13.3 习题课讲解

13.3.1 Fourier级数

解 (1) $f$是奇函数，其Fourier级数是正弦级数，计算可得$$\begin{array}{}\text{(1)}& b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\sin nx\mathrm{d}x=-\frac{2}{n\pi }{\int }_0^{\pi }x(\cos nx{)}^{\prime }\mathrm{d}x=-\frac{2\pi \cos n\pi }{n\pi }+\frac{2}{n\pi }{\int }_0^{\pi }\cos nx\mathrm{d}x=\frac{(-1{)}^{n-1}}{n}\end{array}$$ 所以$$\begin{array}{}\text{(2)}& f(x)\sim S(x):=\sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin nx\end{array}$$

由于$f$ Riemann可积，从而平方可积，故$f$的Fourier级数在区间$[-\pi ,\pi ]$上平方收敛于$f$。考虑级数$$\begin{array}{}\text{(3)}& \sum _{n=1}^{+\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n}\sin nx=-\mathrm{\Im }\sum _{n=1}^{+\mathrm{\infty }}\frac{{(-{\mathrm{e}}^{\mathrm{e}x})}^n}{n}\end{array}$$ 设$\delta \in (0,\pi )$，当$x\in [\delta -\pi ,\pi -\delta ]$时，计算可得$$\begin{array}{}\text{(4)}& \left|\sum _{n=1}^N{(-{\mathrm{e}}^{\mathrm{e}x})}^n\right|=\left|\frac{-{\mathrm{e}}^{\mathrm{e}x}-{(-{\mathrm{e}}^{\mathrm{e}x})}^N}{1+{\mathrm{e}}^{\mathrm{e}x}}\right|\le \frac{2}{|1+{\mathrm{e}}^{\mathrm{e}x}|}\end{array}$$ 即${(-{\mathrm{e}}^{\mathrm{e}x})}^n$的部分和有界、$\frac{1}{n}$单调趋于$0$，由Dirichlet判别法知$\sum _{n=1}^{+\mathrm{\infty }}\frac{{(-{\mathrm{e}}^{\mathrm{e}x})}^n}{n}$在区间$[\delta -\pi ,\pi -\delta ]$上一致收敛。

综上，$f$的Fourier级数在$[-\pi ,\pi ]$上逐点收敛、在$(-\pi ,\pi )$上内闭一致收敛到$f$。因此，和函数$S$在$(-\pi ,\pi )$上连续且有$S(x)=f(x)$，另有$S(-\pi )=S(\pi )=0$，即$$\begin{array}{}\text{(5)}& S(x)=\{\begin{array}{ll}f(x),& x\in (-\pi ,\pi )\\ 0,& x=\pm \pi \end{array}\end{array}$$

(2) 由Parseval等式可得$$\begin{array}{}\text{(6)}& \frac{2}{3}{\pi }^2=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2\mathrm{d}x=\frac{A_0^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}(A_n^2+B_n^2)=\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{n^2}⟹\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}\end{array}$$

(3) 令$x=\frac{\pi }{2}$可得$$\begin{array}{}\text{(7)}& \frac{\pi }{2}=\sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin \frac{n\pi }{2}=\sum _{k=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{k-1}}{2k-1}⟹\sum _{k=1}^{+\mathrm{\infty }}\frac{(-1{)}^{k-1}}{2k-1}=\frac{\pi }{4}\end{array}$$ $◻$

