DifferentialEquationsTimeless

DifferentialEquationsTimeless - Econ 204 Differential...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Econ 204 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 ) = 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 | ), then there is an interval ( a, b ) containing t 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 | r, | t t | r } U 1 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 ([ t , t + ]) : ( 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 continuous on R . Observe that if ( z ( s ) , s ) R for all s [ t , t + ], 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 , t + ] } K sup {| z ( s ) x ( s ) | : s [ t , t + ] } k z x k 2 2 Therefore, k I ( z ) I ( x ) k k z...
View Full Document

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.

Page1 / 30

DifferentialEquationsTimeless - Econ 204 Differential...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online