例 9.45 求流函数 $\psi(x, y)$ 对于不可压缩流动，其在点 $\mathbf{q}$ 处的速度 $(x, y)$ 为

$$(-y/(x^2 + y^2), x/(x^2 + y^2))$$

解 根据流函数的定义，我们有

$$u(x, y) = -\frac{\partial \psi}{\partial y} \quad \text{and} \quad v(x, y) = \frac{\partial \psi}{\partial x}$$

前提是

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$$

这里我们有

$$u = \frac{-y}{x^2 + y^2} \quad \text{and} \quad v = \frac{x}{x^2 + y^2}$$

因此

$$\frac{\partial u}{\partial x} = \frac{2xy}{(x^2 + y^2)^2} \quad \text{and} \quad \frac{\partial v}{\partial y} = -\frac{2yx}{(x^2 + y^2)^2}$$

证实了

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$$

积分

$$\frac{\partial \psi}{\partial y} = -u(x, y) = \frac{y}{x^2 + y^2}$$

对 y 求偏导，保持 x 不变，得到

$$\psi(x, y) = \frac{1}{2} \ln(x^2 + y^2) + g(x)$$

对 x 求偏导得到

$$\frac{\partial \psi}{\partial x} = \frac{x}{x^2 + y^2} + \frac{dg}{dx}$$

由于已知

$$\frac{\partial \psi}{\partial x} = v(x, y) = \frac{x}{x^2 + y^2}$$

我们有

$$\frac{dg}{dx} = 0$$

$$g(x) = C$$