Partial Differential Equations/The Fourier transform

Partial Differential Equations
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In this chapter, we introduce the Fourier transform. The Fourier transform transforms functions into other functions. It can be used to solve certain types of linear differential equations.

Definition and calculation rulesEdit

Definition 8.1:

Let  . Then the Fourier transform of   is defined as follows:

 

We recall that   is integrable     is integrable.

Now we're ready to prove the next theorem:

Theorem 8.2: The Fourier transform of an integrable   is well-defined.

Proof: Since   is integrable, lemma 8.2 tells us that   is integrable. But

 

, and therefore   is integrable. But then,   is integrable, which is why

 

has a unique complex value, by definition of integrability. 

Theorem 8.3: Let  . Then the Fourier transform of  ,  , is bounded.

Proof:

  

Once we have calculated the Fourier transform   of a function  , we can easily find the Fourier transforms of some functions similar to  . The following calculation rules show examples how you can do this. But just before we state the calculation rules, we recall a definition from chapter 2, namely the power of a vector to a multiindex, because it is needed in the last calculation rule.

Definition 2.6:

For a vector   and a  -dimensional multiindex   we define  ,   to the power of  , as follows:

 

Now we write down the calculation rules, using the following notation:

Notation 8.4:

We write

 

to mean the sentence 'the function   is the Fourier transform of the function  '.

Theorem 8.5:

Let   be the Fourier transform of  . Then the following calculation rules hold:

  1.   for arbitrary  
  2.   for arbitrary  
  3.   for arbitrary  

If additionally   is the Fourier transform of  , we have

4.  

Proof: To prove the first rule, we only need one of the rules for the exponential function (and the symmetry of the standard dot product):

1.

 

For the next two rules, we apply the general integration by substitution rule, using the diffeomorphisms   and  , which are bijections from   to itself.

2.

 

3.

 

4.

 

 

The Fourier transform of Schwartz functionsEdit

In order to proceed with further rules for the Fourier transform which involve Schwartz functions, we first need some further properties of Schwartz functions.

Theorem 8.6:

Let   be a Schwartz function and let  . Then the function

 

is a Schwartz function as well.

Proof:

Let  . Due to the general product rule, we have:

 

We note that for all   and  ,   equals to   to some multiindex power. Since   is a Schwartz function, there exist constants   such that:

 

Hence, the triangle inequality for   implies:

  

Theorem 8.7:

Every Schwartz function is integrable.

Proof:

We use that if the absolute value of a function is almost everywhere smaller than the value of an integrable function, then the first function is integrable.

Let   be a Schwartz function. Then there exist   such that for all  :

 

The latter function is integrable, and integrability of   follows. 

Now we can prove all three of the following rules for the Fourier transform involving Schwartz functions.

Theorem 8.8:

If   is the Fourier transform of a function   in the Schwartz space  , in addition to the rules in theorem 8.4, also the following rules hold:

  1.   for arbitrary  
  2.   for arbitrary  

If additionally   is the Fourier transform of a  , then

3.  

Proof:

1.

For the first rule, we use induction over  .

It is clear that the claim is true for   (then the rule states that the Fourier transform of   is the Fourier transform of  ).

We proceed to the induction step: Let  , and assume that the claim is true for all   such that  . Let   such that  . We show that the claim is also true for  .

Remember that we have  . We choose   such that   (this is possible since otherwise  ), define

 
 

and obtain

 

by Schwarz' theorem, which implies that one may interchange the order of partial derivation arbitrarily.

Let   be an arbitrary positive real number. From Fubini's theorem and integration by parts, we obtain:

 

Due to the dominated convergence theorem (with dominating function  ), the integral on the left hand side of this equation converges to

 

as  . Further, since   is a Schwartz function, there are   such that:

 

Hence, the function within the large parentheses in the right hand sinde of the last line of the last equation is dominated by the   function

 

and hence, by the dominated convergence theorem, the integral over that function converges, as  , to:

 

From the uniqueness of limits of real sequences we obtain 1.

2.

We use again induction on  , note that the claim is trivially true for  , assume that the claim is true for all   such that  , choose   such that   and   such that   and define  .

Theorems 8.6 and 8.7 imply that

  • for all  ,   and
  • for all  ,  .

Further,

  •   exists for all  .

Hence, Leibniz' integral rule implies:

 

3.

  

Theorem 8.9:

Let  , and let   be the Fourier transform of  . Then  .

Proof:

Let   be two arbitrary  -dimensional multiindices, and let  . By theorem 8.6   is a Schwartz function as well. Theorem 8.8 implies:

 

By theorem 8.3,   is bounded. Since   were arbitrary, this shows that  . 

