3.4 Eigenvalue Method

In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Suppose we have such a system

$$\vec{x}' = P \vec{x}, \nonumber$$

where $P$ is a constant square matrix. We wish to adapt the method for the single constant coefficient equation by trying the function $e^{\lambda t}$. However, $\vec{x}$ is a vector. So we try $\vec{x}=\vec{v}e^{\lambda t}$, where $\vec{v}$ is an arbitrary constant vector. We plug this $\vec{x}$ into the equation to get

$$\underbrace{\lambda \vec{v} e^{\lambda t}}_{{\vec{x}}'} = \underbrace{P\vec{v} e^{\lambda t}}_{P\vec{x}}. \nonumber$$

We divide by $e^{\lambda t}$ and notice that we are looking for a scalar $\lambda$ and a vector $\vec{x}$ that satisfy the equation

$$\lambda \vec{v}= P \vec{v}. \nonumber$$

To solve this equation we need a little bit more linear algebra, which we now review.

Eigenvalues and Eigenvectors of a Matrix

Let $A$ be a constant square matrix. Suppose there is a scalar $\lambda$ and a nonzero vector $\vec{v}$ such that

$$A \vec{v}= \lambda \vec{v}. \nonumber$$

We then call $\lambda$ an eigenvalue of $A$ and $\vec{x}$ is said to be a corresponding eigenvector.

Note that for an $n \times n$ matrix, the polynomial we get by computing $det(A-\lambda I)$ will be of degree $n$, and hence we will in general have $n$ eigenvalues. Some may be repeated, some may be complex.

To find an eigenvector corresponding to an eigenvalue $\lambda$, we write

$$(A - \lambda I)\vec{v}= \vec{0},\nonumber$$

and solve for a nontrivial (nonzero) vector $\vec{v}$. If $\lambda$ is an eigenvalue, there will be at least one free variable, and so for each distinct eigenvalue $\lambda$, we can always find an eigenvector

3.4.2Eigenvalue Method with Distinct Real Eigenvalues

We have the system of equations

$$\vec{x}'=P\vec{x}. \nonumber$$

We find the eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ of the matrix $P$, and corresponding eigenvectors $\vec{x}_1, \vec{x}_2, \ldots, \vec{x}_n.$ Now we notice that the functions $\vec{v}_1e^{\lambda_1t},\vec{v}_2e^{\lambda_2t}, \ldots, \vec{v}_ne^{\lambda_nt}$ are solutions of the system of equations and hence $$\vec{x}=c_1 \vec{v}_1e^{\lambda_1t}+ c_2 \vec{v}_2e^{\lambda_2t} + \cdots + c_n \vec{v}_ne^{\lambda_nt}$$ is a solution.

The corresponding fundamental matrix solution is

$$X(t)= [\vec{v}_1e^{\lambda_1t}~~~ \vec{v}_2e^{\lambda_2t} ~~~ \cdots ~~~ \vec{v}_ne^{\lambda_nt}]. \nonumber$$

Complex Eigenvalues

A matrix might very well have complex eigenvalues even if all the entries are real. For example, suppose that we have the system

$$\vec{x}'= \left[ \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right] \vec{x}. \nonumber$$

Let us compute the eigenvalues of the matrix $P=\left[ \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right].$

$$det(P - \lambda I) =det \left( \left[ \begin{array}{cc} 1- \lambda & 1 \\ -1 & 1- \lambda \end{array} \right] \right)=(1-\lambda)^2 + 1=\lambda^2-2\lambda+2= 0. \nonumber$$

Thus $\lambda = 1 \pm i.$ The corresponding eigenvectors are also complex. First take $\lambda = 1 - i,$

$$\begin{align}\begin{aligned} (P-(1-i)I)\vec{v}&=\vec{0}, \\ \left[ \begin{array}{cc} i & 1 \\ -1 & i \end{array} \right] \vec{v}&=\vec{0}.\end{aligned}\end{align} \nonumber$$

The equations $iv_1+v_2=0$ and $-v_1+iv_2=0$ are multiples of each other. So we only need to consider one of them. After picking $v_2=1$, for example, we have an eigenvector $\vec{v}=\left[ \begin{array}{c}i\\1 \end{array} \right]$. In similar fashion we find that $\left[ \begin{array}{c}-i\\1 \end{array} \right]$ is an eigenvector corresponding to the eigenvalue $1+i$.

We could write the solution as

$$\vec{x}=c_1\left[ \begin{array}{c} i \\ 1 \end{array} \right]e^{(1-i)t}+c_2\left[ \begin{array}{c} -i \\ 1 \end{array} \right]e^{(1+i)t}= \left[ \begin{array}{c} c_1ie^{(1-i)t}-c_2ie^{(1+i)t} \\ c_1e^{(1-i)t}+c_2e^{(1+i)t} \end{array} \right]. \nonumber$$

We would then need to look for complex values $c_1$ and $c_2$ to solve any initial conditions. It is perhaps not completely clear that we get a real solution. We could use Euler’s formula and do the whole song and dance we did before, but we will not. We will do something a bit smarter first.

