Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 4 ppt

Solution 48.7
C
1
Extremals. Without loss of generality, we take the vertical line to be the y axis. We will consider x
1
, y
1
> 1. With
ds =

1 + (y

)
2
dx we extremize the integral,

x
1
0
√
y

1 + (y

)
2
dx.
Since the Lagrangian is independent of x, we know that the Euler differential equation has a first integral.
d
dx

F
y

− F
y
= 0
y

F
y

y
+ y

F
y

y

− F
y
= 0
d
dx
(y

F
y

− F ) = 0

y

F
y

− F = const
For the given Lagrangian, this is
y

√
y
y


1 + (y

)
2
−
√
y

1 + (y

)
2
= const,
(y

)

2
√
y −
√
y(1 + (y

)
2
) = const

1 + (y

)
2
,
√
y = const

1 + (y

)
2
y = const is one solution. To find the others we solve for y

and then solve the differential equation.
y = a(1 + (y

)
2
)

y

= ±

y − a
a
dx =

a
y − a
dy
2094
±x + b = 2

a(y − a)
y =
x
2
4a
±
bx
2a
+
b
2
4a
+ a
The natural boundary condition is
F
y




x=0
=
√
yy


1 + (y

)
2




x=0
= 0,
y

(0) = 0
The extremal that satisfies this boundary condition is
y =
x
2
4a
+ a.
Now we apply y(x
1

) = y
1
to obtain
a =
1
2

y
1
±

y
2
1
− x
2
1

for y
1
≥ x
1
. The value of the integral is

x
1
0


x

2
4a
+ a

1 +

x
2a

2

dx =
x
1
(x
2
1
+ 12a
2
)
12a
3/2
.
By denoting y
1
= cx
1
, c ≥ 1 we have
a =
1

2

cx
1
± x
1
√
c
2
− 1

The values of the integral for these two values of a are
√
2(x
1
)
3/2
−1 + 3c
2
± 3c
√
c
2
− 1
3(c ±
√
c
2
− 1)
3/2

.
2095
The values are equal only when c = 1. These values, (divided by
√
x
1
), are plotted in Figure 48.1 as a function of c.
The former and latter are fine and coarse dashed lines, respectively. The extremal with
a =
1
2

y
1
+

y
2
1
− x
2
1

has the smaller performance index. The value of the integral is
x
1
(x
2
1
+ 3(y

1
+

y
2
1
− x
2
1
)
2
3
√
2(y
1
+

y
2
1
− x
2
1
)
3
.
The function y = y
1
is an admissible extremal for all x
1

. The value of the integral for this extremal is x
1
√
y
1
which
is larger than the integral of the quadratic we analyzed before for y
1
> x
1
.
1.2
1.4
1.6
1.8 2
2.5
3
3.5
4
Figure 48.1:
Thus we see that
ˆy =
x
2
4a
+ a, a =
1
2

y

1
+

y
2
1
− x
2
1

2096
is the extremal with the smaller integral and is the minimizing curve in C
1
for y
1
≥ x
1
. For y
1
< x
1
the C
1
extremum
is,
ˆy = y
1
.
C
1

p
Extremals. Consider the parametric form of the Lagrangian.

t
1
t
0

y(t)

(x

(t))
2
+ (y

(t))
2
dt
The Euler differential equations are
d
dt
f
x

− f
x
= 0 and
d
dt

f
y

− f
y
= 0.
If one of the equations is satisfied, then the other is automatically satisfied, (or the extremal is straight). With either
of these equations we could derive the quadratic extremal and the y = const extremal that we found previously. We
will find one more extremal by considering the first parametric Euler di fferential equation.
d
dt
f
x

− f
x
= 0
d
dt


y(t)x

(t)

(x

(t))
2
+ (y


(t))
2

= 0

y(t)x

(t)

