设 u 是调和函数，但不恒为常数，则 u^2 不是调和函数。A. 对B. 错

本题考查调和函数的定义及性质，解题思路是先明确调和函数的定义，再根据已知条件对$u^2$求二阶偏导数，最后判断$u^2$是否满足调和函数的条件。

1. 明确调和函数的定义：
   若函数$u(x,y)$在区域$D$内具有二阶连续偏导数，且满足拉普拉斯方程$\Delta u=\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}} = 0$，则称$u(x,y)$为区域$D$内的调和函数。
   已知$u$是调和函数，则$\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}} = 0$。
2. 对$u^2$求二阶偏导数：
   • 先求$\frac{\partial(u^2)}{\partial x}$，根据复合函数求导法则$(f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)$，可得$\frac{\partial(u^2)}{\partial x}=2u\frac{\partial u}{\partial x}$。
   • 再对$\frac{\partial(u^2)}{\partial x}$求关于$x$的偏导数，根据乘积的求导法则$(uv)^\prime = u^\prime v + uv^\prime$，可得：
     $\frac{\partial^{2}(u^2)}{\partial x^{2}}=\frac{\partial}{\partial x}(2u\frac{\partial u}{\partial x})=2(\frac{\partial u}{\partial x})^2 + 2u\frac{\partial^{2}u}{\partial x^{2}}$。
   • 同理，$\frac{\partial(u^2)}{\partial y}=2u\frac{\partial u}{\partial y}$，$\frac{\partial^{2}(u^2)}{\partial y^{2}}=\frac{\partial}{\partial y}(2u\frac{\partial u}{\partial y})=2(\frac{\partial u}{\partial y})^2 + 2u\frac{\partial^{2}u}{\partial y^{2}}$。
3. 计算$\Delta(u^2)$：
   $\Delta(u^2)=\frac{\partial^{2}(u^2)}{\partial x^{2}}+\frac{\partial^{2}(u^2)}{\partial y^{2}}=2((\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2)+2u(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}})$。
   因为$\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}} = 0$，所以$\Delta(u^2)=2((\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2)$。
4. 判断$u^2$是否为调和函数：
   由于$u$不恒为常数，所以$\frac{\partial u}{\partial x}$和$\frac{\partial u}{\partial y}$不同时为$0$，那么$(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2\gt0$，从而$\Delta(u^2)=2((\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2)\gt0$，不满足调和函数的定义，所以$u^2$不是调和函数。