calculus of variations summer term 2016 · = const 2xu (1 +u2)2 = 0 y0+u = 0 (8:1) c daria...

Calculus of VariationsSummer Term 2016

Lecture 8

Universität des Saarlandes

31. Mai 2016

c© Daria Apushkinskaya (UdS) Calculus of variations lecture 8 31. Mai 2016 1 / 24

Purpose of Lesson

Purpose of Lesson:

To discuss Newton’s aerodynamic problem.

To discuss necessary and sufficient conditions for extrema

To introduce the Legendre condition


Purpose of Lesson Newton’s aerodynamic problem

Newton’s aerodynamic problem

Problem 8-1 (Newton’s aerodynamic problem)Finding the profile of a body that gives the minimal (aerodynamicor hydrodynamic) resistance to the motion is one of the firstproblems in the theory of the calculus of variations.

Consider finding the optimal shape of a rocket’s nose cone inorder that it creates the least resistance when passing through air.

In 1685 Sir Isaac Newton studied this problem and presented avery simple model to compute the resistance of a body movingthrough an inviscid and incompressible medium.



Sir Isaac Newton (1642-1727)

In his words "..If in a rare medium, consisting of equal particles freelydisposed at equal distances from each other, a globe and a cylinderdescribed on equal diameter move with equal velocities in the directionof the axis of the cylinder, (then) the resistance of the globe will be halfas great as that of the cylinder. ...I reckon that this proposition will benot without application in the building of ships.."

(I.Newton, Principia Mathematica)



Problem 8-1 (Newton’s aerodynamic problem)Assumptions:

1 As the rocket may rotate along its length, the nose cone must becircularly symmetric, and so we reduce the problem to one ofdetermining the optimal profile of the nose cone.

2 The rocket’s nose cone must have radius R at its base, and lengthL, and its shape should be convex.

3 Air is thin, and composed of perfectly elastic particles.

4 Particles will bounce off the nose cone with equal speed, and equalangle of reflection and incidence.

5 We ingnore tangential friction.

6 We ignore „non-Newtonian“ affects such as those fromcompression of the air.

These assumptions are realistic for high-altitude, supersonic flight.




!

! RL

!

! L/2R! > 5 : 1

Variational Methods & Optimal Control: lecture 06 – p.17/32

θ

y

R

L

θ

θ

v

v

x


Force= ma

! m=! a=

a= v! s= 2vsin2θ.

2vm= 1,

Force= sin2θ=1

1+ cot2θ=

11+ y"2

.


!= 1/(1+ y"2)

!! x

2πxdx.

!

F{y} =! R

0

x1+ y"2

dx,

! y(0) = L y(R) = 0y" # 0 y"" $ 0





Force = ma

m = mass

a = acceleration = change in velocity

a = 2v sin2 θ

Scale constants so that2vm = 1

and thenForce = sin2 θ =

11 + cot2 θ

=1

1 + y ′2




Previous calculation gives force per particle

=1

1 + y ′2

Need to integrate over surface area

Surface area at radius x is

2πxdx

Scaling to remove irrelevant constants, the functional describingthe resistance

J[y ] =

R∫0

x1 + y ′2

dx ,

subject to y(0) = L and y(R) = 0 and y ′ 6 0 and y ′′ > 0.




Problem 8-1 (Newton’s aerodynamic problem)Find extremal of ”air resistance”

J[y ] =

R∫0

x1 + y ′2

dx ,

subject to y(0) = L and y(R) = 0 and y ′ 6 0 and y ′′ > 0.



The Euler-Lagrange equation is

∂F∂y− d

dx

(∂F∂y ′

)= − d

dx

(2xy ′

(1 + y ′2)2

)= 0.

Rearranging we get

2xy ′ = C(

1 + y ′2)2

which isn’t much fun to solve directly.

Alternative: define a new variable u, and constrain it

u = −y ′

Add Lagrange multiplier λ(x) to the functional

H[y ,u, λ] =

∫ R

0

(x

1 + u2 + λ(y ′ + u)

)dx .



