Partial Differential Equations/The transport equation

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

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

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

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

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

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

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

$\forall (x,y)\in O\times B:|\partial _{x}f(x,y)|\leq |g(y)|{\text{ and }}\int _{B}|g(y)|dy<\infty$

hold, then

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

We will omit the proof.

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

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

solves the inhomogenous $d$-dimensional transport equation

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

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

Proof:

1.

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

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

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

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

we will later in this proof show to be equal to

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

which exists because

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

just consists of the derivatives

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

2.

We show that

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

in three substeps.

2.1

We show that

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

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

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

We choose

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

so that we have

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

By the multi-dimensional chain rule, we obtain

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

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

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

and on the other hand

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

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

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

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

2.3

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

Initial value problem

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

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

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

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

Proof:

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

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

We define $w:=u-v$. Then

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

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

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

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

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

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

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

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

Exercises

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

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

   is equal to

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

   and therefore exists.

2. Let $g\in {\mathcal {C}}^{1}(\mathbb {R} ^{d})$ and $\mathbf {v} \in \mathbb {R} ^{d}$. Calculate $\partial _{t}g(x+\mathbf {v} t)$.

3. Find the unique solution to the initial value problem

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

Sources

• Martin Brokate (2011/2012), Partielle Differentialgleichungen, Vorlesungsskript (PDF) (in German) {{citation}}: Check date values in: |year= (help)