56方程组与高阶方程的情形ppt课件

1、1第六节第六节 方程组与高阶方程方程组与高阶方程的情形的情形2 2009, Henan Polytechnic University25.6.1 一阶方程组一阶方程组一阶微分方程组的一般形式为：一阶微分方程组的一般形式为：= = = = )(,.),(,()(.)(,.),(,()(1111xyxyxfxyxyxyxfxymmmm初值初值0002020101)(,.,)(,)(mmyxyyxyyxy= = = =将问题记作向量形式，令：将问题记作向量形式，令： = = = = = =001011.,.,.mmmyyyfffyyy = = = 00)(),()(yxyyxfxy前述所有公式皆前述

2、所有公式皆适用于向量形式。适用于向量形式。3 2009, Henan Polytechnic University3以两个方程构成的方程组为例：以两个方程构成的方程组为例： = = = = = = 0000)(),()(),(zxzzyxgzyxyzyxfy设设 为节点上的近似解，为节点上的近似解，则有改进的则有改进的EulerEuler格式为格式为 iiizyiihxx,);, 3 , 2 , 1(0= = = =),(1iiiiizyxhfyy = = ),(1iiiiizyxhgzz = = 预告：预告： ),(),(21111 = =iiiiiiiizyxfzyxfhyy ),(),(

3、21111 = =iiiiiiiizyxgzyxghzz校正：校正： 4 2009, Henan Polytechnic University4例例 用改进的用改进的EulerEuler法求解初值问题法求解初值问题 = = = = = = = = 2)0(1)0(zzyxzyzxyy2.00 x取步长取步长h=0.1h=0.1，保留六位小数。，保留六位小数。 解解: : 改进的改进的EulerEuler法公式为法公式为),(1iiiiizyxhfyy = = ),(1iiiiizyxhgzz = = 预告：预告： ),(),(21111 = =iiiiiiiizyxfzyxfhyy ),(),

4、(21111 = =iiiiiiiizyxgzyxghzz校正：校正： 5 2009, Henan Polytechnic University5 = = = = iiiiiiiiiizyxzzzyxyy1.0)(1.011 = = = = 111111105. 0)()(05. 0iiiiiiiiiiiiiiiizyxzyxzzzyxzyxyy由初值由初值 , ,计算得计算得 2)0(, 1)0(00= = = = =zzyy = = =050000.2800000.011zy = = = = 046951. 2)1 . 0(801500. 0)1 . 0(11zzyy = = =09099

5、2.2604820.022zy = = = = 088216. 2)2 . 0(604659. 0)2 . 0(22zzyy6 2009, Henan Polytechnic University6相应的四阶龙格相应的四阶龙格库塔格式经典格式为库塔格式经典格式为 = = = = )22(6)22(64321143211LLLLhzzKKKKhyyiiii7 2009, Henan Polytechnic University7 = = = = = = = = = = = = = = ),(),()2,2,()2,2,()2,2,()2,2,(),(),(33143314222132221311

6、2121121211hLzhKyxgLhLzhKyxfKLhzKhyxgLLhzKhyxfKLhzKhyxgLLhzKhyxfKzyxgLzyxfKiiiiiiiiiiiiiiiiiiiiiiii式中式中 8 2009, Henan Polytechnic University8 这是一步法，利用节点 上的值 ， ，ixiyiz = = = = )22(6)22(6432114321LLLLhzzKKKKhyyiiii44332211,LKLKLKLK由上式顺序计算由上式顺序计算然后代入然后代入11, iizy1 ix即可求得即可求得 节点上节点上的的9 2009, Henan Polytec

7、hnic University9),()21,2()21,2(),()22(61342312143211kyhxhfkkyhxhfkkyhxhfkyxhfkkkkkyynniinniinniinniiiiiiinin = = = = = = = = = 多个方程的多个方程的Runge-KuttaRunge-Kutta形式可写为：形式可写为：10 2009, Henan Polytechnic University105.6.2 化高阶方程组为一阶方程组化高阶方程组为一阶方程组= = = = = = = 10)1(1000)1()()(,.,)(,)(),.,(nnnnaxyaxyaxyyyyx

8、fy化作一阶微分方程组求解。化作一阶微分方程组求解。引入新变量引入新变量)1(21,., = = = = =nnyyyyyy = = = = = = ),.,(.1121nnnnyyxfyyyyy初值条件为：初值条件为：10102001)(.)()( = = = =nnaxyaxyaxy即可将即可将n n阶方程化为如下的一阶方程组阶方程化为如下的一阶方程组 11 2009, Henan Polytechnic University11例如例如, ,二阶微分方程的初值问题二阶微分方程的初值问题 = = = = = = 0000)(,)(),(yxyyxyyyxfy在引入新的变量在引入新的变量 后

9、后, ,即化为一阶方程组初值问题即化为一阶方程组初值问题: :yz = = = = = = = = 0000)(,)(,),(yxzyxyzyzyxfz上式为一个一阶方程组的初值问题。上式为一个一阶方程组的初值问题。应用四阶龙格应用四阶龙格- -库塔公式得库塔公式得 12 2009, Henan Polytechnic University12 = = = = )22(6)22(64321143211LLLLhzzKKKKhyyiiii = = = = = = = = = = = = = = ),()2,2,(2)2,2,(2),(3314342221323112121211hLzhKyxfL

10、hLzKLhzKhyxfLLhzKLhzKhyxfLLhzKzyxfLzKiiiiiiiiiiiiiiii13 2009, Henan Polytechnic University13消去消去 ，上式简化为：，上式简化为： )4 , 3 , 2 , 1( = =iKi = = = = )22(6)(64321132121LLLLhzzLLLhhzyyiiiii = = = = = = = ),2,()2,42,()2,2,(),(3221421221312121hLzLhhzyxfLLhzLhzhyxfLLhzzhyxfLzyxfLiiiiiiiiiiiiiii上述方法同样可以用来处理三阶或更

11、高阶的微分方上述方法同样可以用来处理三阶或更高阶的微分方程或方程组的初值问题程或方程组的初值问题 14 2009, Henan Polytechnic University14例例 求解下列二阶微分方程的初值问题求解下列二阶微分方程的初值问题 = = = = = 1)0(, 0)0(yyxyy10 x取步长取步长h=0.1 h=0.1 解解: :先作变换：令先作变换：令 ，代入上式，得一阶方程组，代入上式，得一阶方程组 yz = = = = = = = = 1)0(, 0)0(,zyzyxzz用四阶龙格用四阶龙格- -库塔方法求解计算：库塔方法求解计算：取步长取步长 , , , , ,1 .

12、0= =h00= =x00= =y10= =z15 2009, Henan Polytechnic University15 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =2105. 1)105. 11 . 01 () 1 . 00()()(1105. 1105. 11 . 01105. 1) 1 . 121 . 01 ()21 . 00()2()2(055. 11 . 121 . 0121 . 1) 121 . 01 ()21 . 00()2()2(05. 1121 . 012101),(130043042003203100210200000101hLzhxLhLzKLhzhxLLhzKLhzhxLLhzKxzzyxfLzK0= =i时时 16 2009, Henan Polytechnic University16 = = = = = = = = = = =1104. 1)2105. 1105.