https://r2.erweima.ai/mathbot/image/17188724439019frfmwdm.png

要解决这个问题，我们首先需要找出笛卡尔坐标系和极坐标系之间的导数转换关系，然后用这些关系来简化并解决所给的一阶偏微分方程 \(x w_y - y w_x = 0\)。

### 1. 极坐标和笛卡尔坐标之间的导数关系

在极坐标中，\(x\) 和 \(y\) 的定义是：

\[
x = r \cos\theta
\]
\[
y = r \sin\theta
\]

我们需要找到 \(\frac{\partial}{\partial x}\)，\(\frac{\partial}{\partial y}\) 和 \(\frac{\partial}{\partial r}\)，\(\frac{\partial}{\partial \theta}\) 之间的关系。我们通过链式法则可以得到：

\[
\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}
\]
\[
\frac{\partial}{\partial y} = \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}
\]

利用以下导数关系：

\[
r = \sqrt{x^2 + y^2} \quad \Rightarrow \quad \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r}, \quad \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{y}{r}
\]

\[
\theta = \arctan\left(\frac{y}{x}\right) \quad \Rightarrow \quad \frac{\partial \theta}{\partial x} = -\frac{y}{x^2 + y^2} = -\frac{y}{r^2}, \quad \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2} = \frac{x}{r^2}
\]

将这些导数关系代入，得到：

\[
\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} - \frac{y}{r^2} \frac{\partial}{\partial \theta}
\]
\[
\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} + \frac{x}{r^2} \frac{\partial}{\partial \theta}
\]

### 2. 简化偏微分方程

给定的偏微分方程是：

\[
x w_y - y w_x = 0
\]

我们现在将 \( w_x \) 和 \( w_y \) 转换到极坐标系中：

\[
w_x = \frac{\partial w}{\partial x} = \frac{x}{r} \frac{\partial w}{\partial r} - \frac{y}{r^2} \frac{\partial w}{\partial \theta}
\]
\[
w_y = \frac{\partial w}{\partial y} = \frac{y}{r} \frac{\partial w}{\partial r} + \frac{x}{r^2} \frac{\partial w}{\partial \theta}
\]

代入原方程：

\[
x \left( \frac{y}{r} \frac{\partial w}{\partial r} + \frac{x}{r^2} \frac{\partial w}{\partial \theta} \right) - y \left( \frac{x}{r} \frac{\partial w}{\partial r} - \frac{y}{r^2} \frac{\partial w}{\partial \theta} \right) = 0
\]

展开并简化：

\[
\frac{xy}{r} \frac{\partial w}{\partial r} + \frac{x^2}{r^2} \frac{\partial w}{\partial \theta} - \frac{xy}{r} \frac{\partial w}{\partial r} + \frac{y^2}{r^2} \frac{\partial w}{\partial \theta} = 0
\]

\[
\frac{x^2 + y^2}{r^2} \frac{\partial w}{\partial \theta} = 0
\]

因为 \( x^2 + y^2 = r^2 \)，所以这个方程变为：

\[
\frac{r^2}{r^2} \frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial \theta} = 0
\]

这意味着 \( w \) 不依赖于 \(\theta\)，即 \( w \) 仅为 \( r \) 的函数。

所以，偏微分方程的解形式为 \( w = f(r) \)，其中 \( f \) 是关于 \( r \) 的任意函数。

请问你需要更详细的解释或有其他问题吗？你可以问以下问题：

1. 如何求解具体的偏微分方程？
2. 其他坐标系下的导数关系是什么？
3. 偏微分方程的物理应用有哪些？
4. 如何使用极坐标系简化复杂的数学问题？
5. 泰勒级数在偏微分方程中的应用？

**提示：** 使用坐标变换可以简化许多数学问题，特别是在解决偏微分方程时非常有用。

Learn how to solve the first-order partial differential equation \( x w_y - y w_x = 0 \) using polar coordinates. This problem involves deriving and applying transformations between Cartesian and polar coordinates, and solving using partial derivatives. Suitable for advanced undergraduate mathematics students.

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