(x

(t))
2
+ (y

(t))
2
= const
Note that x(t) = const is a solution. Thus the extremals are of the three forms,
x = const,
y = const,
y =
x
2
4a
+
bx
2a
+

b
2
4a
+ a.
2097
The Erdmann corner conditions require that
F
y

=
√
yy


1 + (y

)
2
,
F −y

F
y

=
√
y

1 + (y


)
2
−
√
y(y

)
2

1 + (y

)
2
=
√
y

1 + (y

)
2
are continuous at corners. There can be corners only if y = 0.
Now we piece the three forms together to obtain C
1
p
extremals that satisfy the Erdmann corner conditions. The
only possibility that is not C
1
is the extremal that is a horizontal line from (0, 0) to (x
1

, 0) and then a vertical line from
(x
1
, y
1
). The value of the integral for this extremal is

y
1
0
√
t dt =
2
3
(y
1
)
3/2
.
Equating the performance indices of the quadratic extremum and the piecewise smooth extremum,
x
1
(x
2
1
+ 3(y
1
+

y

2
1
− x
2
1
)
2
3
√
2(y
1
+

y
2
1
− x
2
1
)
3
=
2
3
(y
1
)
3/2
,
y

1
= ±x
1

3 ± 2
√
3
√
3
.
The only real positive solution is
y
1
= x
1

3 + 2
√
3
√
3
≈ 1.46789 x
1
.
The piecewise smooth extremal has the smaller performance index for y
1
smaller than this value and the quadratic
extremal has the smaller performance index for y
1
greater than this value.

The C
1
p
extremum is the piecewise smooth extremal for y
1
≤ x
1

3 + 2
√
3/
√
3
and is the quadratic extremal for y
1
≥ x
1

3 + 2
√
3/
√
3.
2098
Solution 48.8
The shape of the rope will b e a catenary between x
1
and x
2
and be a vertically hanging segment after that. Let the

length of the vertical segment be z. Without loss of generality we take x
1
= y
2
= 0. The potential energy, (relative to
y = 0), of a length of rope ds in 0 ≤ x ≤ x
2
is mgy = ρgy ds. The total potential energy of the vertically hanging
rope is m(cen ter of mass)g = ρz(−z/2)g. Thus we seek to minimize,
ρg

x
2
0
y ds −
1
2
ρgz
2
, y(0) = y
1
, y(x
2
) = 0,
subject to the isoperimetric constraint,

x
2
0
ds − z = L.

Writing the arc-length differential as ds =

1 + (y

)
2
dx we minimize
ρg

x
2
0
y

1 + (y

)
2
ds −
1
2
ρgz
2
, y(0) = y
1
, y(x
2
) = 0,
subject to,


x
2
0

1 + (y

)
2
dx − z = L.
Consider the more general problem of finding functions y(x) and numbers z which extremize I ≡

b
a
F (x, y, y

) dx+
f(z) subject to J ≡

b
a
G(x, y, y

) dx + g(z) = L.
Suppose y(x) and z are the desired solutions and form the comparison families, y(x) +
1
η
1
(x) +
2
η

2
(x), z + 
1
ζ
1
+

2
ζ
2
. Then, there exists a constant such that
∂
∂
1
(I + λJ)



1
,
2
=0
= 0
∂
∂
2
(I + λJ)




1
,
2
=0
= 0.
2099
These equations are

b
a

d
dx
H
,y

− H
y

η
1
dx + h

(z)ζ
1
= 0,
and

b
a


d
dx
H
,y

− H
y

η
2
dx + h

(z)ζ
2
= 0,
where H = F + λG and h = f + λg. From this we conclude that
d
dx
H
,y

− H
y
= 0, h

(z) = 0
with λ determined by
J =


b
a
G(x, y, y

) dx + g(z) = L.
Now we apply these results to our problem. Since f(z) = −
1
2
ρgz
2
and g(z) = −z we have
−ρgz − λ = 0,
z = −
λ
ρg
.
It was shown in class that the solution of the Euler differential equation is a family of catenaries,
y = −
λ
ρg
+ c
1
cosh

x − c
2
c
1

.

