x2f(x) − xf(x)2

)dx.



∫ 1

0

(x2f(x) − xf(x)2

)dx.

I(f) is a real-valued functionon the space C of continuousfunctions on the interval [0, 1]. f

y

C

y = I(f)



∫ 1

0

(x2f(x) − xf(x)2

)dx.


y

C

y = I(f)

C is an infinite-dimensional vector space (i.e. equipped withaddition and scalar multiplication, hence a notion of “direction”).



∫ 1

0

(x2f(x) − xf(x)2

)dx.


y

C

y = I(f)

C is an infinite-dimensional vector space (i.e. equipped withaddition and scalar multiplication, hence a notion of “direction”).

I(f) depends continuously (even differentiably) on f ∈ C.


Directional derivative of I(f) on the function space C:

Pick a function f ∈ C

f

C



Pick a function f ∈ C and a “direction” — say, the direction ofthe function g ∈ C.

f λg

f + λgC



Pick a function f ∈ C and a “direction” — say, the direction ofthe function g ∈ C. How quickly does I(f) change as we move fin the direction of g?

f λg

f + λgC



Pick a function f ∈ C and a “direction” — say, the direction ofthe function g ∈ C. How quickly does I(f) change as we move fin the direction of g?

f λg

f + λgC

The rate of change of I(f) in the g-direction is the directional orGâteaux derivative :

DgI(f) = limλ→0

I(f + λg) − I(f)

λ= d

dλI(f + λg)∣∣∣λ=0

Application to the Putnam Problem:

For I(f) =

∫ 1

0

(x2f(x) − xf(x)2

)dx this is an easy calculation:


For I(f) =

∫ 1

0

(x2f(x) − xf(x)2


DgI(f) = ddλI(f + λg)

∣∣∣λ=0

=d

dλ

[ ∫ 1

0x2

(f(x) + λg(x)

)− x

(f(x) + λg(x)

)2dx

]

λ=0


For I(f) =

∫ 1

0

(x2f(x) − xf(x)2



∣∣∣λ=0

=d

dλ

[ ∫ 1

0x2

(f(x) + λg(x)

)− x

(f(x) + λg(x)

)2dx

]

λ=0

= d

dλ

»

R

1

0x2

`

f(x) − xf(x)2´

dx + λR

1

0

`

x2g(x) − 2xf(x)g(x)´

dx − λ2R

1

0xg(x)2 dx

–

λ=0


For I(f) =

∫ 1

0

(x2f(x) − xf(x)2



∣∣∣λ=0

=d

dλ

[ ∫ 1

0x2

(f(x) + λg(x)

)− x

(f(x) + λg(x)

)2dx

]

λ=0

= d

dλ

»

R

1

0x2

`

f(x) − xf(x)2´

dx + λR

1

0

`


dx − λ2R

1

0xg(x)2 dx

–

λ=0

=

[ ∫ 1

0

(x2g(x) − 2xf(x)g(x)

)− 2λ

∫ 1

0

xg(x)2 dx

]

λ=0


For I(f) =

∫ 1

0

(x2f(x) − xf(x)2



∣∣∣λ=0

=d

dλ

[ ∫ 1

0x2

(f(x) + λg(x)

)− x

(f(x) + λg(x)

)2dx

]

λ=0

= d

dλ

»

R

1

0x2

`

f(x) − xf(x)2´

dx + λR

1

0

`


dx − λ2R

1

0xg(x)2 dx

–

λ=0

=

[ ∫ 1

0

(x2g(x) − 2xf(x)g(x)

)− 2λ

∫ 1

0

xg(x)2 dx

]

λ=0

=

∫ 1

0

(x2

− 2xf(x))g(x) dx.


So the directional derivative of I(f) in the “direction” of thefunction g is:

DgI(f) =

∫ 1

0

(x2

− 2xf(x))g(x) dx



DgI(f) =

∫ 1

0

(x2

− 2xf(x))g(x) dx

If this derivative is to be zero for every direction g, then

x2− 2xf(x) = 0



DgI(f) =

∫ 1

0

(x2

− 2xf(x))g(x) dx

If this derivative is to be zero for every direction g, then

x2− 2xf(x) = 0 =⇒ f(x) =

x

2

PROCLAMATION:

“Since it is known with certainty that there is scarcely anythingwhich more greatly excites noble and ingenious spirits to laborswhich lead to the increase of knowledge than to propose difficultand at the same time useful problems through the solution ofwhich, as by no other means, they may attain to fame and buildfor themselves eternal monuments among prosperity; so Ishould expect to deserve the thanks of the mathematical world if. . . I should bring before the leading analysts of this age someproblem up which as upon a touchstone they could test theirmethods, exert their powers, and, in case they brought anythingto light, could communicate with us in order that everyone mightpublicly receive his deserved praise from us.”

