Partial Differential Equations/The transport equation

Partial Differential Equations
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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.

SolutionEdit

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.

Definition 2.1:

Let   be a function and  . We say that   is   times continuously differentiable iff all the partial derivatives

 

exist and are continuous. We write  .

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 problemEdit

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  . 

ExercisesEdit

  1. Let   and  . Using Leibniz' integral rule, show that for all   the derivative

     

    is equal to

     

    and therefore exists.

  2. Let   and  . Calculate  .
  3. Find the unique solution to the initial value problem

     .

SourcesEdit

Partial Differential Equations
 ← Introduction and first examples The transport equation Test functions →