解 直接计算可得$$\begin{array}{}\text{(8)}& \begin{array}{rl}{A}_n& =\frac{1}{\pi }{\int }_0^{2\pi }f(-x+L)\cos nx\mathrm{d}x=-\frac{1}{\pi }{\int }_L^{-2\pi +L}f(t)\cos n(L-t)\mathrm{d}t\\ & =\frac{1}{\pi }{\int }_{L-2\pi }^{L}f(t)\cos nL\cos t+f(t)\sin nL\sin t\mathrm{d}t=\cos nL\cdot a_n+\sin nL\cdot b_n\\ B_n& =\frac{1}{\pi }{\int }_0^{2\pi }f(-x+L)\sin nx\mathrm{d}x=-\frac{1}{\pi }{\int }_L^{-2\pi +L}f(t)\sin n(L-t)\mathrm{d}t\\ & =\frac{1}{\pi }{\int }_{L-2\pi }^{L}f(t)\sin nL\cos t-f(t)\cos nL\sin t\mathrm{d}t=\sin nL\cdot a_n-\cos nL\cdot b_n\end{array}\end{array}$$ 这表明可以直接利用换元得到Fourier级数展开，即$$\begin{array}{}\text{(9)}& \begin{array}{rl}f(-x+L)& \sim \frac{a_0}{2}+\sum _{n=1}^{+\mathrm{\infty }}(a_n\cos n(L-x)+b_n\sin n(L-x))\\ & =\frac{a_0}{2}+\sum _{n=1}^{+\mathrm{\infty }}[(a_n\cos nL+b_n\sin nL)\cos nx+(-b_n\cos nL+a_n\sin nL)\sin nx]\end{array}\end{array}$$ $◻$

解 当$-\pi <x<L-\pi$时，$-\pi <L-x-2\pi <L-\pi <0$，故有$$\begin{array}{}\text{(10)}& f(x)+f(L-x)=x+f(L-x-2\pi )=x+L-x-2\pi =L-2\pi \end{array}$$ 当$L-\pi <x<\pi$时，$-\pi <L-\pi <L-x<\pi$，故有$$\begin{array}{}\text{(11)}& f(x)+f(L-x)=x+L-x=L\end{array}$$ 所以$g(x)=f(x)+f(L-x)$是分段常值函数，其表达式为$$\begin{array}{}\text{(12)}& g(x)=\{\begin{array}{ll}L-2\pi ,& -\pi <x<L-\pi \\ L,& L-\pi <x<\pi \end{array}\end{array}$$ 其Fourier级数对函数满足线性，所以$$\begin{array}{}\text{(13)}& \begin{array}{rl}g(x)& \sim \sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin nx+\sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin n(L-x)\\ & =\sum _{n=1}^{+\mathrm{\infty }}(\frac{2(-1{)}^{n-1}\sin nL}{n}\cos nx+\frac{2(-1{)}^{n-1}(1-\cos nL)}{n}\sin nx)\end{array}\end{array}$$ $◻$

解 (1) $f$是偶函数，其Fourier级数是余弦级数，计算可得$$\begin{array}{}\text{(14)}& \begin{array}{rl}{a}_0& =\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\mathrm{d}x=\frac{2}{\pi }{\int }_0^{\pi }x\mathrm{d}x=\pi \\ a_n& =\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos nx\mathrm{d}x=\frac{2}{\pi }{\int }_0^{\pi }x\cos nx\mathrm{d}x=\\ & ={x\frac{2\sin nx}{n\pi }|}_0^{\pi }-\frac{2}{\pi }{\int }_0^{\pi }\frac{\sin nx}{n}\mathrm{d}x=\frac{(-1{)}^n-1}{n^2\pi }\{\begin{array}{ll}0,& n\in \mathrm{even}\\ -\frac{4}{n^2\pi },& n\in \mathrm{odd}\end{array}\end{array}\end{array}$$ 所以$$\begin{array}{}\text{(15)}& f(x)\sim \frac{\pi }{2}-\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{(2n-1{)}^2\pi }\cos (2n-1)x\end{array}$$ 上述Fourier级数一致收敛，而$f$是分段${\mathcal{C}}^1$的连续函数，所以它的Fourier级数一致收敛到$f$自身，即$$\begin{array}{}\text{(16)}& f(x)=\frac{\pi }{2}-\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{(2n-1{)}^2\pi }\cos (2n-1)x\end{array}$$

(2) 令$x=0$可得$$\begin{array}{}\text{(17)}& 0=\frac{\pi }{2}-\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{(2n-1{)}^2\pi }⟹\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{(2n-1{)}^2}=\frac{{\pi }^2}{8}\end{array}$$

(3) 由Parseval等式可得$$\begin{array}{}\text{(18)}& \frac{2}{3}{\pi }^2=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2\mathrm{d}x=\frac{A_0^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}(A_n^2+B_n^2)=\frac{{\pi }^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}\frac{16}{(2n-1{)}^4{\pi }^2}⟹\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{(2n-1{)}^4}=\frac{{\pi }^4}{96}\end{array}$$

