Introduction and first examples[edit]
What is a partial differential equation?[edit]
Let
be a natural number, and let
be an arbitrary set. A partial differential equation on
looks like this:

is an arbitrary function here, specific to the partial differential equation, which goes from
to
, where
is a natural number. And a solution to this partial differential equation on
is a function
satisfying the above logical statement. The solutions of some partial differential equations describe processes in nature; this is one reason why they are so important.
Multiindices[edit]
In the whole theory of partial differential equations, multiindices are extremely important. Only with their help we are able to write down certain formulas a lot briefer.
Definitions 1.1:
A
-dimensional multiindex is a vector
, where
are the natural numbers and zero.
If
is a multiindex, then its absolute value
is defined by

If
is a
-dimensional multiindex,
is an arbitrary set and
is sufficiently often differentiable, we define
, the
-th derivative of
, as follows:

Types of partial differential equations[edit]
We classify partial differential equations into several types, because for partial differential equations of one type we will need different solution techniques as for differential equations of other types. We classify them into linear and nonlinear equations, and into equations of different orders.
Definition 1.3:
Let
. We say that a partial differential equation has
-th order iff
is the smallest number such that it is of the form

First example of a partial differential equation[edit]
Now we are very curious what practical examples of partial differential equations look like after all.
Theorem and definition 1.4:
If
is a differentiable function and
, then the function

solves the one-dimensional homogenous transport equation

Proof: Exercise 2.
We therefore see that the one-dimensional transport equation has many different solutions; one for each continuously differentiable function in existence. However, if we require the solution to have a specific initial state, the solution becomes unique.
Theorem and definition 1.5:
If
is a differentiable function and
, then the function

is the unique solution to the initial value problem for the one-dimensional homogenous transport equation

Proof:
Surely
. Further, theorem 1.4 shows that also:

Now suppose we have an arbitrary other solution to the initial value problem. Let's name it
. Then for all
, the function

is constant:

Therefore, in particular

, which means, inserting the definition of
, that

, which shows that
. Since
was an arbitrary solution, this shows uniqueness.
In the next chapter, we will consider the non-homogenous arbitrary-dimensional transport equation.
Exercises[edit]
- Have a look at the definition of an ordinary differential equation (see for example the Wikipedia page on that) and show that every ordinary differential equation is a partial differential equation.
- Prove Theorem 1.4 using direct calculation.
- What is the order of the transport equation?
- Find a function
such that
and
.
Sources[edit]
The transport equation[edit]
In the first chapter, we had already seen the one-dimensional transport equation. In this chapter we will see that we can quite easily generalise the solution method and the uniqueness proof we used there to multiple dimensions. Let
. The inhomogenous
-dimensional transport equation looks like this:

, where
is a function and
is a vector.
Solution[edit]
The following definition will become a useful shorthand notation in many occasions. Since we can use it right from the beginning of this chapter, we start with it.
Before we prove a solution formula for the transport equation, we need a theorem from analysis which will play a crucial role in the proof of the solution formula.
Theorem 2.2: (Leibniz' integral rule)
Let
be open and
, where
is arbitrary, and let
. If the conditions
- for all
, 
- for all
and
,
exists
- there is a function
such that

hold, then

We will omit the proof.
Theorem 2.3:
If
,
and
, then the function

solves the inhomogenous
-dimensional transport equation

Note that, as in chapter 1, that there are many solutions, one for each continuously differentiable
in existence.
Proof:
1.
We show that
is sufficiently often differentiable. From the chain rule follows that
is continuously differentiable in all the directions
. The existence of

follows from the Leibniz integral rule (see exercise 1). The expression

we will later in this proof show to be equal to
,
which exists because

just consists of the derivatives

2.
We show that

in three substeps.
2.1
We show that

This is left to the reader as an exercise in the application of the multi-dimensional chain rule (see exercise 2).
2.2
We show that

We choose

so that we have

By the multi-dimensional chain rule, we obtain

But on the one hand, we have by the fundamental theorem of calculus, that
and therefore

and on the other hand

, seeing that the differential quotient of the definition of
is equal for both sides. And since on the third hand

