Closed Form Solutions by the Direct Method

$\newenvironment {prompt}{}{} \newcommand {\Sec }[2]{\section {#1}} \newcommand {\trademark }[0]{{R\!\!\!\!\!\bigcirc }} \newcommand {\EXER }[0]{\section *{Exercises}} \newcommand {\CEXER }[0]{} \newcommand {\TEXER }[0]{} \newcommand {\R }[0]{\mbox {$\Bbb {R}$}} \newcommand {\C }[0]{\mbox {$\Bbb {C}$}} \newcommand {\Z }[0]{\mbox {$\Bbb {Z}$}} \newcommand {\N }[0]{\mbox {$\Bbb {N}$}} \newcommand {\D }[0]{\mbox {{\bf D}}} \newcommand {\setmin }[0]{\;\mbox {--}\;} \newcommand {\Matlab }[0]{{M\small {AT\-LAB}} } \newcommand {\Matlabp }[0]{{M\small {AT\-LAB}}} \newcommand {\computer }[0]{\Matlab Instructions} \newcommand {\half }[0]{\mbox {$\frac {1}{2}$}} \newcommand {\compose }[0]{\raisebox {.15ex}{\mbox {{\scriptsize $\circ $}}}} \newcommand {\AND }[0]{\quad \mbox {and}\quad } \newcommand {\vect }[2]{\left (\begin {array}{c} #1_1 \\ \vdots \\ #1_{#2}\end {array}\right )} \newcommand {\mattwo }[4]{\left (\begin {array}{rr} #1 & #2\\ #3 &#4\end {array}\right )} \newcommand {\mattwoc }[4]{\left (\begin {array}{cc} #1 & #2\\ #3 &#4\end {array}\right )} \newcommand {\vectwo }[2]{\left (\begin {array}{r} #1 \\ #2\end {array}\right )} \newcommand {\vectwoc }[2]{\left (\begin {array}{c} #1 \\ #2\end {array}\right )} \newcommand {\inv }[0]{^{-1}} \newcommand {\CC }[0]{{\cal C}} \newcommand {\CCone }[0]{\CC ^1} \newcommand {\Span }[0]{{\rm span}} \newcommand {\rank }[0]{{\rm rank}} \newcommand {\trace }[0]{{\rm tr}} \newcommand {\RE }[0]{{\rm Re}} \newcommand {\IM }[0]{{\rm Im}} \newcommand {\nulls }[0]{{\rm null\;space}} \newcommand {\dps }[0]{\displaystyle } \newcommand {\arraystart }[0]{\renewcommand {\arraystretch }{1.8}} \newcommand {\arrayfinish }[0]{\renewcommand {\arraystretch }{1.2}} \newcommand {\Start }[1]{\vspace {0.08in}\noindent {\bf Section~\ref {#1}}} \newcommand {\exer }[1]{\noindent {\bf \ref {#1}}} \newcommand {\ans }[0]{} \newcommand {\matthree }[9]{\left (\begin {array}{rrr} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9\end {array}\right )} \newcommand {\cvectwo }[2]{\left (\begin {array}{c} #1 \\ #2\end {array}\right )} \newcommand {\cmatthree }[9]{\left (\begin {array}{ccc} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9\end {array}\right )} \newcommand {\vecthree }[3]{\left (\begin {array}{r} #1 \\ #2 \\ #3\end {array}\right )} \newcommand {\cvecthree }[3]{\left (\begin {array}{c} #1 \\ #2 \\ #3\end {array}\right )} \newcommand {\cmattwo }[4]{\left (\begin {array}{cc} #1 & #2\\ #3 &#4\end {array}\right )} \newcommand {\thehelp }[0]{\thesection .\arabic {equation}} \newcommand {\matlabEquation }[0]{\let \oldtheequation \theequation \renewcommand {\theequation }{\oldtheequation *}\begin {equation}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$ $% This file was *autogenerated* from closedFormSolutionsByTheDirectMethod.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %039a8f7e5ad3a1b02e9d1d6ae1b419f4% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

In Section ?? we showed in detail how solutions to planar systems of constant coefficient differential equations with distinct real eigenvalues are found. This method was just reviewed in Section ?? where we saw that the crucial step in solving these systems of differential equations is the step where we find two linearly independent solutions. In this section we discuss how to find these two linearly independent solutions when the eigenvalues of the coefficient matrix are either complex or real and equal.