(4) 设级数收敛于$S$，则有$$\begin{array}{}\text{(19)}& S=\frac{1}{2^4}(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\cdots )+(\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\cdots )=\frac{S}{2^4}+\frac{{\pi }^4}{96}⟹S=\frac{{\pi }^4}{90}\end{array}$$ $◻$

解 正弦级数由奇函数展开得到，所以做周期奇延拓，即$$\begin{array}{}\text{(20)}& f(x)=x^2\mathrm{sgn}x=\{\begin{array}{ll}{x}^2,& 0\le x\le \pi \\ -x^2,& -\pi <x<0\end{array}\end{array}$$ 在区间$(-\pi ,\pi )$上$f^{\prime }(x)=2|x|$，由上题可知$$\begin{array}{}\text{(21)}& f^{\prime }(x)=\pi -\sum _{n=1}^{+\mathrm{\infty }}\frac{8}{(2n-1{)}^2\pi }\cos (2n-1)x\end{array}$$ $\mathrm{\forall }x\in (-\pi ,\pi )$，逐项积分可得$$\begin{array}{}\text{(22)}& \begin{array}{rl}f(x)& =f(0)+{\int }_0^x[\pi -\sum _{n=1}^{+\mathrm{\infty }}\frac{8}{(2n-1{)}^2\pi }\cos (2n-1)t]\mathrm{d}t\\ & =\pi x-\sum _{n=1}^{+\mathrm{\infty }}\frac{8}{(2n-1{)}^3\pi }\sin (2n-1)x=\pi x-\sum _{n=1}^{+\mathrm{\infty }}\frac{4(1-(-1{)}^n)}{n^3\pi }\sin nx\end{array}\end{array}$$ 把$x$在区间$(-\pi ,\pi )$上展成Fourier级数，得到$$\begin{array}{}\text{(23)}& x=\sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin nx,x\in (-\pi ,\pi )\end{array}$$ 所以$$\begin{array}{}\text{(24)}& f(x)=x^2\mathrm{sgn}x\sim \sum _{n=1}^{+\mathrm{\infty }}[\frac{2\pi (-1{)}^{n-1}}{n}-\frac{4(1-(-1{)}^n)}{n^3\pi }]\sin nx,x\in (-\pi ,\pi )\end{array}$$ 关于收敛性，参考例13.3.1可得级数第一项在$(-\pi ,\pi )$上内闭一致收敛，级数第二项在$[-\pi ,\pi ]$上一致绝对收敛，故$f$的Fourier级数在$(-\pi ,\pi )$上内闭一致收敛到$f$，即$f$的Fourier级数在$(-\pi ,\pi )$上连续；在$[-\pi ,\pi ]$上逐点收敛到$0$。 $◻$

13.3.2 *正交多项式

解 (1) 微分方程可化为$$\begin{array}{}\text{(43)}& (1-x^2)y^{″}-2x{y}^{\prime }+l(l+1)y=0\end{array}$$ 方程两边求$n-2$阶导数可得$$\begin{array}{}\text{(44)}& (1-x^2)y^{(n)}+(n-2)(-2x)y^{(n-1)}+\frac{(n-2)(n-3)}{2}(-2)y^{(n-2)}-2x{y}^{(n-1)}-(n-2)2y^{(n-2)}+l(l+1)y^{(n-2)}=0\end{array}$$ 令$x=0$可得$$\begin{array}{}\text{(45)}& \{\begin{array}{l}{y}^{″}(0)+l(l+1)y(0)=0\\ y^{(n)}(0)+[l(l+1)-(n-2)(n-1)]y^{(n-2)}(0)=0\end{array}\end{array}$$

