Applications of Eigenvalues and Eigenvectors

Section 4.4 Applications of Eigenvalues and Eigenvectors

We will develop a simple model of population growth in this section and show an example where eigenvalues and eigenvectors play an important role.

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Consider a population of a single species, perhaps a bacteria, which varies over time. Let \(p_n\) be the population at the \(n\)th time unit (day, month, year, \ldots). A simple model will be that the population at the next time will be proportional to the current population, thus \(p_{n+1} = r p_n\text{.}\)

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If the number of bacteria starting is 50, then \(p_0=50\) and the population doubles every time unit, so \(r=2\text{,}\) then

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\begin{align*} p_1 \amp= 2 p_0 = 100 \\ p_2 \amp = 2p_1 = 200 \\ p_3 \amp = 2p_2 = 400 \end{align*}

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and so on. In general

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\begin{align*} p_1 \amp = r p_0 \\ p_2 \amp = r p_1 = r^2 p_0 \\ p_3 \amp = r p_2 = r^3 p_0 \\ \vdots \amp = \vdots \\ p_n \amp= r^n p_0 \end{align*}

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From this model, the population increases without bound if \(r \gt 1\text{,}\) drops to zero if \(r \lt 1\) and if \(r=1\text{,}\) \(p_n =p_0\text{.}\)

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Subsection 4.4.1 Leslie Population Models

The previous model actually works well for species like single-celled organisms, however, more advanced species will require a more complicated model. Consider a species, like some type of bird, that has two stages of life: juvenile and adult. We will also assume that the population of males and females are equal. The juvenile female cannot reproduce and the adult female can. Let \(p_n\) and \(q_n\) be the population of the juveniles and adults females at stage \(n\text{.}\) A simple model for growth is

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\begin{align*} p_{n+1} \amp = k q_n \\ q_{n+1} \amp = \alpha p_n + \beta q_n \end{align*}

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where at each stage the fraction \(\alpha\) of the juveniles survive to adulthood, and the population of the adults will produce a factor \(k\) juveniles. Finally, the fraction \(\beta\) of the adults survive one time unit. This linear system can be written as the matrix equation

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\begin{equation*} \vec{p}_{n+1} = A \vec{p}_n \end{equation*}

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where

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\begin{align*} \vec{p} \amp= \begin{bmatrix} p \\ q \end{bmatrix} \amp A \amp = \begin{bmatrix} 0 \amp k \\ \alpha \amp \beta \end{bmatrix} \end{align*}

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As a specific example, let’s consider

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\begin{equation*} A = \begin{bmatrix} 0 \amp 2 \\ 0.3 \amp 0.5 \end{bmatrix} \end{equation*}

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The first line means that for every adult female, two juvenile females are produced. The second line means that 0.3 (30 percent) of the juveniles survive to adulthood and at each stage 0.5 (50 percent) of the adults survive to the next stage.

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Let’s see what happens to this population for some initial condition. Let’s say that we have 10 female adults and 0 juvenile adults. Then

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\begin{equation*} \vec{p}_1 = A \vec{p}_0 = \begin{bmatrix} 0 \amp 2 \\ 0.3 \amp 0.5 \end{bmatrix} \begin{bmatrix} 0\\ 10 \end{bmatrix} = \begin{bmatrix} 20 \\ 5 \end{bmatrix} \end{equation*}

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In general,

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\begin{equation*} \vec{p}_n = A^n \vec{p}_0 \end{equation*}

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Table 4.4.1. Table of population values

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year	\(p_n\)	\(q_n\)	\(T_n\)	\(p_n/q_n\)	\(T_{n+1}/T_n\)
0	0	10	10	0
1	20	5	25	4.00	2.50
2	10	8	18	1.18	0.74
3	17	7	24	2.34	1.31
4	14	8	22	1.66	0.96
5	17	8	25	2.00	1.13
\(\vdots\)
10	22	12	34	1.87	1.06
11	24	12	36	1.88	1.07
12	25	13	38	1.88	1.06

We’d like to explain what is going on here. As we will show, the eigenvalues are important. The characteristic equation is

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\begin{equation*} \lambda^2 -0.5 \lambda -0.6 = 0 \end{equation*}

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which has the solutions

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\begin{equation*} \lambda = \frac{0.5 \pm \sqrt{0.25 + 2.4}}{2} \approx 1.06, -0.56 \end{equation*}

