Economics 701 Homework #4 Solutions Chapter 3.2 1. Given that V ...

Economics 701
Homework #4
Solutions
Chapter 3.2
T
1. Given that V!y" ! ! !t 2 " y# 2 "dt has an Euler equation whose general solution is
0
y " !t" ! c 1 t " c 2 , find the extremal if y!0" ! 4 and y!2" is free.
If y!0" ! 4 # c 2 ! 4.
Taking the appropriate partial F y# ! 2y# .
This yields the equation 2y# ! 0. Now take the general solution.
y ! c 1 t " 4. y$ !c 1 . By the above equation c 1 ! 0.
This is a flat line with intercept 4.
3. Suppose the problem were exactly as above but y!T" ! 5 and T is free.
Taking the appropriate partials: F y# ! 2y# , y# F y# ! 2y# 2 .
F # y# F y# ! t 2 # y# 2 ! 0.
Again take the general solution. y# 2 ! c 1 2 . T 2 ! c 1 2 . Now since T must be positive, we
ignore the negative root of this equation. # T ! c 1 .
y!T" ! y!c 1 " ! c 1 2 " 4 ! 5. # c 1 ! 1 ! T.
y " !t" ! t " 4; and T " ! 1.
This is a line with slope 1 and intercept 4.
Chapter 5.2
1. ! ! 50K # K 2 . C ! K# 2 " 2K#
F ! e #"t $50K # K 2 # K# 2 # 2K# %
$
max ! F!K,K #
0
,t"dt
This is now a problem just like the ones in Chapter 2.
F K ! e #"t $50 # 2K%; F K # ! e #"t $#2K# # 2%
F K # K ! 0; F K # K# ! e #"t $#2%; F K # t ! "e #"t $2K# " 2%
The Euler equation is:

2
#2e #"t K% " "e #"t $2K# " 2% # e #"t $50 # 2K% ! 0.
#2K% ""$2K# " 2% # $50 # 2K% ! 0
K% # "K# # " " 25 # K ! 0
K% # "K# # K ! " # 25
Begin by finding the homogenous solution:
K% # "K# # K ! 0
characteristic equation: r 2 # "r # 1 ! 0.
r ! " & "2 " 4
. Let r
2
1 ! " " "2 " 4
; Let r
2
2 ! " # "2 " 4
2
The general solution is K ! c 1 e r1t " c 2 e r 2t
For the particular solution conjecture a constant solution.
K' ! 25 # ". Note that this holds.
K " !t" ! c 1 e r 1t " c 2 e r 2t " K' .
Two boundary conditions:
TC: lim t($
F # K# F K # !K!t",K# !t",t" ! 0 # c 1 ! 0. This is because r 1 is positive, and it is the
explosive root.
Let K!0" ! K 0 .
K 0 ! c 2 " 25 # ". # c 2 ! K 0 # 25 " ".
K " !t" ! !K 0 # 25 " "" exp
" # " 2 " 4
2
t " 25 # ".
3. The problem from 5.13 is
$
max )!K" ! ! $!!K" # C!K # "%e #"t dt
0
s.t. K!0" ! K 0 , given.
F K ! e #"t ! $ !K"
F K # ! #e #"t C $ !K# "
F K # K# ! #e #"t C $$ !K# "
F K # K ! 0.
F K # t ! "e #"t C $ !K# "
After dividing through by #e #"t , the Euler equation is:
K% $C $$ !K# "% # "C $ !K# " " ! $ !K" ! 0