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ID:15702782
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页数:36页
时间:2018-08-05
《化工问题的建模与数学分析方法 第6章 习题及答案》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、第六章习题1.用摄动法求解以下三次代数方程证明其三个根的渐进展开式为将以上解与数值解或准确解比较,取和0.1,讨论其结果。解:,(1)①证明:(A)设解具有如下形式的渐进展开式(2)将其代入(1)式,可得(3)比较两边的同幂次项系数,得到:::所求摄动解为:由于方程(1)有三个根,用上述摄动法只能求得一个根,即接近退化解的那个根。(B)做变换,代入方程(1),可得,(4)此时,小参数不在最高次幂项上。设此时解具有如下形式的渐进展开式,(5)将其代入(4)式,可得(6)比较两边的同幂次项系数,得到或所以,可
2、得方程(1)的另外两个解,至此,方程(1)三个根的渐进展开式为:②上述三个渐进解分别与准确解和数值解的比较分析。(1)采用Matlab的符号工具箱,求得方程(1)的三个准确解分别为(其中epsilon表示):x(1)=1/6/epsilon*((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3)+2/((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*e
3、psilon^2)^(1/3)x(2)=-1/12/epsilon*((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3)-1/((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3)+1/2*i*3^(1/2)*(1/6/epsilon*((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/eps
4、ilon)^(1/2))*epsilon^2)^(1/3)-2/((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3))x(3)=-1/12/epsilon*((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3)-1/((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2
5、))*epsilon^2)^(1/3)-1/2*i*3^(1/2)*(1/6/epsilon*((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3)-2/((-108*a+12*3^(1/2)*((-4+27*a^2*epsilon)/epsilon)^(1/2))*epsilon^2)^(1/3))当时,x(1)=1/3*(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)+10/(-135
6、*a+15*(-120+81*a^2)^(1/2))^(1/3)x(2)=-1/6*(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)-5/(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/3*(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)-10/(-135*a+15*(-120+81*a^2)^(1/2))^(1/3))x(3)=-1/6*(-135*a+15*(-120+81*a^2)^(1
7、/2))^(1/3)-5/(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)-1/2*i*3^(1/2)*(1/3*(-135*a+15*(-120+81*a^2)^(1/2))^(1/3)-10/(-135*a+15*(-120+81*a^2)^(1/2))^(1/3))当时,x(1)=1/3*(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3)+100/(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3)x(2)=-1/6
8、*(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3)-50/(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/3*(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3)-100/(-1350*a+150*(-1200+81*a^2)^(1/2))^(1/3))x(3)=-1/6*(-1350*a+150*(
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