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Chris L T521 · #13 August 29th, 2009, 11:03 PM

System of Differential Equations (Part III - Matrix Methods (cont.))

In Part II, we ended with two special cases for the eigenvalues of an n x n matrix system. We now devote an entire post
to the last special case.

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Case III: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real but not distinct.

When $\lambda_1,\lambda_2,\dots,\lambda_n$ were distinct (real or complex), then the general solution of
$\mathbf{x}^{\prime}=\mathbf{Ax}$ took on the form

$\mathbf{x}(t)=c_1\mathbf{v}_1e^{\lambda_1t}+c_2\mathbf{v}_2e^{\lambda_2t}+\dots+c_n\mathbf{v}_ne^{\lambda_nt}$ .

We now consider when the characteristic equation $\left|\mathbf{A}-\lambda\mathbf{I}\right|=0$ doesn't have n
distinct root --> the characteristic equation has at least one repeated root.

In that case, we refer to the eigenvalue as having multiplicity. An eigenvalue is of multiplicity k if it is a k-fold
root of the characteristic equation. If $\lambda$ is of multiplicity k, then there is at least one
eigenvector $\mathbf{v}$ associated with it. However, we may not always be able to find k linearly
independent eigenvectors associated with $\lambda$ (this is referred to as a defect of $\lambda$ ,
which will be discussed later). If we can find k linearly independent eigenvectors associated with $\lambda$ ,
we say that $\lambda$ is complete.

Example 28

Find a general solution of the system

$\mathbf{x}^{\prime}=\begin{bmatrix}9 & 4 & 0\\-6 & -1 & 0\\6 & 4 & 3\end{bmatrix}\mathbf{x}$

The characteristic equation of $\mathbf{A}=\begin{bmatrix}9 & 4 & 0\\-6 & -1 & 0\\6 & 4 & 3\end{bmatrix}$ is

$\begin{vmatrix}9-\lambda & 4 & 0\\-6 & -1-\lambda & 0\\6 & 4 & 3-\lambda\end{vmatrix}=(3-\lambda)\begin{vmatrix}9 -\lambda & 4\\ -6 & -1-\lambda\end{vmatrix}$ $=(3-\lambda)(\lambda^2-8\lambda+15)=(5-\lambda)(3-\lambda)^2=0$

Here, we see that $\lambda_1=5$ and $\lambda_2=3$ with multiplicity 2.

Case I: $\lambda=5$

The eigenvector equation is

$\begin{bmatrix}4 & 4 & 0\\-6 & -6 & 0\\6 & 4 & -2\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin {bmatrix}0\\0\\0\end{bmatrix}$

Thus, we have the following system of equations:

$\left\{\begin{aligned}4v_1+4v_2 & = 0\\-6v_1-6v_2&=0\\6v_1+4v_2-2v_3&=0\end{aligned}\right.$

The first two deduce to $v_2=-v_1$

Now, it follows the third equation can be written as $2v_1-2v_3=0\implies v_3=v_1$ . Thus, if we pick $v_1=1$ , we have the eigenvector $\mathbf{v}_1=\begin{bmatrix}1\\-1\\1\end{bmatrix}$ associated with
$\lambda=5$ .

Case II: $\lambda=3$

The eigenvector equation is

$\begin{bmatrix}6 & 4 & 0\\-6 & -6 & 0\\6 & 4 & 0\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin {bmatrix}0\\0\\0\end{bmatrix}$

Thus, we have a nonzero eigenvector iff $6v_1+4v_2=0\implies v_2=-\tfrac{3}{2}v_1$ . Thus, $v_3$
is arbitrary. So if we pick $v_3=1$ , we can let $v_1=v_2=0$ . Thus, $\mathbf{v}_2=\begin {bmatrix}0\\0\\1\end{bmatrix}$ is an associated eigenvector to $\lambda=3$ . However, there is one more eigenvector!

