8.2 解答

8.2

1.

$${\cal F}_{s}[u_{xx}] = - \omega {\cal F}_{c}[u_{x}]$$

$$= - \omega[-u(0,t) + \omega {\cal F}_{s}[u]]$$

$$= -\omega^2 {\cal F}_{s}[u] = - \omega^2 U_{s}(\omega,t)$$

$${\cal F}_{s}[u_{t}] = \frac{\partial}{\partial t}{\cal F}_{s}[u] = \frac{\partial}{\partial t}U_{s}(\omega ,t)$$

これより， $u_{t} = 2u_{xx}$の両辺にフーリエ変換を施すと

$$\frac{\partial}{\partial t}U_{s}(\omega ,t) = - 2\omega^2 U_{s}(\omega,t)$$

この式は$t$について1階線形なので，

$$U_{s}(\omega,t) = C_{\omega}e^{-2\omega^2 t}$$

を得る．ここで，初期条件 $u(x,0) = 25x$を用いると

$${\cal F}_{s}[u(x,0)] = U_{s}(\omega,0) = C_{\omega}$$

また

$${\cal F}_{s}[u(x,0)] = {\cal F}_{s}[25x] = \frac{2}{\pi}\int_{0}^{4}25x \sin{\omega x}dx$$

$$= \frac{50}{\pi}\left\{\left[\frac{-x\cos{\omega x}}{\omega} \right]_{0}^{4} + \frac{1}{\omega}\int_{0}^{4} \cos{\omega x} dx \right\}$$

$$= \frac{50}{\pi}\left\{\frac{-4 \cos{4 \omega}}{\omega} + \left[\frac{\sin{\omega x}}{\omega^2}\right]_{0}^{4}\right\}$$

$$= \frac{50}{\pi}\left[\frac{-4\cos{4 \omega}}{\omega} + \frac{\sin{4 \omega x}}{\omega^2} \right]$$

よって，

$$U_{s}(\omega,t) = \frac{50}{\pi}\left[\frac{-4\omega \cos{4\omega} + \sin{4\omega}}{\omega^2}\right] e^{-2\omega^2 t}$$

ここでフーリエ反転公式を用いると

$$u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}U_{s}(\omega,t)\sin{\omega x}dx$$

$$= \frac{2}{\pi}\int_{0}^{\infty}\frac{50}{\pi}\left[\frac{-4\omega ... ...omega} + \sin{4\omega}}{\omega^2}\right] e^{-2\omega^2 t}\sin{\omega x} d\omega$$

$$= \frac{100}{\pi^2} \int_{0}^{\infty}\left[\frac{-4\omega \cos{4\omega} + \sin{4\omega}}{\omega^2}\right] e^{-2\omega^2 t}\sin{\omega x} d \omega$$

2.

$${\cal F}[u_{tt}] = \frac{\partial^2}{\partial t^2}U(\omega,t)$$

$${\cal F}[u_{xx}] = (i \omega^2)U(\omega,t)= - \omega^2 U(\omega,t)$$

より偏微分方程式 $u_{tt} = c^2 u_{xx}$は次の常微分方程式に変換される．

$$\frac{d^2}{dt^2}U(\omega,t) + c^2 \omega^2 U(\omega,t) = 0$$

この式は$t$について線形で，特性根は $\pm ic \omega$なので

$$U(\omega,t) = A_{\omega} \cos{c \omega t} + B_{\omega} \sin{c \omega t}$$

と表わせる．ここで $u(x,0) = f(x)$より

$${\cal F}[u(x,0)] = U(\omega,0) = F(\omega) = A_{\omega}$$

また $u_{t}(x,0) = 0$より

$${\cal F}[u_{t}(x,0)] = \frac{\partial}{\partial t}U(\omega,0) = c\omega B_{\omega} = 0$$

