Introduction to Partial Differential Equations
偏微分方程导论

Dongfen Bian $^{1}$ and Emmanuel Grenier $^{1}$
董芬·边 $^{1}$ 和埃马纽埃尔·格雷尼耶 $^{1}$

September 9, 2024 2024 年 9 月 9 日

^{1}

School of Mathematics and Statistics, Beijing Institute of Technology, Beijing, China
北京理工大学数学与统计学院，中国北京

Chapter 1 第一章

Introduction 介绍

1.1 Objectives of these book
1.1 本书的目标

This book is an introduction to the theory of partial differential equations with two main axis
这本书是关于偏微分方程理论的介绍，主要有两个轴心

the computation of simple explicit solutions that can be studied in details in order to understand the properties of the corresponding partial differential equations,
简单显式解的计算可以详细研究，以便理解相应偏微分方程的性质，
the study of the existence, uniqueness and qualitative properties of solutions of simple partial differential equations, relying on methods which do not use explicit solutions. These methods are general and can be applied to many different equations.
对简单偏微分方程解的存在性、唯一性和定性性质的研究，依赖于不使用显式解的方法。这些方法是一般性的，可以应用于许多不同的方程。

The reader should be familiar with calculus, integration theory, implicit function theorem and elementary Fourier analysis.
读者应熟悉微积分、积分理论、隐函数定理和基础傅里叶分析。

1.2 Notations 1.2 符号

1.2.1 Derivatives 1.2.1 导数

R^{n}

, we will use the following notations
在

R^{n}

中，我们将使用以下符号

\partial_{t} = \frac{\partial}{\partial t}, \partial_{j} = \frac{\partial}{\partial x_{j}}

where

j = 1, 2, \dots, n

and

x_{j}

is the

j^{th}

coordinate.
其中

j = 1, 2, \dots, n

和

x_{j}

是

j^{th}

坐标。
In

R

, we will denote the partial derivative with respect to

x

\partial_{x}

.
在

R

中，我们将用

\partial_{x}

表示对

x

的偏导数。

Moreover,

\nabla

is the operator “nabla”
此外，

\nabla

是算子“nabla”

\nabla = (\partial_{1}, \dots, \partial_{n})

In particular, if

V (t, x) = (V_{1} (t, x), \dots, V_{n} (t, x))

, 特别是，如果

V (t, x) = (V_{1} (t, x), \dots, V_{n} (t, x))

，

\nabla \cdot V = \sum_{j = 1}^{n} \partial_{j} V_{j}

is the divergence of

V

是

V

的散度
If

V (t, x) = (V_{1} (t, x), \dots, V_{n} (t, x))

and if

ϕ (t, x)

is a function, then
如果

V (t, x) = (V_{1} (t, x), \dots, V_{n} (t, x))

，并且如果

ϕ (t, x)

是一个函数，那么

(V \cdot \nabla) ϕ = V \cdot \nabla ϕ = \sum_{j = 1}^{n} V_{j} \partial_{j} ϕ

is called the transport of

ϕ

V

.
被称为

V

对

ϕ

的运输。

1.2.2 Functional spaces 1.2.2 功能空间

The measure will always be Lebesgue measure. Let

Ω

R^{n}, R^{+} \times R

, or an open set of

R^{n}

. The space

L^{\infty} (Ω)

is the set of measurable functions such that
该度量将始终是勒贝格测度。设

Ω

为

R^{n}, R^{+} \times R

，或

R^{n}

的开集。空间

L^{\infty} (Ω)

是可测函数的集合，使得

‖ f ‖_{L^{\infty}} = sup_{x \in Ω} | f (x) | < + \infty .

For

1 \leq p < + \infty

, the space

L^{p} (Ω)

is the set of measurable function

f

such that
对于

1 \leq p < + \infty

，空间

L^{p} (Ω)

是可测函数

f

的集合，使得

‖ f ‖_{L^{p}} = {(\int_{Ω} | f (x) |^{p} d x)}^{1 / p} < + \infty

We recall that, for every

p \in [1, \infty], L^{p} (Ω)

is complete for the norm

‖ \cdot ‖_{L^{p}}

and that

L^{2} (Ω)

is an Hilbert space with the associated scalar product
我们回忆到，对于每个

p \in [1, \infty], L^{p} (Ω)

在范数

‖ \cdot ‖_{L^{p}}

下是完备的，并且

L^{2} (Ω)

是一个具有相关标量积的希尔伯特空间

⟨ f, g ⟩_{L^{2}} = \int_{Ω} f (x) g (x) d x

1.2.3 Miscellaneous 1.2.3 杂项

f (t, x)

is a function of two (or more) variables,

f (t, \cdot)

denotes the function of

x

defined by

x \to f (t, x)

.
如果

f (t, x)

是一个两个（或更多）变量的函数，

f (t, \cdot)

表示由

x \to f (t, x)

定义的

x

的函数。

Let

a \in R

, then

1_{x ∣ e a}

is the function which equals 1 if

x \leq a

and 0 if

x > a

. Similarly,

1_{x \geq a}

equals 1 if

x \geq a

and 0 if

x < a

. More generally, if

I

an interval, or a set,

1_{I} (x)

equals 1 if

x \in I

and 0 if

x \notin I

.
设

a \in R

，则

1_{x ∣ e a}

是一个函数，当

x \leq a

时等于 1，当

x > a

时等于 0。类似地，

1_{x \geq a}

在

x \geq a

时等于 1，在

x < a

时等于 0。更一般地，如果

I

是一个区间或集合，

1_{I} (x)

在

x \in I

时等于 1，在

x \notin I

时等于 0。

Chapter 2 第二章

Transport equations 运输方程

2.1 A first example in $R$
2.1 第一个例子在 $R$

2.1.1 Introduction 2.1.1 介绍

We first consider the transport equation in dimension 1 with constant speed

c \in R

. Let

ϕ_{0} \in C^{1} (R)

. We want to “transport” the values of

ϕ_{0}

.
我们首先考虑一维常速

c \in R

的传输方程。设

ϕ_{0} \in C^{1} (R)

。我们想要“传输”

ϕ_{0}

的值。

We define the function

ϕ (t, x)

by: the value of

ϕ

at time

t

and position

x + c t

equals the value of

ϕ_{0}

at position

x

, namely
我们通过以下方式定义函数

ϕ (t, x)

：在时间

t

和位置

x + c t

时，

ϕ

的值等于在位置

x

时

ϕ_{0}

的值，即

ϕ (t, x + c t) = ϕ_{0} (x), \forall t \in R^{+}, \forall x \in R

This defines a function

ϕ

by
这通过定义一个函数

ϕ

ϕ (t, x) = ϕ_{0} (x - c t), \forall t \in R^{+}, \forall x \in R

ϕ_{0} \in C^{1} (R)

, then

ϕ \in C^{1} (R^{+} \times R)

. We observe that
如果

ϕ_{0} \in C^{1} (R)

，那么

ϕ \in C^{1} (R^{+} \times R)

。我们观察到

\begin{aligned} \partial_{t} ϕ + c \partial_{x} ϕ & = 0, \forall t \in R^{+}, \forall x \in R \\ ϕ (0, x) & = ϕ_{0} (x), \forall x \in R \end{aligned}

The equation (2.3) is called the transport equation. Equation (2.4) provides an initial data. Let us introduce some vocabulary
方程 (2.3) 被称为输运方程。方程 (2.4) 提供了初始数据。让我们引入一些词汇。

in equation (2.1), we “follow” the values of $ϕ$ . It is called a Lagrangian point of view.
在方程（2.1）中，我们“跟随” $ϕ$ 的值。这被称为拉格朗日视角。
in equation (2.3), we do not “move”. It is called an Eulerian point of view.
在方程（2.3）中，我们不“移动”。这被称为欧拉视角。

2.1.2 Solution of the transport equation
2.1.2 运输方程的解

Theorem 2.1.1. Let

ϕ_{0} \in C^{1} (R)

. Then there exists a unique solution

ϕ \in C^{1} (R^{+} \times R)

to 2.3 [2.4). Moreover,

ϕ

is given by
定理 2.1.1. 设

ϕ_{0} \in C^{1} (R)

。则存在唯一解

ϕ \in C^{1} (R^{+} \times R)

对于 2.3 [2.4)。此外，

ϕ

由以下给出

ϕ (t, x) = ϕ_{0} (x - c t)