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The eigenvectors are

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\begin{align*} \vec{v}_1 \amp = \begin{bmatrix} 1.88 \\ 1 \end{bmatrix} \amp \vec{v}_2 \amp = \begin{bmatrix} -3.54 \\1 \end{bmatrix} \end{align*}

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Assume that \(\lambda_1\) and \(\lambda_2\) are the eigenvalues of \(A\) and \(\vec{v}_1\) and \(\vec{v}_2\) are the eigenvectors. If \(\vec{v}_1\) and \(\vec{v}_2\) are linearly independent, then we can write

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\begin{equation*} \vec{p}_0 = a_1 \vec{v}_1 + a_2 \vec{v}_2 \end{equation*}

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for some real numbers \(a_1\) and \(a_2\text{.}\) Let’s use the matrix equations above to find \(\vec{p}_1\text{,}\)

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\begin{align*} \vec{p}_1 \amp = A \vec{p}_0 = A (a_1 \vec{v}_1 + a_2 \vec{v}_2) \\ \amp = a_1 A \vec{v}_1 + a_2 A \vec{v}_2 \\ \amp = a_1 \lambda_1 \vec{v}_1 + a_2 \lambda_2 \vec{v}_2 \end{align*}

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similarly,

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\begin{align*} \vec{p}_2 \amp = A \vec{p}_1 = A(a_1 \lambda_1 \vec{v}_1 + a_2 \lambda_2 \vec{v}_2)\\ \amp= a_1 \lambda_1 A \vec{v}_1 + a_2 \lambda_2 A \vec{v}_2 \\ \amp = a_1 \lambda_1^2 \vec{v}_1 + a_2 \lambda_2^2 \vec{v}_2 \end{align*}

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Now assume that \(|\lambda_1|>|\lambda_2|\) and factor \(\lambda_1^n\) out from the equation above:

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\begin{equation*} \vec{p}_n = \lambda_1^n \biggl( a_1 \vec{v}_1 + a_2 \frac{\lambda_2^n}{\lambda_1^n} \vec{v}_2 \biggr) \end{equation*}

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as \(n\) increases the ratio \((\lambda_2/\lambda_1)^n\) goes to zero, so

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\begin{equation*} \vec{p}_n \approx a_1 \lambda_1^n \vec{v}_1 \end{equation*}

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Subsection 4.4.2 Second-Order Differential Equations Systems

Consider a mass-spring system:

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Figure 4.4.2. Find a point on a line that minimizes distance
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where \(x_1\) and \(x_2\) are the positions of the masses \(m_1\) and \(m_2\) relative to the unstretched spring positions and \(k_1\) and \(k_2\) are the spring constants of the two springs. The mass spring system above can be modeled using the 2nd order system of differential equations.

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\begin{align*} m_1 x_1 '' \amp = - k_1 x_1 + k_2 (x_2-x_1) \\ m_2 x_2 '' \amp = - k_2 (x_2-x_1) . \end{align*}

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This can be written in matrix form as

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\begin{equation*} M \vec{x}'' = K\vec{x} \end{equation*}

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with

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\begin{align*} M \amp = \begin{bmatrix} m_1 \amp 0 \\ 0 \amp m_2 \end{bmatrix} \amp K \amp =\begin{bmatrix} -(k_1+k_2) \amp k_2 \\ -k_2 \amp k_1 \end{bmatrix} \end{align*}

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where \(M\) is the mass matrix and \(K\) is called the stiffness matrix, which can take other forms depending on the connectedness of the spring system. Since \(M\) is nonsingular, we can find \(M^{-1}\) and write the system above as:

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\begin{equation*} \vec{x}'' = A \vec{x} \end{equation*}

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where \(A = M^{-1} K\text{.}\)

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Subsection 4.4.3 Solutions of \(\vec{x}''= A\vec{x}\)

Let \(v\) be an eigenvector of \(A\text{,}\) with eigenvalue \(\lambda = \alpha^2\text{,}\) then \(\vec{x} =\vec{v}(c_1e^{\alpha t} + c_2 e^{-\alpha t})\) is a solution.