$1^{\circ }$ 取$y(0)=1,y^{\prime }(0)=0$可得$$\begin{array}{}\text{(46)}& \begin{array}{rl}{y}^{\prime }(0)& =\cdots =y^{(2k-1)}(0)=\cdots =0\\ y^{(2)}(0)& =-l(l+1)\\ y^{(2k)}(0)& =\prod _{i=1}^k[(2i-1)(2i-2)-l(l+1)],k\ge 1\end{array}\end{array}$$ 此时得到幂级数解$$\begin{array}{}\text{(47)}& y=1+\sum _{k=1}^{+\mathrm{\infty }}\frac{\prod _{i=1}^k[(2i-1)(2i-2)-l(l+1)]}{(2k)!}{x}^{2k}\end{array}$$ 如果$l$是非负偶数，则这个幂级数解是多项式；如果$l$不是非负偶数，则由d’Alembert判别法，该幂级数的收敛半径为$$\begin{array}{}\text{(48)}& R={\left(\underset{k\to \mathrm{\infty }}{lim}\left|\frac{a_k}{a_{k-1}}\right|\right)}^{-1}={\left(\underset{k\to \mathrm{\infty }}{lim}\left|\frac{(2k-1)(2k-2)-l(l+1)}{2k(2k-1)}\right|\right)}^{-1}=1\end{array}$$

$2^{\circ }$ 取 $y(0)=0,y^{\prime }(0)=1$可得$$\begin{array}{}\text{(49)}& \begin{array}{rl}{y}^{(2)}(0)& =\cdots =y^{(2k)}(0)=0\\ y^{(3)}(0)& =2-l(l+1)\\ y^{(2k+1)}(0)& =\prod _{j=1}^k[2j(2j-1)-l(l+1)],k\ge 1\end{array}\end{array}$$ 于是得到幂级数解$$\begin{array}{}\text{(50)}& y=x+\sum _{k=1}^{+\mathrm{\infty }}\frac{\prod _{j=1}^k[2j(2j-1)-l(l+1)]}{(2k+1)!}{x}^{2k+1}\end{array}$$ 如果$l$是正奇数，则这个幂级数是多项式；如果$l$不是正奇数，则用d’Alembert比值判别法求得该幂级数收敛半径$R=1$。

(2) 由(1)可知$$\begin{array}{}\text{(51)}& y=\{\begin{array}{ll}1+\sum _{k=1}^{l/2}\frac{\prod _{i=1}^k[(2i-1)(2i-2)-l(l+1)]}{(2k)!}{x}^{2k},& l\in \mathrm{even}\\ x+\sum _{k=1}^{(l-1)/2}\frac{\prod _{j=1}^k[2j(2j-1)-l(l+1)]}{(2k+1)!}{x}^{2k+1},& l\in \mathrm{odd}\end{array}\end{array}$$ 前6个多项式为$$\begin{array}{}\text{(52)}& \begin{array}{rl}l=0& :y=1\\ l=1& :y=x\\ l=2& :y=1-3x^2\\ l=3& :y=x+\frac{2-12}{6}{x}^3=x-\frac{5}{3}{x}^3\\ l=4& :y=1-10x^2+\frac{35}{3}{x}^4\\ l=5& :y=x-\frac{14}{3}{x}^3+\frac{21}{5}{x}^5\end{array}\end{array}$$

(3) 由分部积分可得$$\begin{array}{}\text{(53)}& \begin{array}{rl}& {\int }_{-1}^1-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}f}{\mathrm{d}x}\right]g(x)\mathrm{d}x={\left[(1-x^2)\frac{\mathrm{d}f}{\mathrm{d}x}\right]g(x)|}_{-1}^1+{\int }_{-1}^1\left[(1-x^2)\frac{\mathrm{d}f}{\mathrm{d}x}\right]g^{\prime }(x)\mathrm{d}x\\ & ={f(x)\left[(1-x^2)\frac{\mathrm{d}g}{\mathrm{d}x}\right]|}_{-1}^1+{\int }_{-1}^1-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}g}{\mathrm{d}x}\right]f(x)\mathrm{d}x={\int }_{-1}^1-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}g}{\mathrm{d}x}\right]f(x)\mathrm{d}x\end{array}\end{array}$$ 即$$\begin{array}{}\text{(54)}& Ay=-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}y}{\mathrm{d}x}\right]\end{array}$$ 是内积${\int }_{-1}^{1}f(x)g(x)\mathrm{d}x$下的对称线性变换，Legendre多项式是这个对称线性变换的特征函数，对应于不同的特征值，所以彼此正交。事实上，若$P_m$和$P_n$是Legendre多项式，次数$m\ne n$，则$$\begin{array}{}\text{(55)}& \begin{array}{rl}& m(m+1){\int }_{-1}^1{P}_m(x)P_n(x)\mathrm{d}x={\int }_{-1}^1-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}{P}_m}{\mathrm{d}x}\right]P_n(x)\mathrm{d}x\\ & ={\int }_{-1}^1-\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}{P}_n}{\mathrm{d}x}\right]P_m(x)\mathrm{d}x=n(n+1){\int }_{-1}^1{P}_m(x)P_n(x)\mathrm{d}x\end{array}\end{array}$$ 所以$$\begin{array}{}\text{(56)}& {\int }_{-1}^1{P}_m(x)P_n(x)\mathrm{d}x=0\end{array}$$ 可以证明：在平方可积函数空间中，Legendre多项式是完备的，即任何平方可积函数可以用Legendre多项式写成平方平均收敛的级数。从而可以建立基于Legendre多项式的Fourier级数理论。 $◻$

