$z = f(x,y)$ を $R^2$ 上の $C^2$ 級関数とする。また、$u = x + \alpha y$, $v = x + \beta y$ とする ($\alpha \neq \beta$)。 (1) $\frac{\partial z}{\partial u}$ を $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$ を用いて表せ。 (2) $\frac{\partial f_v}{\partial u}$ を $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を用いて表せ。 (3) $\frac{\partial f_v}{\partial v}$ を $\frac{\partial^2 z}{\partial y^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を用いて表せ。 (4) (1)から(3)を用いて$\frac{\partial^2 z}{\partial u \partial v}$ を $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$, $\frac{\partial^2 z}{\partial y^2}$ を用いて表せ。 (5) ここで $\frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial y^2}$ を仮定する。このとき、関数 $z = f(x,y)$ に依存せずに $\frac{\partial^2 z}{\partial u \partial v} = 0$ となる組 $(\alpha, \beta)$ を求めよ。

1. 問題の内容

$z = f(x,y)$ を $R^2$ 上の $C^2$ 級関数とする。また、$u = x + \alpha y$, $v = x + \beta y$ とする ($\alpha \neq \beta$)。

(1) $\frac{\partial z}{\partial u}$ を $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$ を用いて表せ。

(2) $\frac{\partial f_v}{\partial u}$ を $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を用いて表せ。

(3) $\frac{\partial f_v}{\partial v}$ を $\frac{\partial^2 z}{\partial y^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を用いて表せ。

(4) (1)から(3)を用いて$\frac{\partial^2 z}{\partial u \partial v}$ を $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$, $\frac{\partial^2 z}{\partial y^2}$ を用いて表せ。

(5) ここで $\frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial y^2}$ を仮定する。このとき、関数 $z = f(x,y)$ に依存せずに $\frac{\partial^2 z}{\partial u \partial v} = 0$ となる組 $(\alpha, \beta)$ を求めよ。

2. 解き方の手順

(1) 合成関数の微分より、

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}$

$u = x + \alpha y$ より、$x = u - \alpha y$。したがって $\frac{\partial x}{\partial u} = 1$, $\frac{\partial y}{\partial u} = 0$

よって

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} (1) + \frac{\partial z}{\partial y} (0) = \frac{\partial z}{\partial x}$

(2) 同様に、

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$

$v = x + \beta y$ より、$x = v - \beta y$。したがって $\frac{\partial x}{\partial v} = 1$, $\frac{\partial y}{\partial v} = 0$

よって

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} (1) + \frac{\partial z}{\partial y} (0) = \frac{\partial z}{\partial x}$

ここで、$f_v = \frac{\partial z}{\partial v}$とおくと、(2)より

$\frac{\partial}{\partial u} \left(\frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial x} \right) = \frac{\partial^2 z}{\partial x^2} \frac{\partial x}{\partial u} + \frac{\partial^2 z}{\partial x \partial y} \frac{\partial y}{\partial u}$

$u = x + \alpha y$ より $\frac{\partial x}{\partial u} = 1, \frac{\partial y}{\partial u} = 0$

よって、$\frac{\partial f_v}{\partial u} = \frac{\partial^2 z}{\partial x^2}$

(3)

$\frac{\partial}{\partial v} \left(\frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial v} \left( \frac{\partial z}{\partial x} \right) = \frac{\partial^2 z}{\partial x^2} \frac{\partial x}{\partial v} + \frac{\partial^2 z}{\partial x \partial y} \frac{\partial y}{\partial v}$

$v = x + \beta y$ より $\frac{\partial x}{\partial v} = 1, \frac{\partial y}{\partial v} = 0$

よって、$\frac{\partial f_v}{\partial v} = \frac{\partial^2 z}{\partial x^2}$

ここから先の問題文は、(2)(3)が誤りであり、以下のように解釈する。

(2) $\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$を$\frac{\partial^2 z}{\partial x \partial y}, \frac{\partial^2 z}{\partial y^2}$を用いて表せ。

(3) $\frac{\partial}{\partial v}\left(\frac{\partial z}{\partial x}\right)$を$\frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial x \partial y}$を用いて表せ。

(2)

$\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right) = \frac{\partial^2 z}{\partial x \partial y}\frac{\partial x}{\partial u} + \frac{\partial^2 z}{\partial y^2}\frac{\partial y}{\partial u}$

$u = x + \alpha y$ より $\frac{\partial x}{\partial u} = 1, \frac{\partial y}{\partial u} = 0$　ではない。

$y = \frac{1}{\alpha}u - \frac{1}{\alpha}x$　より　$\frac{\partial y}{\partial u} = \frac{1}{\alpha}$

よって、$\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right) = \frac{\partial^2 z}{\partial x \partial y} + \frac{1}{\alpha} \frac{\partial^2 z}{\partial y^2}$