, the second part of the second part of the proof is finished.
2.3
We add
and
together, use the linearity of derivatives and see that the equation is satisfied.
Initial value problem[edit]
Theorem and definition 2.4:
If
and
, then the function

is the unique solution of the initial value problem of the transport equation

Proof:
Quite easily,
. Therefore, and due to theorem 2.3,
is a solution to the initial value problem of the transport equation. So we proceed to show uniqueness.
Assume that
is an arbitrary other solution. We show that
, thereby excluding the possibility of a different solution.
We define
. Then

Analogous to the proof of uniqueness of solutions for the one-dimensional homogenous initial value problem of the transport equation in the first chapter, we define for arbitrary
,

Using the multi-dimensional chain rule, we calculate
:

Therefore, for all
is constant, and thus

, which shows that
and thus
.
Exercises[edit]
- Let
and
. Using Leibniz' integral rule, show that for all
the derivative
is equal to

and therefore exists.
- Let
and
. Calculate
.
Find the unique solution to the initial value problem
.
Sources[edit]
Test functions[edit]
Motivation[edit]
Before we dive deeply into the chapter, let's first motivate the notion of a test function. Let's consider two functions which are piecewise constant on the intervals
and zero elsewhere; like, for example, these two:
Let's call the left function
, and the right function
.
Of course we can easily see that the two functions are different; they differ on the interval
; however, let's pretend that we are blind and our only way of finding out something about either function is evaluating the integrals
and 
for functions
in a given set of functions
.
We proceed with choosing
sufficiently clever such that five evaluations of both integrals suffice to show that
. To do so, we first introduce the characteristic function. Let
be any set. The characteristic function of
is defined as

With this definition, we choose the set of functions
as

It is easy to see (see exercise 1), that for
, the expression

equals the value of
on the interval
, and the same is true for
. But as both functions are uniquely determined by their values on the intervals
(since they are zero everywhere else), we can implement the following equality test:

This obviously needs five evaluations of each integral, as
.
Since we used the functions in
to test
and
, we call them test functions. What we ask ourselves now is if this notion generalises from functions like
and
, which are piecewise constant on certain intervals and zero everywhere else, to continuous functions. The following chapter shows that this is true.
Bump functions[edit]
In order to write down the definition of a bump function more shortly, we need the following two definitions:
Now we are ready to define a bump function in a brief way:
These two properties make the function really look like a bump, as the following example shows:
The standard mollifier

in dimension

Example 3.4: The standard mollifier
, given by

, where
, is a bump function (see exercise 2).
Schwartz functions[edit]
As for the bump functions, in order to write down the definition of Schwartz functions shortly, we first need two helpful definitions.
Now we are ready to define a Schwartz function.
Definition 3.7:
We call
a Schwartz function iff the following two conditions are satisfied:


By
we mean the function
.

Example 3.8:
The function

is a Schwartz function.
Theorem 3.9:
Every bump function is also a Schwartz function.
This means for example that the standard mollifier is a Schwartz function.
Proof:
Let
be a bump function. Then, by definition of a bump function,
. By the definition of bump functions, we choose
such that

, as in
, a set is compact iff it is closed & bounded. Further, for
arbitrary,

Convergence of bump and Schwartz functions[edit]
Now we define what convergence of a sequence of bump (Schwartz) functions to a bump (Schwartz) function means.
Definition 3.11:
We say that the sequence of Schwartz functions
converges to
iff the following condition is satisfied:

Theorem 3.12:
Let
be an arbitrary sequence of bump functions. If
with respect to the notion of convergence for bump functions, then also
with respect to the notion of convergence for Schwartz functions.
Proof:
Let
be open, and let
be a sequence in
such that
with respect to the notion of convergence of
. Let thus
be the compact set in which all the
are contained. From this also follows that
, since otherwise
, where
is any nonzero value
takes outside
; this would contradict
with respect to our notion of convergence.
In
, ‘compact’ is equivalent to ‘bounded and closed’. Therefore,
for an
. Therefore, we have for all multiindices
:

Therefore the sequence converges with respect to the notion of convergence for Schwartz functions.
The ‘testing’ property of test functions[edit]
In this section, we want to show that we can test equality of continuous functions
by evaluating the integrals
and 
for all
(thus, evaluating the integrals for all
will also suffice as
due to theorem 3.9).
But before we are able to show that, we need a modified mollifier, where the modification is dependent of a parameter, and two lemmas about that modified mollifier.
Definition 3.13:
For
, we define
.
Lemma 3.14:
Let
. Then
.
Proof:
From the definition of
follows
.
Further, for

