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.

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.

We will omit the proof.

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.

Initial value problem
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
\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}$$.
 * 1) Let $$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)$$.
 * 3) Find the unique solution to the initial value problem $$\begin{cases}