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## Adomian Decomposition Method for a fractional power

### Zulfikar Ahmed<zulfikar.moinuddin.ahmed@gmail.com>

2:37 PM (44 minutes ago)

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Dear Cutiest,

While you sort out your fears regarding manic communications and relations and their ships etc.  I am trying to nail down the Adomian Decomposition Method which is pretty widespread.  The various nonlinear partial differential equations solved by this method.  This is not very deep but requires some skills.

For a (fractional) partial differential equation with a nonlinear term Nu, for example Nu = u^alpha or whatever, the key issue is the computation of the so-called Adomian polynomial, A_0, A_1, A_2, …  What are these polynomials?

The idea is that you want to solve a partial differential equation say of type Lu + Ru + Nu = g, very very general.  If Nu were a linear operator, solving this is classical.  George Adomian came up with a general method of solving these with the following conditions:

(a)  Easily invertible L, the highest order linear differential operator
(b)  The final solution will be a series expansion \sum_{k=0}^\infty u_k

(c)  The solution will be iterative with u_0, u_1, u_2, u_3, … determined stepwise where the nonlinear contribution Nu will be approximated at each step by a polynomial A_k( u_0, …, u_k) instead of a linear combination.

So the central issue is to determine these polynomials.  Formally the polynomials have a nice looking formula but computing this formula requires some skill:

A_k(u_0,…,u_k) = 1/k! d^k/dz^k N(\sum_k z^k u_k)|_{z=0}

The explicit calculation of for a specific form of N of this formula such as Nu = u^alpha requires calculus skills. So this is the problem of the moment, calculus skills from tenth grade … determining these Adomian Polynomials … not very deep mathematics but useful.

Why would we care?  We would care because we’re interested in nonlinear FORCING term ‘F’ in our model equations for volatility.  We want to be able to numerically simulate solutions of fractional diffusions with nonlinearities:

((d/dt)^alpha + (-Laplacian)^beta)u = F(u)

where F(u) might be quite general.  Our F(u) is Nu in the setting.  The attached paper is recent where some Iranians solve some fpde using this method and we want to be able to gain some skills for this.

Attachments area

### Zulfikar Ahmed<zulfikar.moinuddin.ahmed@gmail.com>

3:21 PM (0 minutes ago)

 to namewithdrawn
Let’s do the exercise of computing the Adomian polynomials of a fractional power directly by dividing the problem into two steps.  We use the chain rule for derivatives

A_0 = u_0^alpha

For A_1,

d/dz (u_0 + u_1 z + u_2 z^2 + …)^alpha = alpha (u_0 + u_1 z + ….)^{alpha-1} ( u_1 + 2*u_2*z + …)

from which we get, setting z=0 that

A_1 = alpha*u_0^{alpha-1}*u_1

Then we have to use the product rule for derivatives for A_2:

d^2/dz^2 (u_0 + u_1 z + u_2 z^2 + …) = alpha * d/dz (u_0+u_1 z + …)^{alpha-1} (u_1 + 2 u_2 z + …) + alpha * (u_0+u_1 z + …)^{alpha-1} d/dz(u_1 + 2 u_2 z + …)

so

A_2 = alpha*(alpha-1) u_1^{alpha-2} u_1^2 + alpha*u_0^{alpha-1}* 2*u_2
Well that’s already quite messy, but it’s not deep just messy.