Solutions Manual Applied Mathematics, 3rd Edition

3.3. THE SIMPLEST PROBLEM 45

2. (a) The Euler equation is

Ly − d Ly = d (2y /x3) = 0
dx dx

Thus y(x) = Ax4 + B

(b) The Euler equation is y − y = ex

The general solution is

y(x) = Aex + Be−x + x ex
2

3. The Euler equation is

Ly − d Ly = fy 1 + (y )2 − d f y = 0.
dx dx 1 + (y )2

Taking the total derivative and then multiplying by 1 + (y )2 gives

fy − y fx − f y + f (y )2y = 0
1 + (y )2

This reduces to fy
1 + (y
fy −y fx − )2 = 0

4. The Euler equation is − d Ly =0
or dx

y =c
x 1 + (y )2

Solving for y (take the positive square root since, by the boundary con-
ditions, we want y > 0), separating variables, and integrating yields

y(x) = 1 c2x2 dx + k
− c2x2

Then make the substitution u = 1 − c2x2 to perform the integration. We

obtain 1
c
y(x) = 1 − c2x2 + k

Applying the boundary conditions to determine the constants finally leads

to

y(x) = − 5 − x2 + 2

which is an arc of a circle.

46 CHAPTER 3. CALCULUS OF VARIATIONS

5. Notice that, by expanding, combining terms, and using the Euler equation,

b [2pY h + 2qY h + ph 2 + qh2]dx

J(Y + h) − J(Y ) =

a

b

> [2pY h + 2qY h]dx

a

b

= 2pY h|ab − [(2pY ) h − 2qY h]dx

a

= 0.

The last two lines follow from integration by parts and the fact that Y
satisfies the Euler equation

(pY ) − qY = 0.

6. Let h ∈ C2. Then

bb

δJ(y, h) = K(s, t)[y(s)h(t) + h(s)y(t)]ds dt

aa

bb

+2 y(t)h(t)dt − 2 h(t)f (t)dt

aa

Now, using the symmetry of K and interchanging the order of integration
allows us to rewrite the first integral as

bb

2 K(s, t)y(s)h(t)dsdt

aa

Then

b b

δJ(y, h) = 2 K(s, t)y(s)ds + y(t) − f (t) h(t)dt

a a

Thus b

K(s, t)y(s)ds + y(t) − f (t) = 0

a

This is a Fredholm integral equation (see Chapter 4) for y.

7. The Euler equation is −((1 + x)y ) = 0
or y = c1/(1 + x)
Integrating again y(x) = c1 ln(1 + x) + c2

3.3. THE SIMPLEST PROBLEM 47

If y(0) = 0, y(1) = 1 then we get

y(x) = ln(1 + x)
ln 2

If the boundary condition at x = 1 is changed to y (1) = 0, then the
extremal is y(x) ≡ 0.

8. In each case we minimize the functional T (y) given in Example 3.18 on
page 174.

(a) When n = kx the Lagrangian is independent of y, and therefore the
Euler equation reduces to

x 1 + y 2 = C.

Separating variables and integrating gives

y= C1 − x2 dx + C2 .
x2

The right hand side can be integrated using a trigonometric substi-
tution.

(b) The integrand is independent of x.

(c) The integrand is independent of x.

(d) This problem is similar to the brachistochrone problem (see Example
3.17).

9. Observe that

Lt − d (L − y Ly ) = Lt − dL +y dLy +y Ly
dt dt dt

= Lt − Lt − Lyy − Ly y +y dLy + y Ly
dt

= −y (Ly − d Ly )
dt

10. The minimal surface of revolution is found by minimizing

b

J(y) = 2πy 1 + (y )2dx

a

Because the integrand does not depend explicitly on x, a first integral is

L − y Ly = c

Upon expanding and simplifying, this equation leads to

dy = k2y2 − 1
dx

48 CHAPTER 3. CALCULUS OF VARIATIONS

Now separate variables and integrate while using the fact that

d cosh−1 u = √1 .
du u2 − 1

11. The Euler equation becomes
x2y + 2xy − y = 0

which is a Cauchy-Euler equation. Its solution is

√√

y(x) = Ax(−1+ 5)/2 + Bx(−1− 5)/2

(recall that a Cauchy-Euler equation can be solved by trying power func-
tions, y = xm for some m).

12. To find Euler’s equation, use the fact that L is independent of x and
therefore L − y Ly = C.

13. Using the Euler equation it is straight forward to see that the extremal

is Y = 0, giving the value J(Y ) = 4. Take, for example, y = sin πx.