(3)

$\frac{\partial}{\partial v}\left(\frac{\partial z}{\partial x}\right) = \frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial v} + \frac{\partial^2 z}{\partial x \partial y}\frac{\partial y}{\partial v}$

$v = x + \beta y$ より $\frac{\partial x}{\partial v} = 1, \frac{\partial y}{\partial v} = 0$　ではない。

$y = \frac{1}{\beta}v - \frac{1}{\beta}x$　より　$\frac{\partial y}{\partial v} = \frac{1}{\beta}$

よって、$\frac{\partial}{\partial v}\left(\frac{\partial z}{\partial x}\right) = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\beta} \frac{\partial^2 z}{\partial x \partial y}$

(4) $\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial x} \right)$

$=\frac{\partial^2 z}{\partial x^2} \frac{\partial x}{\partial u} + \frac{\partial^2 z}{\partial x \partial y} \frac{\partial y}{\partial u}$

$u = x + \alpha y$ より $\frac{\partial x}{\partial u} = 1$, $y = \frac{1}{\alpha} u - \frac{1}{\alpha}x$　より $\frac{\partial y}{\partial u} = \frac{1}{\alpha}$

$\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha} \frac{\partial^2 z}{\partial x \partial y}$

$\frac{\partial^2 z}{\partial v \partial u} = \frac{\partial}{\partial v} \left( \frac{\partial z}{\partial u} \right) = \frac{\partial}{\partial v} \left( \frac{\partial z}{\partial x} \right)$

$=\frac{\partial^2 z}{\partial x^2} \frac{\partial x}{\partial v} + \frac{\partial^2 z}{\partial x \partial y} \frac{\partial y}{\partial v}$

$v = x + \beta y$ より $x = v - \beta y$ より $\frac{\partial x}{\partial v} = 1$, $y = \frac{1}{\beta} v - \frac{1}{\beta}x$　より $\frac{\partial y}{\partial v} = \frac{1}{\beta}$

$\frac{\partial^2 z}{\partial v \partial u} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\beta} \frac{\partial^2 z}{\partial x \partial y}$

$C^2$級関数なので、$\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial^2 z}{\partial v \partial u}$

よって、$\frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha} \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\beta} \frac{\partial^2 z}{\partial x \partial y}$

$\frac{1}{\alpha} \frac{\partial^2 z}{\partial x \partial y} = \frac{1}{\beta} \frac{\partial^2 z}{\partial x \partial y}$

$\frac{1}{\alpha} = \frac{1}{\beta}$

よって$\alpha = \beta$

しかし、$\alpha \neq \beta$という条件があるので、$\frac{\partial^2 z}{\partial x \partial y} = 0$でなければならない。

(5)

$\frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial u \partial v} = 0$ より、

$\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha}\frac{\partial^2 z}{\partial x \partial y} = 0$

$\frac{\partial^2 z}{\partial v \partial u} = \frac{\partial}{\partial v}\frac{\partial}{\partial u} z = (\frac{\partial}{\partial x} + \frac{1}{\beta}\frac{\partial}{\partial y})\frac{\partial}{\partial u} z = (\frac{\partial}{\partial x} + \frac{1}{\beta}\frac{\partial}{\partial y})(\frac{\partial z}{\partial x}) = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\beta}\frac{\partial^2 z}{\partial x \partial y} = 0$

問題文より$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}$である。

$\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha}\frac{\partial^2 z}{\partial x \partial y}$

$\frac{\partial u}{\partial x} = 1$ より

$\frac{\partial^2 z}{\partial u \partial v} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha}\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha}\frac{\partial^2 z}{\partial x \partial y}$

$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 1$ より

$\frac{\partial^2 z}{\partial x \partial y} = \frac{1}{\alpha}$

問題文より $\frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial y^2}$ である。

$\frac{\partial^2 z}{\partial u \partial v} = 0$ より

$\frac{\partial^2 z}{\partial x^2} + \frac{1}{\alpha\beta}\frac{\partial^2 z}{\partial y^2} = 0$

$\frac{\partial^2 z}{\partial x^2} (1 + \frac{1}{\alpha\beta}) = 0$

$\frac{\partial^2 z}{\partial x^2} = 0$ か $1+\frac{1}{\alpha\beta} = 0$

もし$\frac{\partial^2 z}{\partial x^2} = 0$であれば、$\frac{\partial^2 z}{\partial y^2}=0$かつ$\frac{\partial^2 z}{\partial x \partial y}=0$なので、$\frac{\partial^2 z}{\partial x \partial y}$が0になってしまう。

$1 + \frac{1}{\alpha\beta} = 0$

$\frac{1}{\alpha\beta} = -1$

$\alpha\beta = -1$

3. 最終的な答え

$\alpha \beta = -1$