嘿，咱们接着之前用格林公式证明法向导数积分与拉普拉斯积分关系的思路，一步步推导格林第一恒等式（也就是你给出的包含两个函数(u)和(v)的等式），逻辑和之前完全相通，咱们来拆解一下：

首先明确前提：区域(D)满足格林定理的适用条件，函数(u, v: \mathbb{R}^2 \rightarrow \mathbb{R})足够光滑（保证各阶偏导数存在且连续，符合格林定理的要求），(\partial D)是(D)的边界，(\frac{\partial u}{\partial n})是(u)沿边界外法向(n)的方向导数。

第一步：将边界法向导数积分转化为线积分 #

咱们已经知道法向导数和梯度的关系：(\frac{\partial u}{\partial n} = \langle\nabla u,n\rangle)，结合之前推导的“法向导数积分转线积分”的结论，对(\int\limits_{\partial D} v\frac{\partial u}{\partial n}, dS)做同样的转换：
$$\int\limits_{\partial D} v\frac{\partial u}{\partial n}, dS = \int\limits_{\partial D} v\langle\nabla u,n\rangle , dS = \int\limits_{\partial D} v\left( \frac{\partial u}{\partial x}dy - \frac{\partial u}{\partial y}dx \right)$$

第二步：对转换后的线积分应用格林定理 #

格林定理的核心形式是：
$$\int\limits_{\partial D} Pdx + Qdy = \iint\limits_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dxdy$$
把咱们的线积分对应到这个形式里：令(P = -v\frac{\partial u}{\partial y})，(Q = v\frac{\partial u}{\partial x})。

接下来用乘积法则计算偏导数：

• (\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left( v\frac{\partial u}{\partial x} \right) = \frac{\partial v}{\partial x}\frac{\partial u}{\partial x} + v\frac{\partial^2 u}{\partial x^2})
• (\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left( -v\frac{\partial u}{\partial y} \right) = -\left( \frac{\partial v}{\partial y}\frac{\partial u}{\partial y} + v\frac{\partial^2 u}{\partial y^2} \right))

把这两个结果代入格林定理的右侧：
$$
\begin{align*}
\iint\limits_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dxdy &= \iint\limits_D \left[ \left( \frac{\partial v}{\partial x}\frac{\partial u}{\partial x} + v\frac{\partial^2 u}{\partial x^2} \right) - \left( -\left( \frac{\partial v}{\partial y}\frac{\partial u}{\partial y} + v\frac{\partial^2 u}{\partial y^2} \right) \right) \right] dxdy \
&= \iint\limits_D \left( \frac{\partial v}{\partial x}\frac{\partial u}{\partial x} + v\frac{\partial^2 u}{\partial x^2} + \frac{\partial v}{\partial y}\frac{\partial u}{\partial y} + v\frac{\partial^2 u}{\partial y^2} \right) dxdy \
&= \iint\limits_D v\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) dxdy + \iint\limits_D \left( \frac{\partial u}{\partial x}\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\frac{\partial v}{\partial y} \right) dxdy
\end{align*}
$$
因为拉普拉斯算子(\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2})，上式可以简化为：
$$= \iint\limits_D v\Delta u ,dxdy + \iint\limits_D \left( \frac{\partial u}{\partial x}\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\frac{\partial v}{\partial y} \right) dxdy$$

第三步：移项得到目标恒等式 #

结合第一步的结论，我们有：
$$\int\limits_{\partial D} v\frac{\partial u}{\partial n}, dS = \iint\limits_D v\Delta u ,dxdy + \iint\limits_D \left( \frac{\partial u}{\partial x}\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\frac{\partial v}{\partial y} \right) dxdy$$
把右边的第二个二重积分项移到左侧，就得到了咱们要证明的格林第一恒等式：
$$\iint\limits_D v\Delta u ,dxdy=-\iint\limits_D \left(\frac{\partial u}{\partial x}\cdot \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\cdot \frac{\partial v}{\partial y} \right) ,dx,dy + \int\limits_{\partial D} v\frac{\partial u}{\partial n}, dS$$

整个推导和之前的思路完全一致：先把边界的法向导数积分转成线积分，再用格林定理展开成二重积分，最后整理移项得到结果，核心就是对乘积形式的项用乘积法则展开偏导数～

备注：内容来源于stack exchange，提问作者Paull