{"id":"byWRDkC4M2qIGC9O","question":"考虑如下热传导方程的初边值问题：\n$$\n\\left\\{\\begin{array}{l}\n\\frac{\\partial u}{\\partial t}-\\frac{\\partial^{2} u}{\\partial x^{2}}+u=f(x, t), x \\in(0,1), t \\in(0, T] \\\\\nu(x, 0)=\\varphi(x), x \\in[0,1], \\\\\nu(0, t)=\\alpha(t), u(1, t)=\\beta(t), t \\in[0, T] .\n\\end{array}\\right.\n$$\n\n已知空间步长 $h=1/m$，时间步长 $\\tau=T/n$，网格节点定义为 $x_{i}=i h$（$0 \\leqslant i \\leqslant m$）和 $t_{k}=k \\tau$（$0 \\leqslant k \\leqslant n$）。请设计一个两层隐式差分格式，使其截断误差达到 $O(\\tau^{2}+h^{2})$，并推导该截断误差的具体表达式。","concepts":"偏微分方程数值解法;Crank-Nicolson格式","answer":"差分格式：$\\left\\{\\begin{array}{l}\n\\frac{1}{\\tau}\\left(u_{i}^{k+1}-u_{i}^{k}\\right)-\\frac{1}{2 h^{2}}\\left(u_{i-1}^{k}-2 u_{i}^{k}+u_{i+1}^{k}+u_{i-1}^{k+1}-2 u_{i}^{k+1}+u_{i+1}^{k+1}\\right) \\\\\n\\quad+\\frac{1}{2}\\left(u_{i}^{k}+u_{i}^{k+1}\\right)=f\\left(x_{i}, t_{k+1 / 2}\\right), \\quad 1 \\leqslant i \\leqslant m-1,0 \\leqslant k \\leqslant n-1, \\\\\nu_{i}^{0}=\\varphi\\left(x_{i}\\right), \\quad 1 \\leqslant i \\leqslant m-1, \\\\\nu_{0}^{k}=\\alpha\\left(t_{k}\\right), u_{m}^{k}=\\beta\\left(t_{k}\\right), \\quad 0 \\leqslant k \\leqslant n .\n\\end{array}\\right.$；截断误差：$R_{i}^{k}=\\left[\\frac{1}{24} \\frac{\\partial^{3} u\\left(x_{i}, \\eta_{i}^{k}\\right)}{\\partial t^{3}}-\\frac{1}{8} \\frac{\\partial^{4} u\\left(x_{i}, \\tilde{\\eta}_{i}^{k}\\right)}{\\partial x^{2} \\partial t^{2}}+\\frac{1}{8} \\frac{\\partial^{2} u\\left(x_{i}, \\bar{\\eta}_{i}^{k}\\right)}{\\partial t^{2}}\\right] \\tau^{2} -\\frac{1}{24}\\left[\\frac{\\partial^{4} u\\left(\\xi_{i}^{k}, t_{k}\\right)}{\\partial x^{4}}+\\frac{\\partial^{4} u\\left(\\xi_{i}^{k+1}, t_{k+1}\\right)}{\\partial x^{4}}\\right] h^{2}$","final_answer":"差分格式：$\\left\\{\\begin{array}{l}\n\\frac{1}{\\tau}\\left(u_{i}^{k+1}-u_{i}^{k}\\right)-\\frac{1}{2 h^{2}}\\left(u_{i-1}^{k}-2 u_{i}^{k}+u_{i+1}^{k}+u_{i-1}^{k+1}-2 u_{i}^{k+1}+u_{i+1}^{k+1}\\right) \\\\\n\\quad+\\frac{1}{2}\\left(u_{i}^{k}+u_{i}^{k+1}\\right)=f\\left(x_{i}, t_{k+1 / 2}\\right), \\quad 1 \\leqslant i \\leqslant m-1,0 \\leqslant k \\leqslant n-1, \\\\\nu_{i}^{0}=\\varphi\\left(x_{i}\\right), \\quad 1 \\leqslant i \\leqslant m-1, \\\\\nu_{0}^{k}=\\alpha\\left(t_{k}\\right), u_{m}^{k}=\\beta\\left(t_{k}\\right), \\quad 0 \\leqslant k \\leqslant n .