13 Solutions for Selected Exercises
255
Exercise 11.4
The Riccati equation for this process is given by (11.57), which in our case reads
˙
κ = −q − 2aκ +
b
2
r
κ
2
.
(13.66)
In steady-state, we have ˙
κ = 0 and the equation becomes a quadratic equation for κ
κ
2
− 2
ar
b 2 κ −
qr
b 2 = 0
(13.67)
which has the solutions
κ =
ar
b 2
1 ±
1 +
b 2 q
a 2 r
.
(13.68)
Inserting this value of κ in (11.54), we arrive at the result shown in (11.62).
Exercise 11.5
The cost function from (11.41) is g(x, u) = (qx
2
+ ru
2
)/2 such that the Hamiltonian
becomes
H(x, u, p) =
1
2
qx
2
+ ru
2
+ p
αx
β
+ u
.
(13.69)
Applying Hamilton’s equation yields
˙
p = −
∂H
∂ x
= −
qx + αpβx
β−1
˙
x =
∂H
∂ p
= αx
β
+ u
(13.70)
0 =
∂H
∂u
= ru + p
Solving the last equation for u = −p/r allows us to eliminate u from the second
equation and results in the following set of two coupled non-linear differential equations
˙
p = −qx − αpβx
β−1
˙
x = αx
β
−
1
r
p .
(13.71)