和達三樹《物理用數學》習題1.4.1

==問題==

$f(x)$與$g(x)$為任意函數，試證明
$$u(a, t) = f(x+at) + g(x-at)$$
可看作是
$$\frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}.$$

==解答==

\begin{align*}
\frac{\partial}{\partial t} u(x, t) &= \frac{\partial}{\partial t} \left[ f(x+at) + g(x-at) \right] \\
&= \frac{\partial}{\partial t} f(x+at) + \frac{\partial}{\partial t} g(x-at) \\
&= f'(x+at) \cdot a + g'(x-at) \cdot (-a) \\
&= af'(x+at) - ag'(x-at) \\
\frac{\partial^2}{\partial t^2} u(x, t) &= \frac{\partial}{\partial t} \left[ \frac{\partial}{\partial t} u(x, t) \right] \\
&= \frac{\partial}{\partial t} \left[ af'(x+at) - ag'(x-at) \right] \\
&= \frac{\partial}{\partial t} \left[ af'(x+at) \right] - \frac{\partial}{\partial t} \left[ ag'(x-at) \right] \\
&= a \frac{\partial}{\partial t} f'(x+at) - a \frac{\partial}{\partial t} g'(x-at) \\
&= a f''(x+at) \cdot a - a g''(x-at) \cdot (-a) \\
&= a^2 \left[ f''(x+at) + g''(x-at) \right] \\
\frac{\partial}{\partial x} u(x, t) &=  \frac{\partial}{\partial x} \left[ f(x+at) + g(x-at) \right] \\
&= \frac{\partial}{\partial x} f(x+at) + \frac{\partial}{\partial x} g(x-at) \\
&= f'(x+at) \cdot 1 + g'(x-at) \cdot 1 \\
&= f'(x+at) + g'(x-at) \\
\frac{\partial^2}{\partial x^2} u(x, t)  &= \frac{\partial}{\partial x} \left[ \frac{\partial}{\partial x} u(x, t)  \right] \\
&= \frac{\partial}{\partial x} \left[ f'(x+at) + g'(x-at) \right] \\
&= \frac{\partial}{\partial x} f'(x+at) + \frac{\partial}{\partial x} g'(x-at) \\
&= f''(x+at) \cdot 1 + g''(x-at) \cdot 1 \\
&= f''(x+at) + g''(x-at) \\
\frac{\partial^2 u}{\partial t^2} &= a^2 \left[ f''(x+at) + g''(x-at) \right] \\
&= a^2 \cdot \frac{\partial^2}{\partial x^2} u(x, t).
\end{align*}

(解答結束)