2019年考研数学二第20题
📝 题目
已知函数 $u(x, y)$ 满足 $2 \displaystyle\frac{\partial^{2} u}{\partial x^{2}}-2 \displaystyle\frac{\partial^{2} u}{\partial y^{2}}+3 \displaystyle\frac{\partial u}{\partial x}+3 \displaystyle\frac{\partial u}{\partial y}=0$ ，求 $a, b$ 的值，使得在变换 $u(x, y)= v(x, y) \mathrm{e}^{a x+b y}$ 下，上述等式可化为 $v(x, y)$ 不含一阶偏导数的等式。
💡 答案解析
$\displaystyle\frac{\partial u}{\partial x}=\displaystyle\frac{\partial v}{\partial x} \mathrm{e}^{a x+b y}+a v \mathrm{e}^{a x+b y}, ~ \displaystyle\frac{\partial u}{\partial y}=\displaystyle\frac{\partial v}{\partial y} \mathrm{e}^{a x+b y}+b v \mathrm{e}^{a x+b y}$ ， $\displaystyle\frac{\partial^{2} u}{\partial x^{2}}=\displaystyle\frac{\partial^{2} v}{\partial x^{2}} \mathrm{e}^{a x+b y}+2 a \displaystyle\frac{\partial v}{\partial x} \mathrm{e}^{a x+b y}+a^{2} v \mathrm{e}^{a x+b y}$, $\displaystyle\frac{\partial^{2} u}{\partial y^{2}}=\displaystyle\frac{\partial^{2} v}{\partial y^{2}} \mathrm{e}^{a x+b y}+2 b \displaystyle\frac{\partial v}{\partial y} \mathrm{e}^{a x+b y}+b^{2} v \mathrm{e}^{a x+b y}$, 代人已知等式得
$$ \begin{aligned} & 2 \frac{\partial^{2} v}{\partial x^{2}} \mathrm{e}^{a x+b y}+4 a \frac{\partial v}{\partial x} \mathrm{e}^{a x+b y}+2 a^{2} v \mathrm{e}^{a x+b y}-2 \frac{\partial^{2} v}{\partial y^{2}} \mathrm{e}^{a x+b y}-4 b \frac{\partial v}{\partial y} \mathrm{e}^{a x+b y}-2 b^{2} v \mathrm{e}^{a x+b y}+ \\ & 3 \frac{\partial v}{\partial x} \mathrm{e}^{a x+b y}+3 a v \mathrm{e}^{a x+b y}+3 \frac{\partial v}{\partial y} \mathrm{e}^{a x+b y}+3 b v \mathrm{e}^{a x+b y}=0 \end{aligned} $$
整理得
$$ 2 \frac{\partial^{2} v}{\partial x^{2}}-2 \frac{\partial^{2} v}{\partial y^{2}}+(4 a+3) \frac{\partial v}{\partial x}+(3-4 b) \frac{\partial v}{\partial y}+\left(2 a^{2}-2 b^{2}+3 a+3 b\right) v=0 $$
由题意得 $\left\{\begin{array}{l}4 a+3=0, \\ 3-4 b=0,\end{array}\right.$ 解得 $a=-\displaystyle\frac{3}{4}, b=\displaystyle\frac{3}{4}$ ． （21）【证明】（I）令 $F(x)=\displaystyle\int_{0}^{x} f(t) \mathrm{d} t$ ，则 $F^{\prime}(x)=f(x)$ ， 由拉格朗日中值定理得
$$
1=\int_{0}^{1} f(x) \mathrm{d} x=F(1)-F(0)=F^{\prime}(c)(1-0)=f(c)(0 因为 $f(c)=f(1)=1$ ，所以由罗尔定理，存在 $\xi \in(c, 1) \subset(0,1)$ ，使得 $f^{\prime}(\xi)=0$ ．
（II）令 $\varphi(x)=f(x)+x^{2}$ ，
$\varphi(0)=0, \quad \varphi(c)=f(c)+c^{2}=1+c^{2}, \quad \varphi(1)=2$,
由拉格朗日中值定理，存在 $\eta_{1} \in(0, c), \eta_{2} \in(c, 1)$ ，使得 $$
\begin{gathered}
\varphi^{\prime}\left(\eta_{1}\right)=\frac{\varphi(c)-\varphi(0)}{c}=\frac{1+c^{2}}{c}=c+\frac{1}{c} \\
\varphi^{\prime}\left(\eta_{2}\right)=\frac{\varphi(1)-\varphi(c)}{1-c}=\frac{2-1-c^{2}}{1-c}=1+c
\end{gathered}
$$ 再由拉格朗日中值定理，存在 $\eta \in\left(\eta_{1}, \eta_{2}\right) \subset(0,1)$ ，使得 $$
\varphi^{\prime \prime}(\eta)=\frac{\varphi^{\prime}\left(\eta_{2}\right)-\varphi^{\prime}\left(\eta_{1}\right)}{\eta_{2}-\eta_{1}}=\frac{1+c-c-\frac{1}{c}}{\eta_{2}-\eta_{1}}=\frac{1-\frac{1}{c}}{\eta_{2}-\eta_{1}}<0,
$$ 而 $\varphi^{\prime \prime}(x)=f^{\prime \prime}(x)+2$ ，即 $f^{\prime \prime}(\eta)+2<0$ ，故 $f^{\prime \prime}(\eta)<-2$ ．