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:

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

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

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.

Template:TextBox We will omit the proof.

Template:TextBox 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 $g(x+𝐯t)$ is continuously differentiable in all the directions $t,x_1,\dots ,x_d$. The existence of

${\partial }_{x_n}\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds,n\in \{1,\dots ,d\}$

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

${\partial }_t\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

we will later in this proof show to be equal to

$f(t,x)+𝐯\cdot {\mathrm{\nabla }}_x\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$,

which exists because

${\mathrm{\nabla }}_x\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

just consists of the derivatives

${\partial }_{x_n}\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds,n\in \{1,\dots ,d\}$

2.

We show that

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

in three substeps.

2.1

We show that

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

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

${\partial }_t\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds-𝐯\cdot {\mathrm{\nabla }}_x\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds=f(t,x)(**)$

We choose

$F(t,x):=\underset{0}{\overset{t}{\int }}f(s,x-𝐯s)ds$

so that we have

$F(t,x+𝐯t)=\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

By the multi-dimensional chain rule, we obtain

$\begin{array}{cc}\frac{d}{dt}F(t,x+𝐯t)& =\left(\begin{array}{cccc}{\partial }_{t}F(t,x+𝐯t)& {\partial }_{x_1}F(t,x+𝐯t)& \cdots & {\partial }_{x_d}F(t,x+𝐯t)\end{array}\right)\left(\begin{array}{c}1\\ 𝐯\end{array}\right)\\ & ={\partial }_{t}F(t,x+𝐯t)+𝐯\cdot {\mathrm{\nabla }}_{x}F(t,x+𝐯t)\end{array}$

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

${\partial }_{t}F(t,x+𝐯t)=f(t,x)$

and on the other hand

${\partial }_{x_n}F(t,x+𝐯t)={\partial }_{x_n}\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

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

$\frac{d}{dt}F(t,x+𝐯t)={\partial }_t\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

, 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. $◻$

Initial value problem

Proof:

Quite easily, $u(0,x)=g(x+𝐯\cdot 0)+\underset{0}{\overset{0}{\int }}f(s,x+𝐯(t-s))ds=g(x)$. 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

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

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 $(t,x)\in ℝ\times {ℝ}^d$,

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

Using the multi-dimensional chain rule, we calculate ${\mu }_{{(}^{\prime }t,x)}(\xi )$:

$\begin{array}{ccc}{\mu }_{{(}^{\prime }t,x)}(\xi )& :=\frac{d}{d\xi }w(t-\xi ,x+𝐯\xi )& {\text{ by defs. of the }}^{\prime }\text{ symbol and }\mu \\ & =\left(\begin{array}{cccc}{\partial }_{t}w(t-\xi ,x+𝐯\xi )& {\partial }_{x_1}w(t-\xi ,x+𝐯\xi )& \cdots & {\partial }_{x_d}w(t-\xi ,x+𝐯\xi )\end{array}\right)\left(\begin{array}{c}-1\\ 𝐯\end{array}\right)& \text{chain rule}\\ & =-{\partial }_{t}w(t-\xi ,x+𝐯\xi )+𝐯\cdot {\mathrm{\nabla }}_{x}w(t-\xi ,x+𝐯\xi )& \\ & =0& (*)\end{array}$

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

$\mathrm{\forall }(t,x)\in ℝ\times {ℝ}^d:w(t,x)={\mu }_{(t,x)}(0)={\mu }_{(t,x)}(t)=w(0,x+𝐯t)\stackrel{(**)}{=}0$

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

Exercises

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

   ${\partial }_{x_n}\underset{0}{\overset{t}{\int }}f(s,x+𝐯(t-s))ds$

   is equal to

   $\underset{0}{\overset{t}{\int }}{\partial }_{x_n}f(s,x+𝐯(t-s))ds$

   and therefore exists.

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

3. Find the unique solution to the initial value problem

   $\{\begin{array}{cc}\mathrm{\forall }(t,x)\in ℝ\times {ℝ}^3:& {\partial }_{t}u(t,x)-\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)\cdot {\mathrm{\nabla }}_{x}u(t,x)=t^5+x_1^6+x_2^7+x_3^8\\ \mathrm{\forall }x\in {ℝ}^3:& u(0,x)=x_1^9+x_2^{10}+x_3^{11}\end{array}$.

Sources

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