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Chris L T521 · #12 August 24th, 2009, 12:51 AM

I'm in such the mood to post Part II....so I'll do it now. XD

Systems of Differential Equations (Part II - Matrix Methods)

In part one, we covered basic techniques on how to solve first order system of two (or three) differential equations. What we will discuss in this post are techniques used in solving systems with a larger number of equations, and look at some non-linear systems.

Matrix-Valued Functions

A matrix-valued function is of the form

$\mathbf{x}(t)=\begin{bmatrix}x_(t)\\ x_2(t)\\ \vdots\\ x_n(t)\end{bmatrix}$ or $\mathbf{A}(t)=\begin{bmatrix}a_{11}(t) & a_{12}(t) & \dots & a_{1n}(t)\\ a_{21}(t) & a_{22}(t) & \dots & a_{2n}(t)\\ \vdots & \vdots & \phantom{x}& \vdots\\ a_{m1}(t) & a_{m2}(t) & \dots & a_{mn}(t)\end{bmatrix}$

where each entry is a function of $t$ . Now, $\mathbf{x}(t)$ or $\mathbf{A}(t)$ is differentiable if each entry is differentiable. Thus, we define $\frac{\,d\mathbf{A}}{\,dt}=\left[\frac{\,da_{ij}}{\,dt}\right]$

Let us now look into a popular method (which we will spend the rest of the post discussing) -- the Eigenvalue Method of Homogeneous Systems.

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Eigenvalue Method of Homogeneous Systems

Let us consider the following first order system of $n$ differential equations

$\left\{\begin{aligned}x_1^{\prime} &= a_{11}x_1+a_{12}x_2+\dots+a_{1n}x_n\\x_2^{\prime} &= a_{21}x_1+a_{22}x_2+\dots+a_{2n}x_n\\ &\vdots\\ x_n^{\prime} &= a_{n1}x_1+a_{n2}x_2+\dots+a_{nn}x_n\\\end{aligned}\right.$

It suffices to find $n$ linearly independent solution vectors $\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_n$ such that

$\mathbf{x}(t)=c_1\mathbf{x}_1+c_2\mathbf{x}_2+\dots+c_n\mathbf{x}_n$

is a solution to the general system.

We anticipate the solution vectors to be of the form

$\mathbf{x}(t)=\begin{bmatrix}x_1\\x_2\\x_3\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}v_1e^{\lambda t}\\v_2e^{\lambda t}\\v_3e^{\lambda t}\\\vdots\\v_ne^{\lambda t}\end{bmatrix}=\begin{bmatrix}v_1\\v_2\\v_3\\\vdots\\v_n\end{bmatrix}e^{\lambda t}=\mathbf{v}e^{\lambda t}$

where $\lambda,v_1,v_2,v_3,\dots,v_n$ are appropriate scalar constants.

To expand on this, let us rewrite our general system in matrix form:

$\mathbf{x}^{\prime}=\mathbf{Ax}$

Now, let us substitute the anticipated solution into the differential equation to get

$\left(\mathbf{v}e^{\lambda t}\right)^{\prime}=\mathbf{A}\left(\mathbf{v}e^{\lambda t}\right)\implies \lambda\mathbf{v}e^{\lambda t}=\mathbf{Av}e^{\lambda t}$

Cancelling out $e^{\lambda t}$ , we now have

$\lambda\mathbf{v}=\mathbf{Av}$ .

From this, we see that $\mathbf{x}=\mathbf{v}e^{\lambda t}$ will be a nontrivial solution of $\mathbf{x}^{\prime}=\mathbf{Ax}$ given that $\mathbf{v}\neq\mathbf{0}$ and such that $\mathbf{Av}$ is a scalar multiple of $\mathbf{v}$ .

So ... How do we find $\mathbf{v}$ and $\lambda$ ??

First, we rewrite $\lambda\mathbf{v}=\mathbf{Av}$ as $\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0}$ .

Now we recall from linear algebra, this equation has a nontrivial solution iff

$\det\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$ .

Thus, $\lambda$ is referred to the eigenvalue of $\mathbf{A}$ , and $\mathbf{v}$ is the associated eigenvector.

We also define $\det\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$ to be the characteristic equation of $\mathbf{A}$ .

Now, we lay out the steps of the eigenvalue method:

1. First solve the characteristic equation for the eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ of the matrix $\mathbf{A}$ .