By finding these two linearly independent solutions we will find both the general solution of the system of differential equations $\dot {X}=CX$ and a method for solving the initial value problem

$\begin{equation} \label{e:eqnA} \begin{array}{rcl} \dps\frac{dX}{dt} & = & CX\\ X(0) & = & X_0. \end{array} \end{equation}$ We assume that $C$ is a $2\times 2$ matrix with eigenvalues $\lambda _1$ and $\lambda _2$ . When needed, we denote the associated eigenvectors by $v_1$ and $v_2$ .

Real Distinct Eigenvalues

We have discussed the case when $\lambda _1\neq \lambda _2\in \R$ on several occasions. For completeness we repeat the result. The general solution is:

$\begin{equation} \label{E:RD2} X(t) = \alpha_1 e^{\lambda_1 t}v_1 + \alpha_2 e^{\lambda_2 t}v_2. \end{equation}$ The initial value problem is solved by finding real numbers $\alpha _1$ and $\alpha _2$ such that $X_0 = \alpha _1 v_1 + \alpha _2 v_2.$ See Section ?? for a detailed discussion with examples.

Complex Conjugate Eigenvalues

Suppose that the eigenvalues of $C$ are complex, that is, suppose that $\lambda _1= \sigma +i\tau$ with $\tau \neq 0$ is an eigenvalue of $C$ with eigenvector $v_1=v+iw$ , where $v,w\in \R ^2$ . We claim that $X_1(t)$ and $X_2(t)$ , where

$\begin{equation} \label{E:CC1} \begin{array}{rcl} X_1(t) & = & e^{\sigma t}(\cos(\tau t)v -\sin(\tau t)w)\\ X_2(t) & = & e^{\sigma t}(\sin(\tau t)v +\cos(\tau t)w), \end{array} \end{equation}$ are solutions to (??) and that the general solution to (??) is: $\begin{equation} \label{E:CC2} X(t) = \alpha_1 X_1(t) + \alpha_2 X_2(t), \end{equation}$ where $\alpha _1, \alpha _2$ are real scalars.

There are several difficulties in deriving (??) and (??); these difficulties are related to using complex numbers as opposed to real numbers. In particular, in the derivation of (??) we need to define the exponential of a complex number, and we begin by discussing this issue.

Euler’s Formula

We find complex exponentials by using Euler’s celebrated formula:

$\begin{equation} \label{E:Euler} e^{i\theta} = \cos\theta + i\sin\theta \end{equation}$ for any real number $\theta$ . A justification of this formula is given in Exercise ??. Euler’s formula allows us to differentiate complex exponentials, obtaining the expected result:

$\begin{eqnarray*} \frac{d}{dt}e^{i\tau t} & = & \frac{d}{dt}(\cos(\tau t) + i\sin(\tau t))\\ & = & \tau(-\sin(\tau t) + i\cos(\tau t)) \\ & = & i\tau(\cos(\tau t) + i\sin(\tau t))\\ & = & i\tau e^{i\tau t}. \end{eqnarray*}$

Euler’s formula also implies that

$\begin{equation} \label{E:ecis} e^{\lambda t} = e^{\sigma t + i\tau t} = e^{\sigma t}e^{i\tau t} = e^{\sigma t}(\cos(\tau t) + i\sin(\tau t)), \end{equation}$ where $\lambda = \sigma +i\tau$ . Most importantly, we note that $\begin{equation} \label{E:eldiff} \frac{d}{dt}e^{\lambda t} = \lambda e^{\lambda t}. \end{equation}$ We use (??) and the product rule for differentiation to verify (??) as follows:

