Lewy‘s Equation | 关于 H. Lewy 的微分方程

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#Lewy‘s Equation

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注：本文为 “Lewy‘s Equation” 相关译文。
英文引文，机翻未校。
如有内容异常，请看原文。

ON THE DIFFERENTIAL EQUATIONS OF H. LEWY

关于 H. Lewy 的微分方程

W. B. SMITH-WHITE
(received 10 October 1960)

1. Introduction

1. 引言

It is known that the theory of Cauchy’s problem for differential equations with two independent variables is reducible to the corresponding problem for systems of quasi-linear equations.
众所周知，含两个自变量的微分方程的柯西问题理论可简化为拟线性方程组的相应问题。

The reduction is carried further, by means of the theory of characteristics, to the case of systems of equations of the special form first considered by H. Lewy [1].
借助特征线理论，可进一步将其简化为 H. Lewy [1] 首次考虑的特殊形式方程组的情形。

The simplest case is that of the pair of equations
最简单的情形是如下方程组：
$\left.\begin{array}{l} a_{11} \frac{\partial z_{1}}{\partial x} + a_{12} \frac{\partial z_{2}}{\partial x} = 0 \\ a_{21} \frac{\partial z_{1}}{\partial y} + a_{22} \frac{\partial z_{2}}{\partial y} = 0 \end{array}\right\}\tag{1}$
where the $a_{i j}$ depend on $z_{1}$ and $z_{2}$ .
其中 $a_{ij}$ 依赖于 $z_1$ 和 $z_2$ 。

The problem to be considered is that of finding functions $z_{1}(x, y)$ , $z_{2}(x, y)$ which satisfy (1) and which take prescribed values on $x + y = 0$ .
需要考虑的问题是寻找满足方程组 (1) 且在直线 $x + y = 0$ 上取给定值的函数 $z_1(x, y)$ 和 $z_2(x, y)$ 。

For brevity we shall say that a function $\cdots)$ of the arguments $\cdots$ is of class $C^{(n)}[a, b, c, \cdots]$ or simply of class $C^{(n)}$ , when all the partial derivatives of $f$ with respect to these variables of order $\leqq n$ exist and are continuous.
为简洁起见，若函数 $\cdots)$ 关于其自变量 $\cdots$ 的所有阶数 $\leqq n$ 的偏导数均存在且连续，则称该函数为 $C^{(n)}[a, b, c, \cdots]$ 类函数，或简称为 $C^{(n)}$ 类函数。

If on the line $\lambda$ , $-\lambda$ the Cauchy data are $C^{(2)}[\lambda]$ , if the coefficients $a_{i j}$ are $C^{(2)}[z_{1}, z_{2}]$ and if
若在直线 $\lambda$ 、 $-\lambda$ 上的柯西数据为 $C^{(2)}[\lambda]$ 类，系数 $a_{ij}$ 为 $C^{(2)}[z_1, z_2]$ 类，且满足
$\Delta = \left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right| \neq 0$

Lewy showed
Lewy 证明了：

(I) that there exists a solution-pair $z_{1}$ , $z_{2}$ of equations (1) of class $C^{(1)}[x, y]$ defined near $x = y = 0$ and which takes the prescribed values on $x + y = 0$ , and for which the “mixed” derivatives exist and are continuous.
(I) 存在方程组 (1) 的一组解 $z_1, z_2$ ，其为在 $x = y = 0$ 附近有定义的 $C^{(1)}[x, y]$ 类函数，在 $x + y = 0$ 上取给定值，且其“混合”导数存在且连续。

$\frac{\partial^2 z_1}{\partial x \partial y}, \quad \frac{\partial^2 z_2}{\partial x \partial y}\tag{2}$

(II) that the solution just described is unique.
(II) 上述解是唯一的。

Lewy’s method consists in replacing the differential equations by difference equations and then establishing the validity of a limiting process.
Lewy 的方法是将微分方程替换为差分方程，然后证明极限过程的合理性。

Other writers [2] have obtained the same results under the weaker hypothesis that the Cauchy data are $C^{(1)}[\lambda]$ on the line $\lambda$ , $-\lambda$ .
其他学者 [2] 在更弱的假设下（即直线 $\lambda$ 、 $-\lambda$ 上的柯西数据为 $C^{(1)}[\lambda]$ 类）得到了相同的结果。

But the result on uniqueness is conditional on the existence and continuity of the derivatives (2).
但唯一性结果依赖于导数 (2) 的存在性和连续性。

This involves a condition that is awkward to express for the derived uniqueness theorems for more general equations.
这导致在推导更一般方程的唯一性定理时，需要引入一个难以表述的条件。