Therefore, and since

, we have:


In order to prove the next lemma, we need the following theorem from integration theory:
Theorem 3.15: (Multi-dimensional integration by substitution)
If
are open, and
is a diffeomorphism, then

We will omit the proof, as understanding it is not very important for understanding this wikibook.
Lemma 3.16:
Let
. Then
.
Proof:


Now we are ready to prove the ‘testing’ property of test functions:
Theorem 3.17:
Let
be continuous. If
,
then
.
Proof:
Let
be arbitrary, and let
. Since
is continuous, there exists a
such that

Then we have

Therefore,
. An analogous reasoning also shows that
. But due to the assumption, we have

As limits in the reals are unique, it follows that
, and since
was arbitrary, we obtain
.
Remark 3.18:
Let
be continuous. If
,
then
.
Proof:
This follows from all bump functions being Schwartz functions, which is why the requirements for theorem 3.17 are met.
Exercises[edit]
Let
and
be constant on the interval
. Show that

- Prove that the standard mollifier as defined in example 3.4 is a bump function by proceeding as follows:
Prove that the function

is contained in
.
Prove that the function

is contained in
.
- Conclude that
.
- Prove that
is compact by calculating
explicitly.
- Let
be open, let
and let
. Prove that if
, then
and
.
- Let
be open, let
be bump functions and let
. Prove that
.
- Let
be Schwartz functions functions and let
. Prove that
is a Schwartz function.
- Let
, let
be a polynomial, and let
in the sense of Schwartz functions. Prove that
in the sense of Schwartz functions.
Distributions[edit]
Distributions and tempered distributions[edit]
Theorem 4.3:
Let
be a tempered distribution. Then the restriction of
to bump functions is a distribution.
Proof:
Let
be a tempered distribution, and let
be open.
1.
We show that
has a well-defined value for
.
Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression

makes sense for every
.
2.
We show that the restriction is linear.
Let
and
. Since due to theorem 3.9
and
are Schwartz functions as well, we have

due to the linearity of
for all Schwartz functions. Thus
is also linear for bump functions.
3.
We show that the restriction of
to
is sequentially continuous. Let
in the notion of convergence of bump functions. Due to theorem 3.11,
in the notion of convergence of Schwartz functions. Since
as a tempered distribution is sequentially continuous,
.
The convolution[edit]
The convolution of two functions may not always exist, but there are sufficient conditions for it to exist:
Theorem 4.5:
Let
such that
and let
and
. Then for all
, the integral

has a well-defined real value.
Proof:
Due to Hölder's inequality,
.
We shall now prove that the convolution is commutative, i. e.
.
Proof:
We apply multi-dimensional integration by substitution using the diffeomorphism
to obtain
.
Lemma 4.7:
Let
be open and let
. Then
.
Proof:
Let
be arbitrary. Then, since for all

and further
,
Leibniz' integral rule (theorem 2.2) is applicable, and by repeated application of Leibniz' integral rule we obtain
.
Regular distributions[edit]
In this section, we shortly study a class of distributions which we call regular distributions. In particular, we will see that for certain kinds of functions there exist corresponding distributions.
Two questions related to this definition could be asked: Given a function
, is
for
open given by

well-defined and a distribution? Or is
given by

well-defined and a tempered distribution? In general, the answer to these two questions is no, but both questions can be answered with yes if the respective function
has the respectively right properties, as the following two theorems show. But before we state the first theorem, we have to define what local integrability means, because in the case of bump functions, local integrability will be exactly the property which
needs in order to define a corresponding regular distribution:
Now we are ready to give some sufficient conditions on
to define a corresponding regular distribution or regular tempered distribution by the way of

or
:
Theorem 4.11:
Let
be open, and let
be a function. Then

is a regular distribution iff
.
Proof:
1.
We show that if
, then
is a distribution.
Well-definedness follows from the triangle inequality of the integral and the monotony of the integral:

In order to have an absolute value strictly less than infinity, the first integral must have a well-defined value in the first place. Therefore,
really maps to
and well-definedness is proven.
Continuity follows similarly due to

