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若三对角矩阵A按行严格对角占优则求解三对角方程组Ax=f的追赶法是数值稳定的．

大学 2022-05-20 03:25:55 方程组,矩阵,数值已帮助人

若三对角矩阵A按行严格对角占优，则求解三对角方程组Ax=f的追赶法是数值稳定的．

热搜：n阶方阵a与对角矩阵相似的充要条件

答案解析

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设有舍入误差时由追赶法计算得到的诸量为 $latex.action?latex=XHdpZGV0aWxkZXtsfV9p44CBXHdpZGV0aWxkZXt1fV9p44CBXHdpZGV0aWxkZXt5fV9p44CBXHdp%0D%0AZGV0aWxkZXt4fV9p$ ，它们彼此之间具有下列关系：
$latex.action?latex=XGJlZ2lue2FycmF5fXtjfVx3aWRldGlsZGV7bH1fMT1iXzHvvIxcd2lkZXRpbGRle3l9XzE9XGZy%0D%0AYWN7Zl8xfXtcd2lkZXRpbGRle2x9XzF9K1xkZWx0YSBfMe%2B8jFx3aWRldGlsZGV7dX1fe2ktMX09%0D%0AXGZyYWN7Y197aS0xfX17XHdpZGV0aWxkZXtsfV97aS0xfX0rXHhpIF97aS0xfe%2B8jFxcXHdpZGV0%0D%0AaWxkZXtsfV9pPWJfaS1hX2lcd2lkZXRpbGRle3V9X3tpLTF9K1xlcHNpbG9uIF9p77yMXHdpZGV0%0D%0AaWxkZXt5fV9pPVxmcmFje2ZfaS1hX2lcd2lkZXRpbGRle3l9X3tpLTF9fXtcd2lkZXRpbGRle2x9%0D%0AX2l9K1xkZWx0YSBfaeOAgChpPTIsMyxcY2RvdHMgLG4pXFxcd2lkZXRpbGRle3h9X249XHdpZGV0%0D%0AaWxkZXt5fV9uK1xldGEgX27vvIxcd2lkZXRpbGRle3h9X2k9XHdpZGV0aWxkZXt5fV9pLVx3aWRl%0D%0AdGlsZGV7dX1faVx3aWRldGlsZGV7eH1fe2krMX0rXGV0YSBfaeOAgChpPW4tMSxcY2RvdHMgLDIs%0D%0AMSlcZW5ke2FycmF5fVxyaWdodFx9$ (3.7)
其中出现的量εi、δi、ξi、ηi是近似计算中产生的误差．
假定|εi|≤ε，|δi|≤ε，|ξi|≤ε，|ηi|≤ε，且ε是充分小的正数(要求每一步的计算误差较小)．引进记号
$latex.action?latex=XHdpZGV0aWxkZXtifV8xPWJfMQ%3D%3D$ ， $latex.action?latex=XHdpZGV0aWxkZXtmfV8xPWZfMStcd2lkZXRpbGRle2x9XzEoXGRlbHRhIF8xK1xldGEgXzEp$ ，
$latex.action?latex=XGxlZnRce1xiZWdpbnthcnJheX17Y31cd2lkZXRpbGRle2N9X3tpLTF9PWNfe2ktMX0rXHdpZGV0%0D%0AaWxkZXtsfV97aS0xfVx4aSBfe2ktMX3vvIxcd2lkZXRpbGRle2F9X2k9YV9p77yMXHdpZGV0aWxk%0D%0AZXtifV9pPWJfaStcZXBzaWxvbiBfaVxcXHdpZGV0aWxkZXtmfV9pPWZfaStcd2lkZXRpbGRle2x9%0D%0AX2koXGRlbHRhIF9pK1xldGEgX2kpK2FfaVxldGEgX3tpLTF944CAKGk9MiwzLFxjZG90cyAsbilc%0D%0AZW5ke2FycmF5fQ%3D%3D$
将式(3.7)改写为
$latex.action?latex=XHdpZGV0aWxkZXtsfV8xPVx3aWRldGlsZGV7Yn1fMQ%3D%3D$ ， $latex.action?latex=XGhhdHt5fV8xPVx3aWRldGlsZGV7eX1fMStcZXRhIF8xPVxmcmFje1x3aWRldGlsZGV7Zn1fMX17%0D%0AXHdpZGV0aWxkZXtsfV8xfQ%3D%3D$