In fact these theorems as stated in [2], [3] are not proven.
事实上，[2]、[3] 中所述的这些定理并未得到证明。

To make this point clear let us consider Monge’s equation.
为阐明这一点，我们考虑蒙日方程。
$\tag{3}$

Here
其中
$\frac{\partial^{2} z}{\partial x^{2}}, \quad s = \frac{\partial^{2} z}{\partial x \partial y}, \quad t = \frac{\partial^{2} z}{\partial y^{2}}.$

We write also
我们还定义
$\frac{\partial z}{\partial x}, \quad q = \frac{\partial z}{\partial y}.$

In (3) the coefficients $\cdots$ will depend on $x, y, z, p, q$ and we suppose them to be functions of class $C^{(2)}$ .
在方程 (3) 中，系数 $\cdots$ 依赖于 $x, y, z, p, q$ ，且我们假设它们为 $C^{(2)}$ 类函数。

Five $C^{(1)}$ functions $x(\lambda)$ , $y(\lambda)$ , $z(\lambda)$ , $p(\lambda)$ , $q(\lambda)$ form a “strip” if identically in $\lambda$ .
五个 $C^{(1)}$ 类函数 $x(\lambda)$ , $y(\lambda)$ , $z(\lambda)$ , $p(\lambda)$ , $q(\lambda)$ 若对所有 $\lambda$ 恒满足条件(4)，则构成一个“带”。
$\, dx + q \, dy\tag{4}$

Cauchy’s problem for the equation (3) is to find a solution $z (x, y)$ of class $C^{(2)}$ which “contains” this strip.
方程 (3) 的柯西问题是寻找一个包含该“带”的 $C^{(2)}$ 类解 $z (x, y)$ 。

For equation (3), the strip is “regular” if
对于方程 (3)，若满足
$\, dy^2 - 2H \, dx \, dy + B \, dx^2 \neq 0, \tag{5}$
and is “hyperbolic” if
则该“带”为“正则”的；若满足
$\quad H^{2} - A B > 0 \tag{6}$

Set $\sqrt{H^{2} - A B}$ ; we may suppose that the square root is chosen so that $w$ does not vanish on the strip.
令 $\sqrt{H^2 - A B}$ ；我们可假设所选的平方根使得 $w$ 在该“带”上不恒为零。
Then
此时有
$\begin{align*} A B + w^{2} &= 2 H w \tag{7} \\ (w^{2} - A B)^{2} &= 4 w^{2}(H^{2} - A B) > 0 \tag{8} \end{align*}$

Lewy’s characteristic equations belonging to the equation (3) may be written down in the form
方程 (3) 对应的 Lewy 特征方程可表示为
$\left.\begin{array}{ll} w x_{\alpha} - A y_{\alpha} = 0, & w y_{\beta} - B x_{\beta} = 0, \\ w p_{\alpha} + B q_{\alpha} - C y_{\alpha} = 0, & A p_{\beta} + w q_{\beta} - C x_{\beta} = 0, \\ z_{\alpha} - p x_{\alpha} - q y_{\alpha} = 0, & z_{\beta} - p x_{\beta} - q y_{\beta} = 0 \end{array}\right\}\tag{9}$

where the suffixes denote partial derivatives:
其中下标表示偏导数：
$x_{\alpha} = \frac{\partial x}{\partial \alpha}, \text{ etc.}$

This system of equations may be discussed in just the same way as the pair (1).
该方程组可按与方程组 (1) 相同的方式进行讨论。

Under the assumptions indicated the coefficients in (9) are $C^{(2)}[x, y, z, p, q]$ .
在上述假设下，方程组 (9) 中的系数为 $C^{(2)}[x, y, z, p, q]$ 类函数。

It has been shown [2] that the system of equations (9) have a unique solution $x(\alpha, \beta), \cdots, q(\alpha, \beta)$ of class $C^{(1)}[\alpha, \beta]$ which reduce to $x(\lambda), \cdots, q(\lambda)$ on $\alpha = \lambda$ , $\beta = -\lambda$ and for which all the derivatives
[2] 中已证明，方程组 (9) 存在唯一的 $C^{(1)}[\alpha, \beta]$ 类解 $x(\alpha, \beta), \cdots, q(\alpha, \beta)$ ，该解在 $\alpha = \lambda$ 、 $\beta = -\lambda$ 上退化为 $x(\lambda), \cdots, q(\lambda)$ ，且所有导数
$x_{\alpha \beta}, \cdots, q_{\alpha \beta}\tag{10}$
exist and are continuous.
均存在且连续。