Proof. First we note that

ϕ (t, x) = ϕ_{0} (x - c t)

satisfies
证明。首先我们注意到

ϕ (t, x) = ϕ_{0} (x - c t)

满足

\partial_{t} ϕ = - c \partial_{x} ϕ_{0} = - c \partial_{x} ϕ

Thus

ϕ

is a solution. It is

C^{1}

since

ϕ_{0}

C^{1}

.
因此

ϕ

是一个解决方案。它是

C^{1}

，因为

ϕ_{0}

是

C^{1}

。
Now, if

ϕ (t, x)

is a solution, let us define
现在，如果

ϕ (t, x)

是一个解，我们定义为

ψ (t, x) = ϕ (t, x + c t)

Then 然后

\partial_{t} ψ (t, x) = \partial_{t} ϕ (t, x) + c \partial_{x} ϕ (t, x) = 0

thus

ψ (t, x)

does not depend on

t

因此

ψ (t, x)

不依赖于

t

ψ (t, x) = ψ (0, x) = ϕ (0, x) = ϕ_{0} (x)

As a consequence, 因此，

ϕ (t, x + c t) = ϕ_{0} (x)

hence 因此

ϕ (t, x) = ϕ_{0} (x - c t)

Thus the solution is unique.
因此，解是唯一的。

2.1.3 Well-posedness in the sense of Hadamard
2.1.3 哈达玛德意义下的良好适定性

Definition 2.1.2. A mathematical problem is said to be well-posed in the sense of Hadamard if
定义 2.1.2。如果一个数学问题在哈达玛德的意义上被称为良好提出，则

it has a solution, 它有一个解决方案，
its solution is unique, 它的解决方案是独特的，
its solution continuously depends on the data (parameters, initial data).
其解决方案持续依赖于数据（参数、初始数据）。

The transport equation in 1D with constant velocity is well-posed in the sense of Hadamard. The previous Theorem gives the existence and uniqueness of the solution
一维常速传输方程在哈达玛尔意义上是良构的。前面的定理给出了解的存在性和唯一性。

Let us study the continuity with respect to the initial data. Let us assume that

ϕ_{n, 0} \to ϕ_{\infty, 0}

C^{1} (R)

in the sense:
让我们研究关于初始数据的连续性。我们假设

ϕ_{n, 0} \to ϕ_{\infty, 0}

在

C^{1} (R)

中的意义是：

sup_{x \in R} | ϕ_{n, 0} (x) - ϕ_{\infty, 0} (x) | + sup_{x \in R} | \partial_{x} ϕ_{n, 0} (x) - \partial_{x} ϕ_{\infty, 0} (x) | \to 0

n \to + \infty

. Let

ϕ_{n}

be the solutions of the transport equation with initial data

ϕ_{n, 0}

and let

ϕ_{\infty}

be he solution of the transport equation with initial data

ϕ_{\infty, 0}

. We want to prove that

ϕ_{n} (t, x) \to ϕ_{\infty} (t, x)

C^{1} (R \times R)

in the sense
作为

n \to + \infty

。设

ϕ_{n}

为具有初始数据

ϕ_{n, 0}

的输运方程的解，设

ϕ_{\infty}

为具有初始数据

ϕ_{\infty, 0}

的输运方程的解。我们想证明

ϕ_{n} (t, x) \to ϕ_{\infty} (t, x)

在

C^{1} (R \times R)

的意义上。

\begin{aligned} sup_{x \in R, t \in R} | ϕ_{n} (t, x) - ϕ_{\infty} (t, x) | & + sup_{x \in R, t \in R} | \partial_{x} ϕ_{n} (t, x) - \partial_{x} ϕ_{\infty} (t, x) | \\ + sup_{x \in R, t \in R} | \partial_{t} ϕ_{n} (t, x) - \partial_{t} ϕ_{\infty} (t, x) | \to 0 \end{aligned}

n \to + \infty

. Let 作为

n \to + \infty

。让

ψ_{n} (t, x) = ϕ_{n} (t, x) - ϕ_{\infty} (t, x)

and 和

ψ_{n, 0} (x) = ϕ_{n, 0} (x) - ϕ_{\infty, 0} (x)

Then by linearity 然后通过线性性

\partial_{t} ψ_{n} + c \partial_{x} ψ_{n} = 0

and

ψ_{n, 0} \to 0

C^{1} (R)

n \to + \infty

. We have
和

ψ_{n, 0} \to 0

在

C^{1} (R)

作为

n \to + \infty

。我们有

ψ_{n} (t, x) = ψ_{n, 0} (x - c t)

In particular, 特别是，

sup_{t, x} | ψ_{n} (t, x) | ≲ sup_{x} | ψ_{n, 0} (x) | \to 0

Moreover, 此外，

\partial_{x} ψ_{n} (t, x) = \partial_{x} ψ_{n, 0} (x - c t)

thus 因此

sup_{t, x} | \partial_{x} ψ_{n} (t, x) | ≲ sup_{x} | \partial_{x} ψ_{n, 0} (x) | \to 0

and similarly for

\partial_{t} ψ_{n}

. Thus

ψ_{n} \to 0

C^{1}

, as

n \to + \infty

, which ends the study.
并且对于

\partial_{t} ψ_{n}

也是如此。因此

ψ_{n} \to 0

在

C^{1}

中，作为

n \to + \infty

，这结束了研究。
Remark. The solution is also continuous in

c

, but the proof is a little more delicate.
备注。该解在

c

处也是连续的，但证明稍微复杂一些。

2.1.4 Notion of weak solutions
2.1.4 弱解的概念

We note that 我们注意到

ϕ (t, x) = ϕ_{0} (x - c t)

has a meaning even if

ϕ_{0}

in not

C^{1} (R)

. It may only be

L_{loc}^{1}

for instance.
即使

ϕ_{0}

不在

C^{1} (R)

中，它也有意义。它可能只是

L_{loc}^{1}

，例如。

Can we say that ϕ is a solution of (2.3) if ϕ_{0} \in L_{loc}^{1} ?

We have to change the notion of solution, since

ϕ

is not differentiable. Instead of looking at precise values at a precise point

(t, x)

, we look at averages. This idea is at the starting point of the theory of the distributions, discovered by Laurent Schwartz (Fields medal, 1950).
我们必须改变解决方案的概念，因为

ϕ

是不可微的。我们不再关注在精确点

(t, x)

的精确值，而是关注平均值。这个想法是分布理论的起点，该理论由洛朗·施瓦茨（菲尔兹奖，1950 年）发现。

Let us make a preliminary computation. Let

ψ \in C^{1} (R^{+} \times R)

be a function with compact support. Let

ϕ_{0}

C^{1}

and let

ϕ \in C^{1}

be the corresponding solution of (2.3). Then
让我们进行初步计算。设

ψ \in C^{1} (R^{+} \times R)

是一个具有紧支撑的函数。设

ϕ_{0}

为

C^{1}

，并且设

ϕ \in C^{1}

为(2.3)的相应解。然后

\int_{R^{+}} \int_{R} ψ (t, x) [\partial_{t} ϕ (t, x) + c \partial_{x} ϕ (t, x)] d x d t = 0

We integrate by parts in

x

. There is no boundary term since the support of

ψ

is compact (i.e.

ψ

vanishes for large

t

and

x

)
我们在

x

中分部积分。由于

ψ

的支撑是紧致的（即

ψ

在大

t

和

x

时消失），因此没有边界项。

\int_{R^{+}} \int_{R} ψ (t, x) \partial_{x} ϕ (t, x) d x d t = - \int_{R^{+}} \int_{R} \partial_{x} ψ (t, x) ϕ (t, x) d x d t

We integrate by parts in

t

. There is a boundary term at

t = 0

,
我们在

t

处分部积分。在

t = 0

处有一个边界项，

\begin{aligned} \int_{R} \int_{R^{+}} ψ (t, x) \partial_{t} ϕ (t, x) d t d x = & - \int_{R} \int_{R^{+}} \partial_{t} ψ (t, x) ϕ (t, x) d t d x \\ - \int_{R} ψ (0, x) ϕ (0, x) d x . \end{aligned}

Hence 因此

\int_{R^{+}} \int_{R} ϕ (t, x) [\partial_{t} ψ (t, x) + c \partial_{x} ψ (t, x)] d x d t + \int_{R} ψ (0, x) ϕ_{0} (x) d x = 0 .