$\begin{eqnarray*} \frac{d}{dt}e^{\lambda t} & = & \frac{d}{dt}\left(e^{\sigma t}e^{i\tau t}\right)\\ & = & \left(\sigma e^{\sigma t}\right)e^{i\tau t} + e^{\sigma t}\left(i\tau e^{i\tau t}\right)\\ & = & (\sigma+i\tau)e^{\sigma t + i\tau t} \\ & = & \lambda e^{\lambda t}. \end{eqnarray*}$

Verification that (??) is the General Solution

A complex vector-valued function $X(t)=X_1(t)+iX_2(t)\in \C ^n$ consists of a real part $X_1(t)\in \R ^n$ and an imaginary part $X_2(t)\in \R ^n$ . For such functions $X(t)$ we define $\dot {X} = \dot {X}_1+i\dot {X}_2$ and $CX = CX_1 + iCX_2.$ To say that $X(t)$ is a solution to $\dot {X}=CX$ means that

$\begin{equation} \label{E:X1X2} \dot{X}_1+i\dot{X}_2 = \dot{X} = CX = CX_1 + iCX_2. \end{equation}$

The complex vector-valued function $X(t)$ is a solution to $\dot {X}=CX$ if and only if the real and imaginary parts are real vector-valued solutions to $\dot {X}=CX$ .

Proof: Equating the real and imaginary parts of (??) implies that $\dot {X}_1 = CX_1 \AND \dot {X}_2 = CX_2.$ $\blacksquare$

It follows from Lemma ?? that finding one complex-valued solution to a linear differential equation provides us with two real-valued solutions. Identity (??) implies that $X(t) = e^{\lambda _1 t}v_1$ is a complex-valued solution to (??). Using Euler’s formula we compute the real and imaginary parts of $X(t)$ , as follows.

$\begin{eqnarray*} X(t) & = & e^{(\sigma+i\tau)t}(v+iw) \\ & = & e^{\sigma t} (\cos(\tau t)+i\sin(\tau t))(v+iw)\\ & = & e^{\sigma t}(\cos(\tau t)v-\sin(\tau t)w)+ ie^{\sigma t}(\sin(\tau t)v+\cos(\tau t)w). \end{eqnarray*}$

Since the real and imaginary parts of $X(t)$ are solutions to $\dot {X}=CX$ , it follows that the real-valued functions $X_1(t)$ and $X_2(t)$ defined in (??) are indeed solutions.

Returning to the case where $C$ is a $2\times 2$ matrix, we see that if $X_1(0)=v$ and $X_2(0)=w$ are linearly independent, then Corollary ?? implies that (??) is the general solution to $\dot {X}=CX$ . The linear independence of $v$ and $w$ is verified using the following lemma.

Let $\lambda _1=\sigma +i\tau$ with $\tau \neq 0$ be a complex eigenvalue of the $2\times 2$ matrix $C$ with eigenvector $v_1=v+iw$ where $v,w\in \R ^2$ . Then $\begin{equation} \label{e:complexcoord} \begin{array}{rcl} Cv & = & \sigma v - \tau w \\ Cw & = & \tau v + \sigma w. \end{array} \end{equation}$ and $v$ and $w$ are linearly independent vectors.

Proof: By assumption $Cv_1=\lambda _1v_1$ , that is, $\begin{equation} \label{E:viw} C (v+iw) = (\sigma+i\tau)(v+iw) = (\sigma v - \tau w) + i(\tau v + \sigma w). \end{equation}$ Equating real and imaginary parts of (??) leads to the system of equations (??). Note that if $w=0$ , then $v\neq 0$ and $\tau v = 0$ . Hence $\tau =0$ , contradicting the assumption that $\tau \neq 0$ . So $w\neq 0$ .
Note also that if $v$ and $w$ are linearly dependent, then $v=\alpha w$ . It then follows from the previous equation that $Cw = (\tau \alpha +\sigma )w.$ Hence $w$ is a real eigenvector; but the eigenvalues of $C$ are not real and $C$ has no real eigenvectors. $\blacksquare$

An Example with Complex Eigenvalues

Consider an example of an initial value problem for a linear system with complex eigenvalues. Let

$\begin{equation} \label{e:complexexample} \frac{dX}{dt} = \mattwo{-1}{2}{-5}{-3} X = CX, \end{equation}$ and $X_0=\vectwo {1}{1}.$