よって，

$$U(\omega,t) = F(\omega)\cos{c \omega t}$$

を得る．ここで，反転公式を用いて，$u(x,t)$を求めると，

$$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} U(\omega,t)e^{i \omega x} d\omega$$

$$= \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)\cos{c \omega t}e^{i \omega x} d\omega$$

$$= \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)\left(\frac{e^{ic\omega t} + e^{-ic \omega t}}{2} \right) e^{i \omega x} d\omega$$

$$= \frac{1}{4\pi}\int_{-\infty}^{\infty} F(\omega)\left(e^{i \omega(x +ct)} + e^{i\omega(x - ct} \right) d\omega$$

$$= \frac{1}{2}\left(\frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)... ...ega + \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega(x - ct)} d \omega \right)$$

$$= \frac{1}{2}(f(x+ct) + f(x-ct))$$

3.

$${\cal F}_{c}[u_{xx}] = - u_{x}(0,t) + \omega {\cal F}_{s}[u_{x}]$$

$$= - f(t) - \omega^2 {\cal F}_{c}[u] = -f(t) - \omega^2 U_{c}(\omega,t)$$

$${\cal F}_{c}[u_{tt}] = \frac{\partial^2}{\partial t^2}U_{c}(\omega,t)$$

これより， $u_{tt} = c^2 u_{xx}$の両辺にフーリエ変換を施すと

$$\frac{\partial^2}{\partial t^2}U_{c}(\omega ,t) + c^2 \omega^2 U_{c}(\omega,t) = -c^2 f(t)$$

この式は$t$について1階線形なので，余関数は

$$U_{cc}(\omega,t) = A_{\omega} \cos{c \omega t} + B_{\omega} \sin{c \omega t}$$

また，特殊解は

$$U_{cp} = v_{1}\cos{c \omega t} + v_{2}\sin{c \omega t}$$

定数変化法を用いると

$$v_{1}' \cos{c \omega t} + v_{2}' \sin{c \omega t} = 0$$

$$v_{1}' (-c \omega \sin{c \omega t}) + v_{2}' (c \omega \cos{c \omega t} = -c^2 f(t)$$

と表わせる．これより，

$$v_{1}' = \frac{\left\vert\begin{array}{cc} 0 & \sin{c \omega t}\... ...mega t} \end{array} \right\vert} = \frac{c^2 f(t) \sin{c \omega t}}{c \omega}$$

よって，

$$v_{1} = \frac{1}{\omega}\int_{}^{t} f(x)\sin{c \omega x} dx$$

同様にして，

$$v_{2} = -\frac{1}{\omega}\int_{}^{t} f(x)\cos{c \omega x} dx$$

を得る．これより，

$$U_{c}(\omega,t) = U_{cc}(\omega,t) + U_{cp}(\omega,t)$$

$$= \left(A_{\omega} + \frac{1}{\omega}\int_{}^{t} f(x)\sin{c \omega x} dt\right)\cos{c \omega t}$$

$$+ \left(B_{\omega} - \frac{1}{\omega}\int_{}^{t} f(x)\cos{c \omega x} dt\right)\sin{c \omega t}$$

ここで， $u(x,0) = 0$より

$${\cal F}_{c}[u(x,0)] = U_{c}(\omega,0) = A_{\omega} = 0$$

$${\cal F}_{c}[u_{t}(x,0)] = \frac{\partial}{\partial t}U_{c}(\omega,0) = -\frac{1}{\omega} f(0) + c\omega B_{\omega} = 0$$

これより，

$$B_{\omega} = \frac{f(0)}{c \omega^2}$$

したがって，

$$U_{c}(\omega,t) = \frac{f(0))}{c\omega^2} \sin{c \omega t}$$

ここで，反転公式を用いて，$u(x,t)$を求めると，

$$u(x,t) = \frac{2}{\pi}\int_{0}^{\infty} U_{c}(\omega,t)\cos{ \omega x} d\omega$$

$$= \frac{2f(0)}{c\pi}\int_{0}^{\infty} \frac{\sin{c \omega t} \cos{c \omega t}}{\omega^2} d\omega$$