— Johann Bernoulli, Acta Eruditorum, June 1696

Brachistochrone Problem:

Problem: Bead falls from rest along frictionless wire. Find shapeof wire to minimize transit time.

t = 0

t =?


Problem: Bead falls from rest along frictionless wire. Find shapeof wire to minimize transit time.

t = 0

t =?

x

y

y = y(x)

(x0, y0)


x

y

y = y(x)

(x0, y0)

dx

ds

arc length: ds =√

1 + y′(x)2 dx

speed: v =√

−2gy(x)


x

y

y = y(x)

(x0, y0)

dx

ds

arc length: ds =√

1 + y′(x)2 dx

speed: v =√

−2gy(x)

Minimize total transit time:

T (y) =

∫ds

v


x

y

y = y(x)

(x0, y0)

dx

ds

arc length: ds =√

1 + y′(x)2 dx

speed: v =√

−2gy(x)

Minimize total transit time:

T (y) =

∫ds

v=

∫ x0

0

√

1 + y′(x)2√

−2gy(x)dx

Hanging Cable Problem:

Problem: Flexible cable is supported at endpoints. Findequilibrium shape of cable.



y

x

y = y(x)



y

x

y = y(x)

ds

dx

x0

arc length: ds =√

1 + y′(x)2 dx

potential energy: dP = −y(x) ds = −y(x)√

1 + y′(x)2 dx



y

x

y = y(x)

ds

dx

x0

arc length: ds =√

1 + y′(x)2 dx


1 + y′(x)2 dx

Minimize total potential energy:

P (y) =

∫

dP



y

x

y = y(x)

ds

dx

x0

arc length: ds =√

1 + y′(x)2 dx


1 + y′(x)2 dx

Minimize total potential energy:

P (y) =

∫

dP = −

∫ x0

0y(x)

√

1 + y′(x)2 dx

Navigation Problem:

Problem: Boat cross river at constant speed relative to water.Find route to minimize transit time from A to B.

B

A

Navigation Problem:


y = y(x)

x

y

v(x)

(x0, y0)

Navigation Problem:


y = y(x)

x

y

v(x)

(x0, y0)

Total transit time along route y = y(x):

T (y) =

∫ x0

0

[

α(x)2√

1 + α(x)2y′(x)2 −(α(x)2v(x)y′(x)

]

dx

where α(x) =(1 − v(x)2

)−1/2

Profit Maximization Problem:

Problem: A company uses a resource at a rate R(t) to producewidgets at a rate y

(R(t)

). It sells the widgets for p dollars each.



(R(t)


The resource costs w per unit; a cost c(R′(t)

)is associated with

adjusting the use of the resource.



(R(t)


The resource costs w per unit; a cost c(R′(t)

)is associated with

adjusting the use of the resource.

The company wants to find R(t) to maximize the total profit oversome time interval:

P (R) =

∫ b

a

[

p · y(R(t)

)

︸︷︷︸

revenue

− w · R(t)︸︷︷︸

cost of resource

− c(R′(t)

)

︸︷︷︸

adjustment cost

]

dt

(while perhaps also satisfying various constraints or targets).

A Collection of Optimization Problems:

brachistochrone: T (y) =

∫ x0

0

√

1 + y′(x)2√

−2gy(x)dx

hanging cable: P (y) = −

∫ x0

0y(x)

√

1 + y′(x)2 dx

river navigation: T (y) =

∫ x0

0

[

α(x)2√

1 + α(x)2y′(x)2

−

(α(x)2v(x)y′(x)

]

dx

profit: P (R) =

∫ b

apy

(R(t)

)− wR(t) − c

(R′(t)

)dt

A Collection of Optimization Problems:

brachistochrone: T (y) =

∫ x0

0

√

1 + y′(x)2√

−2gy(x)dx

hanging cable: P (y) = −

∫ x0

0y(x)

√

1 + y′(x)2 dx

river navigation: T (y) =

∫ x0

0

[

α(x)2√

1 + α(x)2y′(x)2

−

(α(x)2v(x)y′(x)

]

dx

profit: P (R) =

∫ b

apy

(R(t)

)− wR(t) − c

(R′(t)

)dt

In general: I(y) =

∫ b

aF

(x, y(x), y′(x)

