Mathematical Models in Finance and Industry Math Problem

Mathematical Models in Finance and Industry
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Question one
qDx,t=-k∂μ∂x(x, t) with the k>0 which is the diffusion coefficient
To make this equation readable we denote qDx,t by μx,t such that
μx,t=-k∂μ∂x(x, t)
Then
ddtabμx,t dx=abμx,t dt
=-(q (b,t)-q(a,t)=-abqA,xx,t+qDxx,tdx
=-abμνxx,t-kμxxx,tdx
Implying that
=>abμxx,t-kμxxx,t+μνxx,tdx……………i
Equation (i) only when all intervals [a, b] and for all t>0.
Integrating equation (i) we get
μxx,t-kμxxx,t+μνxx,t+cμ=0 For all xϵR and for all t>0
Question two
* Ljn=μxj,tn+1-μ(xj,tn)∆t+μxj+1,tn-μ(xj,tn)h
Expansion using Taylor expansion we get
|Ljn|=∆t2μxj,tn+1-μ(xj,tn)∆t+Ch(μxj+1,tn-μxj,tn)h, expansion using Taylor expansion theorem
Simplifying the R.H.S Expressions we get
|Ljn|=∆t2μxj,tn+1-μ(xj,tn)∆t+h2(μxj+1,tn-μxj,tn)h…………………(i)
= ∆tμxj,tn+1-μxj,tn+h (μxj,tn+1-μxj,tn……………ii
, simplifying the two expressions in equation above
=( ∆t+h) (μxj,tn+1-μxj,tn) factoring the ( ∆t+h)
But μxj,tn+1-μxj,tn=CL
Replacing μxj,tn+1-μxj,tn with CL from the equation( i) we get
|Ljn|=CL( ∆t+h) Hence proven.
* To show that ejn=μμj-tn-μj satisfiesejn+1-ejn∆t+aej+1-ejh
We start at n=o by finding out the iterations of ejn
ej1=ej0-a∆tej0-ej-10h, j∈Z
ej2=ej1-a∆tej1-ej-11h, j∈Z
Up to
ejn+1=ejn-a∆tejn-ej-1nh, j∈Z and n>1
The exact solution for this expression
uμj-tn=e0x-at
Hence it satisfies ejn+1-ejn∆t+aej+1-ejh
Using induction on n to prove
supj∈Z| ejn|≤n∆tCL( ∆t+h)
|ejn|≤|(1-a∆th)ejn|+|a∆thμj-1n dropping the sup and expanding the R.H.S expression
≤(1-a∆th)|ejn|+a∆th|μj-1n| taking the absolute values of ejn+a∆th|μj-1n
The equation obtained is the same as
=a∆thsupj∈Z|ejn|
=supj∈Z|ejn|≤supj∈Z|ej0| equating to the original equation
But supj∈Z|ej0|= n...