ME254/ECE271C
Winter ’09
Solutions to HW 4
1.
Note: These solutions were written several years back, so the terminology may not exactly match that
being used in class this quarter, but you should be able to figure things out from context. The technique
also uses a short cut that’s somewhat special to this problem.
Letting ˙
x
1
=
x
2
, ˙
x
2
=
u
, we have the calculus of variations problem:
min
x
1
,x
2
Z
1=
t
f
0
( ˙
x
2
)
2
dt
3
˙
x
1
=
x
2
x
1
(
t
f
)
2
+
x
2
(
t
f
)
2
= 1
x
1
(0) =

2
√
2
, x
2
(0) = 5
√
2
Let the Lagrange multiplier function associated with the differential constraint be
λ
. Then, we seek
an extremal for:
L
:= ( ˙
x
2
)
2
+
λ
(
t
) [ ˙
x
1

x
2
]
.
The EL equations are:
2¨
x
2
=

λ
˙
λ
=
0
→
λ
(
t
) =
c
1
Hence,
2¨
x
2
=

c
1
⇒
˙
x
2
=

1
2
c
1
t
+
c
2
x
2
(
t
) =

1
4
c
1
t
2
+
c
2
t
+ 5
√
2
.
(1)
Since ˙
x
1
=
x
2
, we have
x
1
(
t
) =

1
12
c
1
t
3
+
1
2
c
2
t
2
+ 5
√
2
t

2
√
2
.
(2)
From the target set we have
x
1
(1)
2
+
x
2
(1)
2
= 1
⇒

1
4
c
1
+
c
2
+ 5
√
2
2
+

1
12
c
1
+
1
2
c
2
+ 3
√
2
2
= 1
(3)
which is one of the equations to be solved to determine
c
1
and
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 Winter '09
 Bamieh
 Quadratic equation, Elementary algebra, Boundary value problem, NBC

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