$u=f(x, y)$ は2回微分可能で、2次偏導関数は全て連続である。$x+y=e^{s+t}$, $x-y=e^{s-t}$ の時、 $$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = e^{-2s}\left(\frac{\partial^2 u}{\partial s^2} - \frac{\partial^2 u}{\partial t^2}\right)$$ が成り立つことを示す。

1. 問題の内容

$u=f(x, y)$ は2回微分可能で、2次偏導関数は全て連続である。$x+y=e^{s+t}$, $x-y=e^{s-t}$ の時、

$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = e^{-2s}\left(\frac{\partial^2 u}{\partial s^2} - \frac{\partial^2 u}{\partial t^2}\right)$$

が成り立つことを示す。

2. 解き方の手順

まず、$x$ と $y$ を $s$ と $t$ で表す。

$x+y = e^{s+t}$ と $x-y = e^{s-t}$ を足すと $2x = e^{s+t} + e^{s-t}$ より

$$x = \frac{1}{2}(e^{s+t} + e^{s-t})$$

$x+y = e^{s+t}$ から $x-y = e^{s-t}$ を引くと $2y = e^{s+t} - e^{s-t}$ より

$$y = \frac{1}{2}(e^{s+t} - e^{s-t})$$

次に、偏微分の連鎖律を用いて、$\frac{\partial}{\partial x}$ と $\frac{\partial}{\partial y}$ を $\frac{\partial}{\partial s}$ と $\frac{\partial}{\partial t}$ で表す。

$$\frac{\partial}{\partial s} = \frac{\partial x}{\partial s}\frac{\partial}{\partial x} + \frac{\partial y}{\partial s}\frac{\partial}{\partial y}$$

$$\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial y}{\partial t}\frac{\partial}{\partial y}$$

ここで

$$\frac{\partial x}{\partial s} = \frac{1}{2}(e^{s+t} + e^{s-t}) = x$$

$$\frac{\partial x}{\partial t} = \frac{1}{2}(e^{s+t} - e^{s-t}) = y$$

$$\frac{\partial y}{\partial s} = \frac{1}{2}(e^{s+t} - e^{s-t}) = y$$

$$\frac{\partial y}{\partial t} = \frac{1}{2}(e^{s+t} + e^{s-t}) = x$$

よって

$$\frac{\partial}{\partial s} = x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}$$

$$\frac{\partial}{\partial t} = y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}$$

したがって

$$\frac{\partial^2}{\partial s^2} = \left(x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}\right)\left(x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}\right) = x^2\frac{\partial^2}{\partial x^2} + xy\frac{\partial^2}{\partial x \partial y} + yx\frac{\partial^2}{\partial y \partial x} + y^2\frac{\partial^2}{\partial y^2} + \frac{\partial x}{\partial s}\frac{\partial}{\partial x} + \frac{\partial y}{\partial s}\frac{\partial}{\partial y} = x^2\frac{\partial^2}{\partial x^2} + 2xy\frac{\partial^2}{\partial x \partial y} + y^2\frac{\partial^2}{\partial y^2} + x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}$$

$$\frac{\partial^2}{\partial t^2} = \left(y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}\right)\left(y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}\right) = y^2\frac{\partial^2}{\partial x^2} + yx\frac{\partial^2}{\partial x \partial y} + xy\frac{\partial^2}{\partial y \partial x} + x^2\frac{\partial^2}{\partial y^2} + \frac{\partial y}{\partial t}\frac{\partial}{\partial x} + \frac{\partial x}{\partial t}\frac{\partial}{\partial y} = y^2\frac{\partial^2}{\partial x^2} + 2xy\frac{\partial^2}{\partial x \partial y} + x^2\frac{\partial^2}{\partial y^2} + x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}$$

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

$x^2 - y^2 = (x+y)(x-y) = e^{s+t}e^{s-t} = e^{2s}$ より

$$\frac{\partial^2 u}{\partial s^2} - \frac{\partial^2 u}{\partial t^2} = e^{2s}\left(\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2}\right)$$

よって

$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = e^{-2s}\left(\frac{\partial^2 u}{\partial s^2} - \frac{\partial^2 u}{\partial t^2}\right)$$

3. 最終的な答え

$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = e^{-2s}\left(\frac{\partial^2 u}{\partial s^2} - \frac{\partial^2 u}{\partial t^2}\right)$$