2. Attempt to find $n$ linearly independent eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_n$ associated with the eigenvalues.

3. If step 2 is possible (it may not always be!), we have $n$ linearly independent solutions $\mathbf{x}_1=\mathbf{v}_1e^{\lambda_1t}, \mathbf{x}_2=\mathbf{v}_2e^{\lambda_2t},\dots,\mathbf{x}_n=\mathbf{v}_ne^{\lambda_nt}$ . Thus, $\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+\dots+c_n\mathbf{x}_n(t)$ is the general solution of $\mathbf{x}^{\prime}=\mathbf{Ax}$

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Let us now go through two special cases (each illustrated by an example):

Case I: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real and distinct.

Let us start with an example.

Example 26

Find a general solution for the system

$\left\{\begin{aligned}x_1^{\prime} & = 4x_1 + 2x_2\\ x_2^{\prime} &= 3x_1-x_2\end{aligned}\right.$

To solve this, let us rewrite the system in matrix form:

$\mathbf{x}^{\prime}=\begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}\mathbf{x}$

It follows that the characteristic equation is

$\begin{vmatrix}4-\lambda & 2 \\ 3 & -1-\lambda\end{vmatrix}=-\left(4-\lambda\right)\left(1+\lambda\right)-6=\lambda^2-3\lambda-10=0$

Thus, $\lambda^2-3\lambda-10=0\implies\left(\lambda-5\right)\left(\lambda+2\right)=0\implies \lambda_1=-2$ and $\lambda_2=5$ .

Now that we have the eigenvalues, let us try to find the eigenvectors.

Note that the eigenvector equation in this case is

$\begin{bmatrix}4-\lambda & 2 \\ 3 & -1-\lambda\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

Case I: $\lambda=-2$ .

Here, the eigenvector equation becomes

$\begin{bmatrix}6& 2 \\ 3 & 1\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

This gives us the linear system

$\left\{\begin{aligned}6v_1+2v_2 & =0\\ 3v_1 + v_2 &= 0\end{aligned}\right.$ .

It is evident that there are infinitely many solutions. So what now? What we usually do is pick a simple value. So for example, if $v_1=1$ , we have $v_2=-3$ .

Therefore, $\mathbf{v}_1=\begin{bmatrix}1\\-3\end{bmatrix}$ is the eigenvector associated to $\lambda_1=-2$ . Thus, $\mathbf{x}_1(t)=\begin{bmatrix}1\\-3\end{bmatrix}e^{-2t}$ is a solution to the general equation.

Case II: $\lambda=5$ .

Here, the eigenvector equation becomes

$\begin{bmatrix}-1& 2 \\ 3 & -6\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

This gives us the linear system

$\left\{\begin{aligned}-v_1+2v_2 & =0\\ 3v_1 - 6v_2 &= 0\end{aligned}\right.$ .

It is evident that there are infinitely many solutions. So what now? What we usually do is pick a simple value. So for example, if $v_2=1$ , we have $v_1=2$ .

Therefore, $\mathbf{v}_2=\begin{bmatrix}2\\1\end{bmatrix}$ is the eigenvector associated to $\lambda_2=5$ . Thus, $\mathbf{x}_2(t)=\begin{bmatrix}2\\1\end{bmatrix}e^{5t}$ is a solution to the general equation.

It is easy to show that $e^{-2t}$ and $e^{5t}$ are linearly independent (via Wronskian).

Now, by the principle of superposition, it follows that

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

satisfies $\mathbf{x}^{\prime}=\begin{bmatrix}4&2\\3&-1\end{bmatrix}\mathbf{x}$

(Written in scalar form, the solutions would be $\color{red}\boxed{\mathbf{x}_1(t)=c_1e^{-2t}+2c_2e^{5t}}$ and $\color{red}\boxed{\mathbf{x}_2(t)=-3c_1e^{-2t}+c_2e^{5t}}$ )

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Case II: $\lambda_1,\lambda_2,\dots,\lambda_n$ are complex.

Prelim Theory

We are after real valued solutions (it will turn out to be real and imaginary parts of the general solution). When complex eigenvalues pop up, they always appear in conjugate pairs (i.e. $\lambda=p+qi$ and $\bar{\lambda}=p-qi$ ).