$latex.action?latex=XGxlZnRce1xiZWdpbnthcnJheX17Y31cd2lkZXRpbGRle3V9X3tpLTF9PVxmcmFje1x3aWRldGls%0D%0AZGV7Y31fe2ktMX19e1x3aWRldGlsZGV7bH1fe2ktMX1977yMXHdpZGV0aWxkZXtsfV9pPVx3aWRl%0D%0AdGlsZGV7Yn1faS1cd2lkZXRpbGRle2F9X2lcd2lkZXRpbGRle3V9X3tpLTF9XFxcaGF0e3l9X2k9%0D%0AXHdpZGV0aWxkZXt5fV9pK1xldGEgX2k9XGZyYWN7XHdpZGV0aWxkZXtmfV9pLVx3aWRldGlsZGV7%0D%0AYX1faVxoYXR7eX1fe2ktMX19e1x3aWRldGlsZGV7bH1fe2l9feOAgChpPTIsMyxcY2RvdHMgLG4p%0D%0AXGVuZHthcnJheX0%3D$
将上式与求三对角方程组Ax=f的追赶法计算公式进行比较可得， $latex.action?latex=XHdpZGV0aWxkZXt4fV9p$ (i=1，2，…，n)应当是扰动三对角方程组 $latex.action?latex=XHdpZGV0aWxkZXtBfVx3aWRldGlsZGV7eH09XHdpZGV0aWxkZXtmfQ%3D%3D$ 的精确解．
由于三对角矩阵A按行严格对角占优，所以存在正数ρ和δ，使得
$latex.action?latex=XGxlZnRce1xiZWdpbnthcnJheX17Y31cZGVsdGEgXGdlcSB8Yl8xfFxnZXEgfGNfMXwrXHJobyBc%0D%0AXFxkZWx0YSBcZ2VxIHxiX2l8XGdlcSB8YV9pfCt8Y19pfCtccmhvIOOAgChpPTIsXGNkb3RzICxu%0D%0ALTEpXFxcZGVsdGEgXGdlcSB8Yl9ufFxnZXEgfGFfbnwrXHJob1xlbmR7YXJyYXl9$ (3.8)
即A满足引理的条件．显然有|ai|＜δ，|ci|＜δ．根据式(3.7)的改写形式可得
$latex.action?latex=fFx3aWRldGlsZGV7Y31fMS1jXzF8PXxcd2lkZXRpbGRle2x9XzFceGkgXzF8PXxiXzFceGkgXzF8%0D%0AXGxlcSBcZGVsdGEgXGVwc2lsb24g$ ，
$latex.action?latex=fFx3aWRldGlsZGV7Zn1fMS1mXzF8PXxcd2lkZXRpbGRle2x9XzEoXGRlbHRhIF8xK1xldGEgXzEp%0D%0AfD18Yl8xKFxkZWx0YSBfMStcZXRhIF8xKXxcbGVxIDJcZGVsdGEgXGVwc2lsb24g$ ．
因为ε充分小，所以可要求ε满足
$latex.action?latex=XGVwc2lsb24gXGxlcSBcZnJhY3t8Yl8xfC18Y18xfH17fGJfMXx9$ ， $latex.action?latex=XGVwc2lsb24gXGxlcSBcZnJhY3tccmhvIH17M1xkZWx0YSArMX0%3D$ ，ε≤δ
于是有
$latex.action?latex=fFx3aWRldGlsZGV7dX1fMXw9XGxlZnR8IFxmcmFje1x3aWRldGlsZGV7Y31fMX17XHdpZGV0aWxk%0D%0AZXtsfV8xfSBccmlnaHR8PVxsZWZ0fCBcZnJhY3tjXzErYl8xXHhpIF8xfXtiXzF9IFxyaWdodHxc%0D%0AbGVxIFxmcmFje3xjXzF8K3xiXzF8XGVwc2lsb24gfXt8Yl8xfH1cbGVxIDE%3D$
假设 $latex.action?latex=fFx3aWRldGlsZGV7dX1fe2ktMX18XGxlcSAx$ ，则有