The characteristic polynomial for the matrix $C$ is: $p_C(\lambda ) = \lambda ^2 +4\lambda + 13,$ whose roots are $\lambda _1 = -2+3i$ and $\lambda _2 = -2-3i$ . So $\sigma = -2 \AND \tau = 3.$ An eigenvector corresponding to the eigenvalue $\lambda _1$ is $v_1 = \vectwoc {2}{-1+3i} = \vectwo {2}{-1}+i\vectwo {0}{3}=v+iw.$ It follows from (??) that $\begin {array}{rcl} X_1(t) & = & e^{-2t}(\cos (3t)v -\sin (3t)w)\\ X_2(t) & = & e^{-2t}(\sin (3t)v +\cos (3t)w), \end {array}$ are solutions to (??) and $X=\alpha _1X_1+\alpha _2X_2$ is the general solution to (??). To solve the initial value problem we need to find $\alpha _1,\alpha _2$ such that $X_0 = X(0) = \alpha _1X_1(0) + \alpha _2X_2(0) = \alpha _1 v + \alpha _2 w,$ that is, $\vectwo {1}{1} = \alpha _1\vectwo {2}{-1} + \alpha _2\vectwo {0}{3}.$ Therefore, $\alpha _1 = \frac {1}{2}$ and $\alpha _2=\frac {1}{2}$ and

$\begin{equation} \label{e:complexexampleans} X(t) = e^{-2t}\vectwoc{\cos(3t)+\sin(3t)}{\cos(3t)-2\sin(3t)}. \end{equation}$

Real and Equal Eigenvalues

There are two types of $2\times 2$ matrices that have real and equal eigenvalues — those that are scalar multiples of the identity and those that are not. An example of a $2\times 2$ matrix that has real and equal eigenvalues is

$\begin{equation} \label{E:equalex} A = \mattwoc{\lambda_1}{1}{0}{\lambda_1}, \quad \lambda_1\in\R. \end{equation}$ The characteristic polynomial of $A$ is $p_A(\lambda ) = \lambda ^2 - \trace (A)\lambda + \det (A) = \lambda ^2 -2\lambda _1\lambda + \lambda _1^2 = (\lambda -\lambda _1)^2.$ Thus the eigenvalues of $A$ both equal $\lambda _1$ .

Only One Linearly Independent Eigenvector

An important fact about the matrix $A$ in (??) is that it has only one linearly independent eigenvector. To verify this fact, solve the system of linear equations $Av = \lambda _1v.$ In matrix form this equation is $0 = (A-\lambda _1I_2)v = \mattwo {0}{1}{0}{0}v.$ A quick calculation shows that all solutions are multiples of $v_1=e_1=(1,0)^t$ .

In fact, this observation is valid for any $2\times 2$ matrix that has equal eigenvalues and is not a scalar multiple of the identity, as the next lemma shows.

Let $A$ be a $2\times 2$ matrix. Suppose that $A$ has two linearly independent eigenvectors both with eigenvalue $\lambda _1$ . Then $A=\lambda _1 I_2$ .

Proof: Let $v_1$ and $v_2$ be two linearly independent eigenvectors of $A$ , that is, $Av_j = \lambda _1 v_j$ . It follows from linearity that $Av=\lambda _1 v$ for any linear combination $v=\alpha _1v_1+\alpha _2v_2$ . Since $v_1$ and $v_2$ are linearly independent and $\dim (\R ^2)=2$ , it follows that $\{v_1,v_2\}$ is a basis of $\R ^2$ . Thus, every vector $v\in \R ^2$ is a linear combination of $v_1$ and $v_2$ . Therefore, $A$ is $\lambda _1$ times the identity matrix. $\blacksquare$

Generalized Eigenvectors

Suppose that $C$ has exactly one linearly independent real eigenvector $v_1$ with real eigenvalue $\lambda _1$ . We call $w_1$ a generalized eigenvector of $C$ it satisfies the system of linear equations

$\begin{equation} \label{e:Cw=lw+va} (C-\lambda_1I_2)w_1 = v_1. \end{equation}$

The matrix $A$ in (??) has a generalized eigenvector. To verify this point solve the linear system $(A-\lambda _1I_2)w_1 = \mattwo {0}{1}{0}{0}w_1 = v_1 = \vectwo {1}{0}$ for $w_1=e_2$ . Note that for this matrix $A$ , $v_1=e_1$ and $w_1=e_2$ are linearly independent. The next lemma shows that this observation about generalized eigenvectors is always valid.