\n\\end{array}\\right.$；截断误差：$R_{i}^{k}=\\left[\\frac{1}{24} \\frac{\\partial^{3} u\\left(x_{i}, \\eta_{i}^{k}\\right)}{\\partial t^{3}}-\\frac{1}{8} \\frac{\\partial^{4} u\\left(x_{i}, \\tilde{\\eta}_{i}^{k}\\right)}{\\partial x^{2} \\partial t^{2}}+\\frac{1}{8} \\frac{\\partial^{2} u\\left(x_{i}, \\bar{\\eta}_{i}^{k}\\right)}{\\partial t^{2}}\\right] \\tau^{2} -\\frac{1}{24}\\left[\\frac{\\partial^{4} u\\left(\\xi_{i}^{k}, t_{k}\\right)}{\\partial x^{4}}+\\frac{\\partial^{4} u\\left(\\xi_{i}^{k+1}, t_{k+1}\\right)}{\\partial x^{4}}\\right] h^{2}$","explanation":"1. 取时间中间点 $t_{k+1/2}=(t_k+t_{k+1})/2$，在点 $(x_i, t_{k+1/2})$ 处考虑原方程：$\\frac{\\partial u(x_i, t_{k+1/2})}{\\partial t}-\\frac{\\partial^2 u(x_i, t_{k+1/2})}{\\partial x^2}+u(x_i, t_{k+1/2})=f(x_i, t_{k+1/2})$。\n2. 对各项进行 Taylor 展开：\n - 时间导数项：$\\frac{\\partial u(x_i, t_{k+1/2})}{\\partial t}=\\frac{1}{\\tau}[u(x_i, t_{k+1})-u(x_i, t_k)]-\\frac{\\tau^2}{24}\\frac{\\partial^3 u(x_i, \\eta_i^k)}{\\partial t^3}$，其中 $t_k<\\eta_i^k<t_{k+1}$。\n - 空间二阶导数项：$\\frac{\\partial^2 u(x_i, t_{k+1/2})}{\\partial x^2}=\\frac{1}{2}[\\frac{\\partial^2 u(x_i, t_k)}{\\partial x^2}+\\frac{\\partial^2 u(x_i, t_{k+1})}{\\partial x^2}]-\\frac{\\tau^2}{8}\\frac{\\partial^4 u(x_i, \\tilde{\\eta}_i^k)}{\\partial x^2 \\partial t^2}$，进一步将空间二阶导数用中心差商近似：$\\frac{1}{2h^2}[u(x_{i-1}, t_k)-2u(x_i, t_k)+u(x_{i+1}, t_k)+u(x_{i-1}, t_{k+1})-2u(x_i, t_{k+1})+u(x_{i+1}, t_{k+1})]-\\frac{h^2}{24}[\\frac{\\partial^4 u(\\xi_i^k, t_k)}{\\partial x^4}+\\frac{\\partial^4 u(\\xi_i^{k+1}, t_{k+1})}{\\partial x^4}]-\\frac{\\tau^2}{8}\\frac{\\partial^4 u(x_i, \\tilde{\\eta}_i^k)}{\\partial x^2 \\partial t^2}$，其中 $x_{i-1}<\\xi_i^k, \\xi_i^{k+1}<x_{i+1}$，$t_k<\\tilde{\\eta}_i^k<t_{k+1}$。\n - $u$ 项：$u(x_i, t_{k+1/2})=\\frac{1}{2}[u(x_i, t_k)+u(x_i, t_{k+1})]-\\frac{\\tau^2}{8}\\frac{\\partial^2 u(x_i, \\bar{\\eta}_i^k)}{\\partial t^2}$，其中 $t_k<\\bar{\\eta}_i^k<t_{k+1}$。\n3. 将上述展开式代入方程，忽略截断误差，结合初边值条件 $u(x_i, 0)=\\varphi(x_i)$（$1\\leqslant i\\leqslant m-1$）、$u(0, t_k)=\\alpha(t_k)$ 和 $u(1, t_k)=\\beta(t_k)$（$0\\leqslant k\\leqslant n$），得到差分格式。\n4. 整理截断误差项，得到 $R_i^k$ 的表达式。","batch":"graduate_data_12000_1127","type":"解答题"}