$latex.action?latex=fFx3aWRldGlsZGV7bH1fe2l9fD18XHdpZGV0aWxkZXtifV9pLVx3aWRldGlsZGV7YX1faVx3aWRl%0D%0AdGlsZGV7dX1fe2ktMX18PXxiX2krXGVwc2lsb24gX2ktYV9pXHdpZGV0aWxkZXt1fV97aS0xfXxc%0D%0AbGVxIHxiX2l8K3xcZXBzaWxvbiBfaXwrfGFfaeKAllx3aWRldGlsZGV7dX1fe2ktMX18XGxlcSAz%0D%0AXGRlbHRhIA%3D%3D$
$latex.action?latex=fFx3aWRldGlsZGV7bH1fe2l9fFxnZXEgfGJfaXwtfGFfaeKAllx3aWRldGlsZGV7dX1fe2ktMX18%0D%0ALXxcZXBzaWxvbiBfaXxcZ2VxIHxiX2l8LXxhX2l8LVxlcHNpbG9uIFxnZXEgfGNfaXwrXHJobyAt%0D%0AXGVwc2lsb24g$
$latex.action?latex=fFx3aWRldGlsZGV7Y31faS1jX2l8PXxcd2lkZXRpbGRle2x9X2lceGkgX2l8XGxlcSAzXGRlbHRh%0D%0AIFxlcHNpbG9uIA%3D%3D$
$latex.action?latex=fFx3aWRldGlsZGV7Zn1faS1mX2l8PXxcd2lkZXRpbGRle2x9X2koXGRlbHRhIF9pK1xldGEgX2kp%0D%0AK2FfaVxldGEgX3tpLTF9fFxsZXEgfFx3aWRldGlsZGV7bH1faXwofFxkZWx0YSBfaXwrfFxldGEg%0D%0AX2l8KSt8YV9p4oCWXGV0YSBfe2ktMX18XGxlcSA3XGRlbHRhIFxlcHNpbG9uIA%3D%3D$
$latex.action?latex=fFx3aWRldGlsZGV7dX1faXw9XGZyYWN7fFx3aWRldGlsZGV7Y31faXx9e3xcd2lkZXRpbGRle2x9%0D%0AX2l8fVxsZXEgXGZyYWN7fGNfaXwrM1xkZWx0YSBcZXBzaWxvbiB9e3xjX2l8K1xyaG8gLVxlcHNp%0D%0AbG9uIH1cbGVxIDE%3D$
根据归纳法原理可得
$latex.action?latex=fFx3aWRldGlsZGV7Y31faS1jX2l8XGxlcSBcd2lkZXRpbGRle1xlcHNpbG9uIH0%3D$ (i=1，2，…，n-1)
$latex.action?latex=fFx3aWRldGlsZGV7Zn1faS1mX2l8XGxlcSBcd2lkZXRpbGRle1xlcHNpbG9uIH0%3D$ (i=1，2，…，n)
其中 $latex.action?latex=XHdpZGV0aWxkZXtcZXBzaWxvbiB9PW1heFxsZWZ0XHsgXGVwc2lsb24gLDdcZGVsdGEgXGVwc2ls%0D%0Ab24gIFxyaWdodFx9$ ．注意到
$latex.action?latex=fFx3aWRldGlsZGV7Yn1fMS1iXzF8PTA%3D$ ， $latex.action?latex=fFx3aWRldGlsZGV7Yn1faS1iX2l8PXxcZXBzaWxvbiBfaXxcbGVxIFx3aWRldGlsZGV7XGVwc2ls%0D%0Ab24gfQ%3D%3D$ (i=1，2，…，n)
$latex.action?latex=fFx3aWRldGlsZGV7YX1faS1hX2l8PTA%3D$ (i=2，3，…，n)
由引理可得 $latex.action?latex=4oCWXHdpZGV0aWxkZXt4fS144oCWX3tcaW5mdHkgfVxsZXEgTVxlcHNpbG9uIA%3D%3D$ ，其中 $latex.action?latex=XHdpZGV0aWxkZXtcZXBzaWxvbiB9$ 是充分小的正数，M是与n无关的常数．这表明追赶法是数值稳定的．
证毕．