Now we solve this new problem with three dependent variables(y ,u, λ) of x . So, we expect three Euler-Lagrange equations.

The Euler-Lagrange equations

∂H∂y− d

dx

(∂H∂y ′

)= 0

∂H∂u− d

dx

(∂H∂u′

)= 0

∂H∂λ− d

dx

(∂H∂λ′

)= 0

gives the DEsλ = const

λ− 2xu(1 + u2)2 = 0

y ′ + u = 0

(8.1)




If λ = 0, then for x > 0 we get u = 0, and hence y = const .

If λ 6= 0, then the second equation in (8.1) implies

x(u) =Cu

(1 + u2)2 = C(

1u

+ 2u + u3)

for C constant.




From the last equation in (8.1) (which we insisted on at the start),we get

dydx

= −u.

Now note that from the chain rule

dydu

=dydx

dxdu

= −udxdu

= C(

1u− 2u − 3u3

)which we can integrate with respect to u to get

y(u) = const − C(− ln u + u2 +

34

u4).




Thus, we have

x(u) = C(

1u

+ 2u + u3)

=Cu

(1 + u2

)2> 0 for all u


34

u4)

End-point conditions take the form

y(u1) = L y(u2) = 0

x(u1) = x1 x(u2) = R

but we don’t know x1, u1 or u2.





34

u4).

At u1 we have y(u1) = L. For convenience we write

y(u) = L− C(−A− ln u + u2 +

34

u4).

So, at u1 we get

L = L− C(− ln u1 − A + u2

1 +34

u41

)0 = −C

(− ln u1 − A + u2

1 +34

u41

)A = − ln u1 + u2

1 +34

u41



Thus, we get

y(u) = L− C(−A− ln u + u2 +

34

u4)

x(u) =Cu

(1 + u2)2

Now at u2 we have x(u2) = R and y(u2) = 0. So

L = C(−A− ln u2 + u2

2 +34

u42

)R =

Cu2

(1 + u2

2

)2(8.2)

Dividing the first equation in (8.2) by the second one, we get.....



LR

= u2

(−A− ln u2 + u2

2 +34

u42

)(1 + u2

2)−2 (8.3)

The function on the RHS of (8.3) is increasing. So, we can solve(8.3) numerically and obtain a value for u2.

We can find C using x(u2) = R.

R =Cu2

(1 + u2

2

)2

C =u2R

(1 + u22)2

All we need to know now is u1, which gives us A and x(u1) whichgives us u2, which gives us C.




ddx

∂ f∂y!

" ∂ f∂y

=ddx

2xy!

(1+ y!2)2= 0

c

2xy!

(1+ y!2)2= c.

2xy! = c(1+ y!2)2


x(u) = c!1u

+2u+u3"

=cu(1+u2)2

y(u) = L" c!" lnu" 7

4+u2+

34u4

"

dydx

=dydu

dudx

=dydu

/dxdu

= "u


!

0 0.5 10

0.5

1

1.5

2

0 0.5 10

0.2

0.4

0.6

0.8

1

1.2


Variational Methods & Optimal Control: lecture 06 – p.24/32c© Daria Apushkinskaya (UdS) Calculus of variations lecture 8 31. Mai 2016 18 / 24


Alternatives: cylinder (R = L = 1)

y ′ = 0

J[y ] =

1∫0

xdx =12

y! = 0

F{y} =! R

0xdx

=R2

2

R= 1

F = 1/2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.500


y! = "L/RF{y} =

! R

0

x1+(L/R)2

dx

=R2

2(1+(L/R)2)

R= L= 1

F = 1/4

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


R= L= 1

x2+ y2 = 1y! = "x/y

= "x/!1" x2

F{y} =! 1

0

x1+ y!2

dx

=! 1

0x(1" x2)dx

=14

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


a

y! =

"0 x# a"L/(R"a) x$ a

F{y} =! a

0xdx+

! R

a

x1+ y!2

dx

=a2L2+R2(R"a)22(L2+(R"a)2)

a

a=(L2+2R2)"L%L2+4R2

2R0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

F = 0.191




Alternatives: sphere (R = L = 1)

x2 + y2 = 1

y ′ = −xy

= − x√1− x2

J[y ] =

1∫0

x1 + y ′2

dx =

1∫0

x(1− x2)dx =14

y! = 0

F{y} =! R

0xdx

=R2

2

R= 1

F = 1/2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.500


y! = "L/RF{y} =

! R

0

x1+(L/R)2

dx

=R2

2(1+(L/R)2)

