Fourier Series For Function Not Centred at Zero

In summary, the problem involves solving a differential equation using separation of variables, with two coordinates and a given boundary condition. The solution involves an exponential function and a Fourier Series expansion, but the issue arises when the right hand side contains both cosine and sine terms. The question is how to determine the coefficients in this case and if there is a theory for shifting a function to obtain a Fourier Series in a different domain.
  • #1
modulus
127
3

Homework Statement


I was working on a problem where I had been given a differential equation to be solved using separation of variables. Two coordinates: a time coordinate and a single spatial coordinate (1-D problem).

Homework Equations


The domain for the spatial part was [0, L].
Given spatial function at t=0 (boundary condition). Call it Uo(x).

The Attempt at a Solution


I ended up with an exponential function in the time coordinate multiplied by an expansion that looked exactly like the Fourier Series of the spatial part. So, when I put t=0, I end up with Uo(x) on the left of the equality and what should be its Fourier Series on the right.

Under usual circumstances, I could extend the function to the domain [-L, L] by simply assigning Uo(x) (resp. its negative) to the domain [-L,0] to get an even (resp. odd) function and a cosine (resp. sine) Fourier series.

But the problem is that the right hand side has both cos and sin terms!

So, how do I determine the coefficients of in my solution now?

Is there any theory that can allow me to take a function centred at a point other than zero and shift it to get a Fourier Series for it in the relevant domain? From what I've tried out, I think it can be done, but it would involve making the arguments of the sines and cosines in the series2n*pi(x-a)/L where a is the point around which my function is centred.
Edit: The spatial boundary conditions (other than the one mentioned above, that was a boundary condition for the time coordinate) force the Fourier Series spatial solution to retain both cosine and sine terms.
 
Last edited:
Physics news on Phys.org
  • #2
modulus said:

Homework Statement


I was working on a problem where I had been given a differential equation to be solved using separation of variables. Two coordinates: a time coordinate and a single spatial coordinate (1-D problem).

Homework Equations


The domain for the spatial part was [0, L].
Given spatial function at t=0 (boundary condition). Call it Uo(x).

The Attempt at a Solution


I ended up with an exponential function in the time coordinate multiplied by an expansion that looked exactly like the Fourier Series of the spatial part. So, when I put t=0, I end up with Uo(x) on the left of the equality and what should be its Fourier Series on the right.

Under usual circumstances, I could extend the function to the domain [-L, L] by simply assigning Uo(x) (resp. its negative) to the domain [-L,0] to get an even (resp. odd) function and a cosine (resp. sine) Fourier series.

But the problem is that the right hand side has both cos and sin terms!

So, how do I determine the coefficients of in my solution now?

Is there any theory that can allow me to take a function centred at a point other than zero and shift it to get a Fourier Series for it in the relevant domain? From what I've tried out, I think it can be done, but it would involve making the arguments of the sines and cosines in the series2n*pi(x-a)/L where a is the point around which my function is centred.
Edit: The spatial boundary conditions (other than the one mentioned above, that was a boundary condition for the time coordinate) force the Fourier Series spatial solution to retain both cosine and sine terms.

It would be good if you gave us more equations and less prose so we understand better what is really going on.
 

Related to Fourier Series For Function Not Centred at Zero

What is a Fourier series for a function not centred at zero?

A Fourier series is a mathematical representation of a periodic function using a sum of sine and cosine functions. Typically, these functions are centred at zero, but in some cases, the function may be centred at a different point.

How is a Fourier series for a function not centred at zero different from a regular Fourier series?

The main difference is in the formula used to calculate the coefficients of the series. In a regular Fourier series, the coefficient formula involves integrals from -L to L, where L is half the period of the function. In a Fourier series for a function not centred at zero, the coefficient formula involves integrals from c-L to c+L, where c is the centre of the function.

Can a Fourier series for a function not centred at zero still be used to approximate the function?

Yes, a Fourier series for a function not centred at zero can still be used to approximate the function. However, the accuracy of the approximation may be affected by the location of the centre of the function.

What are the applications of Fourier series for functions not centred at zero?

Some applications of Fourier series for functions not centred at zero include analyzing non-periodic signals, studying non-symmetric functions, and solving boundary value problems in physics and engineering.

Are there any limitations or drawbacks to using a Fourier series for a function not centred at zero?

One limitation is that the coefficient formula may become more complicated and difficult to calculate. Additionally, the accuracy of the approximation may be affected by the location of the centre of the function. For some functions, it may be more appropriate to use other methods of approximation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
397
  • Calculus and Beyond Homework Help
Replies
3
Views
440
  • Calculus and Beyond Homework Help
Replies
6
Views
471
  • Calculus and Beyond Homework Help
Replies
5
Views
330
  • Calculus and Beyond Homework Help
Replies
1
Views
306
  • Calculus and Beyond Homework Help
Replies
6
Views
961
  • Calculus and Beyond Homework Help
Replies
1
Views
362
  • Calculus and Beyond Homework Help
Replies
5
Views
501
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Back
Top