If we pick $v_3=0$ , we can pick $v_1$ and $v_2$ such that we don't have the zero vector. So if we take $v_2=2$ , we see that $v_3=-3$ . Thus, $\mathbf{v}_3=\begin{bmatrix}2\\-3\\0\end{bmatrix}$ is the eigenvector associated with $\lambda=3$ .

Therefore, the general solution is

$\color{red}\boxed{\mathbf{x}(t)=c_1\begin{bmatrix}1\\-1\\1\end{bmatrix}e^{5t}+c_2\begin{bmatrix}0\\0\\1\end{bmatrix}e^{3t}+c_3\begin{bmatrix}2\\-3\\0\end{bmatrix}e^{3t}}$

Remark: With regards to the two eigenvectors for $\lambda=3$ , the fact that $v_2=-\tfrac{3}{2}v_1$ is worth taking note of. The eigenvector can be rewritten as

$\mathbf{v}=\begin{bmatrix}v_1\\-\frac{3}{2}v_1\\v_3\end{bmatrix}=v_3\begin{bmatrix}0\\0\\1\end{bmatrix}+\tfrac{1}{2}v_1\begin{bmatrix}2\\-3\\0\end{bmatrix}=v_3\mathbf{v}_2-\tfrac{1}{2}v_1\mathbf{v}_3$

Thus, we could replace $\mathbf{v}$ for the eigenvector and still get the same answer we did when considering both eigenvectors. This tells us that we don't have to worry about making the right choice -- its just advisible that we pick the simplest one.

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Defective Eigenvalues

We start this section with an example.

Example 29

Consider the coefficient matrix $\mathbf{A}=\begin{bmatrix}1 &-3\\3 & 7\end{bmatrix}$ .

The characteristic equation is $\begin{vmatrix}1-\lambda & -3\\ 3 & 7-\lambda\end{vmatrix}=\lambda^2-8\lambda+16=\left(\lambda-4\right)^2=0$ .

Thus, $\lambda=4$ is an eigenvalue of multiplicity two.

Now, the eigenvector equation is

$\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

Thus, it follows that our system of equation is

$\left\{\begin{aligned}-3v_1-3v_2 & = 0\\3v_1 + 3v_2 & = 0\end{aligned}\right.$

Thus, $v_2=-v_1$ .

Thus the eigenvector is of the form $\mathbf{v}=\begin{bmatrix}v_1\\-v_1\end{bmatrix}=v_1\begin{bmatrix}1\\-1\end{bmatrix}$ .

This implies that all eigenvectors associated with $\lambda=4$ will be a constant multiple of $\begin{bmatrix}1\\-1\end{bmatrix}$ . Therefore, there is only one linearly independent eigenvector associated with $\lambda=4$ , making $\lambda=4$ incomplete.

The eigenvalue in the above example is incomplete, or defective.

Now, if an eigenvalue $\lambda$ has $p<k$ linearly independent eigenvectors, then $d=k-p$ is the number of missing eigenvectors - the defect of the defective eigenvalue $\lambda$ .

In Example 29, the defect would be $d=2-1=1$ .

What we do now is consider a way to solve a system of differential equations given the defect $d=1$ .

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Case IV: $\lambda$ has multiplicity two and is defective.

Suppose that $\lambda$ has one linearly independent eigenvector, implying that $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}$ is the only solution (that we know of) to $\mathbf{x}^{\prime}=\mathbf{Ax}$ .

However, we hope to find a second solution of the form $\mathbf{x}_2(t)=\mathbf{v}_2te^{\lambda t}$ . Substituting it into the system, we have

$\mathbf{v}_2e^{\lambda t}+\lambda\mathbf{v}_2te^{\lambda t}=\mathbf{Av}_2te^{\lambda t}$ .

Since the coefficients of $e^{\lambda t}$ and $te^{\lambda t}$ need to balance, it follows from the above equation that $\mathbf{v}_2=\mathbf{0}$ and consequently, $\mathbf{x}_2(t)\equiv\mathbf{0}$ .