One can find c
1
and c
2
in terms of λ by applying the end conditions y(0) = y
1
and y(x
2
) = 0. Then the expression for
y(x) and z = −λ/ρg are substituted into the isoperimetric constraint to determine λ.
Consider the special case that (x
1
, y
1
) = (0, 0) and (x
2
, y
2
) = (1, 0). In this case we can use the fact that
y(0) = y(1) to solve for c
2
and write y in the form
y = −
λ
ρg
+ c
1
cosh

x − 1/2

c
1

.
2100
Applying the condition y(0) = 0 would give us the algebraic-transcendental equation,
y(0) = −
λ
ρg
+ c
1
cosh

1
2c
1

= 0,
which we can’t solve in closed form. Since we ran into a dead end in applying the boundary condition, we turn to the
isoperimetric constraint.

1
0

1 + (y

)
2
dx − z = L


1
0
cosh

x − 1/2
c
1

dx − z = L
2c
1
sinh

1
2c
1

− z = L
With the isoperimetric constraint, the algebraic-transcendental equation and z = −λ/ρg we now have
z = −c
1
cosh

1
2c
1

,
z = 2c
1

sinh

1
2c
1

− L.
For any fixed L, we can numerically solve for c
1
and thus obtain z. You can derive that there are no solutions unless
L is greater than about 1.9366. If L is smaller than this, the rope would slip off the pin. For L = 2, c
1
has the values
0.4265 and 0.7524. The larger value of c
1
gives the smaller potential energy. The position of the end of the rope is
z = −0.9248.
Solution 48.9
Using the method of Lagrange multipliers, we look for stationary values of

c
0
((y

)
2
+ λy
2
) dx,
δ


c
0
((y

)
2
+ λy
2
) dx = 0.
2101
The Euler differential equation is
d
dx
F
(
, y

) − F
,y
= 0,
d
dx
(2y

) − 2λy = 0.
Together with the homogeneous boundary conditions, we have the problem
y

− λy = 0, y(0) = y(c) = 0,

which has the solutions,
λ
n
= −

nπ
c

2
, y
n
= a
n
sin

nπx
c

, n ∈ Z
+
.
Now we determine the constants a
n
with the moment of inertia constraint.

c
0
a
2
n

sin
2

nπx
c

dx =
ca
2
n
2
= A
Thus we have the extremals,
y
n
=

2A
c
sin

nπx
c

, n ∈ Z
+
.
The drag for these extremals is
D =
2A

c

c
0

nπ
c

2
cos
2

nπx
c

dx =
An
2
π
2
c
2
.
We see that the drag is mini mum for n = 1. The shape f or minimum drag is
ˆy =

2A
c
sin


nπx
c

.
2102
Solution 48.10
Consider the general problem of determining the stationary values of the quantity ω
2
given by
ω
2
=

b
a
F (x, y, y

, y

) dx

b
a
G(x, y, y

, y

) dx
≡
I

J
.
The variation of ω
2
is
δω
2
=
JδI − IδJ
J
2
=
1
J

δI −
I
J
δJ

=
1
J

δI − ω
2
δJ

.
The the values of y and y


are specified on the boundary, then the variations of I and J are
δI =

b
a

d
2
dx
2
F
,y

−
d
dx
F
,y

+ F
,y

δy dx, δJ =

b
a

d
2

dx
2
G
,y

−
d
dx
G
,y

+ G
,y

δy dx
Thus δω
2
= 0 becomes

b
a

d
2
dx
2
H
,y

−

d
dx
H
,y

+ H
,y

δy dx

b
a
G dx
= 0,
where H = F −ω
2
G. A necessary condition for an extremum is
d
2
dx
2
H
,y

−
d
dx
H
,y


+ H
,y
= 0 where H ≡ F −ω
2
G.
For our problem we have F = EI(y

)
2
and G = ρy so that the extremals are solutions of
d
2
dx
2

EI
dy
dx

− ρω
2
y = 0,
2103
With homogeneous boundary conditions we have an eigenvalue problem with deflections modes y
n
(x) and corresponding
natural frequencies ω
n
.
Solution 48.11

We assume that v
0
> w(x, y, t) so that the problem has a solution for any end point. The crossing time is
T =

l
0

˙
X(t)

−1
dx =
1
v
0

l
0
sec α(t) dx.
Note that
dy
dx
=
w + v
0
sin α
v
0
cos α

=
w
v
0
sec α + tan α
=
w
v
0
sec α +
√
sec
2
α − 1.
We solve this relation for sec α.