Definitions 8.10:

We define the Fourier transform on the Schwartz space to be the function

 .

Theorem 8.9 assures that this function really maps to  . Furthermore, we define the inverse Fourier transform on the Schwartz space to be the function

 .

This function maps to   since  .

Both the Fourier transform and the inverse Fourier transform are sequentially continuous:

Theorem 8.11:

Let   and let   be a sequence of Schwartz functions such that  . Then   and  , both in the sense of Schwartz function convergence as defined in definition 3.11.

Proof:

1. We prove  .

Let  . Due to theorem 8.8 1. and 2. and the linearity of derivatives, integrals and multiplication, we have

 .

As in the proof of theorem 8.3, we hence obtain

 .

Due to the multi-dimensional product rule,

 .

Let now   be arbitrary. Since   as defined in definition 3.11, for each   we may choose   such that

 .

Further, we may choose   such that

 .

Hence follows for  :

 

Since   was arbitrary, we obtain  .

2. From 1., we deduce  .

If   in the sense of Schwartz functions, then also   in the sense of Schwartz functions, where we define

  and  .

Therefore, by 1. and integration by substitution using the diffeomorphism  ,  . 

In the next theorem, we prove that   is the inverse function of the Fourier transform. But for the proof of that theorem (which will be a bit long, and hence to read it will be a very good exercise), we need another two lemmas:

Lemma 8.12:

If we define the function

 ,

then   and  .

Proof:

1.  :

We define

 .

By the product rule, we have for all  

 .

Due to 1. of theorem 8.8, we have

 ;

from 2. of theorem 8.8 we further obtain

 .

Hence,   is constant. Further,

 .

2.  :

By substitution using the diffeomorphism  ,

 . 

For the next lemma, we need example 3.4 again, which is why we restate it:

Example 3.4: The standard mollifier  , given by

 

, where  , is a bump function (see exercise 3.2).

Lemma 8.13:

Let  , and for each   define  . Then   in the sense of Schwartz functions.

Proof:

Let   be arbitrary. Due to the generalised product rule,

 .

By the triangle inequality, we may hence deduce

 .

Since both   and   are Schwartz functions (see exercise 3.2 and theorem 3.9), for each   we may choose   such that

  and  .

Further, for each  , we may choose   such that

 .

Let now   be arbitrary. We choose   such that for all  

 .

Further, we choose   such that

 .

This is possible since

 

due to our choice of  .

Then we choose   such that for all  

 .

Inserting all this in the above equation gives   for  . Since  ,   and   were arbitrary, this proves   in the sense of Schwartz functions. 

Theorem 8.14:

Let  . Then   and  .

Proof:

1. We prove that if   is a Schwartz function vanishing at the origin (i. e.  ), then  .

So let   be a Schwartz function vanishing at the origin. By the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral, we have

 .

Defining  ,

 ,

and multiplying both sides of the above equation by  , we obtain

 .

Since by repeated application of Leibniz' integral rule for all  

 ,

all the   are bump functions (due to theorem 4.15 and exercise 3.?), and hence Schwartz functions (theorem 3.9). Hence, by theorem 8.8 and the linearity of the Fourier transform (which follows from the linearity of the integral),

 .

Hence,

 .

Let  . By Fubini's theorem, the fundamental theorem of calculus and since   is a bump function, we have

 .

If we let  , theorem 8.11 and lemma 8.13 give the claim.

2. We deduce from 1. that if   is an arbitrary Schwartz function, then  .

As in lemma 8.12, we define

 .

Let now   be any Schwartz function. Then   is also a Schwartz function (see exercise 3.?). Further, since  , it vanishes at the origin. Hence, by 1.,

 .

Further, due to lemma 8.12 and the linearity of the Fourier transform,

 .

3. We deduce from 2. that if   is a Schwartz function and   is arbitrary, then   (i. e.  .

Let   and   be arbitrary. Due to the definition of  ,

 .

Further, if we define  ,

 .

Hence, by 2.,

 .

4. We deduce from 3. that for any Schwartz function   we have  .

Let   and   be arbitrary. Then we have

 . 

The Fourier transform of tempered distributionsEdit

Definition 8.15:

Let   be a tempered distribution. We define

 .

Theorem 8.16:

  is a tempered distribution.

Proof:

1. Sequential continuity follows from the sequential continuity of   and   (theorem 8.11) and that the composition of two sequentially continuous functions is sequentially continuous again.

2. Linearity follows from the linearity of   and   and that the composition of two linear functions is linear again. 

Definition 8.17:

Let   be a tempered distribution. We define

 .

ExercisesEdit

SourcesEdit

Partial Differential Equations
 ← The heat equation The Fourier transform The wave equation →