204Lecture142009 - Economics 204 Lecture 14Thursday, August...

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Unformatted text preview: Economics 204 Lecture 14Thursday, August 13, 2009 Revised 8/13/09, revisions indicated by ** and Sticky Notes Differential Equations Existence and Uniqueness of Solutions Definition 1 A differential equation is an equation of the form y ( t ) = F ( y ( t ) , t ) where F : U R n where U is an open subset of R n R . An initial value problem is a differential equation combined with an initial condition 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 | 1 then there is an interval ( a, b ) containing t such that the solution is unique on ( a, b ) . Proof: We consider only the case in which F is Lipschitz. Since U is open, we may choose r > 0 such that R = { ( y, t ) : | y y | r, | t t | r } U Since F is continuous, we may find M R ** such that | F ( y, t ) | M for all ( y, t ) R . 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 : ( 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 continu- ous on R . Observe that if ( z ( s ) , s ) R for all s [ t , t + ], 2 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 K sup {| z ( s ) x ( s ) | : s [ t , t + ] } k z x k 2 Therefore,...
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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.

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204Lecture142009 - Economics 204 Lecture 14Thursday, August...

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