This formula does not involve derivatives of

ϕ

. Thus, the left hand side of 2.5) has a meaning even if

ϕ

is only

L_{l o c}^{1} (R^{+} \times R)

. We take this formula as the definition of a new notion of solutions.
这个公式不涉及

ϕ

的导数。因此，即使

ϕ

仅为

L_{l o c}^{1} (R^{+} \times R)

，2.5)的左侧也有意义。我们将这个公式视为新解的定义。

Definition 2.1.3. Let

ϕ_{0} \in L_{l o c}^{1} (R)

. We say that

ϕ \in L_{loc}^{1} (R^{+} \times R)

is a weak solution of (2.3 2.4) if, for every function

ψ \in C^{1} (R^{+} \times R)

with compact support,
定义 2.1.3. 设

ϕ_{0} \in L_{l o c}^{1} (R)

。我们称

ϕ \in L_{loc}^{1} (R^{+} \times R)

为 (2.3 2.4) 的弱解，如果对于每个具有紧支撑的函数

ψ \in C^{1} (R^{+} \times R)

，

\int_{R^{+}} \int_{R} ϕ (t, x) [\partial_{t} ψ (t, x) + c \partial_{x} ψ (t, x)] d x d t + \int_{R} ψ (0, x) ϕ_{0} (x) d x = 0 .

The function $ψ$ is called a “test function”. Test functions are $C^{1}$ and with compact support. The notion of “weak solution” is linked to the particular equation under study.
函数 $ψ$ 被称为“测试函数”。测试函数是 $C^{1}$ 并具有紧支撑。 “弱解”的概念与所研究的特定方程相关。
If $ϕ \in C^{1}$ is a solution in the usual sense, then $ϕ$ is called a “strong solution”, or a “classical solution”.
如果 $ϕ \in C^{1}$ 是通常意义上的解，那么 $ϕ$ 被称为“强解”或“经典解”。

2.1.5 An example of weak solution
2.1.5 弱解的一个例子

Let

ϕ_{0} (x) = 1_{x \geq 0}

. Let us show that
让

ϕ_{0} (x) = 1_{x \geq 0}

。让我们证明。

ϕ (t, x) = 1_{x \geq c t}

is a weak solution of (2.32.4). Let

ψ \in C^{1} (R^{+} \times R)

with compact support. Then we have
是(2.32.4)的弱解。设

ψ \in C^{1} (R^{+} \times R)

具有紧支撑。然后我们有

\begin{aligned} \int_{R^{+}} \int_{R} ϕ (t, x) [\partial_{t} ψ (t, x) + c \partial_{x} ψ (t, x)] d x d t \\ = & \int_{0}^{+ \infty} \int_{0}^{x / c} \partial_{t} ψ (t, x) d t d x + c \int_{0}^{+ \infty} \int_{c t}^{+ \infty} \partial_{x} ψ (t, x) d x d t \\ = & \int_{0}^{+ \infty} ψ (\frac{x}{c}, x) d x - \int_{0}^{+ \infty} ψ (0, x) d x - c \int_{0}^{+ \infty} ψ (t, c t) d t \\ = & - \int_{0}^{+ \infty} ψ (0, x) d x . \end{aligned}

Thus

ϕ

is a weak solution.
因此

ϕ

是一个弱解。

2.1.6 Link between weak and strong solutions
2.1.6 弱解与强解之间的联系

Theorem 2.1.4. If

ϕ \in C^{1} (R^{+} \times R)

is a strong solution of (2.3 2.4), then it is also a weak solution of (2.3 2.4).
定理 2.1.4。如果

ϕ \in C^{1} (R^{+} \times R)

是 (2.3 2.4) 的强解，那么它也是 (2.3 2.4) 的弱解。

ϕ \in L_{loc}^{1} (R^{+} \times R)

is a weak solution of (2.3.2.4) and if moreover

ϕ \in C^{1} (R^{+} \times R)

, then

ϕ

is a strong solution of (2.3 (2.4).
如果

ϕ \in L_{loc}^{1} (R^{+} \times R)

是(2.3.2.4)的弱解，并且如果

ϕ \in C^{1} (R^{+} \times R)

，那么

ϕ

是(2.3(2.4))的强解。

Proof. If

ϕ \in C^{1}

, then we do again the computations of section 2.1.4 which gives (2.6), namely

ϕ

is a weak solution.
证明。如果

ϕ \in C^{1}

，那么我们再次进行 2.1.4 节的计算，得到(2.6)，即

ϕ

是一个弱解。

ϕ

is a weak solution and if moreover

ϕ \in C^{1}

, then we can integrate by parts (2.6), which gives
如果

ϕ

是一个弱解，并且如果

ϕ \in C^{1}

，那么我们可以通过分部积分(2.6)进行积分，这给出

\int_{R^{+}} \int_{R} ψ (t, x) [\partial_{t} ϕ (t, x) + c \partial_{x} ϕ (t, x)] d x d t = 0

We recall that if a function

f (t, x) \in C^{0}

satisfies
我们回忆起，如果一个函数

f (t, x) \in C^{0}

满足

\int_{R^{+}} \int_{R} ψ (t, x) f (t, x) d x d t = 0

for any

ψ \in C^{1} (R^{+} \times R)

, then

f

is identically 0 (c.f. lectures on the Lebesgue integral). Thus
对于任何

ψ \in C^{1} (R^{+} \times R)

，则

f

恒等于 0（参见关于勒贝格积分的讲座）。因此

\partial_{t} ϕ (t, x) + c \partial_{x} ϕ (t, x) = 0

which means that

ϕ

is a strong solution.
这意味着

ϕ

是一个强溶液。
Thus 因此

Strong solution $\to$ weak solution,
强解 $\to$ 弱解，
Weak solution and regularity $\to$ strong solution.
弱解和正则性 $\to$ 强解。

2.1.7 Existence and uniqueness of weak solutions
2.1.7 弱解的存在性和唯一性

Let us now prove the existence and uniqueness of a weak solution.
现在让我们证明弱解的存在性和唯一性。
Theorem 2.1.1. Let

ϕ_{0} \in L_{loc}^{1} (R^{+} \times R)

. Then there exists a unique weak solution

ϕ \in L_{loc}^{1} (R^{+} \times R)

to
定理 2.1.1. 设

ϕ_{0} \in L_{loc}^{1} (R^{+} \times R)

。则存在唯一的弱解

ϕ \in L_{loc}^{1} (R^{+} \times R)

使得

\partial_{t} ϕ + c \partial_{x} ϕ = 0

with initial data

ϕ_{0}

. 带有初始数据

ϕ_{0}

。
Proof. Let us begin with the existence part and let us check that

ϕ (t, x) =

ϕ_{0} (x - c t)

is a weak solution. Let

ψ

C^{1} (R^{+} \times R)

with compact support. We have
证明。我们先从存在性部分开始，检查

ϕ (t, x) =

ϕ_{0} (x - c t)

是否是一个弱解。设

ψ

为具有紧支撑的

C^{1} (R^{+} \times R)

。我们有

\int_{R^{+} \times R} ϕ (t, x) [\partial_{t} ψ + c \partial_{x} ψ] d x d t = \int_{R^{+} \times R} ϕ_{0} (x - c t) [\partial_{t} ψ + c \partial_{x} ψ] d x d t

We make the change of variables

y = x - c t

in the previous equation, which leads to
我们在前面的方程中进行变量变换

y = x - c t

，这导致了

\begin{aligned} \int_{R^{+} \times R} ϕ_{0} (y) [\partial_{t} ψ (t, y + c t) + c \partial_{x} ψ (t, y + c t)] d y d t \\ = \int_{R^{+} \times R} ϕ_{0} (y) \frac{d}{d t} [ψ (t, y + c t)] d y d t = - \int_{R} ϕ_{0} (y) ψ (0, y) d y \end{aligned}

which ends the proof of existence.
这结束了存在性的证明。

Let us turn to the uniqueness. Let

ϕ_{1}

and

ϕ_{2}

be two weak solutions. Then, by linearity,

ϕ = ϕ_{1} - ϕ_{2}

is a weak solution with initial data 0 .
让我们转向唯一性。设

ϕ_{1}

和

ϕ_{2}

是两个弱解。那么，由于线性性质，

ϕ = ϕ_{1} - ϕ_{2}

是一个初始数据为 0 的弱解。

If we prove that 如果我们证明了

\int_{R^{+} \times R^{2}} ϕ (t, x) θ (t, x) d x d t = 0

for every function

θ \in C^{1} (R^{+} \times R)

with compact support, then we obtain that

ϕ = 0

, which will prove the uniqueness of the solution.
对于每个具有紧支撑的函数

θ \in C^{1} (R^{+} \times R)