Then J(y) > 4. Hence, Y = 0 does not give a local maximum. Because

the extremals are only necessary conditions, there is no guarantee that

Y = 0 provides a minimum either. Note also that, for any y, we have
1
J(y) = 0 [y2 + (y − 2)2] dx ≥ 0.

14. Substitution of r into the integral gives the variational problem

T

E(y) = 2 e−βt αy − y dt → max.

0

The Euler equation is

√α1 y−−βy − d √αy1− y = 0.
dt

This simplifies to

y + (α + 2β − 2)y + 2α(1 − β)y = 0,

which is a linear equation with constant coefficients.

3.4 Generalizations

Exercises, page 184
1. (a) The Euler equations are
8y1 − y2 = 0, 2y2 − y1 = 0.

3.4. GENERALIZATIONS 49

Eliminating y2 gives y1(4) − 16y1 = 0.

The characteristic equation is m4 − 16 = 0, which has roots m =

±2, ±2i. Therefore the solution is

y1(x) = ae2x + be−2x + c cos 2x + d sin 2x,

where a, b, c, d are arbitrary constants. Then y2 = 0.5y1 , and the four
constants a, b, c, d can be computed from the boundary conditions.

(b) The Euler equation is y(4) = 0,

and thus the extremals are

y(x) = a + bx + cx2 + dx3.

The four constants a, b, c, d can be computed from the boundary con-
ditions.
(c) The Euler equation is y(4) − 2y + y = 0.
(d) The Euler equation is y(4) = 0.
(e) The Euler equation is y(4) − y(3) − y − y = 0.

2. The Euler equation for J(y) = L(x, y, y , y )dx is

Ly − (Ly ) + (Ly ) = 0

If Ly = 0 then clearly

Ly − (Ly ) = const.

If Lx = 0, then expand all the derivatives and use the Euler equation to

show d
dx
(L − y (Ly − (Ly )−y Ly )) = 0

3. The Euler equation is

LM − d LM = 2a(aM − M − b) + d 2(aM − M ) = 0,
dx dx

or
M − a2M = −ab.

The general solution is

M (t) = Aeat + Be−at + b .
a

The left boundary condition is M (0) = M0; the right boundary condition
is the natural boundary condition LM = 0 at t = T , or

aM (T ) − M (T ) = b.

50 CHAPTER 3. CALCULUS OF VARIATIONS

4. The two Euler equations (expanded out) are

Ly y y + Ly z z =0
Lz y y + Ly z z =0

It is given that the determinant of the coefficient matrix of this system is
nonzero. Therefore the only solution is y = z = 0, which gives linear
functions for y and z.

5. (a) The natural boundary condition is y (1) + y(1) = 0. The extremals
are
y(x) = aex + be−x

. The boundary conditions force a + b = 1 and a = 0, so y(x) = e−x.

(b) The Euler equation is

y + 2y + y = 0,

giving extremals

y(x) = ae−x + bxe−x.

The boundary conditions are y(0) = 1 and y (3) = 0.
(c) The Euler equation is

x2y + 2xy + 1 y = 0,
4

which is a Cauchy-Euler equation. Assuming solutions of the form
y = tm gives the characteristic equation

m(m − 1) + 2m + 1 = 0,
4

which has a real double root m = − 1 . Thus the general solution is
2

y(x) = a √1x + b √1x ln x.

The given boundary condition is y(1) = 1; the natural boundary
condition at x = e is y (e) = 0. One finds from these two conditions
that a = b = 1.

(d) The Euler equation is

y − 2y = −1.

(e) The extremals are

y(x) = ax + b.

The boundary conditions are y(0) = 1 and y (1) + y(1) = 0.

3.5. THE CANONICAL FORMALISM 51

6. The natural boundary condition is
Ly (b, y(b), y (b)) + G (y(b)) = 0.

7. The Lagrangian is L = 1 − k2 + y 2 − ky , where 0 < k < 1. It does not
depend on x, and therefore a first integral is

y + k = C.
1 − k2 − y 2

Solving gives y(x) = Ax + B. Now y(0) = 0 forces B = 0 and y(x) = Ax.
Now apply the natural boundary condition Ly = 0 at x = b. We get

y +k= √ A + k = 0.

1 − k2 − y 2 1 − k2 − A

This gives

A=− k2(1 − k2) .
1 + k2

8. The natural boundary condition is (see Problem 6)

9y (2) + 9y(2) = 2.