The uniqueness established depends upon this last condition (10).
所证明的唯一性依赖于最后的条件 (10)。

If for instance the system (9) were satisfied by functions of class $C^{(1)}[\alpha, \beta]$ which satisfy also the conditions on $\alpha = \lambda$ , $\beta = -\lambda$ , but for which the derivatives (10) do not exist, then these functions would yield a solution of the regular hyperbolic Cauchy problem different from the one whose existence has been established.
例如，若方程组 (9) 存在另一组 $C^{(1)}[\alpha, \beta]$ 类解，其满足 $\alpha = \lambda$ 、 $\beta = -\lambda$ 上的条件，但导数 (10) 不存在，则该解将对应正则双曲型柯西问题的一个与已证明存在性的解不同的解。

This possibility is not excluded by the discussion of [1], [2], [3].
[1]、[2]、[3] 中的讨论并未排除这种可能性。

Lewy recognises this effectively in a footnote to his paper.
Lewy 在其论文的一个脚注中明确指出了这一点。

The proof of the statement just made may be had in a few lines.
上述结论的证明可简述如下。

If $\dot{x} = d x / d \lambda$ etc., we have
若令 $\dot{x} = d x / d \lambda$ 等，则有
$\dot{x} = x_{\alpha} - x_{\beta}, \quad \dot{y} = y_{\alpha} - y_{\beta}.$

On $\alpha + \beta = 0$ , using (9)，
在 $\alpha + \beta = 0$ 上，利用 (9) 可得，

$\begin{aligned} \frac{\partial(x, y)}{\partial(\alpha, \beta)} \cdot \left(w^{2} - A B\right) & = \left|\begin{array}{ll} x_{\alpha} & y_{\alpha} \\ x_{\beta} & y_{\beta} \end{array}\right| \cdot \left|\begin{array}{ll} w & -B \\ -A & w \end{array}\right| \\[1em]&= \left|\begin{array}{ll} 0 & -B \dot{x} + w \dot{y} \\ w \dot{x} - A \dot{y} & 0 \end{array}\right| \\[1em] & = w\left(A \dot{y}^{2} - 2 H \dot{x} \dot{y} + B \dot{x}^{2}\right). \end{aligned}$

Hence by (8),(5)
因此，由 (8),(5) 有
$\frac{\partial(x, y)}{\partial(\alpha, \beta)} = \frac{A \dot{y}^{2} - 2 H \dot{x} \dot{y} + B \dot{x}^{2}}{2 \sqrt{H^{2} - A B}} \neq 0.$

This implies also $y_{\alpha} \neq 0$ .
这也意味着 $y_{\alpha} \neq 0$ 。

We may therefore express $z, p, q$ as functions of $x, y$ of class $C^{(1)}$ .
因此，我们可将 $z, p, q$ 表示为 $x, y$ 的 $C^{(1)}$ 类函数。

Then
此时
$z_{\alpha} d \alpha + z_{\beta} d \beta = \left(p x_{\alpha} + q y_{\alpha}\right) d \alpha + \left(p x_{\beta} + q y_{\beta}\right) d \beta = p d x + q d y.$

So $\partial z / \partial x$ , $\partial z / \partial y$ and $z$ is of class $C^{(2)}[x, y]$ .
因此， $\partial z / \partial x$ 、 $\partial z / \partial y$ ，且 $z$ 为 $C^{(2)}[x, y]$ 类函数。

Finally, from equation (9),
最后，由方程 (9) 可得
$\begin{aligned} 0 & = w^{2} p_{\alpha} + w B q_{\alpha} - w C y_{\alpha} \\ & = w^{2}\left(r x_{\alpha} + s y_{\alpha}\right) + w B\left(s x_{\alpha} + t y_{\alpha}\right) - w C y_{\alpha} \\ & = A w y_{\alpha} r + \left(w^{2} + A B\right) y_{\alpha} s + B w y_{\alpha} t - C w y_{\alpha} \\ & = w y_{\alpha}[A r + 2 H s + B t - C]. \end{aligned}$

Since $y_{\alpha} \neq 0$ , so equation (3) is satisfied.
由于 $y_{\alpha} \neq 0$ ，因此方程 (3) 成立。

It will be seen that, contrary to the statements of the text books, we cannot infer on the basis of anything yet proved, the uniqueness of the solution of Cauchy’s problem for functions $z (x, y)$ of class $C^{(2)}$ simply.
可以看出，与教科书所述相反，仅基于已证明的结论，我们无法推出 $C^{(2)}$ 类函数 $z (x, y)$ 对应的柯西问题解的唯一性。