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\begin{align*} \vec{x}'' \amp = \alpha^2 \vec{v} (c_1 e^{\alpha t}+ c_2 e^{-\alpha t}) \\ \amp = A \vec{v}(c_1e^{\alpha t} + c_2 e^{-\alpha t}) \end{align*}

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if \(\lambda \lt 0\text{,}\) then the \(\lambda = - \omega^2\) and the solution is:

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\begin{equation*} \vec{x} = \vec{v} (a \cos \omega t + b \sin \omega t) \end{equation*}

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If \(\lambda = 0\) is an eigenvalue of \(A\) with corresponding eigenvector \(\vec{v}_0\text{,}\) then the part of the solution associate with this is:

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\begin{equation*} \vec{x}(t) = (a_0 + b_0 t) \vec{v}_0 \end{equation*}

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Theorem 4.4.3.

If the \(n \times n\) matrix \(A\) has distinct negative eigenvalues \(-\omega_1^2, -\omega_2^2, \ldots, -\omega_n^2\) with associated real eigenvectors, then a general solution to

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\begin{equation*} \vec{x''} = A x \end{equation*}

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is given by

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\begin{equation*} \vec{x} = \sum_{i=1}^n (a_i \cos \omega_i t + b_i \sin \omega_i t)\vec{v}_i \end{equation*}

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where \(a_i\) and \(b_i\) arbitrary constants. If \(\lambda = 0\) is an eigenvalue of \(A\) with corresponding eigenvector \(\vec{v}_0\text{,}\) then the part of the solution associate with this is:

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\begin{equation*} \vec{x}(t) = (a_0 + b_0 t) \vec{v}_0 \end{equation*}

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Example 4.4.4.

Consider the mass-spring system above with \(m_1=1\text{,}\) \(m_2=2\text{,}\) \(k_1 = 1\text{,}\) and \(k_2 = 2\text{,}\) then the mass matrix and stiffness matrix are:

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\begin{align*} M \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 2 \end{bmatrix} \amp K \amp = \begin{bmatrix} -3 \amp 2 \\ -2 \amp 4 \end{bmatrix} \end{align*}

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resulting in the system:

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\begin{equation*} \vec{x}'' = \begin{bmatrix} -3 \amp 2 \\ -1 \amp 2 \end{bmatrix} \vec{x} \end{equation*}

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The eigenvalues and eigenvectors are:

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\begin{align*} \lambda_1 \amp = -4 \amp \vec{v}_1 \amp = (-2,1)^T \\ \lambda_2 \amp = -1 \amp \vec{v}_2 \amp = (1,1)^T \end{align*}

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Therefore the solution is:

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\begin{equation} \vec{x} = (a_1 \cos 2t + b_1 \sin 2t) \begin{bmatrix} -2 \\ 1 \end{bmatrix} + (a_2 \cos t + b_2 \sin t) \begin{bmatrix} 1 \\1 \end{bmatrix}\tag{4.4.1} \end{equation}

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Next, let’s find the values of the \(a\)’s and \(b\)’s if \(x_1=1\) and \(x_2=2\) when \(t=0\) or

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\begin{equation*} \vec{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{equation*}

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and \(\vec{x}'(0)=\vec{0}\text{.}\) First use the initial value \(\vec{x}(0)\) in ((4.4.1)),

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\begin{equation*} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = a_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + a_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{equation*}

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and this is true when \(a_1=1/3\) and \(a_2=5/3\text{.}\) To find \(b_1\) and \(b_2\text{,}\) differentiate ((4.4.1)),

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\begin{align*} \vec{x}'(t) \amp = (-2a_1 \sin 2t + 2b_1 \cos 2t ) \begin{bmatrix} -2 \\ 1 \end{bmatrix} + (-a_2 \sin t + b_2 \cos t) \begin{bmatrix} 1 1 \end{bmatrix} \end{align*}

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and evaluate it at \(t=0\)

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\begin{equation*} \vec{x}'(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 2b_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + 2b_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{equation*}

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or \(b_1=0\) and \(b_2=0\text{.}\) The solution with the given initial condition,

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\begin{equation*} \vec{x}(t) = \frac{1}{3} \cos 2t \begin{bmatrix} -2 \\ 1 \end{bmatrix} + \frac{5}{3} \cos t \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{equation*}

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The value of \(x_1\) and \(x_2\text{,}\) the position of the two masses are the first and second components. A plot of these two are:

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Figure 4.4.5. Solution of a second-order differential equation.
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