多変数の関数高次導関数、テイラーの定理 2変数、偏微分、等式

学習環境

解析入門(中) (松坂和夫数学入門シリーズ 5) (松坂和夫 (著)、岩波書店)の第14章(多変数の関数)、14.2(高次導関数、テイラーの定理)、問題7の解答を求めてみる。

\begin{array}{l} s = x + a y \\ t = x - a y \end{array}

とおくと、

\frac{\partial z}{\partial x} = (\frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial g}{\partial t} \frac{\partial t}{\partial x}, \frac{\partial f}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial g}{\partial t} \frac{\partial t}{\partial y}) \cdot (1, 1)

= \frac{\partial f}{\partial s} + \frac{\partial g}{\partial t} + a \frac{\partial f}{\partial s} - a \frac{\partial g}{\partial t}

\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial s^{2}} + \frac{\partial^{2} g}{\partial t^{2}} + a^{2} \frac{\partial^{2} f}{\partial s^{2}} + a^{2} \frac{\partial^{2} g}{\partial t^{2}}

\frac{\partial z}{\partial y} = (\frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial g}{\partial t} \frac{\partial t}{\partial x}, a \frac{\partial f}{\partial s} - a \frac{\partial y}{\partial t}) \cdot (a, - a)

= a (\frac{\partial f}{\partial s} + \frac{\partial g}{\partial t}) - a^{2} (\frac{\partial f}{\partial s} - \frac{\partial g}{\partial t})

\frac{\partial^{2} z}{\partial y^{2}} = a^{2} (\frac{\partial^{2} f}{\partial s^{2}} + \frac{\partial^{2} g}{\partial t^{2}}) + a^{4} (\frac{\partial^{2} f}{\partial s^{2}} + \frac{\partial^{2} f}{\partial t^{2}})

よって、

a^{2} \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} z}{\partial y^{2}}

（証明終）

\frac{\partial^{2} z}{\partial s \partial t} = 0

を満たす関数は、

z = f (s) + g (t)

と表される。

\begin{array}{l} s = x + a y \\ t = x - a y \end{array}

とおけば、

z = f (x + a y) + g (x - a y)

a^{2} \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} z}{\partial y^{2}}