BEEM103 –Optimization Techniques for Economists
Dieter Balkenborg
Departments of Economics
University of Exeter
Class Exercises Week 6 - Solutions
Exercise 1 Find the solutions of the form e
x•
t
2x_ + x = 0
Solution 1 For such a solution x_ = e t , x• =
x•
A solution is given by
2x_ + x =
to the homogenous equation
2
e
t
and hence
2 +1 e
t
=(
2
1)2 e
t
= 1, i.e. x (t) = et .
Exercise 2 Find a solution to the inhomogeneous equation
x•
2x_ + x = 3
Solution 2 A solution is given by x (t) = 3.
Exercise 3 Describe all solutions to this equation.
Solution 3 Because 2 2 + 1 has the double root = 1 it follows from Theorem 6.3.1
in the textbook by Sydsæter, Hammond, Seierstad and Strøm that the general solution
to the homogeneous equation x• 2x_ + x = 0 is (A + Bt) et . The same book discusses two
pages later that the general solution to the inhomogeneous solution is
(A + Bt) et + 3
Exercise 4 Find a solution with x (0) = 0, x (1) = 1
Solution 4
x (0) = A + 3 = 0 () A = 3
x (1) = ( 3 + B) e + 3 = 2 ()
3+B =
e
1
() B =
e
1
+3
Exercise 5 Find the solutions to the homogenous equation
x• + x = 0
Solution 5 The characteristic equation 2 + 1 has no real solution. By Theorem 6.3.1
in the textbook by Sydsæter, Hammond, Seierstad and Strøm the general solution is
x = A cos t + B sin t
Exercise 6 Minimize
Z
1
p
1 + x_ 2 dt
0
subject to the boundary conditions x (0) = 4, x (1) = 2:
Solution 6 Let F =
p
1 + x_ 2 . We have
1
1
@F
1
@F
= 0,
=
1 + x_ 2 2 2x_ = 1 + x_ 2 2 x_
@x
@ x_
2
@ 2F
@ 2F
=
=0
@ x@t
_
@ x@x
_
3
1
@ 2F
1
=
1 + x_ 2 2 2x_ x_ + 1 + x_ 2 2
@ x@
_ x_
2
1
x_ 2
1
1
= p
+1 = p
>0
2
2
2
1 + x_
1 + x_
1 + x_ 1 + x_ 2
The Euler equation is hence in this case
1
1
0= p
x• () x• = 0
2
1 + x_ 1 + x_ 2
The general solution to x• = 0 is x = At + B. To …t the boundary conditions we must
have
x (0) = B = 4
x (1) = A + 4 = 2 , A =
Thus x (t) =
2
2t + 4
Exercise 7 Minimize
Z
1
x2 + 2txx_ + t2 x_ 2 dt
0
subject to the boundary conditions x (0) = 0, x (1) = 2:
Solution 7
@F
= 2x + 2tx,
_
@x
@ 2F
= 2x + 4tx,
_
@ x@t
_
@F
= 2tx + 2t2 x_
@ x_
@ 2F
@ 2F
= 2t,
= 2t2
@ x@x
_
@ x@
_ x_
The Euler equation is
2x + 2tx_
0
2x + 4tx_ + 2tx_ + 2t2 x•
4tx_ + 2t2 x•
2
() x• =
x_ for t 6= 0
t
=
=
Let u = x.
_ Then the last equation becomes
du
=
dt
2
2
u
t
This is a separable di¤erential equation which we can solve as follows
1
du
Z u
1
du
u
ln juj
ln juj
juj
=
=
=
=
=
2
dt
t
Z
2
dt
t
2 ln jtj + C
ln (jtj) 2 + C = ln t
t 2 or u =
t 2
2
where = ln C.
Next
dx
= u=
dt
Z
x =
t
2
t 2 dt =
t
1
+D
The only function possible which could in addition satisfy x (0) = 0 would be obtained
for = 0 (otherwise the function would not be de…ned at zero) and A = 0, i.e., we would
have x (t) = 0 for all t. However, x (2) = 2 would then not be possible. Hence we cannot
…nd a solution and suspect that none exists.
Exercise 8 Find the Euler equation for the Ramsey problem
Z T
U f (K (t)) K_ (t) e rt dt
0
subject to the boundary conditions x (0) = 4, x (1) = 2: Use the relative rate of change
_
_
consumption C=C
where C = f (K) K.
Solution 8 For this and the next exercise see Sydsæter, Hammond, Seierstad and Strøm
Section 8.4:
Exercise 9 Solve the equation in the case f (K) = bK; U (C) = C 1 v = (1
3
v).