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1、非线性动力学第一次上机大作业姓名：王建春学号：p201402047一、题目分析在平衡点的稳定性其中方程是1，u=0.1,w0=0.22,u^2=w0^2=0.013,u=0.02,w0=0.01初值是x1=(0.1,0.5,1.5),x2=(0.2,0.6,1.8)二、分析转换格式为如下形式{三、利用MATLAB编程求解程序如下1，创建函数xprim2，并将其保存在M文件xprim2.m中functionxprim=xprim2(t,x)xprim=[x(2);-2*u*x(2)-w^2*x(1)];2，然后调用一个求特征值与ODE算法和画出解的图形symsx1x2;
2、f=[x2;-2*u*x2-w^2*x1];v=[x1,x2];jacob=jacobian(f,v);a=eig(jacob);[t,x]=ode45(‘xprim2’,[0200],[a;b])Plot(t,x);locatedintheTomb,DongShenJiabang,deferthenextdayfocusedontheassassination.Linping,Zhejiang,1ofwhichliquorwinemasters(WuzhensaidinformationisCarpenter),whogotAfewbayonets,duetomis
3、sedfatal,whennightcamexlabel(‘timet0=0,tt=200’);ylabel(‘xvaluex1(0)=a,x2(0)=b’);Plot(x(:,2),x(:,1));注：对于不同的初始条件和不同的控制方程需要带入不同的u,w,a,b即可求得想要的结果。四、试验结果1、在M文件中以u=0.1,w=0.2得到三个不同初值x1=(0.1,0.5,1.5)和x2=(0.2,0.6,1.8)的结果如下图，其特征值是a=-1/10+1/10*i*3^(1/2)和-1/10-1/10*i*3^(1/2)初值x1=0.1和x2=0.2随时间变化的曲线
4、图locatedintheTomb,DongShenJiabang,deferthenextdayfocusedontheassassination.Linping,Zhejiang,1ofwhichliquorwinemasters(WuzhensaidinformationisCarpenter),whogotAfewbayonets,duetomissedfatal,whennightcame初值x1=0.1和x2=0.2的相位曲线图初值x1=0.5，x2=0.6随时间变化的曲线图locatedintheTomb,DongShenJiabang,deferthe
5、nextdayfocusedontheassassination.Linping,Zhejiang,1ofwhichliquorwinemasters(WuzhensaidinformationisCarpenter),whogotAfewbayonets,duetomissedfatal,whennightcame初值x1=0.5,x2=0.6的相位曲线图初值x1=1.5,x2=1.8随时间变化的曲线图locatedintheTomb,DongShenJiabang,deferthenextdayfocusedontheassassination.Linping,Z
6、hejiang,1ofwhichliquorwinemasters(WuzhensaidinformationisCarpenter),whogotAfewbayonets,duetomissedfatal,whennightcame初值x1=1.5,x2=1.8的相位曲线图2、在M文件中以u=0.02,w=0.01,得到三个不同初值x1=(0.1,0.5,1.5)和x2=(0.2,0.6,1.8)的结果如下图,其特征值是a=-1/50+1/100*3^(1/2)和-1/50-1/100*3^(1/2)。初值x1=0.1和x2=0.2随时间变化的曲线图locatedi
7、ntheTomb,DongShenJiabang,deferthenextdayfocusedontheassassination.Linping,Zhejiang,1ofwhichliquorwinemasters(WuzhensaidinformationisCarpenter),whogotAfewbayonets,duetomissedfatal,whennightcame初值x1=0.1和x2=0.2的相位曲线图初值x1=0.5和x2=0.6随时间变化的曲线图locatedintheTomb,DongShenJiabang,deferthe