y

−
w
v
0
sec α

2
= sec
2
α − 1
(y


)
2
− 2
w
v
0
y

sec α +
w
2
v
2
0
sec
2
α = sec
2
α − 1
(v
2
0
− w
2
) sec
2
α + 2v
0
wy


sec α − v
2
0
((y

)
2
+ 1) = 0
sec α =
−2v
0
wy

±

4v
2
0
w
2
(y

)
2
+ 4(v
2
0
− w
2
)v

2
0
((y

)
2
+ 1)
2(v
2
0
− w
2
)
sec α = v
0
−wy

±

v
2
0
((y

)
2
+ 1) − w
2
(v
2

0
− w
2
)
2104
Since the steering angle satisfies −π/2 ≤ α ≤ π/2 only the positive solution is relevant.
sec α = v
0
−wy

+

v
2
0
((y

)
2
+ 1) − w
2
(v
2
0
− w
2
)
Time Independent Current. If we make the assumption that w = w(x, y) then we can write the crossing time
as an integral of a function of x and y.
T (y) =


l
0
−wy

+

v
2
0
((y

)
2
+ 1) − w
2
(v
2
0
− w
2
)
dx
A necessary condition for a minimum is δT = 0. The Euler differential equation for this problem is
d
dx
F
,y

− F

,y
= 0
d
dx

1
v
2
0
− w
2

−w +
v
2
0
y


v
2
0
((y

)
2
+ 1) − w
2

−

w
y
(v
2
0
− w
2
)
2

w(v
2
(1 + 2(y

)
2
) − w
2
)

v
2
0
((y

)
2
+ 1) − w
2
− y


(v
2
0
+ w
2
)

By solving this second order differential equation subject to the boundary conditions y(0) = 0, y(l) = y
1
we obtain the
path of minimum crossing time.
Current w = w (x). If the current is only a function of x, then the Euler differential equation can be integrated to
obtain,
1
v
2
0
− w
2

−w +
v
2
0
y


v
2

0
((y

)
2
+ 1) − w
2

= c
0
.
Solving for y

,
y

= ±
w + c
0
(v
2
0
− w
2
)
v
0

1 − 2c
0

w − c
2
0
(v
2
0
− w
2
)
.
Since y(0) = 0, we have
y(x) = ±

x
0
w(ξ) + c
0
(v
2
0
− (w(ξ))
2
)
v
0

1 − 2c
0
w(ξ) − c
2

0
(v
2
0
− (w(ξ))
2
)
.
2105
For any given w(x) we can use the condition y(l) = y
1
to solve for the constant c
0
.
Constant Current. If the current is constant then the Lagrangian is a function of y

alone. The admissible
extremals are straight lines. The solution is then
y(x) =
y
1
x
l
.
Solution 48.12
1. The kinetic energy of the first particle is
1
2
m((α − x)
ˆ

θ)
2
. Its potential energy, relative to the table top, is zero.
The kinetic energy of the second particle is
1
2
mˆx
2
. Its p otential energy, relative to its equilibrium position is
−mgx. The Lagrangian is the difference of kinetic and potential energy.
L = m

˙x
2
+
1
2
(α − x)
2
˙
θ
2
+ gx

The Euler differential equations are the equations of motion.
d
dt
L
, ˙x
− L

x
= 0,
d
dt
L
,
˙
θ
− L
θ
= 0
d
dt
(2m ˙x) + m(α − x)
˙
θ
2
− mg = 0,
d
dt

m(α − x)
2
˙
θ
2

= 0
2¨x + (α − x)
˙

θ
2
− g = 0, (α −x)
2
˙
θ
2
= const
When x = 0,
˙
θ = ω =

g/α. This determines the constant in the equation of motion for θ.
˙
θ =
α
√
αg
(α − x)
2
Now we substitute the expression for
˙
θ into the equation of motion for x.
2¨x + (α − x)
α
3
g
(α − x)
4
− g = 0

