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Note: 2 lectures, §9.3 in [EP], §10.4 in [BD]

You may have noticed by now that an odd function has no cosine terms in the Fourier series and an even function has no sine terms in the Fourier series. This observation is not a coincidence. Let us look at even and odd periodic function in more detail.

Recall that a function is odd if . A function is even if . For example, is even and is odd. Similarly the function is even if is even and odd when is odd.

Exercise 4.4.1: Take two functions and and deﬁne their product . a) Suppose both are odd, is odd or even? b) Suppose one is even and one is odd, is odd or even? c) Suppose both are even, is odd or even?

If and are both odd, then is odd. Similarly for even functions. On the other hand, if is odd and even, then we cannot say anything about the sum . In fact, the Fourier series of any function is a sum of an odd (the sine terms) and an even (the cosine terms) function.

In this section we consider odd and even periodic functions. We have previously deﬁned the -periodic extension of a function deﬁned on the interval . Sometimes we are only interested in the function on the range and it would be convenient to have an odd (resp. even) function. If the function is odd (resp. even), all the cosine (resp. sine) terms will disappear. What we will do is take the odd (resp. even) extension of the function to and then extend periodically to a -periodic function.

Take a function deﬁned on . On deﬁne the functions

Extend and to be -periodic. Then is called the odd periodic extension of , and is called the even periodic extension of .Example 4.4.1: Take the function deﬁned on . Figure 4.11 shows the plots of the odd and even extensions of .

Let be an odd -periodic function. We write the Fourier series for . First, we compute the coeﬃcients (including ) and get

That is, there are no cosine terms in the Fourier series of an odd function. The integral is zero because is an odd function (product of an odd and an even function is odd) and the integral of an odd function over a symmetric interval is always zero. The integral of an even function over a symmetric interval is twice the integral of the function over the interval . The function is the product of two odd functions and hence is even.

We now write the Fourier series of as

Similarly, if is an even -periodic function. For the same exact reasons as above, we ﬁnd that and

The formula still works for , in which case it becomes

The Fourier series is then

An interesting consequence is that the coeﬃcients of the Fourier series of an odd (or even) function can be computed by just integrating over the half interval . Therefore, we can compute the Fourier series of the odd (or even) extension of a function by computing certain integrals over the interval where the original function is deﬁned.

Theorem 4.4.1. Let be a piecewise smooth function deﬁned on . Then the odd extension of has the Fourier series

where

The even extension of has the Fourier series

where

The series is called the sine series of and the series is called the cosine series of . We often do not actually care what happens outside of . In this case, we pick whichever series ﬁts our problem better.

It is not necessary to start with the full Fourier series to obtain the sine and cosine series. The sine series is really the eigenfunction expansion of using the eigenfunctions of the eigenvalue problem , , . The cosine series is the eigenfunction expansion of using the eigenfunctions of the eigenvalue problem , , . We could have, therefore, gotten the same formulas by deﬁning the inner product

and following the procedure of § 4.2. This point of view is useful, as we commonly use a speciﬁc series that arose because our underlying question led to a certain eigenvalue problem. If the eigenvalue problem is not one of the three we covered so far, you can still do an eigenfunction expansion, generalizing the results of this chapter. We will deal with such a generalization in chapter 5.

Example 4.4.2: Find the Fourier series of the even periodic extension of the function for .

We want to write

where

and

Note that we have “detected” the continuity of the extension since the coeﬃcients decay as . That is, the even extension of has no jump discontinuities. It does have corners, since the derivative, which is an odd function and a sine series, has jumps; it has a Fourier series whose coeﬃcients decay only as .

Explicitly, the ﬁrst few terms of the series are

Exercise 4.4.3: a) Compute the derivative of the even extension of above and verify it has jump discontinuities. Use the actual deﬁnition of , not its cosine series! b) Why is it that the derivative of the even extension of is the odd extension of ?

Fourier series ties in to the boundary value problems we studied earlier. Let us see this connection in more detail.

Suppose we have the boundary value problem for ,

for the Dirichlet boundary conditions , . By using the Fredholm alternative (Theorem 4.1.2) we note that as long as is not an eigenvalue of the underlying homogeneous problem, there exists a unique solution. Note that the eigenfunctions of this eigenvalue problem are the functions . Therefore, to ﬁnd the solution, we ﬁrst ﬁnd the Fourier sine series for . We write also as a sine series, but with unknown coeﬃcients. We substitute the series for into the equation and solve for the unknown coeﬃcients. If we have the Neumann boundary conditions , , we do the same procedure using the cosine series.

Let us see how this method works on examples.

Example 4.4.3: Take the boundary value problem for ,

where on , and satisfying the Dirichlet boundary conditions , . We write as a sine series

where

We write as

We plug in to obtain

Therefore,

or

We have thus obtained a Fourier series for the solution

Example 4.4.4: Similarly we handle the Neumann conditions. Take the boundary value problem for ,

where again on , but now satisfying the Neumann boundary conditions , . We write as a cosine series

where

and

We write as a cosine series

We plug in to obtain

Therefore, , for even () and for odd we have

or

The Fourier series for the solution is

Exercise 4.4.4: Take deﬁned on . a) Sketch the plot of the even periodic extension of . b) Sketch the plot of the odd periodic extension of .

Exercise 4.4.5: Find the Fourier series of both the odd and even periodic extension of the function for . Can you tell which extension is continuous from the Fourier series coeﬃcients?

Exercise 4.4.6: Find the Fourier series of both the odd and even periodic extension of the function for .

where on . a) Solve for the Dirichlet conditions . b) Solve for the Neumann conditions .

for on . a) Solve for the Dirichlet conditions . b) Solve for the Neumann conditions .

where . Write the solution as a Fourier series, where the coeﬃcients are given in terms of .

Exercise 4.4.11: Let for . Let be the odd periodic extension. Compute , , , , , , . Note: Do not compute using the sine series.

Exercise 4.4.101: Let on . a) Find the Fourier series of the even periodic extension. b) Find the Fourier series of the odd periodic extension.

Exercise 4.4.102: Let on . a) Find the Fourier series of the even periodic extension. b) Find the Fourier series of the odd periodic extension.