问题及解答

是否存在可微函数 $u(x,y)$, 使得 $u'_x=xy$, $u'_y=-x^2y$ ?

假设存在这样的函数 $u(x,y)$, 满足

\[
\left\{
\begin{aligned}
\frac{\partial u}{\partial x}&=xy,\\
\frac{\partial u}{\partial y}&=-x^2y.
\end{aligned}
\right.
\]

则

\[
\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial}{\partial y}\biggl(\frac{\partial u}{\partial x}\biggr)=\frac{\partial}{\partial y}(xy)=x,
\]

而

\[
\frac{\partial^2 u}{\partial y\partial x}=\frac{\partial}{\partial x}\biggl(\frac{\partial u}{\partial y}\biggr)=\frac{\partial}{\partial x}(-x^2y)=-2xy,
\]

两者虽然存在但不相等, 这说明 $\frac{\partial^2 u}{\partial x\partial y}$ 和 $\frac{\partial^2 u}{\partial y\partial x}$ 都不是连续的.

\[
u(x,y)=\int\frac{\partial u}{\partial x}\mathrm{d}x+\varphi(y),
\]

这里 $\varphi(y)$ 是关于 $y$ 的可微函数. 将 $\dfrac{\partial u}{\partial x}=xy$ 代入, 得

\[
u(x,y)=\int xy\mathrm{d}x+\varphi(y)=\frac{1}{2}x^2 y+\varphi(y).\tag{*}
\]

于是

\[
\frac{\partial u}{\partial y}=\frac{1}{2}x^2+\varphi'(y),
\]

又 $\dfrac{\partial u}{\partial y}=-x^2y$, 故

\[
\varphi'(y)=-x^2y-\frac{1}{2}x^2.
\]

从而

\[
\varphi(y)=\int(-x^2y-\frac{1}{2}x^2)\mathrm{d}y=-\frac{1}{2}x^2 y^2-\frac{1}{2}x^2y+C.
\]

代入 $(*)$, 得

\[
u(x,y)=-\frac{1}{2}x^2 y^2+C.
\]

但 $u$ 不满足最初的条件 $u'_x=xy$.