# 英国代考|MATH0008 Applied Mathematics 1应用数学 1

Consider the harmonic oscillator with mass $m=1$, spring constant $k=2$, and damping coefficient $b=2 \sqrt{2}$. The second-order equation that models the motion of the oscillator is
$$\frac{d^{2} y}{d t^{2}}+2 \sqrt{2} \frac{d y}{d t}+2 y=0$$
and its associated system is
$$\frac{d \mathbf{Y}}{d t}=\mathbf{B Y}, \quad \text { where } \mathbf{B}=\left(\begin{array}{cc} 0 & 1 \ -2 & -2 \sqrt{2} \end{array}\right), \quad \mathbf{Y}=\left(\begin{array}{l} y \ v \end{array}\right)$$
and $v=d y / d t$. The eigenvalues of this system are the roots of the characteristic polynomial
$$\operatorname{det}(\mathbf{B}-\lambda \mathbf{I})=(0-\lambda)(-2 \sqrt{2}-\lambda)+2=\lambda^{2}+2 \sqrt{2} \lambda+2$$
Using the quadratic formula, we have
$$\lambda=\frac{-2 \sqrt{2} \pm \sqrt{8-8}}{2}$$

and consequently $-\sqrt{2}$ is a repeated eigenvalue.
Given the eigenvalue $\lambda=-\sqrt{2}$, we determine the eigenvectors $\mathbf{V}=(y, v)$ by solving $\mathbf{B V}=\lambda \mathbf{V}$, which is
$$\left(\begin{array}{cc} 0 & 1 \ -2 & -2 \sqrt{2} \end{array}\right)\left(\begin{array}{l} y \ v \end{array}\right)=-\sqrt{2}\left(\begin{array}{l} y \ v \end{array}\right)$$
So the eigenvectors lie on the line $v=-\sqrt{2} y$. For instance, one convenient eigenvector is $\mathbf{V}=(1,-\sqrt{2})$. The phase portrait for this system is shown in Figure 3.34. All solutions approach the origin as $t$ increases, and all solutions are tangent to the line of eigenvectors as they approach the origin.

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