Let $C$ be a $2\times 2$ matrix with both eigenvalues equal to $\lambda _1$ and with one linearly independent eigenvector $v_1$ .

(a) Let $w_1$ be a generalized eigenvector of $C$ , then $v_1$ and $w_1$ are linearly independent.

(b) Let $w$ be any vector such that $v_1$ and $w$ are linearly independent. Then $w$ is a nonzero scalar multiple of a generalized eigenvector of $C$ .

Proof

(a) If $v_1$ and $w_1$ were linearly dependent, then $w_1$ would be a multiple of $v_1$ and hence an eigenvector of $C$ . But $C-\lambda _1I_2$ applied to an eigenvector is zero, which is a contradiction. Therefore, $v_1$ and $w_1$ are linearly independent.

(b) Let $w$ be any vector that is linearly independent of the eigenvector $v_1$ . It follows that $\{v_1,w\}$ is a basis for $\R ^2$ ; hence

$\begin{equation} \label{E:Cw1} Cw = \alpha v_1 + \beta w \end{equation}$ for some scalars $\alpha$ and $\beta$ . If $\alpha =0$ , then $w$ is an eigenvector of $C$ , contradicting the assumption that $C$ has only one linearly independent eigenvector. Therefore, $\alpha \neq 0$ .

We claim that $\beta =\lambda _1$ and we prove the claim by showing that $\beta$ is an eigenvalue of $C$ . Hence $\beta$ must equal $\lambda _1$ since both eigenvalues of $C$ equal $\lambda _1$ . To see that $\beta$ is an eigenvalue, define the nonzero vector $u = \alpha v_1 +(\beta -\lambda _1)w$ and compute $Cu = \lambda _1 \alpha v_1 + (\beta -\lambda _1)(\alpha v_1+\beta w) = \beta u.$ So $u$ is an eigenvector of $C$ with eigenvalue $\beta$ . It now follows from (??) that $(C-\lambda _1I_2)w = \alpha v_1.$ Therefore, $w_1=\frac {1}{\alpha }w$ is a generalized eigenvector of $C$ . $\blacksquare$

Independent Solutions to Differential Equations with Equal Eigenvalues

In the equal eigenvalue one eigenvector case, we claim that the general solution to $\dot {X}=CX$ is:

$\begin{equation} \label{e:exp1eva} X(t) = e^{\lambda_1 t}\left(\alpha v_1 +\beta (w_1+tv_1)\right), \end{equation}$ where $v_1$ is an eigenvector of $C$ and $w_1$ is the generalized eigenvector.

Since $v_1$ is an eigenvector of $C$ with eigenvalue $\lambda _1$ , the function $X_1(t)=e^{\lambda _1 t}v_1$ is a solution to $\dot {X}=CX$ . Suppose that we can show that $X_2(t)=e^{\lambda _1 t}(w_1+tv_1)$ is also a solution to $\dot {X}=CX$ . Then (??) is the general solution, since $X_1(0)=v_1$ and $X_2(0)=w_1$ are linearly independent by Lemma ??(a). Apply Theorem ??.

To verify that $X_2(t)$ is a solution (that is, that $\dot {X}_2=CX_2$ ), calculate $\dot {X}_2(t) = \lambda _1 e^{\lambda _1 t}(w_1+tv_1) + e^{\lambda _1 t}v_1= e^{\lambda _1 t}(\lambda _1 w_1 + v_1 +t\lambda v_1)$ and $CX_2(t) = e^{\lambda _1 t}(Cw_1+tCv_1) = e^{\lambda _1 t} ((v_1+\lambda _1 w_1)+t\lambda _1 v_1),$ using (??). Note that $X(0)=\alpha v_1 + \beta w_1$ , so $\alpha$ and $\beta$ are found by solving $X_0= \alpha v_1 + \beta w_1$ .