We claim that we did not have to look for a second eigenvector (nor for the second eigenvalue). All complex eigenvalues come in pairs (because the matrix $P$ is real).

First a small side note. The real part of a complex number $z$ can be computed as $\frac{z+\bar{z}}{2}$, where the bar above $z$ means $\overline{a+ib}=a-ib$. This operation is called the complex conjugate. If $a$ is a real number, then $\bar{a}=a$. Similarly we can bar whole vectors or matrices by taking the complex conjugate of every entry. If a matrix $P$ is real, then $\bar{P}=P$. We note that $\overline{P \vec{x}}=\bar{P} \bar{\vec{x}}=P \bar{\vec{x}}.$. Also the complex conjugate of $0$ is still $0$, therefore,

$$\vec{0}=\overline{\vec{0}}= \overline{(P- \lambda I) \vec{v }} = (P- \bar{\lambda} I) \bar{\vec{v }}. \nonumber$$

So if $\vec{v}$ is an eigenvector corresponding to the eigenvalue $\lambda=a+ib$, then $\bar{\vec{v}}$ is an eigenvector corresponding to the eigenvalue $\bar{\lambda}=a-ib$.

Suppose that $a+ib$ is a complex eigenvalue of $P$, and $\vec{v}$ is a corresponding eigenvector. Then

$$\vec{x}_1=\vec{v}e^{(a+ib)t} \nonumber$$

is a solution (complex valued) of $\vec{x}'=P\vec{x}$. Euler’s formula shows that $\overline{e^{a+ib}}=e^{a-ib}$, and so

$$\vec{x}_2=\overline{\vec{x_1}}=\bar{\vec{v}}e^{(a+ib)t} \nonumber$$

is also a solution. As $\vec{x_{1}}$ and $\vec{x_{2}}$ are solutions, the function

$$\vec{x}_3 = {\rm{Re~}} \vec{x}_1= {\rm{Re~}} \vec{v}e^{(a+ib)t}=\frac{\vec{x}_1+\overline{\vec{x_1}}}{2}=\frac{\vec{x}_1+\vec{x_2}}{2}=\frac{1}{2}\vec{x_{1}}+\frac{1}{2}\vec{x_{2}} \nonumber$$

is also a solution. And $\vec{x}_3$ is real-valued! Similarly as ${\rm{Im~}}z= \frac{z-\bar{z}}{2i}$ is the imaginary part, we find that

$$\vec{x}_4= {\rm{Im~}} \vec{x}_1= \frac{\vec{x}_1-\overline{\vec{x_1}}}{2i}=\frac{\vec{x}_1-\vec{x_2}}{2i} \nonumber$$

is also a real-valued solution. It turns out that $\vec{x}_3$ and $\vec{x}_4$ are linearly independent. We will use Euler’s formula to separate out the real and imaginary part.

Returning to our problem,

$$\vec{x}_1 = \begin{bmatrix} i \\ 1 \end{bmatrix} e^{(1-i)t} = \begin{bmatrix} i \\ 1 \end{bmatrix} \left( e^t \cos t - i e^t \sin t \right) = \begin{bmatrix} i e^t \cos t + e^t \sin t \\ e^t \cos t - i e^t \sin t \end{bmatrix} = \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} + i \begin{bmatrix} e^t \cos t \\ - e^t \sin t \end{bmatrix}. \nonumber$$

Then

$${\rm{Re~}} \vec{x}_1= \left[ \begin{array}{c} e^t \sin t \\ e^t\cos t \end{array} \right], \quad\text{and}\quad {\rm{Im~}} \vec{x}_1= \left[ \begin{array}{c} e^t \cos t \\ -e^t\sin t \end{array} \right], \nonumber$$

are the two real-valued linearly independent solutions we seek.

The general solution is

$$\vec{x}=c_1 \left[ \begin{array}{c} e^t \sin t \\ e^t\cos t \end{array} \right] + c_2 \left[ \begin{array}{c} e^t \cos t \\ -e^t\sin t \end{array} \right]= \left[ \begin{array}{c} c_1 e^t \sin t +c_2 e^t \cos t \\ c_1 e^t \cos t -c_2 e^t \sin t \end{array} \right].\nonumber$$

This solution is real-valued for real $c_1$ and $c_2$. At this point, we would solve for any initial conditions we may have to find $c_{1}$ and $c_{2}$.

Let us summarize the discussion as a theorem.

For each pair of complex eigenvalues $a+ib$ and $a-ib$, we get two real-valued linearly independent solutions. We then go on to the next eigenvalue, which is either a real eigenvalue or another complex eigenvalue pair. If we have $n$ distinct eigenvalues (real or complex), then we end up with $n$ linearly independent solutions. If we had only two equations $(n=2)$ as in the example above, then once we found two solutions we are finished, and our general solution is

$$\vec{x}=c_{1}\vec{x}_{1}+c_{2}\vec{x}_{2}=c_{1}\left(\text{Re }\vec{v}e^{(a+ib)t}\right) +c_{2}\left(\text{Im }\vec{v}e^{(a+ib)t}\right). \nonumber$$

We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. When we have repeated eigenvalues, matters get a bit more complicated and we will look at that situation in Section 3.7.

Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.