13.3.3 *Fourier变换的应用

解 对定解问题作关于$x$的Fourier变换可得$$\begin{array}{}\text{(58)}& \{\begin{array}{ll}{\hat{u}}_t=-a^2{\xi }^2\hat{u}+\hat{f}(\xi ,t),& \xi \in \mathbb{R},t>0\\ \hat{u}(\xi ,0)=\hat{\phi }(\xi ),& \xi \in \mathbb{R}\end{array}\end{array}$$ 解得$$\begin{array}{}\text{(59)}& \hat{u}(\xi ,t)=\hat{\phi }(\xi ){\mathrm{e}}^{-a^2{\xi }^{2}t}+{\int }_0^t\hat{f}(\xi ,\tau ){\mathrm{e}}^{-a^2{\xi }^2(t-\tau )}\mathrm{d}\tau \end{array}$$ 作Fourier逆变换可得$$\begin{array}{}\text{(60)}& u(x,t)={\mathcal{F}}^{-1}[\hat{u}(\cdot ,t)](x)\end{array}$$ 因此$$\begin{array}{}\text{(61)}& \begin{array}{rl}u(x,t)& ={\mathcal{F}}^{-1}[\hat{\phi }(\xi ){\mathrm{e}}^{-a^2{\xi }^{2}t}](x)+{\int }_0^t{\mathcal{F}}^{-1}[\hat{f}(\xi ,\tau ){\mathrm{e}}^{-a^2{\xi }^2(t-\tau )}](x)\mathrm{d}\tau \\ & =\phi (x)\ast \frac{1}{2a\sqrt{\pi t}}{\mathrm{e}}^{-\frac{x^2}{4a^{2}t}}+{\int }_0^{t}f(x,\tau )\ast \frac{1}{2a\sqrt{\pi (t-\tau )}}{\mathrm{e}}^{-\frac{x^2}{4a(t-\tau )}}\mathrm{d}\tau \\ & =\frac{1}{2a\sqrt{\pi t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi (y){\mathrm{e}}^{-\frac{(x-y{)}^2}{4a^{2}t}}\mathrm{d}y+{\int }_0^t\mathrm{d}\tau {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{f(y,\tau )}{2a\sqrt{\pi (t-\tau )}}{\mathrm{e}}^{-\frac{(x-y{)}^2}{4a^2(t-\tau )}}\mathrm{d}y\end{array}\end{array}$$ 当$f=0$时，对应的齐次方程的解为$$\begin{array}{}\text{(62)}& u(x,t)=\frac{1}{2a\sqrt{\pi t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi (y){\mathrm{e}}^{-\frac{(x-y{)}^2}{4a^{2}t}}\mathrm{d}y\end{array}$$ $◻$