An Example with Equal Eigenvalues

Consider the system of differential equations

$\begin{equation} \label{e:shearexample} \frac{dX}{dt} = \mattwo{1}{-1}{9}{-5} X \end{equation}$ with initial value $X_0 = \vectwo {2}{3}.$ The characteristic polynomial for the matrix $C=\mattwo {1}{-1}{9}{-5}$ is $p_C(\lambda ) = \lambda ^2 + 4\lambda +4 = (\lambda + 2)^2.$ Thus $\lambda _1=-2$ is an eigenvalue of multiplicity two. Since $C$ is not a multiple of the identity matrix, it must have precisely one linearly independent eigenvector $v_1$ . This eigenvector is found by solving the equation $0 = (C-\lambda _1I_2)v_1 = (C+2I_2)v_1 = \mattwo {3}{-1}{9}{-3}v_1$ for $v_1=\vectwo {1}{3}.$ To find the generalized eigenvector $w_1$ , we solve the system of linear equations $(C-\lambda _1I_2)w_1=(C+2I_2)w_1=\mattwo {3}{-1}{9}{-3}w_1= v_1=\vectwo {1}{3}$ by row reducing the augmented matrix $\left (\begin {array}{rr|r} 3 & -1 & 1\\ 9 & -3 & 3 \end {array}\right )$ to obtain $w_1 = \vectwo {1}{2}.$

We may now apply (??) to find the general solution to (??) $X(t) = e^{-2t}\left (\alpha v_1 +\beta (w_1+tv_1)\right ).$ We solve the initial value problem by solving $\vectwo {2}{3} = X_0 = X(0) = \alpha v_1 +\beta w_1 = \mattwo {1}{1}{3}{2}\vectwo {\alpha }{\beta }$ for $\alpha =-1$ and $\beta =3$ . So the closed form solution to this initial value problem is

$\begin{eqnarray*} X(t) & = & e^{-2t}\left(-v_1 + 3(w_1+tv_1)\right)\\ & = & e^{-2t}\left(-\vectwo{1}{3}+3\vectwoc{1+t}{2+3t}\right)\\ & = & e^{-2t}\vectwo{2+3t}{3+9t}. \end{eqnarray*}$

There is a simpler method for finding this solution — a method that does not require solving for either the eigenvector $v_1$ or the generalized eigenvector $w_1$ that we will discuss later. See Section ??.

Exercises

Justify Euler’s formula (??) as follows. Recall the Taylor series

$\begin{eqnarray*} e^x & = & 1 + x + \frac{1}{2!}x^2 + \cdots + \frac{1}{n!}x^n + \cdots\\ \cos x & = & 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \cdots + (-1)^n \frac{1}{(2n)!}x^{2n} + \cdots \\ \sin x & = & x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 + \cdots + (-1)^n \frac{1}{(2n+1)!}x^{2n+1} + \cdots. \end{eqnarray*}$

Now evaluate the Taylor series $e^{i\theta }$ and separate into real and imaginary parts.

In modern language De Moivre’s formula states that $e^{ni\theta } = \left (e^{i\theta }\right )^n.$ In Exercises ?? – ?? use De Moivre’s formula coupled with Euler’s formula (??) to determine trigonometric identities for the given quantity in terms of $\cos \theta$ , $\sin \theta$ , $\cos \varphi$ , $\sin \varphi$ .

$\cos (\theta +\varphi )$ .

$\sin (3\theta )$ .

In Exercises ?? – ?? compute the general solution for the given system of differential equations.

$\dps \frac {dX}{dt} = \mattwo {-1}{-4}{2}{3} X$ .

$\dps \frac {dX}{dt} = \mattwo {8}{-15}{3}{-4} X$ .

$\dps \frac {dX}{dt} = \mattwo {5}{-1}{1}{3} X$ .

$\dps \frac {dX}{dt} = \mattwo {-4}{4}{-1}{0} X$ .