, where
is the compact set in which all the supports of
and
are contained (remember: The existence of a compact set such that all the supports of
are contained in it is a part of the definition of convergence in
, see the last chapter. As in the proof of theorem 3.11, we also conclude that the support of
is also contained in
).
Linearity follows due to the linearity of the integral.
2.
We show that
is a distribution, then
(in fact, we even show that if
has a well-defined real value for every
, then
. Therefore, by part 1 of this proof, which showed that if
it follows that
is a distribution in
, we have that if
is a well-defined real number for every
,
is a distribution in
.
Let
be an arbitrary compact set. We define

is continuous, even Lipschitz continuous with Lipschitz constant
: Let
. Due to the triangle inequality, both

and

, which can be seen by applying the triangle inequality twice.
We choose sequences
and
in
such that
and
and consider two cases. First, we consider what happens if
. Then we have
.
Second, we consider what happens if
:

Since always either
or
, we have proven Lipschitz continuity and thus continuity. By the extreme value theorem,
therefore has a minimum
. Since
would mean that
for a sequence
in
which is a contradiction as
is closed and
, we have
.
Hence, if we define
, then
. Further, the function

has support contained in
, is equal to
within
and further is contained in
due to lemma 4.7. Hence, it is also contained in
. Since therefore, by the monotonicity of the integral

,
is indeed locally integrable.
Theorem 4.12:
Let
, i. e.

Then

is a regular tempered distribution.
Proof:
From Hölder's inequality we obtain
.
Hence,
is well-defined.
Due to the triangle inequality for integrals and Hölder's inequality, we have

Furthermore
.
If
in the notion of convergence of the Schwartz function space, then this expression goes to zero. Therefore, continuity is verified.
Linearity follows from the linearity of the integral.
Equicontinuity[edit]
We now introduce the concept of equicontinuity.
So equicontinuity is in fact defined for sets of continuous functions mapping from
(a set in a metric space) to the real numbers
.
Proof:
In order to prove uniform convergence, by definition we must prove that for all
, there exists an
such that for all
.
So let's assume the contrary, which equals by negating the logical statement
.
We choose a sequence
in
. We take
in
such that
for an arbitrarily chosen
and if we have already chosen
and
for all
, we choose
such that
, where
is greater than
.
As
is sequentially compact, there is a convergent subsequence
of
. Let us call the limit of that subsequence sequence
.
As
is equicontinuous, we can choose
such that
.
Further, since
(if
of course), we may choose
such that
.
But then follows for
and the reverse triangle inequality:

Since we had
, the reverse triangle inequality and the definition of t

, we obtain:

Thus we have a contradiction to
.
Proof: We have to prove equicontinuity, so we have to prove
.
Let
be arbitrary.
We choose
.
Let
such that
, and let
be arbitrary. By the mean-value theorem in multiple dimensions, we obtain that there exists a
such that:

The element
is inside
, because
is convex. From the Cauchy-Schwarz inequality then follows:


The generalised product rule[edit]
Definition 4.16:
If
are two
-dimensional multiindices, we define the binomial coefficient of
over
as
.
We also define less or equal relation on the set of multi-indices.
Definition 4.17:
Let
be two
-dimensional multiindices. We define
to be less or equal than
if and only if
.
For
, there are vectors
such that neither
nor
. For
, the following two vectors are examples for this:

This example can be generalised to higher dimensions (see exercise 6).
With these multiindex definitions, we are able to write down a more general version of the product rule. But in order to prove it, we need another lemma.
Lemma 4.18:
If
and
, where the
is at the
-th place, we have

for arbitrary multiindices
.
Proof:
For the ordinary binomial coefficients for natural numbers, we had the formula
.
Therefore,


This is the general product rule:
Theorem 4.19:
Let
and let
. Then

Proof:
We prove the claim by induction over
.
1.
We start with the induction base
. Then the formula just reads

, and this is true. Therefore, we have completed the induction base.
2.
Next, we do the induction step. Let's assume the claim is true for all
such that
. Let now
such that
. Let's choose
such that
(we may do this because
). We define again
, where the
is at the
-th place. Due to Schwarz' theorem and the ordinary product rule, we have