，我们可以得到

ϕ = 0

，这将证明解的唯一性。

Let

θ

be given. It remains to construct

ψ \in C^{1}

with compact support such that
设

θ

已知。接下来需要构造

ψ \in C^{1}

具有紧支撑，使得

\partial_{t} ψ + c \partial_{x} ψ = θ

Let

f (t, x) = ψ (t, x + c t)

. Then 让

f (t, x) = ψ (t, x + c t)

。然后

\partial_{t} f (t, x) = θ (t, x + c t)

We take 我们采取

f (t, x) = - \int_{t}^{+ \infty} θ (s, x + c s) d s

We then have 我们接下来有

ψ (t, x) = f (t, x - c t)

Then

ψ (t, x)

is a solution of (2.7) with compact support. We have
那么

ψ (t, x)

是具有紧支撑的 (2.7) 的解。我们有

\int_{R^{+} \times R^{-}} ϕ (t, x) θ (t, x) d x d t = \int_{R^{+} \times R} ϕ (t, x) [\partial_{t} ψ (t, x) + c \partial_{x} ψ (t, x)] d x d t = 0

since

ϕ

is a weak solution with

ϕ (0, x) = 0

, which ends the proof.
由于

ϕ

是一个弱解，且满足

ϕ (0, x) = 0

，这就结束了证明。

2.1.8 Qualitative properties
2.1.8 质量属性

In PDEs, a “qualitative property” is a property of the equation, like its sign, estimates on its maximum, its behavior in long time, namely, any property of the solution which allows to describe it.
在偏微分方程中，“定性属性”是方程的一个属性，比如它的符号、对其最大值的估计、在长时间内的行为，即任何允许描述解的属性。

For instance, if the support of

ψ_{0}

is compact, then for every

t

, the support of

ϕ (t, x)

is compact, as stated in the next proposition.
例如，如果

ψ_{0}

的支撑是紧致的，那么对于每个

t

，

ϕ (t, x)

的支撑也是紧致的，如下一个命题所述。

Proposition 2.1.5. Let

M > 0

and let us assume that

ϕ_{0} \in C^{1}

and vanishes outside

[- M, M]

. Then

ϕ (t, \cdot)

vanishes outside

[- M + c t, + M + c t]

.
命题 2.1.5. 设

M > 0

，并假设

ϕ_{0} \in C^{1}

在

[- M, M]

之外消失。那么

ϕ (t, \cdot)

在

[- M + c t, + M + c t]

之外消失。

Proof. As

ϕ (t, x) = ϕ_{0} (x - c t)

, we have

ϕ (t, x) = 0

| x - c t | \geq M

, which gives the proposition.
证明。如

ϕ (t, x) = ϕ_{0} (x - c t)

，我们有

ϕ (t, x) = 0

如果

| x - c t | \geq M

，这给出了命题。

This proof however completely relies on the explicit solution of 2.3 2.4 and can not be generalized to other kinds of equations. It is thus interesting to find another proof, which does not rely on this explicit solution. The proof will be more “robust” and can be applied to many other equations where there is no explicit solution.
然而，这个证明完全依赖于 2.3 和 2.4 的显式解，无法推广到其他类型的方程。因此，寻找另一个不依赖于这个显式解的证明是很有趣的。这个证明将更加“稳健”，可以应用于许多没有显式解的其他方程。

Let us prove the following weaker result, without using the explicit formula: if for any

t, ϕ (t, x)

goes to 0 as

| x | \to + \infty

, then

ϕ (t, \cdot)

is supported in

[- M + c t, M + c t]

.
让我们证明以下较弱的结果，而不使用显式公式：如果对于任何

t, ϕ (t, x)

，当

| x | \to + \infty

趋向于 0 时，

ϕ (t, \cdot)

在

[- M + c t, M + c t]

上是有支撑的。

Let us introduce 让我们介绍一下

I (t) = \int_{x \geq M + c t} | ϕ (t, x) |^{2} d x

Then

I (t)

C^{1}

and
然后

I (t)

是

C^{1}

和

\begin{aligned} I^{'} (t) & = 2 \int_{x \geq M + c t} ϕ (t, x) \partial_{t} ϕ (t, x) d x - c | ϕ (t, M + c t) |^{2} \\ = - 2 c \int_{x \geq M + c t} ϕ (t, x) \partial_{x} ϕ (t, x) d x - c | ϕ (t, M + c t) |^{2} \\ = c | ϕ (t, M + c t) |^{2} - c | ϕ (t, M + c t) |^{2} = 0 \end{aligned}

Thus,

I (t)

is a constant. As

I (0) = 0

by assumption,

I (t) = 0

for every

t

, which implies that

ϕ (t, x) = 0

for every

t \geq 0

and

x \geq M + c t

. We then repeat the argument on

x < - M + c t

.
因此，

I (t)

是一个常数。根据假设，

I (0) = 0

对于每个

t

，这意味着

ϕ (t, x) = 0

对于每个

t \geq 0

和

x \geq M + c t

。然后我们在

x < - M + c t

上重复这个论证。

This kind of method is called an “energy method”. We define an “energy”

I (t)

and try to get informations on it. Such techniques are of constant use in PDEs.
这种方法被称为“能量方法”。我们定义一个“能量”

I (t)

，并试图获取关于它的信息。这种技术在偏微分方程中常常使用。

Similar ideas can be used to prove the uniqueness of compactly supported solutions. Namely, let

ϕ_{1}

and

ϕ_{2}

be two solutions of (2.3) 2.4 which go to 0 at inifinity for each time

t \geq 0

. Then

ϕ = ϕ_{1} - ϕ_{2}

goes to 0 as

| x | \to + \infty

and satisy
类似的思想可以用来证明紧支撑解的唯一性。即，设

ϕ_{1}

和

ϕ_{2}

是(2.3) 2.4 的两个解，对于每个时间

t \geq 0

，它们在无穷远处趋向于 0。那么

ϕ = ϕ_{1} - ϕ_{2}

在

| x | \to + \infty

时也趋向于 0，并满足

\partial_{t} ϕ + c \partial_{x} ϕ = 0

with initial data

ϕ (0, x) = 0

, for every

x \in R

. We have
初始数据为

ϕ (0, x) = 0

，对于每个

x \in R

。我们有

\partial_{t} \int_{R} | ϕ (t, x) |^{2} d x = 2 \int_{R} ϕ \partial_{t} ϕ d x = - 2 c \int_{R} ϕ \partial_{x} ϕ d x = - c \int_{R} \partial_{x} (ϕ^{2}) d x = 0

,
thus

ϕ (t, x) = 0

for every

t \geq 0

and

x \in R

, which gives uniqueness.
因此，对于每个

t \geq 0

和

x \in R

，这赋予了唯一性。
Let us now turn to another property: the transport equation satisfies the following “maximum principle”
现在让我们转向另一个性质：传输方程满足以下“极大值原理”

Proposition 2.1.6. Let

ϕ

be a classical solution of 2.3.2.4). Then, if

ϕ_{0} (x) \leq M

for every

x \in R

,
命题 2.1.6. 设

ϕ

是 2.3.2.4) 的一个经典解。那么，如果对每个

x \in R

都有

ϕ_{0} (x) \leq M

。

ϕ (t, x) \leq M, \forall t \geq 0, \forall x \in R

ϕ_{0} (x) \geq 0

for every

x \in R

, then

ϕ (t, x) \geq 0

for every

t \geq 0

and

x \in R

.
如果

ϕ_{0} (x) \geq 0

对于每个

x \in R

，那么

ϕ (t, x) \geq 0

对于每个

t \geq 0

和

x \in R

。
Proof. The proof of this proposition is obvious, since, for any

t

and

x

,
证明。这个命题的证明显而易见，因为对于任何

t

和

x

，

ϕ (t, x) = ϕ_{0} (x - c t) \leq M

This proof again completely relies on the explicit solution of 2.3 2.4 and can not be generalized to other kinds of equations. It is thus interesting to find another proof, which does not rely on this explicit solution.
这个证明再次完全依赖于 2.3 和 2.4 的显式解，无法推广到其他类型的方程。因此，寻找一个不依赖于这个显式解的其他证明是很有趣的。