2106
2¨x +

α
3
(α − x)
3
− 1

g = 0
2¨x +

1
(1 − x/α)
3
− 1

g = 0
2. For small oscillations,


x
α


 1. Recall the binomial expansion,
(1 + z)
a
=
∞


n=0

a
n

z
n
, for |z| < 1,
(1 + z)
a
≈ 1 + az, for |z|  1.
We make the approximation,
1
(1 − x/α)
3
≈ 1 + 3
x
α
,
to obtain the linearized equation of motion,
2¨x +
3g
α
x = 0.
This is the equation of a harmonic oscillator with solution
x = a sin


3g2α(t − b)


.
The period of oscillation is,
T = 2π
√
2α3g.
Solution 48.13
We write the equation of motion and boundary conditions,
¨x = U(t) −g, x(0) = ˙x(0) = 0, x(T) = h,
2107
as the first order system,
˙x = 0, x(0) = 0, x(T ) = h,
˙y = U(t) − g, y(0) = 0.
We seek to minimize,
T =

T
0
dt,
subject to the constraints,
˙x − y = 0,
˙y − U(t) + g = 0,

T
0
U
2
(t) dt = k
2
.

Thus we seek extrema of

T
0
H dt ≡

T
0

1 + λ(t)( ˙x − y) + µ(t)( ˙y − U(t) + g) + νU
2
(t)

dt.
Since y is not specified at t = T , we have the natural boundary condition,
H
, ˙y


t=T
= 0,
µ(T ) = 0.
The first Euler differential equation is
d
dt
H
, ˙x
− H
,x
= 0,

d
dt
λ(t) = 0.
2108
We see that λ(t) = λ is constant. The next Euler DE is
d
dt
H
, ˙y
− H
,y
= 0,
d
dt
µ(t) + λ = 0.
µ(t) = −λt + const
With the natural boundary condition, µ(T ) = 0, we have
µ(t) = λ(T − t).
The final Euler DE is,
d
dt
H
,
˙
U
− H
,U
= 0,
µ(t) − 2νU(t) = 0.
Thus we have

U(t) =
λ(T −t)
2ν
.
This is the required thrust function. We use the constraints to find λ, ν and T .
Substituting U(t) = λ(T −t)/(2ν) into the isoperimetric constraint,

T
0
U
2
(t) dt = k
2
yields
λ
2
T
3
12ν
2
= k
2
,
U(t) =
√
3k
T
3/2
(T −t).
The equation of motion for x is

¨x = U(t) −g =
√
3k
T
3/2
(T −t).
2109
Integrating and applying the initial conditions x(0) = ˙x(0) = 0 yields,
x(t) =
kt
2
(3T −t)
2
√
3T
3/2
−
1
2
gt
2
.
Applying the condition x(T ) = h gives us,
k
√
3
T
3/2
−
1

2
gT
2
= h,
1
4
g
2
T
4
−
k
3
T
3
+ ghT
2
+ h
2
= 0.
If k ≥ 4

2/3g
3/2
√
h then this fourth degree polynomial has positive, real solutions for T. With strict inequality, the
minimum time is the smaller of the two positive, real solutions. If k < 4

2/3g
3/2

√
h then there is not enough fuel to
reach the target height.
Solution 48.14
We have ¨x = U(t) where U(t) is the acceleration furnished by the thrust of the vehicles engine. In practice, the engine
will be des igned to operate within certain bounds, say −M ≤ U(t) ≤ M, where ±M is the maximum forward/backward
acceleration. To account for the inequality constraint we write U = M sin V (t) for some suitable V (t). More generally,
if we had φ(t) ≤ U(t) ≤ ψ(t), we could write this as U(t) =
ψ+φ
2
+
ψ−φ
2
sin V (t).
We write the equation of motion as a first order system,
˙x = y, x(0) = a, x(T ) = 0,
˙y = M sin V, y(0) = b, y(T ) = 0.
Thus we minimize
T =

