204Lecture142009

# 204Lecture142009 - Economics 204 Lecture 14–Thursday...

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Unformatted text preview: Economics 204 Lecture 14–Thursday, August 13, 2009 Revised 8/13/09, revisions indicated by ** and Sticky Notes 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 : U → R n where U is an open subset of R n × R . An initial value problem is a differential equation combined with an initial condition 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 ) , (ˆ y, t ) ∈ U , | F ( y, t ) − F (ˆ y, t ) | ≤ K | y − ˆ y | 1 then there is an interval ( a, b ) containing t such that the solution is unique on ( a, b ) . 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 | ≤ r, | t − t | ≤ r } ⊆ U Since F is continuous, we may find M ∈ R ** such that | F ( y, t ) | ≤ M for all ( y, t ) ∈ R . Given the Lipschitz condition, we may assume that | F ( y, t ) − F (ˆ y, t ) | ≤ K | y − ˆ y | for all ( y, t ) , (ˆ y, t ) ∈ R Let δ = min ⎧ ⎪ ⎨ ⎪ ⎩ 1 2 K , r M ⎫ ⎪ ⎬ ⎪ ⎭ We claim the initial value problem has a unique solution on ( t − δ, t + δ ). Let C be the space of continuous functions from [ t − δ, t + δ ] to R n , endowed with the sup norm k f k ∞ = sup {| f ( t ) | : t ∈ [ t − δ, t + δ ] } Let S = { z ∈ C : ( z ( s ) , s ) ∈ R for all s ∈ [ t − δ, t + 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 + Z t t F ( z ( s ) , s ) ds I ( z ) is defined and continuous because F is bounded and continu- ous on R . Observe that if ( z ( s ) , s ) ∈ R for all s ∈ [ t − δ, t + δ ], 2 then | I ( z )( t ) − y | = Z t t F ( z ( s ) , s ) ds ≤ | t − t | max {| F ( y, s ) | : ( y, s ) ∈ R } ≤ δM ≤ r so ( I ( z )( t ) , t ) ∈ R for all t ∈ [ t − δ, t + δ ]. Thus, I : S → S . Given two functions x, z ∈ S and t ∈ [ t − δ, t + δ ], | 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 ≤ δK sup {| z ( s ) − x ( s ) | : s ∈ [ t − δ, t + δ ] } ≤ k z − x k ∞ 2 Therefore,...
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204Lecture142009 - Economics 204 Lecture 14–Thursday...

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