Let

f \in C^{1} (R)

. Then

f (ϕ)

C^{1}

and
让

f \in C^{1} (R)

。那么

f (ϕ)

是

C^{1}

和

f^{'} (ϕ (t, x)) \partial_{t} ϕ + c f^{'} (ϕ (t, x)) \partial_{t} ϕ = 0

thus 因此

\partial_{t} f (ϕ) + c \partial_{x} f (ϕ) = 0

Thus 因此

ψ (t, x) = f (ϕ (t, x))

is also a solution of the transport equation
也是传输方程的一个解

\partial_{t} ψ + c \partial_{x} ψ = 0

Assume now that

f (y) = 0

y \leq M

. Then we have

ψ (0, x) = 0

for every

x

. Using the uniqueness of the solution of the equation, this gives

ψ (t, x) = 0

for every

t

and

x

. Let us take
假设现在如果

y \leq M

，则

f (y) = 0

。那么我们对每个

x

都有

ψ (0, x) = 0

。利用方程解的唯一性，这为每个

t

和

x

提供了

ψ (t, x) = 0

。让我们取

f (y) = (sup (0, y - M))^{2}

Then as

f (ϕ (t, x)) = 0

for every

t

and

x

, this means that

ϕ (t, x) \leq M

everywhere which ends the proof.
那么作为

f (ϕ (t, x)) = 0

对于每个

t

和

x

，这意味着

ϕ (t, x) \leq M

到处都是，这结束了证明。

The proof that

ϕ_{0} (x) \geq 0

for every

x

implies

ϕ (t, x) \geq 0

for ever

t

and

x

is similar.
证明

ϕ_{0} (x) \geq 0

对于每个

x

意味着

ϕ (t, x) \geq 0

对于每个

t

和

x

是类似的。

Remark. The idea to introduce nonlinear functions of the solution, like

f (u)

is called “renomarlization”. It is of wide use in many different kinds of PDEs.
备注。引入解的非线性函数的想法，如

f (u)

，称为“重标定”。它在许多不同类型的偏微分方程中被广泛使用。

2.1.9 Numerical schemes 2.1.9 数值方案

A central question in applications is to compute numerically solutions of PDEs, since most of them have no explicit solutions. It is of course not possible to evaluate a solution everywhere, namely for every time and position, since computers can only handle a finite number of informations. The starting point is thus to construct a “grid” and to “discretize” the space and time.
在应用中，一个核心问题是数值计算偏微分方程的解，因为大多数方程没有显式解。当然，无法在每个时间和位置评估解，因为计算机只能处理有限数量的信息。因此，起点是构建一个“网格”，并对空间和时间进行“离散化”。

Let

h > 0

and

k

be (small) real numbers. We define the points

x_{j}

by
让

h > 0

和

k

是（小）实数。我们通过以下方式定义点

x_{j}

：

x_{j} = j h, j \in Z

and the times

t_{n}

by
和时代

t_{n}

由

t_{n} = n k, n \in N

The couples

{(t_{n}, x_{j})}_{n \in N, j \in Z}

form a “grid” on which we want to evaluate the solution

ϕ

to (2.3 2.4). We will denote by

ϕ_{n}^{j}

a numerical approximation of

ϕ (t_{n}, x_{j})

.
这些夫妇

{(t_{n}, x_{j})}_{n \in N, j \in Z}

形成了一个“网格”，我们希望在其上评估解决方案

ϕ

对(2.3 2.4)。我们将用

ϕ_{n}^{j}

表示

ϕ (t_{n}, x_{j})

的数值近似。

The central question of “numerical analysis” is to find algorithms to evaluate

ϕ_{n}^{j}

which are precise and fast. Of course

ϕ_{n}^{j}

will in general not equal

ϕ (t_{n}, x_{j})

but we want the error
“数值分析”的核心问题是寻找精确且快速的算法来评估

ϕ_{n}^{j}

。当然，

ϕ_{n}^{j}

通常不等于

ϕ (t_{n}, x_{j})

，但我们希望误差

ε_{n}^{j} = ϕ_{n}^{j} - ϕ (t_{n}, x_{j})

to be as small and possible, and we want to compute as fast as possible.
尽可能小，并且我们希望计算尽可能快。
Let us now detail a simple “numerical scheme”. We approximate

\partial_{t} ϕ

and

\partial_{x} ϕ

by
现在让我们详细介绍一个简单的“数值方案”。我们通过以下方式近似

\partial_{t} ϕ

和

\partial_{x} ϕ

：

\partial_{t} ϕ (t_{n}, x_{j}) \approx \frac{ϕ_{n + 1}^{j} - ϕ_{n}^{j}}{k}, \partial_{x} ϕ (t_{n}, x_{j}) \approx \frac{ϕ_{n}^{j} - ϕ_{n}^{j - 1}}{h}

This leads to the following relations
这导致了以下关系

\frac{ϕ_{n + 1}^{j} - ϕ_{n}^{j}}{k} + c \frac{ϕ_{n}^{j} - ϕ_{n}^{j - 1}}{h} = 0

We can start with 我们可以从这里开始

ϕ_{0}^{j} = ϕ_{0} (x_{j})

Then the various

ϕ_{n}^{j}

can be computed by induction, using
然后可以通过归纳法计算各种

ϕ_{n}^{j}

ϕ_{n + 1}^{j} = ϕ_{n}^{j} - c \frac{k}{h} (ϕ_{n}^{j} - ϕ_{n}^{j - 1})

Starting with

{(ϕ_{0}^{j})}_{j \in Z}

, we compute

{(ϕ_{1}^{j})}_{j \in Z}

and iterate the process.
从

{(ϕ_{0}^{j})}_{j \in Z}

开始，我们计算

{(ϕ_{1}^{j})}_{j \in Z}

并迭代这个过程。

Consistence of the numerical scheme
数值方案的一致性

The error of consistence is the error made when we replace

ϕ_{n}^{j}

ϕ (t_{n}, x_{j})

, namely,
一致性错误是指当我们用

ϕ (t_{n}, x_{j})

替换

ϕ_{n}^{j}

时所犯的错误，即，

δ_{n, j} = ϕ (t_{n + 1}, x_{j}) - ϕ (t_{n}, x_{j}) + c \frac{k}{h} [ϕ (t_{n}, x_{j}) - ϕ (t_{n}, x_{j - 1})] .

Using Taylor’s formula, assuming that the derivative of

ϕ_{0}

is uniformly bounded (which implies that the derivatives of

ϕ

are uniformly bounded), we have
使用泰勒公式，假设

ϕ_{0}

的导数是均匀有界的（这意味着

ϕ

的导数是均匀有界的），我们有

\begin{aligned} δ_{n, j} & = k \partial_{t} ϕ (t_{n}, x_{j}) + O (k^{2}) + c \frac{k}{h} [\partial_{x} ϕ (t_{n}, x_{j}) h + O (h^{2})] \\ = O (k^{2}) + O (k h) \end{aligned}

Thus

δ_{n, j}

is small, like

k (h + k)

provided

k

and

h

are small. The numerical scheme is thus said “consistent”.
因此

δ_{n, j}

很小，就像

k (h + k)