3.5 The Canonical Formalism

Exercises, page 196

1. The Hamiltonian is

H(t, y, p) = p2 − q(t)y2
4r(t)

Hamilton’s equations are

y = p , p = 2q(t)y
2r(t)

2. Here we have

J(y) = (t2 + y2)(1 + (y )2)

We find p = Ly = t2 + y2y
which yields 1 + (y )2

(y )2 = t2 p2 p2
+ y2 −

52 CHAPTER 3. CALCULUS OF VARIATIONS

The Hamiltonian simplifies, after some algebra, to

H(t, y, p) = − t2 + y2 − p2

Then Hamilton’s equations are

dy = p, dp = y
dt t2 + y2 − p2 dt t2 + y2 − p2

Dividing the two equations gives

dy = p
dp y

Integrating yields

y2 − p2 = const

These are hyperbola in the yp phase plane.

3. Hamilton’s equations for the pendulum are

θ = p , p = −mgl sin θ
ml2

4. (a) The potential energy is the negative integral of the force, or

V (y) = − (−ω2y + ay2)dy = 1 ω2y2 − 1 ay3
2 3

The Lagrangian is L = 1 m(y )2 −V (y). The Euler equation coincides
2
with Newton’s second law:

my = −ω2y + ay2

(b) The momentum is p = Ly = my , which gives y = p/m. The

Hamiltonian is 1
2
H(y, p) = (p/m)2 + V (y)

which is the kinetic plus the potential energy, or the total energy of
the system. Yes, energy is conserved (L is independent of time).

(c) We have H = E for all time t, so at t = 0 we have

1 (p(0)/m)2 + V (y(0)) = ω2
2 10

If y(0) = 0 we can solve for p(0) to get the momentum at time zero;
but this gives

y (0) = ± ω2/5m

3.5. THE CANONICAL FORMALISM 53

(d) The potential energy has a local maximum at y = ω2/a and is equal

to ω6
6a2
Vmax = V (ω2/a) =

Note that V = 0 at y = 0, 3ω2/2a. If E < Vmax then we obtain
oscillatory motion; if E > Vmax, then the motion is not oscillatory.
Observe that the phase diagram (the solution curves in yp–space, or

phase space), can be found by graphing

p = ± 2m(E − V (y)

for various constants E.

5. The Euler equations are

Fy − d Fy = 0, Fp − d Fp = 0,
dt dt

where F = py − H(t, y, p). Easily we find from these two equations that

−Hy − d p = 0, y − Hp = 0.
dt

6. Here the force is F (t, y) = ket/y2. We can define a potential by V (t, y) =
− F (t, y)dy = ket/y. The Lagrangian is

L(t, y, y ) = m (y )2 − V (t, y)
2

The Euler equation is my − ket/y2 = 0

which is Newton’s second law of motion. The Hamiltonian is

H(t, y, p) = p2 + k et
2m y

which is the total energy. Is energy conserved? We can compute dH/dt

to find dH
dt
= ket/y =0

So the energy is not constant.

7. The kinetic energy is T = m(y )2/2. The force is mg (with a plus sign since

positive distance is measured downward). Thus V (y) = −mgy. Then the
Lagrangian is L = m(y )2/2 + mgy.

8. We have, for example,

Lxi − d Lxi = ∂V − d (mxi ) = 0,
dt ∂xi dt

or

mxi = Fi.

54 CHAPTER 3. CALCULUS OF VARIATIONS

9. We have r = −α and so r(t) = −αt + l. Then the kinetic energy is

T = m ((r )2 + r2(θ )2) = m (α2 + (l − αt)2(θ )2),
2 2

and the potential energy is

V = mgh = mg(l − r cos θ) = mg(l − (l − αt) cos θ.)

The Lagrangian is L = T − V . The Euler equation, or the equation of
motion, is

g sin θ + (l − αt)θ − 2αθ = 0.

One can check that the Hamiltonian is not the same as the total energy;
energy is not conserved in this system. If fact, one can verify that

d (T + V ) = −mgα cos θ = 0.
dt

10. Follow the instructions.

11. The Emden-Fowler equation is

y + 2 y + y5 = 0.
t

Multiply by the integrating factor t2 to write the equation in the form

(t2y ) + t2y5 = 0.

Now we can identify this with the Euler equation:

Ly = −t2y5, Ly = t2y .