To obtain any statement of uniqueness for equation (3) we would need to admit only those solutions of class $C^{(2)}$ such that, when $x, y, z, p, q$ are expressed by means of the characteristic parameters $\alpha, \beta$ , all the mixed derivatives (10) exist and are continuous.
要得到方程 (3) 的唯一性结论，我们只能考虑这样的 $C^{(2)}$ 类解：当 $x, y, z, p, q$ 通过特征参数 $\alpha, \beta$ 表示时，所有混合导数 (10) 均存在且连续。

This is the awkward and unsatisfactory result for Monge’s equation which I referred to at the beginning.
这就是我在开头提到的蒙日方程对应的那个难以接受且不尽人意的结果。

It will be clear also that what we need is an unconditional uniqueness theorem; in the case of the pair of equations (1) we need a proof of uniqueness which is independent of the existence or otherwise of the derivatives (2).
显然，我们需要的是一个无条件的唯一性定理；对于方程组 (1)，我们需要一个不依赖于导数 (2) 是否存在的唯一性证明。

Such a proof will now be given.
下面将给出该证明。

2. The Uniqueness Theorem

2. 唯一性定理

Suppose $z_{1}$ , $z_{2}$ and $\bar{z}_{1}$ , $\bar{z}_{2}$ are two solution-pairs of (1) of class $C^{(1)}[x, y]$ and that on $x + y = 0$ ,
假设 $z_1, z_2$ 和 $\bar{z}_1, \bar{z}_2$ 是方程组 (1) 的两组 $C^{(1)}[x, y]$ 类解，且在 $x + y = 0$ 上满足
$z_{1} = \bar{z}_{1}, \quad z_{2} = \bar{z}_{2}.$

Write $\bar{a}_{i j} = a_{i j}(\bar{z}_{1}, \bar{z}_{2})$ .
令 $\bar{a}_{ij} = a_{ij}(\bar{z}_1, \bar{z}_2)$ 。

Set $u_{1} = z_{1} - \bar{z}_{1}$ , $u_{2} = z_{2} - \bar{z}_{2}$ so that $u_{1}$ , $u_{2}$ both vanish on $x + y = 0$ .
定义 $u_1 = z_1 - \bar{z}_1$ 、 $u_2 = z_2 - \bar{z}_2$ ，则 $u_1$ 和 $u_2$ 在 $x + y = 0$ 上均恒为零。

Now we find
此时可得
$\frac{\partial}{\partial x}\left(a_{11} u_{1}\right) = a_{11} \frac{\partial z_{1}}{\partial x} - \overline{a}_{11} \frac{\partial \overline{z}_{1}}{\partial x} + u_{1} \frac{\partial}{\partial x}\left(a_{11}\right) + \left(\overline{a}_{11} - a_{11}\right) \frac{\partial \overline{z}_{1}}{\partial x}, \\[1em] \frac{\partial}{\partial x}\left(a_{12} u_{2}\right) = a_{12} \frac{\partial z_{2}}{\partial x} - \overline{a}_{12} \frac{\partial \overline{z}_{2}}{\partial x} + u_{2} \frac{\partial}{\partial x}\left(a_{12}\right) + \left(\overline{a}_{12} - a_{12}\right) \frac{\partial \overline{z}_{2}}{\partial x}.$

Adding, and using equations (1),
将两式相加，并利用方程组 (1)，可得
$\begin{aligned} \frac{\partial}{\partial x}\left(a_{11} u_{1} + a_{12} u_{2}\right) & = u_{1} \frac{\partial}{\partial x}\left(a_{11}\right) + u_{2} \frac{\partial}{\partial x}\left(a_{12}\right) + \left(\overline{a}_{11} - a_{11}\right) \frac{\partial \overline{z}_{1}}{\partial x} + \left(\overline{a}_{12} - a_{12}\right) \frac{\partial \overline{z}_{2}}{\partial x}. \end{aligned}$

Now apply the mean value theorem to the differences $\bar{a}_{11} - a_{11}$ , $\bar{a}_{12} - a_{12}$ .
对差值 $\bar{a}_{11} - a_{11}$ 、 $\bar{a}_{12} - a_{12}$ 应用中值定理。

Then we can find a constant $K_{1} > 0$ such that near $x = 0$ , $y = 0$
则存在常数 $K_1 > 0$ ，使得在 $x = 0$ 、 $y = 0$ 附近有
$\left|\frac{\partial}{\partial x}\left(a_{11} u_{1} + a_{12} u_{2}\right)\right| \leqq K_{1}\left(\left|u_{1}\right| + \left|u_{2}\right|\right).$