，前提是

k

和

h

也很小。因此，这个数值方案被称为“一致的”。

Stability 稳定性

A numerical scheme is said “stable” if we can control the growth of norms.
一个数值方案被称为“稳定”，如果我们可以控制范数的增长。
Proposition 2.1.7. Assume that

c > 0

and that

c k / h < 1

, then we have
命题 2.1.7. 假设

c > 0

和

c k / h < 1

，则我们有

sup_{l \in Z} | ϕ_{n + 1}^{l} | \leq sup_{l \in Z} | ϕ_{n}^{l} |

Proof. We have for any

n \geq 0

and

j \in Z

,
证明。对于任何

n \geq 0

和

j \in Z

，

ϕ_{n + 1}^{j} = ϕ_{n}^{j} (1 - c \frac{k}{h}) + c \frac{k}{h} ϕ_{n}^{j - 1}

We note that

1 - c k / h

and

c k / h

positive and less than 1 provided

c > 0

and

c k / h < 1

. Thus
我们注意到

1 - c k / h

和

c k / h

是正数且小于 1，前提是

c > 0

和

c k / h < 1

。因此

| ϕ_{n + 1}^{j} | \leq max (| ϕ_{n}^{j} |, | ϕ_{n}^{j - 1} |)

which ends the proof. 这结束了证明。

Remarks. 备注。

If $c < 0$ , the maximum of $ϕ_{n}^{j}$ may increase at each step and develop “oscillations”. If $c < 0$ , we know that the solution travels “to the left”, thus the evolution of a value at $x_{j}$ depends on the value at $x_{j + 1}$ . A numerical scheme which does not involve $x_{j + 1}$ will fail.
如果 $c < 0$ ， $ϕ_{n}^{j}$ 的最大值可能在每一步增加并产生“振荡”。如果 $c < 0$ ，我们知道解是“向左”移动的，因此在 $x_{j}$ 处的值的演变依赖于在 $x_{j + 1}$ 处的值。一个不涉及 $x_{j + 1}$ 的数值方案将会失败。
The condition $c k / h < 1$ is called a Courant-Friedrichs-Lewy condition (“CFL” condition). It occurs in many different numerical schemes for many different equations. It says that during a time step $k$ , the solution has travelled over a distance $k c$ which is smaller than the step of the grid $h$ .
条件 $c k / h < 1$ 被称为 Courant-Friedrichs-Lewy 条件（“CFL” 条件）。它出现在许多不同方程的许多不同数值方案中。它表示在一个时间步 $k$ 内，解所经过的距离 $k c$ 小于网格步长 $h$ 。
The support of the numerical solution $ϕ_{n}^{j}$ increases by 1 on the right at each time step, namely with a velocity $h / k$ . The velocity of the evolution of the support of $ϕ$ is $c$ . We must have $c < h / k$ , which is the CFL condition.
数值解的支撑在每个时间步长右侧增加 1，即以速度 $h / k$ 。 $ϕ$ 的支撑演化速度为 $c$ 。我们必须满足 $c < h / k$ ，这就是 CFL 条件。

2.1.10 Convergence of the numerical scheme
2.1.10 数值方案的收敛性

Proposition 2.1.8. Let

ϕ_{0} \in C^{1} (R)

with compact support. Let

h_{m}

and

k_{m}

be a sequence of positive real numbers, such that

c k_{m} / h_{m} < 1

. Let

ϕ_{n, m}^{j}

be defined by 2.8 2.9) and let

ϕ

be the solution of 2.3 2.4). Then, for any

T > 0

, there exists a constant

C

such that
命题 2.1.8. 设

ϕ_{0} \in C^{1} (R)

具有紧支撑。设

h_{m}

和

k_{m}

为一列正实数，满足

c k_{m} / h_{m} < 1

。设

ϕ_{n, m}^{j}

由 2.8 2.9) 定义，且

ϕ

为 2.3 2.4) 的解。那么，对于任何

T > 0

，存在一个常数

C

使得

sup_{m \to + \infty} sup_{n k_{m} \leq T, j \in Z} | ϕ_{n, m}^{j} - ϕ (t_{n}, x_{j}) | \leq C h

In other words, the numerical solution converges to the true one with a speed

O (h)

.
换句话说，数值解以速度

O (h)

收敛到真实解。

Proof. Let us fix some

m > 0

and forget the index

m

in the notations. We note that
证明。我们固定一些

m > 0

，并在符号中忽略索引

m

。我们注意到

| δ_{n, j} | \leq C_{0} k (h + k)

for some constant which depends on

T

. Let
对于某个依赖于

T

的常数。让

ε_{n}^{j} = ϕ_{n}^{j} - ϕ (t_{n}, x_{j})

Then 然后

ε_{n + 1}^{j} = ε_{n}^{j} (1 - c \frac{k}{h}) + c \frac{k}{h} ε_{n}^{j - 1} - δ_{n}^{j}

thus, using the CFL condition,
因此，使用 CFL 条件，

sup_{j \in Z} | ε_{n + 1}^{j} | \leq sup_{j \in Z} | ε_{n}^{j} | + C_{0} k (h + k)

This leads to 这导致了

sup_{j \in Z} | ε_{n}^{j} | \leq C_{0} n k (h + k) \leq C_{0} T (h + k) \leq C h

where we have used

n k \leq T

and the CFL condition.
我们使用了

n k \leq T

和 CFL 条件。

Remarks. The previous proof uses only two ingredients: the consistence of the scheme (namely estimates on

δ_{n}^{j}

and the stability of the scheme (estimates on the supremum). The consistence and the stability leads to the convergence of the scheme, which occurs for many different numerical schemes and many different equations. We sum up the proof by saying
备注。之前的证明只使用了两个要素：方案的一致性（即对

δ_{n}^{j}

的估计）和方案的稳定性（对上确界的估计）。一致性和稳定性导致了方案的收敛，这在许多不同的数值方案和许多不同的方程中都发生。我们总结证明如下：

consistence + stability ⟹ convergence.

2.1.11 Boundaries 2.1.11 边界

In applications, we usually are not in the whole space, but only in a bounded domain. Let

a < b

be two real numbers and let us study the transport equation on

[a, b]

. We have to describe what happens at

a

and

b

. If

c > 0

, the flow goes from the left to the right, it “enters” the domain at

a

and “leaves” the domain at

b

. We thus enforce a boundary condition at

x = a

, and nothing at

x = b

. This leads to
在应用中，我们通常不在整个空间中，而只是处于一个有界域中。设

a < b

为两个实数，我们来研究

[a, b]

上的输运方程。我们必须描述在

a

和

b

发生的情况。如果

c > 0

，流动从左向右，它在

a

“进入”该域，在

b

“离开”该域。因此，我们在

x = a

施加边界条件，而在

x = b

则不施加任何条件。这导致了

\begin{matrix} \partial_{t} ϕ + c \partial_{x} ϕ = 0, \forall t \geq 0, \forall x \in [a, b] \\ ϕ (0, x) = ϕ_{0} (x), \forall x \in [a, b] \\ ϕ (t, a) = ϕ_{1} (t), \forall t \geq 0 \end{matrix}

where

ϕ_{0} \in C^{1} (R)

and

ϕ_{1} \in C^{1} (R^{+})

are two given functions.
其中

ϕ_{0} \in C^{1} (R)

和

ϕ_{1} \in C^{1} (R^{+})

是两个给定的函数。
Theorem 2.1.2. Let us assume that
定理 2.1.2. 让我们假设

ϕ_{1} (0) = ϕ_{0} (a), \partial_{t} ϕ_{1}^{'} (0) = c ϕ_{0}^{'} (a)

Then there exists a unique solution

ϕ \in C^{1} (R^{+} \times R)

to 2.12 2.13 (2.14).
那么存在一个唯一解

ϕ \in C^{1} (R^{+} \times R)

对于 2.12 2.13 (2.14)。
Proof. Let 证明。设

ψ (t, x) = ϕ (t, x + c t)

which is defined provided

a \leq x + c t \leq b

. Then, wherever it is defined,
定义为

a \leq x + c t \leq b

。然后，无论它在哪里被定义，

\partial_{t} ψ = \partial_{t} ϕ + c \partial_{x} ϕ = 0

thus

ψ (t, x)

is independent on

t

.
因此

ψ (t, x)

独立于

t

。
Let

t \geq 0

and

x \in [a, b]

. Two cases arise. If

x - c t > a

, then we have
让

t \geq 0

和

x \in [a, b]

。出现两种情况。如果

x - c t > a

，那么我们有

ϕ (t, x) = ϕ (t, x - c t + c t) = ψ (t, x - c t) = ψ (0, x - c t) = ϕ_{0} (x - c t)

On the contrary if

x - c t < a

, then we have, for any

τ

such that

0 \leq τ \leq t

,
相反，如果

x - c t < a

，那么对于任何

τ

，只要

0 \leq τ \leq t

，我们有：

ϕ (t, x) = ϕ (t, x - c t + c t) = ψ (t, x - c t) = ψ (τ, x - c t) = ϕ (τ, x - c t + c τ)

where we choose

τ

such that
我们选择

τ

使得

x - c t + c τ = a

namely 即

τ = t - \frac{x - a}{c}

Thus 因此

ϕ (t, x) = ϕ (τ, a) = ϕ_{1} (t - \frac{x - a}{c}) .

