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Unformatted text preview: Econ 204 Differential Equations In this supplement, we use the methods we have developed so far to study differential equations. 1 Existence and Uniqueness of Solutions Definition 1 A differential equation is an equation of the form y ( t ) = F ( y ( t ) , t ) where U is an open subset of R n R and F : U R n . An initial value problem is a differential equation combined with an initial con dition y ( t ) = y with ( y , t ) U . A solution of the initial value problem is a differentiable function y : ( a, b ) R n such that t ( a, b ), y ( t ) = y and, for all t ( a, b ), dy dt = F ( y ( t ) , t ). The general solution of the differential equation is the family of all solutions for all initial values ( y , t ) U . Theorem 2 Consider the initial value problem y ( t ) = F ( y ( t ) , t ) , y ( t ) = y (1) Let U be an open set in R n R containing ( y , t ) . Suppose F : U R n is continuous. Then the initial value problem has a solution. If, in addition, F is Lipschitz in y on U (i.e. there is a constant K such that for all ( y, t ) , ( y, t ) U ,  F ( y, t ) F ( y, t )  K  y y  ), then there is an interval ( a, b ) containing t such that the solution is unique on ( a, b ) . Proof: We will limit our proof to the case in which F is Lipschitz; for the general case, see Coddington and Levinson [1]. Since U is open, we may choose r > 0 such that R = { ( y, t ) :  y y  r,  t t  r } U 1 Given the Lipschitz condition, we may assume that  F ( y, t ) F ( y, t )  K  y y  for all ( y, t ) , ( y, t ) R . Let = min 1 2 K , r M We claim the initial value problem has a unique solution on ( t , t + ). Let C be the space of continuous functions from [ t , t + ] to R n , endowed with the sup norm k f k = sup { f ( t )  : t [ t , t + ] } Let S = { z C ([ t , t + ]) : ( z ( s ) , s ) R for all s [ t , t + ] } S is a closed subset of the complete metric space C , so S is a complete metric space. Consider the function I : S C defined by I ( z )( t ) = y + Z t t F ( z ( s ) , s ) ds I ( z ) is defined and continuous because F is bounded and continuous on R . Observe that if ( z ( s ) , s ) R for all s [ t , t + ], then  I ( z )( t ) y  = Z t t F ( z ( s ) , s ) ds  t t  max { F ( y, s )  : ( y, s ) R } M r so ( I ( z )( t ) , t ) R for all t [ t , t + ]. Thus, I : S S . Given two functions x, z S and t [ t , t + ],  I ( z )( t ) I ( x )( t )  = y + Z t t F ( z ( s ) , s ) ds y Z t t F ( x ( s ) , s ) ds = Z t t ( F ( z ( s ) , s ) F ( x ( s ) , s )) ds  t t  sup { F ( z ( s ) , s ) F ( x ( s ) , s )  : s [ t , t + ] } K sup { z ( s ) x ( s )  : s [ t , t + ] } k z x k 2 2 Therefore, k I ( z ) I ( x ) k k z...
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at University of California, Berkeley.
 Summer '08
 ANDERSON

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