91影视

Consider a noninvertible $2脳2$matrix A for the system $\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$with two distinct eigenvalues ${位}_1=0$and ${位}_2$is negative.

If $\stackrel{\mathbf{鈫�}}{{\mathbf{v}}_{\mathbf{1}}}$and$\stackrel{\mathbf{鈫�}}{{\mathbf{v}}_{\mathbf{1}}}$are the eigenvector corresponding to the eigenvalues and for the systemlocalid="1659698986072" $\frac{\mathbf{d}\stackrel{\mathbf{鈫�}}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\stackrel{\mathbf{鈫�}}{\mathbf{x}}$then the solution of the system is $\stackrel{\mathbf{鈫�}}{\mathbf{x}}\left(t\right)\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{{\mathbf{位}}_{\mathbf{1}}\mathbf{t}}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{位}}_{\mathbf{2}}\mathbf{t}}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}$.

Using the definition, substitute the value 0 for ${位}_1$in the equation as follows:$\stackrel{鈫�}{x}\left(t\right)=c_1{e}^{{位}_{1}t}{\stackrel{鈫�}{v}}_1+c_2{e}^{{位}_{2}t}{\stackrel{鈫�}{v}}_2$

$$\begin{gathered} \stackrel{鈫�}{x}\left(t\right)=c_1{e}^{{位}_{1}t}{\stackrel{鈫�}{v}}_1+c_2{e}^{{位}_{2}t}{\stackrel{鈫�}{v}}_2 \\ \stackrel{鈫�}{x}\left(t\right)=c_1{e}^{\left(0\right)t}{\stackrel{鈫�}{v}}_1+c_2{e}^{{位}_{2}t}{\stackrel{鈫�}{v}}_2 \\ \stackrel{鈫�}{x}\left(t\right)=c_1{\stackrel{鈫�}{v}}_1+c_2{e}^{{位}_{2}t}{\stackrel{鈫�}{v}}_2 \end{gathered}$$

As ${位}_1=0$and${位}_2$is negative, the trajectory is toward the origin.

Draw the trajectory for the system$\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$with${位}_1=0$and${位}_2$is negative as follows:

Hence, the solution of the system $\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$is $\stackrel{鈫�}{x}\left(t\right)=c_1{\stackrel{鈫�}{v}}_1+c_2{e}^{{位}_{2}t}{\stackrel{鈫�}{v}}_2$where A is a $2脳2$matrix with ${位}_1=0$and ${位}_2$, and the phase portrait of the system $\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$is sketched.