T
0
dt
subject to the constraints,
˙x − y = 0
˙y − M sin V = 0.
2110
Consider
H = 1 + λ(t)( ˙x −y) + µ(t)( ˙y − M sin V ).
The Euler differential equations are

d
dt
H
, ˙x
− H
,x
= 0 ⇒
d
dt
λ(t) = 0 ⇒ λ(t) = const
d
dt
H
, ˙y
− H
,y
= 0 ⇒
d
dt
µ(t) + λ = 0 ⇒ µ(t) = −λt + const
d
dt
H
,
˙
V
− H
,V
= 0 ⇒ µ(t)M cos V (t) = 0 ⇒ V (t) =
π

2
+ nπ.
Thus we see that
U(t) = M sin

π
2
+ nπ

= ±M.
Therefore, if the rocket is to be transferred from its initial state to is sp eci fied final state in minimum time with
a limited source of thrust, (|U| ≤ M), then the engine should operate at full power at all times except possibly for a
finite number of switching times. (Indeed, if some power were not being used, we would expect the transfer would be
speeded up by using the additional power suitably.)
To see how this ”bang-bang” process works, we’ll look at the phase plane. The problem
˙x = y, x(0) = c,
˙y = ±M, y(0) = d,
has the solution
x(t) = c + dt ± M
t
2
2
, y(t) = d ± Mt.
We can eliminate t to get
x = ±
y
2
2M
+ c ∓
d

2
2M
.
These curves are plotted in Figure 48.2.
2111
Figure 48.2:
There is only curve in each case which transfers the initial state to the origin. We will denote these curves γ and Γ,
respectively. Only if the initial point (a, b) lies on one of these two curves can we transfer the state of the system to the
origin along an extremal without switching. If a =
b
2
2M
and b < 0 then this is possible using U(t) = M. If a = −
b
2
2M
and b > 0 then this is possible using U(t) = −M. Otherwise we follow an extremal that intersects the initial position
until this curve intersects γ or Γ. We then follow γ or Γ to the origin.
Solution 48.15
Since the integrand does not explicitly depend on x, the Euler differential equation has the first integral,
F −y

F
y

= const.

y + h

1 + (y


)
2
− y

y

√
y + h

1 + (y

)
2
= const
√
y + h

1 + (y

)
2
= const
y + h = c
2
1
(1 + (y

)
2

)
2112

y + h − c
2
1
= c
1
y

c
1
dy

y + h − c
2
1
= dx
2c
1

y + h − c
2
1
= x −c
2
4c
2
1
(y + h − c

2
1
) = (x − c
2
)
2
Since the extremal passes through the origin, we have
4c
2
1
(h − c
2
1
) = c
2
2
.
4c
2
1
y = x
2
− 2c
2
x (48.6)
Introduce as a parameter the slope of the extremal at the origin; that is, y

(0) = α. Then differentiating (48.6) at
x = 0 yields 4c
2

1
α = −2c
2
. Together with c
2
2
= 4c
2
1
(h − c
2
1
) we obtain c
2
1
=
h
1+α
2
and c
2
= −
2αh
1+α
2
. Thus the equation
of the pencil (48.6) will have the form
y = αx +
1 + α
2

4h
x
2
. (48.7)
To find the envelope of this family we differentiate ( 48.7) with respect to α to obtain 0 = x +
α
2h
x
2
and eliminate α
between this and ( 48.7) to obtain
y = −h +
x
2
4h
.
See Figure 48.3 for a plot of some extremals and the envelope.
All extremals (48.7) lie above the envelope which in ballistics is calle d the parabola of safety. If (m, M) lies outside
the parabola, M < −h +
m
2
4h
, then it cannot be joined to (0, 0) by an extremal. If (m, M) is above the envelope
then there are two candidates. Clearly we rule out the one that touches the envelope because of the occurrence of
conjugate points. For the other extremal, problem 2 shows that E ≥ 0 for all y

. Clearly we can embed this extremal
in an extremal pencil, so Jacobi’s test is satisfied. Therefore the parabola that does not touch the envelope is a strong
minimum.
2113

h
2h
x
-h
y
Figure 48.3: Some Extremals and the Envelope.
Solution 48.16
E = F (x, y, y

