Need help to prove expressions using Fourier Expansion

[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.

 

[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

 

[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 + \Sigmaam*Cos[m*Pi*x/a]

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

Hope this would helps.

Thank you.

 

[hunt_mat]

So:
<br /> 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)<br />

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

 

[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\neq0 and n\neq0:

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\neq0:

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

Case 3: m\neq0 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.