204Lecture142006

# 204Lecture142006 - Economics 204 Lecture 14Thursday Revised...

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Economics 204 Lecture 14–Thursday, August 17, 2006 Revised 8/17/06, Revisions indicated by ** 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 : R n × R R n . An initial value problem is a differ- ential equation combined with an initial condition y ( t 0 ) = y 0 A solution of the initial value problem is a differentiable function y : ( a, b ) R n such that t 0 ( a, b ), y ( t 0 ) = y 0 and, for all t ( a, b ), dy dt = F ( y ( t ) , t ). Theorem 2 Consider the initial value problem y ( t ) = F ( y ( t ) , t ) , y ( t 0 ) = y 0 Let U be an open set in R n × R containing ( y 0 , t 0 ) . 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 ˆ y | then there is an interval ( a, b ) containing t 0 such that the solution is unique on ( a, b ) . 1

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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 0 | ≤ r, | t t 0 | ≤ r } ⊆ U Since F is continuous, we may find M R and r > 0 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 ˆ y | for all ( y, t ) , y, t ) R Let δ = min 1 2 K , r M We claim the initial value problem has a unique solution on ( t 0 δ, t 0 + δ ). Let C be the space of continuous functions from [ t 0 δ, t 0 + δ ] to R n , endowed with the sup norm f = sup {| f ( t ) | : t [ t 0 δ, t 0 + δ ] } Let S = { z C : ( z ( s ) , s ) R for all s [ t 0 δ, t 0 + δ ] } 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 0 + t t 0 F ( z ( s ) , s ) ds I ( z ) is defined and continuous because F is bounded on R . Ob- serve that if ( z ( s ) , s ) R for all s [ t 0 δ, t 0 + δ ], then | I ( z )( t ) y 0 | = t t 0 F ( z ( s ) , s ) ds ≤ | t t 0 | max {| F ( y, s ) | : ( y, s ) R } δM r 2
so ( I ( z )( t ) , t ) R for all t [ t 0 δ, t 0 + δ ]. Thus, I : S S .

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• Fall '08
• Staff
• Economics, Continuous function, Metric space, Taylor’s Theorem, Lipschitz continuity, Lipschitz

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