Appendix: All Solutions to the Problems
867
Similarly, it can be found
d
dt
F ˙
y =
d
2 y
ds 2
ds
dt
,
d
dt
F ˙
z =
d
2 z
ds 2
ds
dt
. Thus the Euler equations
of the extremal function of the auxiliary functional are
⎧
⎪ ⎨
⎪ ⎩
λ(t)ϕ x −
d
2 x
ds 2
ds
dt
= 0
λ(t)ϕ y −
d
2 y
ds 2
ds
dt
= 0
λ(t)ϕ z −
d
2 z
ds 2
ds
dt
= 0
Therefore
d
2 x
ds 2
ds
dt
ϕ x
=
d
2 y
ds 2
ds
dt
ϕ y
=
d
2 z
ds 2
ds
dt
ϕ z
= λ(t)
Quod erat demonstrandum.
Chapter 7: Solutions to the Problems in Variational
Principles
7.1 Let T 1 and T 2 be bounded linear operators of the Hilbert space H to itself, if for
arbitrary x, y ∈ H , there is (T 1 x, y) = (T 2 x, y), prove T 1 = T 2 .
Proof Since (T 1 x, y) = (T 2 x, y), the expression
(T 1 x, y) − (T 2 x, y) = (T 1 x − T 2 x, y) = [(T 1 − T 2 )x, y] = 0
holds for an arbitrary y ∈ H , thus
(T 1 − T 2 )x = 0
holds for an arbitrary x ∈ H , namelyT 1 = T 2 . Quod erat demonstrandum.
7.2 Let the two vectors u and v in an inner product space be orthogonal, prove
u + v
2
= u
2
+ v
2 .
Proof Because u and v are orthogonal, there is
u + v
2
= (u + v, u + v) = (u, u) + 2(u, v) + (v, v) = u
2
+ v
2
Quod erat demonstrandum.
7.3 Find the boundary conditions satisfied that the operator T is a symmetric operator.
(1) T = −
d
dx
x
2 d
dx
, D T : {u(x)|u(x) ∈ C
2
[0, l]};