# Partial Differential Equations/The transport equation

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

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 ${\displaystyle d\in \mathbb {N} }$. The inhomogenous ${\displaystyle d}$-dimensional transport equation looks like this:

${\displaystyle \forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)}$

, where ${\displaystyle f:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} }$ is a function and ${\displaystyle \mathbf {v} \in \mathbb {R} ^{d}}$ is a vector.

## Solution

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 ${\displaystyle f:\mathbb {R} ^{d}\to \mathbb {R} }$ be a function and ${\displaystyle n\in \mathbb {N} }$. We say that ${\displaystyle f}$ is ${\displaystyle n}$ times continuously differentiable iff all the partial derivatives

${\displaystyle \partial _{\alpha }f,\alpha \in \mathbb {N} _{0}^{d}{\text{ and }}|\alpha |\leq n}$

exist and are continuous. We write ${\displaystyle f\in {\mathcal {C}}^{n}(\mathbb {R} ^{d})}$.

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 ${\displaystyle O\subseteq \mathbb {R} }$ be open and ${\displaystyle B\subseteq \mathbb {R} ^{d}}$, where ${\displaystyle d\in \mathbb {N} }$ is arbitrary, and let ${\displaystyle f\in {\mathcal {C}}^{1}(O\times B)}$. If the conditions

• for all ${\displaystyle x\in O}$, ${\displaystyle \int _{B}|f(x,y)|dy<\infty }$
• for all ${\displaystyle x\in O}$ and ${\displaystyle y\in B}$, ${\displaystyle {\frac {d}{dx}}f(x,y)}$ exists
• there is a function ${\displaystyle g:B\to \mathbb {R} }$ such that
${\displaystyle \forall (x,y)\in O\times B:|\partial _{x}f(x,y)|\leq |g(y)|{\text{ and }}\int _{B}|g(y)|dy<\infty }$

hold, then

${\displaystyle {\frac {d}{dx}}\int _{B}f(x,y)dy=\int _{B}{\frac {d}{dx}}f(x,y)}$

We will omit the proof.

Theorem 2.3: If ${\displaystyle f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}$, ${\displaystyle g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})}$ and ${\displaystyle \mathbf {v} \in \mathbb {R} ^{d}}$, then the function

${\displaystyle u:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,u(t,x):=g(x+\mathbf {v} t)+\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

solves the inhomogenous ${\displaystyle d}$-dimensional transport equation

${\displaystyle \forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)}$

Note that, as in chapter 1, that there are many solutions, one for each continuously differentiable ${\displaystyle g}$ in existence.

Proof:

1.

We show that ${\displaystyle u}$ is sufficiently often differentiable. From the chain rule follows that ${\displaystyle g(x+\mathbf {v} t)}$ is continuously differentiable in all the directions ${\displaystyle t,x_{1},\ldots ,x_{d}}$. The existence of

${\displaystyle \partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds,n\in \{1,\ldots ,d\}}$

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

${\displaystyle \partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

we will later in this proof show to be equal to

${\displaystyle f(t,x)+\mathbf {v} \cdot \nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$,

which exists because

${\displaystyle \nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

just consists of the derivatives

${\displaystyle \partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds,n\in \{1,\ldots ,d\}}$

2.

We show that

${\displaystyle \forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)}$

in three substeps.

2.1

We show that

${\displaystyle \partial _{t}g(x+\mathbf {v} t)-\mathbf {v} \cdot \nabla _{x}g(x+\mathbf {v} t)=0~~~~~(*)}$

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

${\displaystyle \partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds-\mathbf {v} \cdot \nabla _{x}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds=f(t,x)~~~~~(**)}$

We choose

${\displaystyle F(t,x):=\int _{0}^{t}f(s,x-\mathbf {v} s)ds}$

so that we have

${\displaystyle F(t,x+\mathbf {v} t)=\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

By the multi-dimensional chain rule, we obtain

{\displaystyle {\begin{aligned}{\frac {d}{dt}}F(t,x+\mathbf {v} t)&={\begin{pmatrix}\partial _{t}F(t,x+\mathbf {v} t)&\partial _{x_{1}}F(t,x+\mathbf {v} t)&\cdots &\partial _{x_{d}}F(t,x+\mathbf {v} t)\end{pmatrix}}{\begin{pmatrix}1\\\mathbf {v} \end{pmatrix}}\\&=\partial _{t}F(t,x+\mathbf {v} t)+\mathbf {v} \cdot \nabla _{x}F(t,x+\mathbf {v} t)\end{aligned}}}

But on the one hand, we have by the fundamental theorem of calculus, that ${\displaystyle \partial _{t}F(t,x)=f(t,x-\mathbf {v} t)}$ and therefore

${\displaystyle \partial _{t}F(t,x+\mathbf {v} t)=f(t,x)}$

and on the other hand

${\displaystyle \partial _{x_{n}}F(t,x+\mathbf {v} t)=\partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

, seeing that the differential quotient of the definition of ${\displaystyle \partial _{x_{n}}}$ is equal for both sides. And since on the third hand

${\displaystyle {\frac {d}{dt}}F(t,x+\mathbf {v} t)=\partial _{t}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

, the second part of the second part of the proof is finished.