Now, if $\mathbf{v}$ is an eigenvector associated with $\lambda$ , such that

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

then taking complex conjugates in the equation gives us

$\left(\mathbf{A}-\bar{\lambda}\mathbf{I}\right)\overline{\mathbf{v}}=\mathbf{0}$

If we take

$\mathbf{v}=\begin{bmatrix}a_1+b_1i\\a_2+b_2i\\\vdots\\a_n+b_ni\end{bmatrix}=\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}+\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}i=\mathbf{a}+\mathbf{b}i$ ,

then $\overline{\mathbf{v}}=\mathbf{a}-\mathbf{b}i$

Therefore, the complex-valued solution associated with $\lambda$ and $\mathbf{v}$ is

$\mathbf{x}(t)=\mathbf{v}e^{\lambda t}=\mathbf{v}e^{\left(p+qi\right)t}=\left(\mathbf{a}+\mathbf{b}i\right)e^{pt}\left[\cos\!\left(qt\right)+\sin\!\left(qt\right)\right]$

Rearranging, we have

$\mathbf{x}(t)=e^{pt}\left[\mathbf{a}\cos\!\left(qt\right)-\mathbf{b}\sin\!\left(qt\right)\right]+ie^{pt}\left[\mathbf{b}\cos\!\left(qt\right)+\mathbf{a}\sin\!\left(qt\right)\right]$ .

Therefore,

$\begin{aligned}\mathbf{x}_1(t)&=\Re\left(\mathbf{x}(t)\right)=e^{pt}\left[\mathbf{a}\cos\!\left(qt\right)-\mathbf{b}\sin\!\left(qt\right)\right]\\\mathbf{x}_2(t)&=\Im\left(\mathbf{x}(t)\right)=e^{pt}\left[\mathbf{b}\cos\!\left(qt\right)+\mathbf{a}\sin\!\left(qt\right)\right]\end{aligned}$

I leave it for you to verify we get the same set of solutions when we check the real and imaginary parts of $\overline{\mathbf{v}}e^{\bar{\lambda}t}$ .

Example 27

Find the general solution of the system

$\begin{aligned}x_1^{\prime} &= 4x_1-3x_2\\ x_2^{\prime}&= 3x_1+4x_2\end{aligned}$

Our coefficient matrix $\mathbf{A}=\begin{bmatrix}4&-3\\3&4\end{bmatrix}$ has the characteristic equation

$\begin{bmatrix}4-\lambda & -3 \\ 3 & 4-\lambda\end{bmatrix}=\left(4-\lambda\right)^2+9=0\implies \lambda=4-3i$ and $\bar{\lambda}=4+3i$ .

Substituting $\lambda=4-3i$ into the eigenvector equation, we have

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

Thus, we have the linear system

$\left\{\begin{aligned}iv_1-v_2 & = 0\\ v_1 + iv_2 & = 0\end{aligned}\right.$

If we take $v_1=1$ , $v_2=i$ . Thus, $\mathbf{v}=\begin{bmatrix}1\\i\end{bmatrix}$ is complex eigenvector associated with $\lambda=4-3i$ .

Now, the corresponding complex solution is

$\mathbf{x}(t)=\begin{bmatrix}1\\i\end{bmatrix}e^{\left(4-3i\right)t}=\begin{bmatrix}1\\i\end{bmatrix}e^{4t}\left(\cos\!\left(3t\right)-i\sin\!\left(3t\right)\right)=e^{4t}\begin{bmatrix}\cos\!\left(3t\right)-i\sin\!\left(3t\right)\\i\cos\!\left(3t\right)+\sin\!\left(3t\right)\end{bmatrix}$

Thus,

$\mathbf{x}_1(t)=\Re\left(\mathbf{x}(t)\right)=e^{4t}\begin{bmatrix}\cos\!\left(3t\right)\\\sin\!\left(3t\right)\end{bmatrix}$ and $\mathbf{x}_2(t)=\Im\left(\mathbf{x}(t)\right)=e^{4t}\begin{bmatrix}-\sin\!\left(3t\right)\\\cos\!\left(3t\right)\end{bmatrix}$

Therefore, a real-valued general solution to $\mathbf{x}^{\prime}=\mathbf{Ax}$ is

$\color{red}\boxed{\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)=e^{4t}\begin{bmatrix}c_1\cos\!\left(3t\right)-c_2\sin\!\left(3t\right)\\c_1\sin\!\left(3t\right)+c_2\cos\!\left(3t\right)\end{bmatrix}}$ .

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I will have to post a Part III for Case III: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real, but not distinct.

I will have that posted sometime tomorrow or the next day.

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