Since that didn't work, let us extend our original idea and replace $\mathbf{v}_2t$ with $\mathbf{v}_1t+\mathbf{v}_2$ . So we suppose now that the second solution will take on the form

$\mathbf{x}_2(t)=\mathbf{v}_1te^{\lambda t}+\mathbf{v_2}e^{\lambda t}$ .

Substituting this into $\mathbf{x}^{\prime}=\mathbf{Ax}$ , we get

$\left(\mathbf{v}_1+\lambda\mathbf{v}_2\right)e^{\lambda t}+\lambda\mathbf{v}_1te^{\lambda t}=\mathbf{Av}_1te^{\lambda t}+\mathbf{Av}_2e^{\lambda t}$

Comparing coefficents of $e^{\lambda t}$ and $te^{\lambda t}$ , we see that

$\mathbf{v}_1+\lambda\mathbf{v}_2=A\mathbf{v}_2\implies \left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v_2}=\mathbf{v_1}$

and

$\lambda\mathbf{v}_1=\mathbf{Av}_1\implies \left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_1=\mathbf{0}$

The second equation confirms that $\mathbf{v}_1$ is an eigenvector for $\lambda$ . Now, it follows that $\mathbf{v}_2$ satisfies the equation

$\left(\mathbf{A}-\lambda\mathbf{I}\right)^2\mathbf{v}_2=\left(\mathbf{A}-\lambda\mathbf{I}\right)\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_2=\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_1=\mathbf{0}$

This tells us that it suffices to find a single solution $\mathbf{v}_2$ to the equation $\left(\mathbf{A}-\lambda\mathbf{I}\right)^2\mathbf{v}_2=\mathbf{0}$ such that $\mathbf{v}_1=\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_2\neq\mathbf{0}$ .

It is always possible to find a solution when the defective eigenvalue $\lambda$ has multiplicity two.

Let us go through an example that illustrates this process.

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Example 30

Find the general solution to the system

$\mathbf{x}^{\prime}=\begin{bmatrix}1 & -3\\ 3 & 7\end{bmatrix}\mathbf{x}$

In example 29, we showed that the characteristic equation produced a defective eigenvalue $\lambda=4$ of multiplicity two.

We now start by calculation $\left(\mathbf{A}-4\mathbf{I}\right)^2$ :

$\left(\mathbf{A}-4\mathbf{I}\right)^2=\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$

Thus, $\left(\mathbf{A}-4\mathbf{I}\right)^2\mathbf{v}_2=\mathbf{0}\implies \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\mathbf{v}_2=\mathbf{0}$ implies that $\mathbf{v}_2$ can be of any (nonzero) form.

So if we take $\mathbf{v}_2=\begin{bmatrix}1\\0\end{bmatrix}$ , then we see that

$\left(\mathbf{A}-4\mathbf{I}\right)\mathbf{v}_2=\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}-3\\3\end{bmatrix}=\mathbf{v}_1$ .

This eigenvector is nonzero, and thus associated with the eigenvalue $\lambda=4$ (Note that this is -3 times the eigenvector we found in example 29).

Therefore, the two solutions to the system are

$\mathbf{x}_1(t)=\mathbf{v}_1e^{4t}=\begin{bmatrix}-3\\3\end{bmatrix}e^{4t}$

and

$\mathbf{x}_2(t)=\left(\mathbf{v}_1t+\mathbf{v}_2\right)e^{4t}=\begin{bmatrix}-3t+1\\3t\end{bmatrix}e^{4t}$

Therefore, the general solution to the system is

$\color{red}\boxed{\mathbf{x}(t)=c_1\begin{bmatrix}-3\\3\end{bmatrix}e^{4t}+c_2\begin{bmatrix}-3t+1\\3t\end{bmatrix}e^{4t}=\begin{bmatrix}-3c_1-3c_2t+c_2\\3c_1+3c_2t\end{bmatrix}e^{4t}}$

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This will conclude the systems of differential equations section of the tutorial.

I will start working on the first of three (or maybe four) posts on Laplace Transforms and their use in IVPs.

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