在理想状态下一根棍子的热传导,配上均匀的边界条件.
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以下解法首先由约瑟夫·傅立叶在他于1822年出版的著作《解析热学》(Théorie analytique de la chaleur)给出.先考虑一维热方程,这可以当作棍子的热传导之模型.方程式如下:$$\ u_t = k u_{xx},\label1\tag1$$其中$u =u(t, x)$是$t$和$x$的二元函数. 假设下述初始条件:$$u(0,x) = f(x) \quad \forall x \in [0,L]\tag2\label2$$ 其中函数$f$是给定的.再配合下述边界条件$$\ u(t,0) = 0 = u(t,L) \quad \forall t > 0 \label3\tag3$$让我们试着找一个非恒等于零的解,使之满足边界条件\eqref{3}并具备以下形式:$$u(t,x) = X(x) T(t)\label4\tag4$$ 这套技术称作分离变数法.现在将$u$代回\eqref{1}:$$\frac{T'(t)}{kT(t)} = \frac{X''(x)}{X(x)}. $$ 由于等式右边只依赖$x$,而左边只依赖$t$,两边都等于某个常数$-\lambda$,于是: $$ \ T'(t) = - \lambda kT(t) \label5\tag5$$ $$ \ X''(x) = - \lambda X(x)\label6\tag6$$ 以下将证明\eqref{6}没有$\lambda\le0$的解:
假设$\lambda\lt0$,则存在实数$B,C$使得$$X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}$$从\eqref{3}得到 $$X(0) = 0 = X(L)$$ 于是有$B= 0 = C$,这蕴含$u$恒等于零. 假设λ = 0,则存在实数$B,C$使得$X(x) = Bx + C. $ 仿上述办法可从等式\eqref{3}推出$u$恒等于零. 因此必然有$\lambda\gt0$,此时存在实数$A,B,C$使得 $$T(t) = A e^{-\lambda k t}$$ $$X(x) = B \sin\left(\sqrt{\lambda} \, x\right) + C \cos\left(\sqrt{\lambda} \, x\right)$$ 从等式\eqref{3}可知$C= 0$,且存在正整数$n$使得 $$\sqrt{\lambda} = n \frac{\pi}{L}$$ 由此得到热方程形如\eqref{4}的解.
一般而言,满足\eqref{1}与\eqref{3}的解相加后仍是满足\eqref{1}与\eqref{3}的解.事实上可以证明满足\eqref{1},\eqref{2},\eqref{3}的解由下述公式给出: $$u(t,x) = \sum_{n = 1}^{+\infty} D_n \left(\sin \frac{n\pi x}{L}\right) e^{-\frac{n^2 \pi^2 kt}{L^2}}$$ 其中 $$D_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \,\mathrm dx$$
求解技巧的推广
上面采用的方法可以推广到许多不同方程.想法是:在适当的函数空间上,算子$u \mapsto u_{xx}$可以用它的特征向量表示.这就自然地导向线性自伴算子的谱理论.
考虑线性算子$Δ u = u_{xx}$,以下函数序列 $$ e_n(x) = \sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L} \quad(n≥ 1)$$ 是Δ的特征向量.诚然:$$ \Delta e_n = -\frac{n^2 \pi^2}{L^2} e_n$$ 此外,任何满足边界条件$f(0)=f(L)=0$的Δ的特征向量都是某个$e_n$.令$L_2(0,L)$表$[0,L]$上全体平方可积函数的向量空间.这些函数$e_n$构成$L_2(0,L)$的一组单位正交基.也就是说,$$ \langle e_n, e_m \rangle = \int_0^L e_n(x) e_m(x)\,\mathrm dx = \left\{ \begin{matrix} 0 & n \neq m \\ 1 & m = n\end{matrix}\right. $$ 最后,序列$\{e_n\}_{n\in\mathbb N}$张出$L_2(0,L)$的一个稠密的线性子空间.这就表明我们实际上已将算子Δ对角化.