R= L= 1

F = 1/4

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


R= L= 1

x2+ y2 = 1y! = "x/y

= "x/!1" x2

F{y} =! 1

0

x1+ y!2

dx

=! 1

0x(1" x2)dx

=14

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


a

y! =

"0 x# a"L/(R"a) x$ a

F{y} =! a

0xdx+

! R

a

x1+ y!2

dx

=a2L2+R2(R"a)22(L2+(R"a)2)

a

a=(L2+2R2)"L%L2+4R2

2R0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

F = 0.191




Alternatives: cone (R = L = 1)

y ′ = − LR

= −1

J[y ] =

1∫0

x2

dx =14

y! = 0

F{y} =! R

0xdx

=R2

2

R= 1

F = 1/2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.500


y! = "L/RF{y} =

! R

0

x1+(L/R)2

dx

=R2

2(1+(L/R)2)

R= L= 1

F = 1/4

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


R= L= 1

x2+ y2 = 1y! = "x/y

= "x/!1" x2

F{y} =! 1

0

x1+ y!2

dx

=! 1

0x(1" x2)dx

=14

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


a

y! =

"0 x# a"L/(R"a) x$ a

F{y} =! a

0xdx+

! R

a

x1+ y!2

dx

=a2L2+R2(R"a)22(L2+(R"a)2)

a

a=(L2+2R2)"L%L2+4R2

2R0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

F = 0.191




Alternatives: frustrum of cone (R = L = 1, corner at a)

y ′ =

0, x 6 a

− 11− a

, x > a, a =

3−√52

J[y ] =

a∫0

xdx +

1∫a

x1 + y ′2

dx =a2 + (1− a)2

2(1 + (1− a)2)

y! = 0

F{y} =! R

0xdx

=R2

2

R= 1

F = 1/2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.500


y! = "L/RF{y} =

! R

0

x1+(L/R)2

dx

=R2

2(1+(L/R)2)

R= L= 1

F = 1/4

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


R= L= 1

x2+ y2 = 1y! = "x/y

= "x/!1" x2

F{y} =! 1

0

x1+ y!2

dx

=! 1

0x(1" x2)dx

=14

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.250


a

y! =

"0 x# a"L/(R"a) x$ a

F{y} =! a

0xdx+

! R

a

x1+ y!2

dx

=a2L2+R2(R"a)22(L2+(R"a)2)

a

a=(L2+2R2)"L%L2+4R2

2R0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

F = 0.191




Alternatives: optimal (R = L = 1)

Optimal profile: J computed numerically

F

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.187


http://en.wikipedia.org/wiki/Nose_cone_design

http://www.info-central.org/?article=125

http://mcfisher.0catch.com/other/mach1/mach1.htm

http://www.if.sc.usp.br/~projetosulfos/artigos/NoseCone_EQN2.PDF

F

0 0.5 10

0.2

0.4

0.6

0.8

1

F1 = 0.187, F2 = 0.240


!

!!

!

"

"

!





Bullets

F

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

F = 0.187


http://en.wikipedia.org/wiki/Nose_cone_design

http://www.info-central.org/?article=125

http://mcfisher.0catch.com/other/mach1/mach1.htm

http://www.if.sc.usp.br/~projetosulfos/artigos/NoseCone_EQN2.PDF

F

0 0.5 10

0.2

0.4

0.6

0.8

1

F1 = 0.187, F2 = 0.240


!

!!

!

"

"

!




calculus of variations summer term 2016 · = const 2xu (1 +u2)2 = 0 y0+u = 0 (8:1) c daria...

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