2 - Larry Karp Notes for Dynamics II General Dynamic...

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P1 max m T t 0 e ± rt U x ( t ), y ( t ) dt s . t . ˙ x ± f ( x ) ² y state equation x t 0 given initial condition x ( T ) ± x T terminal condition Larry Karp Notes for Dynamics September 19, 2001 II. General Dynamic Problem in Resources, Calculus of Variations 1) Description of dynamic optimization problem 2) Statement of necessary condition (Euler equation) 3) Interpretation of Euler Equation 4) Autonomous problems and steady states 5) Derivation of Euler Equation 6) Special functional forms 7) Necessary and sufficient conditions 8) Different boundary conditions for dynamic problems 9)An example: extraction of a non-renewable resource 1) Examples of control problems in natural resources x = stock of resources, x = initial condition 0 y = "harvest" f ( x ) = natural growth U ( x , y ) flow of payoff - the integrand r = discount rate e = discount factor - rt T = final time x = terminal condition T In this problem, x is the state variable, y is the control variable, and dx/dt = f(x) -y is the equation of motion. + +/- fisheries U ( x , y )
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F ( t , x , ˙ x ) / w f ( x ) ² ˙ x ² c x , f ( x ) ² ˙ x e ± rt P2 max m T t 0 F ( t , x , ˙ x ) dt s. t . x 0 given x T given When using the Maximum Principle, we typically retain the control variable y, and we do 1 not eliminate the equation of motion. 2:2 - + fisheries (compet) = p ( t ) y - c ( x , y ) fisheries (monopoly) = p ( y ) @ y - c ( x , y ) fisheries (social planner) = w ( y ) - c ( x , y ) where w N / p ( y ) ____________________ pollution x ± = f ( x ) + g <---contribution to pollution + U ( x,g ) = v ( g ) - D ( x ) benefit damage function function nonrenewable resource f ( x ) / 0 x ± = -y For the C.O.V. problem write integrand as F ( t , x , x ± ) using y = f ( x ) - x ± . For example, for fishery controlled by social planner: I can write P1 as I have eliminated the constraint x ± =f ( x )- y by substitution. I get rid of one variable ( y ) and one constraint and write the problem in terms of x and its derivative dx/dt. It is common to eliminate the control variable (here y) when using COV to solve a dynamic problem. That is, instead of choosing the optimal trajectory for the control variable y (harvest), we explicitly choose dx/dt. 1 2. Necessary Condition
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F x ( t , x ² , ˙ x ² ) ± d dt F ˙ x ( t , x ² , ˙ x ² ) ± F ˙ xt ³ F ˙ xx dx dt ³ F ˙ x ˙ x d 2 x dt 2 ± F ˙ xt ³ F ˙ xx ˙ x ³ F ˙ x ˙ x ¨ x m t 1 t F x ( τ ) d τ ± m t 1 t dF ˙ x ( τ ) ± F ˙ x ( t 1 ) ² F ˙ x ( t ) ³ m t 1 t F x ( τ , x ( τ ) ² , ˙ x ( τ ) ² ) d τ F ˙ x ( t , x ² ( t 1 ), x ² ( t 1 ) ² F ˙ x ( t , x ² t , ˙ x ² ( t ) ) 2:3 (Subscripts indicate partial derivatives.) A necessary condition for P2 is that the optimal x solves * the Euler equation: This is a second order ODE, and we have 2 B.C.'s, x ( ( )= x and x ( T ) = x This is a A two point ( T boundary value problem @ (TPBVP) 3. Interpretation of Euler Equation Interpretation of Euler equation: Integrate, using or, = Benefit over t , t of having Leaving one more fish in sea at t and 1 the extra fish in the stock taking it out at t (reallocating consumption) 1 Mention case where F / 0. This inequality would hold if, for example, a larger stock does x not make it any cheaper to catch fish, and the growth is independent of the stock of fish. The independence of growth on the stock of fish is not plausible. However, the independence of the
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