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DifferentialEquations0808

# DifferentialEquations0808 - Econ 204Summer/Fall 2008...

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Econ 204–Summer/Fall 2008 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 0 ) = y 0 with ( y 0 , t 0 ) U . 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 (1) 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 ) . 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 0 | ≤ r, | t t 0 | ≤ r } ⊆ U 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 1

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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 ([ t 0 δ, t 0 + δ ]) : ( 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 and continuous on R . Observe 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 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 + t t 0 F ( z ( s ) , s ) ds y 0 t t 0 F ( x ( s ) , s ) ds = t t 0 ( F ( z ( s ) , s ) F ( x ( s ) , s )) ds | t t 0 | sup {| F ( z ( s ) , s ) F ( x ( s ) , s ) | : s [ t 0 δ, t 0 + δ ] } δK sup {| z ( s ) x ( s ) | : s [ t 0 δ, t 0 + δ ] } z x 2 Therefore, I ( z ) I ( x ) z x 2 , so I is a contraction. Since S is a complete metric space, I has a unique fixed point y S . Therefore, for all t [ t 0 δ, t 0 + δ ], we have y ( t ) = t t 0 F ( y ( s ) , s ) ds 2
F is continuous, so the Fundamental Theorem of Calculus implies that y ( t ) = F ( y ( t ) , t ) for all t ( t 0 δ, t 0 + δ ). Since we also have y ( t 0 ) = y 0 + t 0 t 0 F ( y ( s ) , s ) ds = y 0 y (restricted to ( t 0 δ, t 0 + δ )) is a solution of the initial value problem. On the other hand, suppose that ˆ y is any solution of the initial value problem on ( t 0 δ, t 0 + δ ). It is easy to check that (ˆ y ( s ) , s ) R for all s ( t 0 δ, t 0 + δ ), so we have | F y ( s ) , s ) | ≤ M ; this implies that ˆ y has a extension to a continuous function (still denoted ˆ y ) in S . Since ˆ y is a solution

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