Last modified on 11 October 2014, at 11:15

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.

SolutionEdit

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 problemEdit

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.

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ExercisesEdit

  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)

SourcesEdit

Partial Differential Equations
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