)dx

General Optimization Problem:

Problem: Find a function y(x) than gives an extreme value of

I(y) =

∫ b

aF

(x, y(x), y′(x)

)dx



I(y) =

∫ b

aF

(x, y(x), y′(x)

)dx

Solution: Consider the directional derivative of I(y) in the“direction” of the function g:

DgI(y) =d

dλI(y + λg)

∣∣∣∣λ=0



I(y) =

∫ b

aF

(x, y(x), y′(x)

)dx

Solution: Consider the directional derivative of I(y) in the“direction” of the function g:

DgI(y) =d

dλI(y + λg)

∣∣∣∣λ=0

=d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0


DgI(y) =d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0


DgI(y) =d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

d

dλF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0


DgI(y) =d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

d

dλF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

∂F

∂y

(x, y + λg, y′ + λg′

)g +

∂F

∂y′(x, y + λg, y′ + λg′

)g′ dx

∣∣∣∣λ=0


DgI(y) =d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

d

dλF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

∂F

∂y

(x, y + λg, y′ + λg′

)g +

∂F

∂y′(x, y + λg, y′ + λg′

)g′ dx

∣∣∣∣λ=0

=

∫ b

a

∂F

∂y

(x, y, y′

)g +

∂F

∂y′(x, y, y′

)g′ dx


DgI(y) =d

dλ

∫ b

aF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

d

dλF

(x, y + λg, y′ + λg′

)dx

∣∣∣∣λ=0

=

∫ b

a

∂F

∂y

(x, y + λg, y′ + λg′

)g +

∂F

∂y′(x, y + λg, y′ + λg′

)g′ dx

∣∣∣∣λ=0

=

∫ b

a

∂F

∂y

(x, y, y′

)g +

∂F

∂y′(x, y, y′

)g′ dx

=

∫ b

a

∂F

∂y

(x, y, y′

)g dx+

∂F

∂y′(x, y, y′

)g

∣∣∣∣

b

a

−

∫ b

a

[d

dx

∂F

∂y′(x, y, y′

)]

g dx


So finally. . .

DgI(y) =

∫ b

a

[∂F

∂y

(x, y, y′

)−

d

dx

∂F

∂y′(x, y, y′

)]

g(x) dx


So finally. . .

DgI(y) =

∫ b

a

[∂F

∂y

(x, y, y′

)−

d

dx

∂F

∂y′(x, y, y′

)]

g(x) dx

Setting DgI(y) = 0 for every direction g, we get theEuler-Lagrange equation (ca. 1750):

∂F

∂y

(x, y, y′

)−

d

dx

∂F

∂y′(x, y, y′

)= 0


So finally. . .

DgI(y) =

∫ b

a

[∂F

∂y

(x, y, y′

)−

d

dx

∂F

∂y′(x, y, y′

)]

g(x) dx

Setting DgI(y) = 0 for every direction g, we get theEuler-Lagrange equation (ca. 1750):

∂F

∂y

(x, y, y′

)−

d

dx

∂F

∂y′(x, y, y′

)= 0

. . . a differential equation for the unknown function y(x).

Back to the Putnam Problem:

Problem: Find y to maximize I(y) =

∫ 1

0

(x2y(x) − xy(x)2

)dx.



∫ 1

0

(x2y(x) − xy(x)2

)dx.

Solution: We have I(y) =

∫ b

aF

(x, y(x), y′(x)

)dx with:

F (x, y, y′) = x2y − xy2



∫ 1

0

(x2y(x) − xy(x)2

)dx.


∫ b

aF

(x, y(x), y′(x)

)dx with:

F (x, y, y′) = x2y − xy2

Apply the Euler-Lagrange equation:

∂F

∂y−

d

dx

∂F

∂y′= 0



∫ 1

0

(x2y(x) − xy(x)2

)dx.


∫ b

aF

(x, y(x), y′(x)

)dx with:

F (x, y, y′) = x2y − xy2


∂F

∂y−

d

dx

∂F

∂y′= 0 =⇒ x2

− 2xy = 0



∫ 1

0

(x2y(x) − xy(x)2

)dx.


∫ b

aF

(x, y(x), y′(x)

)dx with:

F (x, y, y′) = x2y − xy2


∂F

∂y−

d

dx

∂F

∂y′= 0 =⇒ x2

− 2xy = 0 =⇒ y =x

2

The Calculus of Variations - Thompson Rivers …The Calculus of Variations All You Need to Know in...

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Transcript of The Calculus of Variations - Thompson Rivers …The Calculus of Variations All You Need to Know in...