BASIC OPTIMAL SHOCK ISOLATION: SINGLE-DEGREE-OF-FREEDOM SYSTEMS - Injury Biomechanics and Control: Optimal Protection from Impact - page 91

t = V/( 2 U) ( U/V time units after the beginning of the motion) comes to
a complete stop at the initial position x =

0.
To prove that Eqs. (3.135) and (3.136) provide an optimal solution, solve
Eq. (3.128) subject to the conditions of Eqs. (3.129) and (3.130) for an
arbitrary control law u(t) to obtain

⎧
⎨

t
− t ∗

(t − τ)u(τ) dτ for − t ∗ ≤ t< 0 ,

x(t) = (3.139)

+ t

⎩

Vt (t − τ)u(τ) dτ for t ≥ 0 .

− t

∗

Set u(τ ) U on the right-hand side of Eq. (3.139) to obtain a lower

bound for the solution x(t) :

≡−

⎧
⎨

U(t + t ∗ ) 2
2

− for − t ∗ ≤ t< 0 ,

x(t) ≥ (3.140)

⎩

t ∗ ) 2 U(t +

Vt − for t ≥ 0 .

2

The maximum of the right-hand side of this inequality provides a lower
bound for max t x(t) given by

max
t ≥− t ∗

x(t) ≥ p(t ∗ ), (3.141)

where

⎨

V 2
2 U −

V
2 U ,

Vt ∗ for t ∗ <

p(t ∗ ) = (3.142)

⎩

V
2 U .

0 for t ∗ ≥

Use x( 0 ) as an upper bound for min t x(t) , which, from Eq. (3.139), gives

0

min
t ≥− t ∗

x(t) ≤− τu(τ)dτ. (3.143)

− t ∗

In view of the constraint − u(τ ) ≤ U , this inequality shows that an upper

bound for the absolute minimum of x(t) is

U

0

Ut 2

∗

min
t ≥− t

x(t) ≤ τdτ =− . (3.144)

2

∗

− t ∗