DifferentialEquations0808

# - Econ 204Summer/Fall 2008 Differential Equations In this supplement we use the methods we have developed so far to study differential equations 1

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Econ 204–Summer/Fall 2008 Diferential Equations In this supplement, we use the methods we have developed so far to study diFerential equations. 1 Existence and Uniqueness o± Solutions De²nition 1 A diferential equation is an equation of the form y 0 ( t )= F ( y ( t ) ,t )whe re U is an open subset of R n × R and F : U R n .A n initial value problem is a diFerential equation combined with an initial con- dition y ( t 0 y 0 with ( y 0 0 ) U 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 aso lu t ion . 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 ) . Proo±: 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 y 0 r, | t t 0 r }⊆ U Given the Lipschitz condition, we may assume that | F ( ) F y, t ) K | y ˆ y | for all ( ) , y, t ) R .L e t δ =m in ± 1 2 K , r M ² 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 k f k =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 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 . Observe that if ( 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 ± ± ± ± ≤| 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 ) ) 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 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 Z t t 0 F ( y ( s ) ) ds 2
F is continuous, so the Fundamental Theorem of Calculus implies that y 0 ( t )= F ( y ( t ) ,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 ) ,s ) ds = y 0 y (restricted to ( t 0 δ, t 0 + δ )) is a solution of the initial value problem.

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## This note was uploaded on 08/01/2008 for the course ECON 204 taught by Professor Anderson during the Fall '08 term at University of California, Berkeley.

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- Econ 204Summer/Fall 2008 Differential Equations In this supplement we use the methods we have developed so far to study differential equations 1

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