Integrate these two equations to find

L = 1 t2(y )2 − 1 t2 y6 + φ(t)
2 6

12. Multiply y + ay + b = 0 by the integrating factor exp(at) to get
(eaty ) + beat = 0

Now identify the terms in this equation with the terms in the Euler equa-
tion as in Exercise 5.14. Finally we arrive at a Lagrangian

L = eat m (y )2 − by
2

There are many Lagrangians, and we have chosen just one by selecting
the arbitrary functions.

3.6. ISOPERIMETRIC PROBLEMS 55

13. Follow Example 3.27 in the book with m, a, k and y replaced by L, R, C−1
and I, respectively.

14. Multiplying the given equation by exp P (t) makes the sum of the first two
terms a total derivative and we get

d (eP (t)y ) + eP (t)f (y) = 0.
dt

Comparing to the Euler equation we take

Ly = eP (t)f (y), Ly = −eP (t)y .
Integrating the first gives

L = eP (t)F (y) + φ(t, y ),

where φ is arbitrary. Plugging this into the second equation we get

φy = −eP (t)y .

Integrating, 1
2
φ = − y 2eP (t) + ψ(t),

where ψ is arbitrary. Thus,

L = eP (t) F (y) − 1 y 2 + ψ(t).
2

3.6 Isoperimetric Problems

Exercises, page 203

1. Form the Lagrangian L∗ = (y )2 + λy2.

The Euler equation reduces to

y − λy = 0, y(0) = y(π) = 0

The extremals are given by

yn(x) = ± 2/π sin nx, n = 1, 2, 3, . . . .

2. Let L∗ = x2 + (y )2 + λy2. Then the Euler equation for L∗ is

y + λy = 0, y(0) = y(1) = 0.

If λ ≥ 0 then this boundary value problem has only trivial solutions. If
λ < 0, say λ = −β2, then the problem has nontrivial solutions

yn(x) = Bn sin nπx, n = 1, 2, . . . ,

56 CHAPTER 3. CALCULUS OF VARIATIONS

where the Bn are constants. Applying the constraint gives

1

Bn2 sin2 nπx dx = 2.

0

Thus Bn = ±2 for all n. So the extremals are
yn(x) = ±2 sin nπx, n = 1, 2, . . . .

3. The Euler equations become

L∗y1 − d Ly∗1 = 0, Ly∗2 − d Ly∗2 = 0,
dx dx

where L∗ = L + λG.

4. Form the Lagrangian

L∗ = xy − yx + λ (x )2 + (y )2.

Because the Lagrangian does not depend explicitly on t, a first integral is
given by

L∗ − x L∗x − y L∗y = C.

5. The problem is to minimize

1 1 + (y )2dx, y(0) = y(1) = 0

J(y) =

0

subject to the constraint

1

y(x)dx = A.

0

Form the Lagrangian

L∗ = 1 + (y )2 + λy

Because the Lagrangian does not depend explicitly on x, a first integral is
given by

L∗ − y L∗y = c.

Expanding out this equation leads to

y= 1 − (λy − c)2 .
(λy − c)2

Separating variables and integrating, and then using the substitution u =

1 − (λy − c)2, gives 1 √duu = x + c1.
2λ
−

3.6. ISOPERIMETRIC PROBLEMS 57

Thus 1
λ2
(x + c1)2 + (y − c/λ)2 = ,

which is a circle. Now the two boundary conditions and the constraint

give three equations for the constants c, c1, λ.

6. Form the Lagrangian

L∗ = p(y )2 + qy2 + λry2.

Then the Euler equation corresponding to L∗ is

Ly − (Ly ) = 2qy + 2rλy − 2(py ) = 0,

or
(py ) − qy = rλy, y(a) = y(b) = 0

This is a Sturm-Liouville problem for y (see Chapter 4).

7. Solve the constraint equation to obtain

z = g(t, y).

Substitute this into the functional to obtain

bb

W (y) ≡ F (t, y, y ) ≡ L(t, y, g(t, y), y , gt + gyy )dt.

aa

Now form the Euler equation for F . We get

Fy − d Fy = 0,
dt

or, in terms of L,

Ly + Lz gy + Lz (gty + gyy y ) − d (Ly + Lz gy) = 0.
dt

This simplifies to

Ly − d Ly + gy (Lz − d Lz = 0.
dt dt

Also G(t, y, g(t, y)) = 0, and taking the partial with respect to y gives

Gy + Gzgy = 0.

Thus Ly − d Ly Lz − d Lz
dt dt
= .
Gy Gy

Now these two expressions must be equal to the same function of t, that

is d d
dt dt
Ly − Ly = λ(t)Gy, Lz − Lz = λ(t)Gz.