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204Lecture142008Web

# 204Lecture142008Web - Economics 204 Lecture 14Thursday...

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Economics 204 Lecture 14–Thursday, August 14, 2008 Diferential Equations Existence and Uniqueness oF Solutions De±nition 1 A diferential equation is an equation of the form y 0 ( 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 diFerential equation combined with an initial condition y ( t 0 y 0 with ( y 0 0 ) U .A solution of the initial value problem is a diFerentiable 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 ) ). Theorem 2 Consider the initial value problem y 0 ( t F ( y ( t ) ) ,y ( t 0 y 0 (1) Let U be an open set in R n × R containing ( y 0 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 ( ) 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 fnd M R and 0 such that | F ( ) M For all ( ) R . Given the Lipschitz condition, we may assume that | F ( ) F y, t ) K | y ˆ y | For all ( ) , y, t ) R Let δ =m in ± 1 2 K , r M ² 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 ,endowedw iththesupnorm k f k =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 defned by I ( z )( t )= y 0 + Z t t 0 F ( z ( s ) ) ds I ( z ) is defned and continuous because F is bounded and continuous on R .Ob s e rv eth a ti F( z ( s ) ) R For all s [ t 0 δ, t 0 + δ ], then | I ( z )( t ) y 0 | = Z t t 0 F ( z ( s ) ) ds 2
≤| t t 0 | max {| F ( y, s ) | :( ) R } δM r so ( I ( z )( t ) ,t ) R for all t [ t 0 δ, t 0 + δ ]. Thus, I : S S . Given two functions x, z S and t [ t 0 δ, t 0 + δ ], | I ( z )( t ) I ( x )( t ) | = y 0 + Z t t 0 F ( z ( s ) ,s ) ds y 0 Z t t 0 F ( x ( s ) ) ds = Z t t 0 ( F ( z ( s ) ) F ( x ( s ) )) ds t t 0 | sup {| F ( z ( s ) ) F ( x ( s ) ) | : s [ t 0 δ, t 0 + δ ] } δK sup {| z ( s ) x ( s ) | : s [ t 0 δ, t 0 + δ ] } k z x k 2 Therefore, k I ( z ) I ( x ) k k z x k 2 ,so I is a contraction. Since S is a complete metric space, I has a unique Fxed point y S . Therefore, for all t [ t 0 δ, t 0 + δ ], we have y ( t )= y 0 + Z t t 0 F ( y ( s ) ) ds F is continuous, so the ±undamental Theorem of Calculus implies that y 0 ( t F ( y ( t ) ) for all t ( t 0 δ, t 0 + δ ). Since we also have y ( t 0 y 0 + Z t 0 t 0 F ( y ( s ) ) ds = y 0 y (restricted to ( t 0 δ, t 0 + δ )) is a solution of the initial value problem (1). On the other hand, suppose that ˆ y is any solution of the initial value problem (1) on ( t 0 δ, t 0 + δ ). It is easy to check that (ˆ y ( s ) ) R for all s ( t 0 δ, t 0 + δ ), so we have | F y ( s ) ) |≤ M ; 3

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this implies that ˆ y has a extension to a continuous function (still denoted ˆ y )in S .S i n c y is a solution
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204Lecture142008Web - Economics 204 Lecture 14Thursday...

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