Thus, if the solution

ϕ

exists, it is given by 2.16 .
因此，如果解

ϕ

存在，则由 2.16 给出。
It remains to check that

ϕ

is a solution. In order

ϕ

to be

C^{0}

(t, x) =

(0, a)

, we must have
仍需检查

ϕ

是否是一个解。为了

ϕ

在

C^{0}

(t, x) =

(0, a)

，我们必须有

ϕ_{0} (a) = ϕ_{1} (0)

In order

ϕ

to be

C^{1}

(t, x) = (0, a)

, we must have
为了

ϕ

在

C^{1}

时

(t, x) = (0, a)

，我们必须拥有

\partial_{t} ϕ (0, a) = - c \partial_{x} ϕ (0, a)

thus 因此

\partial_{t} ϕ_{1} (0) = - c \partial_{x} ϕ_{0} (a)

The two relations (2.18) and (2.19) are called “compatibility conditions”.
这两个关系（2.18）和（2.19）被称为“兼容性条件”。
Under these two compatibility conditions, it can be checked that

ϕ

satisfies (2.12) when

x > a + c t

, when

x < a + c t

and also when

x = a + c t

.
在这两个兼容性条件下，可以检查到

ϕ

在

x > a + c t

、

x < a + c t

和

x = a + c t

时满足 (2.12)。

2.2 Transport equation in $R^{n}$
2.2 在 $R^{n}$ 中的传输方程

2.2.1 Introduction 2.2.1 引言

Let

n \geq 1

be an integer. Let

V (t, x)

be a vector field on

R^{n}

, namely
设

n \geq 1

为一个整数。设

V (t, x)

为在

R^{n}

上的一个向量场，即

V (t, x) = (V_{1} (t, x), \dots, V_{n} (t, x))

where, for every

1 \leq j \leq n

, 对于每个

1 \leq j \leq n

，

V_{j} \in C^{0} ([0, T], C^{1} (R^{n}))

In this section, we study the transport of quantities by the vector field

V (t, x)

.
在本节中，我们研究由向量场

V (t, x)

运输的量。

Definition 2.2.1. Characteristics of the vector field

V (t, x)

.
定义 2.2.1. 向量场

V (t, x)

的特征。
The solution

X (t; s, x)

to the equations
方程的解

X (t; s, x)

\begin{matrix} \partial_{t} X (t; s, x) = V (t, X (t; s, x)) \\ X (s; s, x) = x \end{matrix}

when it exists, is called the characteristics curve, or integral curves, or characteristics of the vector field

V (t, x)

. Moreover,

X (t; s, x)

is continuous in all its variables.
当它存在时，称为特征曲线，或积分曲线，或向量场

V (t, x)

的特征。此外，

X (t; s, x)

在其所有变量中是连续的。

Remark. The characteristics can be interpreted as follow. Imagine that a particle is at time

s

at the position

x \in R^{n}

and that it evolves according to the velocity field

V (t, x)

, namely that its speed at any time is equal to the value of

V

at it current position. Then its position at time

s

X (t; s, x)

.
备注。特征可以解释如下。想象一个粒子在时间

s

处于位置

x \in R^{n}

，并且它根据速度场

V (t, x)

演变，也就是说它在任何时刻的速度等于其当前位置的

V

值。那么它在时间

s

的位置信息是

X (t; s, x)

。

X (t; s, x)

is the position at time

t

of the particle which at time

s

is at

x

X (t; s, x)

是在时间

t

时粒子的位置，而在时间

s

时粒子位于

x

。
Let us now detail the properties of

X (t; s, x)

.
现在让我们详细说明

X (t; s, x)

的属性。
Lemma 2.2.2. For any

s, t, τ

and

x

, we have
引理 2.2.2。对于任何

s, t, τ

和

x

，我们有

X (τ; s, X (s; t, x)) = X (τ; t, x)

and in particular 特别是

X (t; s, X (s; t, x)) = x

Proof. Note that (2.23) is a consequence of (2.22) by taking

τ = t

. Now both terms of (2.22) are the position at time

τ

of the particle which at time

t

was at

x

.
证明。注意到 (2.23) 是通过取

τ = t

得到 (2.22) 的结果。现在 (2.22) 的两个项是粒子在时间

τ

的位置，而该粒子在时间

t

时位于

x

。

We say that a scalar

ϕ (t, x)

is "transported by the velocity field

V (t, x)

", if its values are constant along a characteristics, namely if, for any

s, t, x

,
我们说标量

ϕ (t, x)

"由速度场

V (t, x)

运输"，如果它的值沿特征线是恒定的，即对于任何

s, t, x

，

ϕ (t, X (t; s, x)) = ϕ (s, X (s; s, x)) = ϕ (s, x)

ϕ (t, x)

C^{1}

t

and

x

, then, differentiating 2.24, we obtain
如果

ϕ (t, x)

在

C^{1}

和

t

中是

x

，那么，对 2.24 进行求导，我们得到

\partial_{t} ϕ + \nabla_{x} ϕ \cdot \partial_{t} X = 0

which leads to the transport equation
这导致了传输方程

\partial_{t} ϕ (t, x) + V (t, x) \cdot \nabla_{x} ϕ (t, x) = 0

This equation describes the transport of a quantity

ϕ

by the vector field

V

.
这个方程描述了量

ϕ

在向量场

V

中的传输。

2.2.2 Resolution of the transport equation
2.2.2 运输方程的求解

In this paragraph, we focus on the resolution of the transport equation
在这一段中，我们专注于运输方程的解

\begin{matrix} \partial_{t} ϕ + V \cdot \nabla_{x} ϕ = 0 \\ ϕ (0, x) = ϕ_{0} (x), \forall x \in R^{d} \end{matrix}

where

ϕ_{0} (x)

is a given

C^{1} (R^{d})

function.
其中

ϕ_{0} (x)

是给定的

C^{1} (R^{d})

函数。
Proposition 2.2.3. Existence and uniqueness of

C^{1}

solutions. Let

ϕ_{0} \in C^{1} (R^{d})

. Let us assume that

V \in C^{0} ([0, T], C^{1} (R^{d}))

and that

V \in L^{\infty} ([0, T] \times R^{n})

. Then the characteristics

X (t; s, x)

are well defined, unique, and

X \in C^{1} ([0, T] \times [0, T] \times R^{d})

. Moreover, there exists a unique solution

ϕ

to 2.27), 2.28) in

C^{1} ([0, T] \times R^{d})

. This solution is given by
命题 2.2.3.

C^{1}

解的存在性和唯一性。设

ϕ_{0} \in C^{1} (R^{d})

。我们假设

V \in C^{0} ([0, T], C^{1} (R^{d}))

和

V \in L^{\infty} ([0, T] \times R^{n})

。那么特征

X (t; s, x)

是良好定义的、唯一的，并且

X \in C^{1} ([0, T] \times [0, T] \times R^{d})

。此外，存在唯一解

ϕ

对于 2.27)、2.28) 在

C^{1} ([0, T] \times R^{d})

中。该解由以下给出：

ϕ (t, x) = ϕ_{0} (X (0; t, x)) .

Proof. Let

t

and

x

be fixed, then 2.20 is an ordinary differential equation in

t

, with an initial condition (2.21) prescribed at time

s

. As

V

C^{1}

x

and

C^{0}

t

, we can apply the Cauchy-Lifschitz theorem with parameters, which provides the existence, uniqueness and smoothness of the characteristics.
证明。设

t

和

x

为固定值，则 2.20 是

t

中的一个常微分方程，初始条件在时间

s

处规定为(2.21)。由于

V

在

C^{1}

中，

C^{0}

在

t

中，我们可以应用带参数的 Cauchy-Lifschitz 定理，这提供了特征的存在性、唯一性和光滑性。

We note that

ϕ

, defined by (2.29) is a solution, since, using (2.22), we have
我们注意到由(2.29)定义的

ϕ

是一个解，因为使用(2.22)，我们有

ϕ (t, X (t; 0, x)) = ϕ_{0} (x)

thus 因此

\partial_{t} ϕ + V \cdot \nabla_{x} ϕ = 0 .