解 对定解问题作关于$\mathit{x}$的Fourier变换可得$$\begin{array}{}\text{(64)}& \{\begin{array}{ll}{\hat{u}}_t=-a^2({\xi }_1^2+\cdots +{\xi }_n^2)\hat{u},& \mathit{\xi }\in {\mathbb{R}}^n,t>0\\ \hat{u}(\mathit{\xi },0)=\hat{\phi }(\mathit{\xi }),& \mathit{\xi }\in {\mathbb{R}}^n\end{array}\end{array}$$ 解得$$\begin{array}{}\text{(65)}& \hat{u}(\mathit{\xi },t)=\hat{\phi }(\mathit{\xi }){\mathrm{e}}^{-a^2({\xi }_1^2+\cdots +{\xi }_n^2)t}\end{array}$$ 作Fourier逆变换可得$$\begin{array}{}\text{(66)}& \begin{array}{rl}u(\mathit{x},t)& ={\mathcal{F}}^{-1}[\hat{u}(\cdot ,t)](\mathit{x})={\mathcal{F}}^{-1}[\hat{\phi }(\mathit{\xi }){\mathrm{e}}^{-a^2({\xi }_1^2+\cdots +{\xi }_n^2)t}](\mathit{x})\\ & =\phi (\mathit{x})\ast {\left(\frac{1}{2\pi \sqrt{at}}\right)}^n{\mathrm{e}}^{-\frac{x_1^2+\cdots +x_n^2}{4a^{2}t}}={\left(\frac{1}{2\pi \sqrt{at}}\right)}^n{\int }_{{\mathbb{R}}^n}\phi (\mathit{y}){\mathrm{e}}^{-\frac{\parallel \mathit{x}-\mathit{y}{\parallel }^2}{4a^{2}t}}{\mathrm{d}}^n\mathit{y}\end{array}\end{array}$$ $◻$

解 对定解问题作关于$\mathit{x}$的Fourier变换可得$$\begin{array}{}\text{(68)}& \{\begin{array}{ll}{\hat{u}}_{tt}=-a^2({\xi }_1^2+\cdots +{\xi }_n^2)\hat{u},& \mathit{\xi }\in {\mathbb{R}}^n,t>0\\ \hat{u}(\mathit{\xi },0)=\hat{\phi }(\mathit{\xi }),\hat{u}(\mathit{\xi },0)=\hat{\psi }(\mathit{\xi }),& \mathit{\xi }\in {\mathbb{R}}^n\end{array}\end{array}$$ 解得$$\begin{array}{}\text{(69)}& \hat{u}(\mathit{\xi },t)=\hat{\phi }(\mathit{\xi })\cos (a\parallel \mathit{\xi }\parallel t)+\frac{\hat{\psi }(\mathit{\xi })}{a\parallel \mathit{\xi }\parallel }\sin (a\parallel \mathit{\xi }\parallel t)\end{array}$$ 作Fourier逆变换可得$$\begin{array}{}\text{(70)}& u(\mathit{x},t)={\mathcal{F}}^{-1}[\hat{u}(\cdot ,t)](\mathit{x},t)\end{array}$$ 对于一般的$n$，上面的逆变换不容易得到。特别地，当$n=1$时，有$$\begin{array}{}\text{(71)}& \begin{array}{rl}\hat{u}(\xi ,t)& =\hat{\phi }(\xi )\cos a\xi t+\frac{\hat{\psi }(\xi )}{a\xi }\sin a\xi t=\hat{\phi }(\xi )\frac{{\mathrm{e}}^{\mathrm{i}\xi at}+{\mathrm{e}}^{-\mathrm{i}\xi at}}{2}+\hat{\psi }(\xi )\frac{{\mathrm{e}}^{\mathrm{i}\xi at}-{\mathrm{e}}^{-\mathrm{i}\xi at}}{2a\mathrm{i}\xi }\end{array}\end{array}$$ 所以$$\begin{array}{}\text{(72)}& \begin{array}{rl}u(x,t)& ={\mathcal{F}}^{-1}[\hat{\phi }(\xi )\frac{{\mathrm{e}}^{\mathrm{i}\xi at}+{\mathrm{e}}^{-\mathrm{i}\xi at}}{2}+\hat{\psi }(\xi )\frac{{\mathrm{e}}^{\mathrm{i}\xi at}-{\mathrm{e}}^{-\mathrm{i}\xi at}}{2a\mathrm{i}\xi }]\\ & =\frac{1}{2}\phi (x)\ast [\delta (x+at)+\delta (x-at)]+\frac{1}{2a}{\int }_{-\mathrm{\infty }}^x\psi (y)\mathrm{d}y\ast [\delta (x+at)+\delta (x-at)]\\ & =\frac{\phi (x+at)+\phi (x-at)}{2}+\frac{1}{2a}{\int }_{x-at}^{x+at}\psi (\xi )\mathrm{d}\xi \end{array}\end{array}$$ 此即d’Alembert公式。 $◻$