hw2-3 - 2 Basic Suppose V S t is the solution of the...

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Unformatted text preview: 2 Basic Options 29 19. Suppose V ( S, t ) is the solution of the problem ∂V ∂t + 1 2 σ 2 ( S ) S 2 ∂ 2 V ∂S 2 + ( r- D ) S ∂V ∂S- rV = 0 , ≤ S, t ≤ T, V ( S, T ) = V T ( S ) , ≤ S. Let ξ = S S + P m , τ = T- t and V ( S, t ) = ( S + P m ) V ( ξ, τ ), where P m is a positive constant. a) Show that V ( ξ, τ ) is the solution of the problem ∂ V ∂τ = 1 2 ¯ σ 2 ( ξ ) ξ 2 (1- ξ ) 2 ∂ 2 V ∂ξ 2 + ( r- D ) ξ (1- ξ ) ∂ V ∂ξ- [ r (1- ξ ) + D ξ ] V , ≤ ξ ≤ 1 , ≤ τ, V ( ξ, 0) = (1- ξ ) P m V T µ P m ξ 1- ξ ¶ , where ¯ σ ( ξ ) = σ µ P m ξ 1- ξ ¶ . b) What are the advantages of reformulating the problem on a finite domain? Solution : a) Noticing S = P m ξ 1- ξ , S + P m = P m 1- ξ , P m S + P m = 1- ξ, dξ dS = P m ( S + P m ) 2 , we have ∂V ∂t = ( S + P m ) ∂ V ∂τ · dτ dt =- ( S + P m ) ∂ V ∂τ =- P m 1- ξ ∂ V ∂τ , ∂V ∂S = V ( ξ, τ ) + ( S + P m ) ∂ V ∂ξ · dξ dS = V ( ξ, τ ) + P m S + P m ∂ V ∂ξ = V ( ξ, τ ) + (1- ξ ) ∂ V ∂ξ , ∂ 2 V ∂S 2 = ∂ V ∂ξ dξ dS- dξ dS ∂ V ∂ξ + (1- ξ ) ∂ 2 V ∂ξ 2 dξ dS = (1- ξ ) 3 P m ∂ 2 V ∂ξ 2 . Substituting these expressions into the equation yields- P m 1- ξ ∂ V ∂τ + 1 2 σ 2 ( ξ ) µ P m ξ 1- ξ ¶ 2 (1- ξ ) 3 P m ∂ 2 V ∂ξ 2 +( r- D ) P m ξ 1- ξ µ V + (1- ξ ) ∂ V ∂ξ ¶- r P m 1- ξ V ( ξ, τ...
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This note was uploaded on 02/11/2010 for the course MATH 6203 taught by Professor Zhu during the Spring '10 term at University of North Carolina Wilmington.

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hw2-3 - 2 Basic Suppose V S t is the solution of the...

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