2.3

We add ${\displaystyle (*)}$ and ${\displaystyle (**)}$ together, use the linearity of derivatives and see that the equation is satisfied. ${\displaystyle \Box }$

## Initial value problem

Theorem and definition 2.4: If ${\displaystyle f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}$ and ${\displaystyle g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})}$, then the function

${\displaystyle u:\mathbb {R} \times \mathbb {R} ^{d}\to \mathbb {R} ,u(t,x):=g(x+\mathbf {v} t)+\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

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

${\displaystyle {\begin{cases}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:&\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x)=f(t,x)\\\forall x\in \mathbb {R} ^{d}:&u(0,x)=g(x)\end{cases}}}$

Proof:

Quite easily, ${\displaystyle u(0,x)=g(x+\mathbf {v} \cdot 0)+\int _{0}^{0}f(s,x+\mathbf {v} (t-s))ds=g(x)}$. Therefore, and due to theorem 2.3, ${\displaystyle u}$ is a solution to the initial value problem of the transport equation. So we proceed to show uniqueness.

Assume that ${\displaystyle v}$ is an arbitrary other solution. We show that ${\displaystyle v=u}$, thereby excluding the possibility of a different solution.

We define ${\displaystyle w:=u-v}$. Then

${\displaystyle {\begin{array}{llll}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:&\partial _{t}w(t,x)-\mathbf {v} \cdot \nabla _{x}w(t,x)&=(\partial _{t}u(t,x)-\mathbf {v} \cdot \nabla _{x}u(t,x))-(\partial _{t}v(t,x)-\mathbf {v} \cdot \nabla _{x}v(t,x))&\\&&=f(t,x)-f(t,x)=0&~~~~~(*)\\\forall x\in \mathbb {R} ^{d}:&w(0,x)=u(0,x)-v(0,x)&=g(x)-g(x)=0&~~~~~(**)\end{array}}}$

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 ${\displaystyle (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}}$,

${\displaystyle \mu _{(t,x)}(\xi ):=w(t-\xi ,x+\mathbf {v} \xi )}$

Using the multi-dimensional chain rule, we calculate ${\displaystyle \mu _{(t,x)}'(\xi )}$:

{\displaystyle {\begin{aligned}\mu _{(t,x)}'(\xi )&:={\frac {d}{d\xi }}w(t-\xi ,x+\mathbf {v} \xi )&{\text{ by defs. of the }}'{\text{ symbol and }}\mu \\&={\begin{pmatrix}\partial _{t}w(t-\xi ,x+\mathbf {v} \xi )&\partial _{x_{1}}w(t-\xi ,x+\mathbf {v} \xi )&\cdots &\partial _{x_{d}}w(t-\xi ,x+\mathbf {v} \xi )\end{pmatrix}}{\begin{pmatrix}-1\\\mathbf {v} \end{pmatrix}}&{\text{chain rule}}\\&=-\partial _{t}w(t-\xi ,x+\mathbf {v} \xi )+\mathbf {v} \cdot \nabla _{x}w(t-\xi ,x+\mathbf {v} \xi )&\\&=0&(*)\end{aligned}}}

Therefore, for all ${\displaystyle (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}}$ ${\displaystyle \mu _{(t,x)}(\xi )}$ is constant, and thus

${\displaystyle \forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}:w(t,x)=\mu _{(t,x)}(0)=\mu _{(t,x)}(t)=w(0,x+\mathbf {v} t){\overset {(**)}{=}}0}$

, which shows that ${\displaystyle w=u-v=0}$ and thus ${\displaystyle u=v}$.${\displaystyle \Box }$

## Exercises

1. Let ${\displaystyle f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}$ and ${\displaystyle \mathbf {v} \in \mathbb {R} ^{d}}$. Using Leibniz' integral rule, show that for all ${\displaystyle n\in \{1,\ldots ,d\}}$ the derivative

${\displaystyle \partial _{x_{n}}\int _{0}^{t}f(s,x+\mathbf {v} (t-s))ds}$

is equal to

${\displaystyle \int _{0}^{t}\partial _{x_{n}}f(s,x+\mathbf {v} (t-s))ds}$

and therefore exists.

2. Let ${\displaystyle g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})}$ and ${\displaystyle \mathbf {v} \in \mathbb {R} ^{d}}$. Calculate ${\displaystyle \partial _{t}g(x+\mathbf {v} t)}$.
3. Find the unique solution to the initial value problem

${\displaystyle {\begin{cases}\forall (t,x)\in \mathbb {R} \times \mathbb {R} ^{3}:&\partial _{t}u(t,x)-{\begin{pmatrix}2\\3\\4\end{pmatrix}}\cdot \nabla _{x}u(t,x)=t^{5}+x_{1}^{6}+x_{2}^{7}+x_{3}^{8}\\\forall x\in \mathbb {R} ^{3}:&u(0,x)=x_{1}^{9}+x_{2}^{10}+x_{3}^{11}\end{cases}}}$.

## Sources

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