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ECON 401 Hartman Autumn 2010 ANSWERS FOR PROBLEM SET VIII 1.a. The Hamiltonian is 1/2 (2 ) bt a Hec k cg k  . We choose c to maximize H , which gives 1 /0 bt a Hc e a c   , and the evolution of is given by 1/2 /( ) Hk k     . The first of these implies that 1 bt a ea c  so that 12 (1 ) bt a bt a be ac e a a c c  . We can substitute into 1/2 () k  to get 1 1 / 2 ) ( ) bt a bt a bt a be ac e a a c c e ac k  , which simplifies to 1/2 ) c ab k c  or 1/2 1 cb k c a  . To use the calculus of variations, solve 1/2 2 kk c g k  for c and substitute into the expression under the integral to get 1/2 0 ) bt a ekg k k d t  . If we let 1/2 (, , ) ) bt a Gtkk e k g k k  and remember that 1/2 2 ck gk k , then 11 / 2 ) bt a Gkea c k , 1 / bt a Gk ea c , and ) bt a bt a dG be ac e a a c c dt k    . The Euler equation requires that Gd G kd t k , and we can substitute to see that the Euler equation is / 2 1 2 ( 1 ) bt a bt a bt a e ac k be ac e a a c c   . The Euler equation simplifies to 1/2 ) c k c or 1/2 1 k c a , which is the same thing we found using the Hamiltonian. The transversality condition is 0.1 1 lim( ) 0 ta t c  . From the c equation, we see that if 0 c , then 1/2 kb  or 2 . In the graph on the next page, the 0 c locus is the vertical line above *2 kk b . If we move horizontally to the right of the line (i.e., if we hold c fixed at any level and increase k ), then c decreases from 0 c to a negative value because 3/2 1 / 2 ) / ( 1 ) 0 c k a     . (Remember that 01 a .) It follows that c is decreasing to the right of the 0 c line and increasing to the left of the line. From the k equation, we see that if 0 k , then 1/2 2 . For this equation, 1/2 / dc dk k and 22 3 / 2 1 / 2 ) 0 dc d k k . It follows that the equation 1/2 2 has a maximum when 2 k  . Now, b  and b  ; it follows that the maximum of the 0 k curve occurs to the right of . Moreover, if 0 k in this equation, then we see that 1/2 2 crosses the vertical axis at g . The 0 k locus has been drawn and labeled in the graph on the next page; note that the maximum for the 0 k curve occurs to the right of the 0 c line. If we move vertically up from the 0 k curve, then k decreases from 0 k to a negative number because /1 0 kc   . It follows that k increases below the 0 k curve and decreases above that curve. The optimal trajectory is the saddle point trajectory indicated in the graph. The steady state at the saddle point is stable along that trajectory.

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