Need help to prove expressions using Fourier Expansion

In summary, based on the given data, the differential equation to be solved is: -\frac{dU^{o}}{dz} (1-\frac{x}{a}) - \frac{dU^{a}}{dz}(\frac{x}{a})
  • #1
kaizen.moto
98
0
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.
 
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  • #2
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
Apologies for the unclear statements.
f1(x) = (1-x/a) can be transformed using Fourier expansion into
bo +[itex]\Sigma[/itex] bm*Cos[m*Pi*x/a] and

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

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

Hope this would helps.

Thank you.
 
Last edited:
  • #4
So:
[tex]
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)
[/tex]

What is the differential equation you're trying to solve?
 
  • #5
The only available data given to me are as follows:
-[itex]\frac{dU^{o}}{dz}[/itex] (1-[itex]\frac{x}{a}[/itex]) - [itex]\frac{dU^{a}}{dz}[/itex]([itex]\frac{x}{a}[/itex])........(1)

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

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


Case 1: m[itex]\neq[/itex]0 and n[itex]\neq[/itex]0:

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

Case 2: m = 0 and n[itex]\neq[/itex]0:

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

Case 3: m[itex]\neq[/itex]0 and n = 0:

eqn (1) = 0


Using Fourier expansion, I could expand the following:

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

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

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

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



Any help is greatly appreciated.
 

Related to Need help to prove expressions using Fourier Expansion

What is a Fourier Expansion?

A Fourier Expansion is a mathematical technique used to represent a function as a sum of infinite trigonometric functions. It is named after mathematician Joseph Fourier and is often used in signal processing and other areas of science and engineering.

How is Fourier Expansion used to prove expressions?

Fourier Expansion can be used to prove expressions by breaking down the original function into simpler trigonometric functions and then manipulating them to match the desired expression. This process involves using Fourier coefficients and properties of trigonometric functions.

What are some common applications of Fourier Expansion?

Fourier Expansion is commonly used in signal processing to analyze and transform signals in the time and frequency domains. It is also used in solving differential equations, heat transfer, and image processing.

What are some challenges in using Fourier Expansion to prove expressions?

One challenge in using Fourier Expansion is determining the appropriate Fourier coefficients for a given function. This often involves complex calculations and can be time-consuming. Additionally, some functions may not have a unique Fourier Expansion, making it difficult to prove certain expressions.

Are there any limitations to using Fourier Expansion?

Yes, Fourier Expansion is limited to functions that are periodic and can be represented as a sum of trigonometric functions. It also assumes that the function is continuous and differentiable, which may not always be the case. Additionally, the convergence of Fourier series may be an issue for some functions.

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