) − F (x, y, p) − (y

− p)F
y

(x, y, p)
= n

1 + (y

)
2
− n

1 + p
2
− (y

− p)
np


1 + p
2
=
n

1 + p
2


1 + (y

)
2

1 + p
2
− (1 + p
2
) − (y

− p)p

=
n

1 + p
2


1 + (y


)
2
+ p
2
+ (y

)
2
p
2
− 2y

p + 2y

p − (1 + py

)

=
n

1 + p
2


(1 + py

)
2

+ (y

− p)
2
− (1 + py

)

≥ 0
2114
The speed of light in an inhomogeneous medium is
ds
dt
=
1
n(x,y
. The time of transit is then
T =

(b,B)
(a,A)
dt
ds
ds =

b
a
n(x, y)

1 + (y


)
2
dx.
Since E ≥ 0, light traveling on extremals follow the time optimal path as long as the extremals do not intersect.
Solution 48.17
Extremals. Since the integrand doe s not depend explicitly on x, the Euler differential equation has the first integral,
F −y

F
,y

= const.
1 + y
2
(y

)
2
− y

−2(1 + y
2
)
(y

)
3
= const
dy


1 + (y

)
2
= const dx
arcsinh(y) = c
1
x + c
2
y = sinh(c
1
x + c
2
)
Jacobi Test. We can see by inspection that no conjugate points exist. Consider the central field through (0, 0),
sinh(cx), (See Figure 48.4).
We can also easily arrive at this conclusion analytically as follows: Solutions u
1
and u
2
of the Jacobi equation are
given by
u
1
=
∂y
∂c
2
= cosh(c

1
x + c
2
),
u
2
=
∂y
∂c
1
= x cosh(c
1
x + c
2
).
Since u
2
/u
1
= x is monotone for all x there are no conjugate points.
2115
-3 -2
-1 1
2 3
-3
-2
-1
1
2
3

Figure 48.4: sinh(cx)
Weierstrass Test.
E = F (x, y, y

) − F (x, y, p) − (y

− p)F
,y

(x, y, p)
=
1 + y
2
(y

)
2
−
1 + y
2
p
2
− (y

− p)
−2(1 + y
2
)
p
3

=
1 + y
2
(y

)
2
p
2

p
3
− p(y

)
2
+ 2(y

)
3
− 2p(y

)
2
p

=
1 + y
2
(y


)
2
p
2

(p − y

)
2
(p + 2y

)
p

For p = p(x, y) bounded away from zero, E is one-signed for values of y

close to p. However, since the factor (p + 2y

)
can have any sign for arbitrary values of y

, the conditions for a strong minimum are not satisfied.
Furthermore, since the extremals are y = sinh(c
1
x + c
2
), the slope function p(x, y) will be of one sign only if the
range of integration is such that we are on a monotonic piece of the sinh. If we span both an increasing and decreasing
section, E changes sign even for weak variations.

2116
Legendre Condition.
F
,y

y

=
6(1 + y
2
)
(y

)
4
> 0
Note that F cannot be represented in a Taylor series for arbitrary values of y

due to the presence of a discontinuity in
F when y

= 0. However, F
,y

y

> 0 on an extremal implies a weak minimum is provided by the extremal.
Strong Variations. Consider

1+y

2
(y

)
2
dx on both an extremal and on the special piecewise continuous variation in
the figure. On P Q we have y

= ∞ with implies that
1+y
2
(y

)
2
= 0 so that there is no contribution to the integral from
P Q.
On QR the value of y

is greater than its value along the extremal P R while the value of y on QR is less than the
value of y along P R. Thus on QR the quantity
1+y
2
(y

)
2
is less than it is on the extremal PR.

QR

1 + y
2
(y

)
2
dx <

P R
1 + y
2
(y

)
2
dx
Thus the weak minimum along the extremal can be weakened by a strong variation.
Solution 48.18
The Euler differential equation is
d
dx
F
,y

− F
,y
= 0.
d
dx
(1 + 2x

2
y

) = 0
1 + 2x
2
y

= const
y

= const
1
x
2
y =
c
1
x
+ c
2
(i) No continuous extremal exists in −1 ≤ x ≤ 2 that satisfies y(−1) = 1 and y(2) = 4.
2117