Fourier Cosine and Sine Transform on fractal space

The fractional Fourier transform has been investigated in a number of papers and has been proved to be very useful in solving engineering problems [1][2][3][4][5][6][7]. It is important to deal with the continuous fractal functions, which are irregular in the real world. Recently, Yang-Fourier transform based on the local fractional calculus was introduced [8] and Yang continued to study this subject [9][10]. The importance of Yang-Fourier transform for fractal functions derives from the fact that this is the only mathematic model which focuses on local fractional continuous functions derived from local fractional calculus. More recently, some model for engineering derived from local fractional derivative was proposed [11][12].

In this section, we start with the following result [9,10]:

where

From (2.2) , the Yang-Fourier transform of f (x) is given by [9,10]

And its Inverse formula of Yang-Fourier’s transforms as follows

Now ,by (2.1) and (2.2) ,we have

(2.5)

Here, we named (2.5) the Yang-Fourier integral formula.

We express the exponential factor E α (i α ω α (x αξ α )) in (2.5) in terms of trigonometric functions on fractal set of fractal dimensionαand use the even and odd nature of the cosine and the sine functions respectively as functions of ω, so that (2.5) can be written as

Hence,we have

This is another version of the Yang-Fourier integral formula.

We now assume that f (x) is an even function and expand the cosine function in (2.6) to obtain

This is called the local fractional Fourier cosine integral formula.

Similarly, for an odd function f (x), we obtain the local fractional Fourier sine integral formula

(2.8)

The local fractional Fourier cosine integral formula (2.7) leads to the local fractional Fourier cosine transform and its inverse defined by

(2.9)

where F α,c is the local fractional Fourier cosine transform operator and F -1 α.c is its inverse operator.

Similarly, the local fractional Fourier sine integral formula (2.8) leads to the local fractional Fourier sine transform and its inverse defined by

where F α,s is the local fractional Fourier cosine transform operator and F -1 α.s is its inverse operator.

(2.14)

By local fractional Fourier cosine transform,we have

(2.15) By (2.15),we obtain

Similarly, we obtain (2.14)

3 Properties of local fractional Fourier Cosine and Sine Transforms

Proof. As a direct application of the local fractional Fourier Cosine transform, we derive the follow identity

Similarly, we obtain (3.2)

Under appropriate conditions, the following properties also hold:

These results can be generalized for the cosine and sine transforms of higher order derivatives of a function.

Proof.Using the definition of the inverse local fractional Fourier cosine transform, we Have

Hence,we obtain

in which the definition of the inverse local fractional Fourier cosine transform is used. This proves (3.8). It also follows from the proof of Theorem 2 that

This proves result (3.9). Putting x = 0 in (3.9), we obtain

This is the Parseval relation for the local fractional Fourier cosine transform.

Similarly, we obtain

which is, by interchanging the order of integration,

in which the inverse local fractional Fourier sine transform is used. Thus, we find

Replacing g(x)) by f (x) gives the Parseval relation for the local fractional Fourier sine transform

(3.13)

Use the local fractional Fourier sine transform to solve the following differential equation:

Since we are interested in positive + region, we can take y(t) to be an odd function and take local fractional Fourier sine transforms. It is clear from its definition that local fractional Fourier sine transform is linear

Using this property and taking local fractional Fourier sine transform of both sides of the differential equation, we have F α,s {y (2α) } -9F α,s {y(t)} = 50F α,s {E α (-2t α )}.

Since F α,s {y (2α) (t)} = -ω 2α F α,s {y(t)} + 2ω α y(0).

-ω 2α F α,s {y(t)} + 2ω α y(0) -9F α,s {y(t)} = 50F α,s {E α (-2t α )}.

which, after collecting terms, becomes (ω 2α + 9)F α,s {y(t)} = -50 2ω α 4 + ω 2α + 2ω α y 0 . ] + y 0 2ω α ω 2α + 9 = (y 0 + 10) 2ω α ω 2α + 9 -10 2ω α (4 + ω 2α ) = (y 0 + 10)F α,s {E α (-3t α )} -10F α,s {E α (-2t α )}.

Taking the inverse local fractional Fourier sine transform, we get the solution y(t) = (y 0 + 10)E α (-3t α ) -10E α (-2t α ).

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