# Partial Differential Equations/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 $d \in \mathbb N$. 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

$\partial_\alpha f, \alpha \in \mathbb N_0^d \text{ and } |\alpha| \le n$

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
$\forall (x, y) \in O \times B : |\partial_x f(x, y)| \le |g(y)| \text{ and } \int_B |g(y)| dy < \infty$

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

$u : \mathbb R \times \mathbb R^d \to \mathbb R, u(t, x) := g(x + \mathbf vt) + \int_0^t f(s, x + \mathbf v(t - s)) ds$

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 $g(x + \mathbf vt)$ 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

$\partial_t g(x + \mathbf vt) - \mathbf v \cdot \nabla_x g(x + \mathbf vt) = 0 ~~~~~ (*)$

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

$F(t, x) := \int_0^t f(s, x - \mathbf vs) ds$

so that we have

$F(t, x + \mathbf vt) = \int_0^t f(s, x + \mathbf v(t - s)) ds$

By the multi-dimensional chain rule, we obtain

\begin{align} \frac{d}{dt} F(t, x + \mathbf vt) &= \begin{pmatrix} \partial_t F (t, x + \mathbf vt) & \partial_{x_1} F (t, x + \mathbf vt) & \cdots & \partial_{x_d} F(t, x + \mathbf vt) \end{pmatrix} \begin{pmatrix} 1 \\ \mathbf v \end{pmatrix} \\ &= \partial_t F (t, x + \mathbf vt) + \mathbf v \cdot \nabla_x F (t, x + \mathbf vt) \end{align}

But on the one hand, we have by the fundamental theorem of calculus, that $\partial_t F (t, x) = f(t, x - \mathbf vt)$ and therefore

$\partial_t F (t, x + \mathbf vt) = f(t, x)$

and on the other hand

$\partial_{x_n} F(t, x + \mathbf vt) = \partial_{x_n} \int_0^t f(s, x + \mathbf v(t - s)) ds$

, seeing that the differential quotient of the definition of $\partial_{x_n}$ is equal for both sides. And since on the third hand

$\frac{d}{dt} F(t, x + \mathbf vt) = \partial_t \int_0^t f(s, x + \mathbf v(t - s)) ds$

, 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.

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## 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

$u : \mathbb R \times \mathbb R^d \to \mathbb R, u(t, x) := g(x + \mathbf vt) + \int_0^t f(s, x + \mathbf v(t - s)) ds$

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{align} \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{align}

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 $u=v$.

$////$

## 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

$\partial_t g(x + \mathbf vt)$

## Sources

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