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3. 设 $u=u(x, y)$ 是单连通区域 $D$ 内的一个调和函数, 证明:
$$
v(x, y):=\int_{\left(x_0, y_0\right)}^{(x, y)}-u_y \mathrm{~d} x+u_x \mathrm{~d} y,\left(x_0, y_0\right) \in D
$$
是其共轭调和函数。
证明
对于 $D$ 内的任意分段光滑闭曲线 $\gamma$, 其围成的区域为 $S$, 根据Green公式有
$$
\oint_\gamma-u_y \mathrm{~d} x+u_x \mathrm{~d} y=\iint_S u_{x x}-\left(-u_{y y}\right) \mathrm{d} x \mathrm{~d} y=\iint_S \Delta u \mathrm{~d} x \mathrm{~d} y=\iint_S 0 \mathrm{~d} x \mathrm{~d} y=0
$$
从而积分
$$
v(x, y):=\int_{\left(x_0, y_0\right)}^{(x, y)}-u_y \mathrm{~d} x+u_x \mathrm{~d} y
$$
与路径无关.
取直线段连接 $\left(x_0, y_0\right)$ 和 $(x, y)$, 设其参数方程为 $\left\{\begin{array}{c}x(t)=x_0+\left(x-x_0\right) t \\ y(t)=y_0+\left(y-y_0\right) t\end{array}, t \in[0,1]\right.$,
于是 $\left\{\begin{array}{l}(x(0), y(0))=\left(x_0, y_0\right) \\ (x(1), y(1))=(x, y)\end{array}\right.$ 且 $\left\{\begin{array}{l}\mathrm{d} x(t)=\left(x-x_0\right) \mathrm{d} t \\ \mathrm{~d} y(t)=\left(y-y_0\right) \mathrm{d} t\end{array}\right.$
从而
$$
v(x, y)=\int_0^1\left[-u_2(x(t), y(t))\left(x-x_0\right)+u_1(x(t), y(t))\left(y-y_0\right)\right] \mathrm{d} t
$$
于是$$
v_y(x, y)=\int_0^1\left[-u_{22}(x(t), y(t))\left(x-x_0\right) t+u_{12}(x(t), y(t))\left(y-y_0\right) t+u_1(x(t), y(t))\right] \mathrm{d} t
$$
根据 $\Delta u(x, y)=0$ 有
$$
\begin{aligned}
v_y(x, y) & =\int_0^1\left[u_{11}(x(t), y(t))\left(x-x_0\right) t+u_{12}(x(t), y(t))\left(y-y_0\right) t+u_1(x(t), y(t))\right] \mathrm{d} t \\
& =\int_0^1\left[t \mathrm{~d} u_1(x(t), y(t))+u_1(x(t), y(t)) \mathrm{d} t\right]=\left.t u_1(x(t), y(t))\right|_0 ^1=u_x(x, y)
\end{aligned}
$$
同理容易验证 $u_y(x, y)=-v_x(x, y)$
由于 $u(x, y) \in C^2(D)$, 故 $u_x, u_y, v_x, v_y \in C^1(D)$, 那么 $v(x, y) \in C^2(D)$,
于是 $f(x+\mathrm{i} y)=u(x, y)+\mathrm{i} v(x, y)$ 可微, 且满足 $\mathrm{C}-\mathrm{R}$ 方程, 故 $f(x+\mathrm{i} y)$ 在 $D$ 上解析,
那么 $v(x, y)$ 是 $u(x, y)$ 的共轭调和函数. |
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