Moreover

ϕ (0, x) = ϕ_{0} (x)

. This solution is

C^{1}

since

X

C^{1}

.
此外

ϕ (0, x) = ϕ_{0} (x)

。这个解决方案是

C^{1}

，因为

X

是

C^{1}

。
Let us turn to the uniqueness. If

ψ

is a solution then

ψ (t, X (t; 0, x))

is constant in

t

and is equal to

ψ (0, X (0; 0, x)) = ϕ_{0} (x)

, thus using (2.22),

ψ = ϕ

.
让我们转向独特性。如果

ψ

是一个解，那么

ψ (t, X (t; 0, x))

在

t

中是常数，并且等于

ψ (0, X (0; 0, x)) = ϕ_{0} (x)

，因此使用(2.22)，

ψ = ϕ

。

2.3 Exercises 2.3 练习

Exercise 2.3.1. Check your knowledge of the Cauchy-Lifschitz theorem with parameters, and read again its proof.
练习 2.3.1。检查你对带参数的柯西-利夫希茨定理的知识，并再次阅读其证明。

Exercise 2.3.2. Let

ϕ_{0} (x) \in C^{1} (R)

be a given bounded function.
练习 2.3.2. 设

ϕ_{0} (x) \in C^{1} (R)

为一个给定的有界函数。

Compute the solution of 计算以下方程的解

\partial_{t} ϕ + \partial_{x} ϕ = 2 ϕ,

with initial data

ϕ_{0}

. 带有初始数据

ϕ_{0}

。
Hint: divide the equation by

ϕ

, and study

\log ϕ - 2 t

.
提示：将方程除以

ϕ

，并研究

\log ϕ - 2 t

。

Same question with 相同的问题与

\partial_{t} ϕ + \partial_{x} ϕ = ϕ^{2} .

Same question with 相同的问题与

\partial_{t} ϕ + \partial_{x} ϕ = t^{2} .

Exrtise 2.3.3. Compute the solution to the following equations
练习 2.3.3。计算以下方程的解。

\begin{array}{crr} \partial_{t} ϕ + \partial_{x} ϕ = 0, & with & ϕ_{0} (x) = \sin x \\ \partial_{t} ϕ + x \partial_{x} ϕ = 0, & with & ϕ_{0} (x) = \sin x, \\ \partial_{t} ϕ + \cos x \partial_{x} ϕ = 0, & with & ϕ_{0} (x) = e^{x} \\ \partial_{t} ϕ + ψ (x) \partial_{x} ϕ = 0, & with & ϕ_{0} (x) = e^{x} . \end{array}

where

ψ (x)

is a given function.
其中

ψ (x)

是一个给定的函数。
Exercise 2.3.4. Compute the solution to the following equations
练习 2.3.4。计算以下方程的解。

\begin{matrix} \partial_{t} ϕ + x \partial_{x} ϕ + y \partial_{y} ϕ = ϕ, with ϕ_{0} (x) = \sin x \cos y \\ \partial_{t} ϕ - y \partial_{x} ϕ + x \partial_{y} ϕ = 0 \end{matrix}

where

ψ (x)

is a given function.
其中

ψ (x)

是一个给定的函数。
Exercise 2.3.5. Let

V (t, x) \in C^{1} (R^{+} \times R^{n})

be a bounded function, whose gradient is also bounded (uniformly in

t

and

x

). Let

ϕ_{0} (x) \in C^{1} (R^{n})

. Prove that there exists a unique solution

ϕ \in C^{1} (R^{+} \times R^{n})

to
练习 2.3.5. 设

V (t, x) \in C^{1} (R^{+} \times R^{n})

是一个有界函数，其梯度在

t

和

x

中也是有界的。设

ϕ_{0} (x) \in C^{1} (R^{n})

。证明存在唯一解

ϕ \in C^{1} (R^{+} \times R^{n})

。

\partial_{t} ϕ + \nabla \cdot (V ϕ) = 0

with initial data

ϕ (0, x) = ϕ_{0} (x)

(for every

x \in R^{n})

.
使用初始数据

ϕ (0, x) = ϕ_{0} (x)

（对于每个

x \in R^{n})

。
Exercise 2.3.6. Using Python or another language, code the numerical scheme. Observe what happens if

c < 0

or if the CFL condition is not satisfied. By taking smaller and smaller

k

and

h

, check the speed of convergence.
练习 2.3.6。使用 Python 或其他语言，编码数值方案。观察如果

c < 0

或 CFL 条件不满足时会发生什么。通过逐渐减小

k

和

h

，检查收敛速度。

Exercise 2.3.7. We consider the numerical scheme
练习 2.3.7。我们考虑数值方案

\frac{ϕ_{n + 1}^{j} - ϕ_{n}^{j}}{k} + c \frac{ϕ_{n}^{j + 1} - ϕ_{n}^{j - 1}}{2 h}

What is its error of consistence ? Is it stable? What is the corresponding CLF? Prove a result similar to Theorem 2.1.8.
它的一致性误差是多少？它稳定吗？相应的 CLF 是什么？证明一个类似于定理 2.1.8 的结果。

Exercise 2.3.8. Let

u_{0} \in C^{1} (R)

and let

f \in C^{1} (R^{+} \times R)

. Prove that there exists a unique classical solution to
练习 2.3.8。设

u_{0} \in C^{1} (R)

和

f \in C^{1} (R^{+} \times R)

。证明存在唯一的经典解。

\partial_{t} u (t, x) + c \partial_{x} u (t, x) = f (t, x)

with initial data

u_{0} (x)

. 带有初始数据

u_{0} (x)

。

Introduction to Partial Differential Equations 偏微分方程导论

Chapter 1 第一章

Introduction 介绍

1.1 Objectives of these book1.1 本书的目标

1.2 Notations 1.2 符号

1.2.1 Derivatives 1.2.1 导数

1.2.2 Functional spaces 1.2.2 功能空间

1.2.3 Miscellaneous 1.2.3 杂项

Chapter 2 第二章

Transport equations 运输方程

2.1 A first example in R R R\mathbb{R}2.1 第一个例子在 R R R\mathbb{R}

2.1.1 Introduction 2.1.1 介绍

2.1.2 Solution of the transport equation2.1.2 运输方程的解

2.1.3 Well-posedness in the sense of Hadamard2.1.3 哈达玛德意义下的良好适定性

2.1.4 Notion of weak solutions2.1.4 弱解的概念

2.1.5 An example of weak solution2.1.5 弱解的一个例子

2.1.6 Link between weak and strong solutions2.1.6 弱解与强解之间的联系

2.1.7 Existence and uniqueness of weak solutions2.1.7 弱解的存在性和唯一性

2.1.8 Qualitative properties2.1.8 质量属性

2.1.9 Numerical schemes 2.1.9 数值方案

Consistence of the numerical scheme数值方案的一致性

Stability 稳定性

Remarks. 备注。

2.1.10 Convergence of the numerical scheme2.1.10 数值方案的收敛性

2.1.11 Boundaries 2.1.11 边界

2.2 Transport equation in R n R n R^(n)\mathbb{R}^{n}2.2 在 R n R n R^(n)\mathbb{R}^{n} 中的传输方程

2.2.1 Introduction 2.2.1 引言

2.2.2 Resolution of the transport equation2.2.2 运输方程的求解

2.3 Exercises 2.3 练习

Introduction to Partial Differential Equations
偏微分方程导论

1.1 Objectives of these book
1.1 本书的目标

2.1 A first example in $R$
2.1 第一个例子在 $R$

2.1.2 Solution of the transport equation
2.1.2 运输方程的解

2.1.3 Well-posedness in the sense of Hadamard
2.1.3 哈达玛德意义下的良好适定性

2.1.4 Notion of weak solutions
2.1.4 弱解的概念

2.1.5 An example of weak solution
2.1.5 弱解的一个例子

2.1.6 Link between weak and strong solutions
2.1.6 弱解与强解之间的联系

2.1.7 Existence and uniqueness of weak solutions
2.1.7 弱解的存在性和唯一性

2.1.8 Qualitative properties
2.1.8 质量属性

Consistence of the numerical scheme
数值方案的一致性

2.1.10 Convergence of the numerical scheme
2.1.10 数值方案的收敛性

2.2 Transport equation in $R^{n}$
2.2 在 $R^{n}$ 中的传输方程

2.2.2 Resolution of the transport equation
2.2.2 运输方程的求解