# Need help to prove expressions using Fourier Expansion

1. Jun 25, 2011

### kaizen.moto

Dear all,

Please help me to prove the Fourier expansion for three different cases as follows. I need some help to show that L.H.S = R.H.S

Case1: when m not equal to 0 and n not equal to 0.

dU^0/dz (1- x/a) - dU^a/dz (x/a) = -(2/m^2 Pi^2)(1- Cos[m*Pi]) [dUn^0/dz - dUn^a/dz]

Case2: when m = 0 and n not equal to 0.

dU^0/dz (1- x/a) - dU^a/dz (x/a) = -1/2 [dUn^0/dz + dUn^a/dz]

Case3: when m not equal to 0 and n = 0.

dU^0/dz (1- x/a) - dU^a/dz (x/a) = 0

where
dU^0/dz = dUn^0/dz Sin[ n*Pi*x/a] and dU^a/dz = dUn^a/dz Sin[ n*Pi*x/a].

The idea is that firstly using Fourier expansion, to expand (1 - x/a), then multiply with dU^0/dz, similarly, apply Fourier expansion for (x/a), then multiply with dU^a/dz.

Thanks for any help.

2. Jun 26, 2011

### hunt_mat

I am not too sure what you're saying here, the function 1-x/a does not have a Fourier transform, can you possibly tex up your question

3. Jun 26, 2011

### kaizen.moto

Apologies for the unclear statements.
f1(x) = (1-x/a) can be transformed using Fourier expansion into
bo +$\Sigma$ bm*Cos[m*Pi*x/a] and

f2(x) = x/a can be transformed using Fourier expansion into
ao + $\Sigma$am*Cos[m*Pi*x/a]

Iam aware that the original equation of f(x) = ao +$\Sigma$am*Cos[m*Pi*x/a] + bm Sin[m*Pi*x/a].

Hope this would helps.

Thank you.

Last edited: Jun 26, 2011
4. Jun 26, 2011

### hunt_mat

So:
$$f(x)=a_{0}+\sum_{m=1}^{\infty}a_{m}\cos\left(\frac{m\pi x}{a}\right) +b_{m}\sin\left(\frac{m\pi x}{a}\right)$$

What is the differential equation you're trying to solve?

5. Jun 26, 2011

### kaizen.moto

The only available data given to me are as follows:
-$\frac{dU^{o}}{dz}$ (1-$\frac{x}{a}$) - $\frac{dU^{a}}{dz}$($\frac{x}{a}$)..................................(1)

U$^{o}$ = U$^{o}_{n}$ *Sin[$\frac{n*π*x}{a}$]....................................(2)

U$^{a}$ = U$^{a}_{n}$ *Sin[$\frac{n*π*x}{a}$]...................................(3)

Case 1: m$\neq$0 and n$\neq$0:

eqn(1) = $\frac{-2}{m^2*π^2}$*[1-cos(m*π)]*[$\frac{dU^{0}_{n}}{dz}$-$\frac{dU^{a}_{n}}{dz}$]

Case 2: m = 0 and n$\neq$0:

eqn(1) = -$\frac{1}{2}$*[$\frac{dU^{0}_{n}}{dz}$+$\frac{dU^{a}_{n}}{dz}$]

Case 3: m$\neq$0 and n = 0:

eqn (1) = 0

Using Fourier expansion, I could expand the following:

($\frac{x}{a}$) $\approx$ a$_{o}$+$\sum$ a$_{m}$*Cos[$\frac{m*Pi*x}{a}$]

$\int$$^{0}_{a}$ ($\frac{x}{a}$)*Cos[$\frac{L*m*x}{a}$] dx=$\sum$$\int$$^{0}_{a}$ a$_{m}$*Cos[$\frac{m*Pi*x}{a}$]*Cos[$\frac{L*m*x}{a}$] dx

$\int$$^{0}_{a}$ ($\frac{x}{a}$) Cos[$\frac{L*m*x}{a}$] dx= $\sum$$\int$$^{0}_{a}$ a$_{m}$*$\frac{1}{2}$[Cos($\frac{m*Pi*x}{a}$-$\frac{L*Pi*x}{a}$)+Cos($\frac{m*Pi*x}{a}$+$\frac{L*Pi*x}{a}$)]dx

Note that L = m, then, Iam stucked up to this point. I have no idea how to proceed next.

Similarly, the above expansion can be used for (1 - $\frac{x}{a}$):

(1-$\frac{x}{a}$) $\approx$ b$_{o}$+$\sum$ b$_{m}$*Cos[$\frac{m*Pi*x}{a}$]

Any help is greatly appreciated.