In just the same way,
同理可得
$\left|\frac{\partial}{\partial y}\left(a_{21} u_{1} + a_{22} u_{2}\right)\right| \leqq K_{1}\left(\left|u_{1}\right| + \left|u_{2}\right|\right).$

Again, if
此外，若令
$a_{11} u_{1} + a_{12} u_{2}, \quad v = a_{21} u_{1} + a_{22} u_{2}$

we find, for a suitable constant $K_{2}$
则对于适当的常数 $K_2$ ，有
$\left.\begin{array}{l} \left|u_{1}\right| = \left|\frac{1}{\Delta}\left(a_{22} u - a_{12} v\right)\right| \leqq K_{2}(|u| + |v|) \\ \left|u_{2}\right| = \left|\frac{1}{\Delta}\left(a_{11} v - a_{21} u\right)\right| \leqq K_{2}(|u| + |v|) \end{array}\right\}.\tag{11}$

Hence if $K = 2 K_{1} K_{2}$
因此，若令 $K = 2 K_1 K_2$ ，则有
$\quad\left|\frac{\partial u}{\partial x}\right| \leqq K(|u| + |v|), \quad\left|\frac{\partial v}{\partial y}\right| \leqq K(|u| + |v|).\tag{12}$

From these inequalities (12) and the fact that $u = 0$ , $v = 0$ on $x + y = 0$ it follows that $u, v$ vanish identically.
由不等式 (12) 以及 $u = 0$ 、 $v = 0$ 在 $x + y = 0$ 上恒成立的事实，可推出 $u$ 和 $v$ 恒为零。

For, choose a constant $G$ such that
事实上，选取常数 $G$ 满足
$\leqq G, \quad|v| \leqq G.$

Considering the part of the plane $x + y > 0$ , set
考虑平面上 $x + y > 0$ 的区域，定义
$\zeta_{n} = \frac{G}{n !}(2 K)^{n}(x + y)^{n}$
so that
则有
$\frac{\partial \zeta_{n}}{\partial x} = \frac{\partial \zeta_{n}}{\partial y} = 2 K \zeta_{n - 1}.$

By induction we find, in $x + y > 0$
通过归纳法可证明，在 $x + y > 0$ 区域内
$\leqq \zeta_{n}, \quad|v| \leqq \zeta_{n}.\tag{13}$

This is true already for $n = 0$ ; if we suppose it holds for $n - 1$ then
当 $n = 0$ 时，上式显然成立；假设当 $n - 1$ 时成立，则
$\begin{aligned} |u| &= \left|\int_{-y}^{x} \frac{\partial u}{\partial x} d x\right| \leqq \int_{-y}^{x} 2 K \zeta_{n - 1} d x \\ &= \int_{-y}^{x} \frac{\partial \zeta_{n}}{\partial x} d x = \zeta_{n} \end{aligned}$
and similarly $\leqq \zeta_{n}$ .
同理可得 $\leqq \zeta_{n}$ 。

Since $\zeta_{n} \to 0$ as $\to \infty$ , so we infer $u = 0$ , $v = 0$ in $x + y > 0$ .
由于当 $\to \infty$ 时， $\zeta_n \to 0$ ，因此在 $x + y > 0$ 区域内 $u = 0$ 、 $v = 0$ 。

It is obvious that the result holds also for $x + y < 0$ .
显然，该结果在 $x + y < 0$ 区域内同样成立。

Now from (11), $u_{1}$ , $u_{2}$ vanish identically, so that $z_{1} = \bar{z}_{1}$ , $z_{2} = \bar{z}_{2}$ and any $C^{(1)}$ solution of the Cauchy problem for (1) is unique.
由 (11) 可知， $u_1$ 和 $u_2$ 恒为零，因此 $z_1 = \bar{z}_1$ 、 $z_2 = \bar{z}_2$ ，即方程组 (1) 的柯西问题的任意 $C^{(1)}$ 类解都是唯一的。

Of course, it now follows from the existence theorem itself, that a $C^{(1)}$ solution necessarily admits the continuous derivatives (2).
当然，由存在性定理本身可推出， $C^{(1)}$ 类解必然存在连续的导数 (2)。

It will be clear that the discussion above extends so as to obtain similar results for a system of $N = m + n$ equations of the form
显然，上述讨论可推广到如下形式的 $N = m + n$ 阶方程组，得到类似的结果：
$\sum_{j=1}^{N} a_{i j} \frac{\partial z_{j}}{\partial x} = 0, \quad i = 1, 2, \cdots, m. \\[1em] \sum_{j=1}^{N} a_{i j} \frac{\partial z_{j}}{\partial y} = 0, \quad i = m + 1, \cdots, N.$