Higher Order Equations

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The order of an ordinary differential equation is the highest order of differentiation that appears in the equation. For instance, the differential equation $\frac {d^3x}{dt^3} + 2\frac {dx}{dt}+ 5tx = \sin t$ has order three.

In this section we begin the study of solutions to differential equations of the form

$\begin{equation} \label{eq:nconst} a_n\frac{d^nx}{dt^n} + \cdots + a_1\frac{dx}{dt}+a_0x = g(t), \end{equation}$ where $a_j\in \R$ are constants with $a_n\neq 0$ and $g(t)$ is a continuous function. These equations are $n^{th}$ order and linear. The constants $a_j$ are called the coefficients of the differential equation (??). After dividing (??) by $a_n$ , we may assume that $a_n=1$ .

In analogy to systems of linear equations, we call (??) homogeneous if $g(t)=0$ for all $t$ . Otherwise the equation is inhomogeneous. In particular, the principle of superposition is valid for the homogeneous equations.

Reduction of Equations to First Order Systems

In principle, we can find solutions to (??) by finding solutions of a corresponding linear system of first order ODEs with constant coefficients. We have previously discussed this reduction in the special case of second order equations to planar systems in Section ??. Let

$\begin{equation} \label{eq:trick} y_1(t)=x(t),\quad y_2(t)=\frac{dx}{dt}(t),\quad \ldots, \quad y_n(t)=\frac{d^{n-1}x}{dt^{n-1}}(t). \end{equation}$ With these functions, (??) (with $a_n=1$ ) can be rewritten as $\frac {dy_n}{dt} + a_{n-1} y_n + \cdots + a_1 y_2 + a_0 y_1 = g(t).$ Hence, (??) is equivalent to $\begin {array}{rcl} \dps \frac {dy_j}{dt}&=&y_{j+1} \hspace {1.6in} (j=1,\ldots ,n-1)\\ \dps \frac {dy_n}{dt}&=& g(t) - a_0y_1 - \cdots - a_{n-1} y_n, \end {array}$ which is a linear constant coefficient system of ODEs. More explicitly, define the $n\times n$ coefficient matrix $Q$ by $\begin{equation} \label{eq:coefma} Q = \left(\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 & \cdots & 1\\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{array}\right), \end{equation}$ and introduce the vectors $Y(t) = (y_1(t),\ldots ,y_n(t))^t \AND G(t) = (0,\ldots ,0,g(t))^t.$ Then we arrive at the system of ODEs $\begin{equation} \label{eq:highrw} \frac{dY}{dt} = QY+G(t). \end{equation}$ Let us summarize.

The function $x(t)$ is a solution to the order $n$ ODE (??) if and only if the vector of functions $Y(t)=(y_1(t),\ldots ,y_n(t))^t$ is a solution of the first order system (??), where $y_1(t) = x(t) \AND y_{j+1}(t)=\dps \frac {d^jx}{dt^j}(t)$ for $j=1,2,\ldots ,n-1$ .

The Homogeneous Equation

Now consider specifically the case when (??) is homogeneous. As before, we assume that $a_n=1$ , and solve

$\begin{equation} \label{eq:nconsthom} \frac{d^nx}{dt^n} + a_{n-1}\frac{d^{n-1}x}{dt^{n-1}} + \cdots + a_1\frac{dx}{dt}+a_0x = 0. \end{equation}$ Proposition ?? implies that each solution of (??) is the first component in a solution to the $n\times n$ system $\begin{equation} \label{eq:highrwh} \frac{dY}{dt} = QY, \end{equation}$ where $Q$ is the matrix defined in (??). From the discussion in Section ??, it follows that we can find all solutions to (??) if we know the eigenvalues and the Jordan block structure of $Q$ . In particular, the first component of solutions to (??) is always just a linear combination of the functions $t^je^{\lambda t}$ where $\lambda$ is an eigenvalue of $Q$ . In the abstract, the only question that remains is which powers $t^j$ can actually occur. Certainly, $j$ must be less than the multiplicity of the eigenvalue $\lambda$ . It also follows from Proposition ?? that there are precisely $n$ linearly independent functions that are solutions to (??).

We can use this abstract information to find a shortcut for solving (??). Suppose that $x(t)=e^{\lambda t}$ . Then we can compute the left hand side of (??) with this $x(t)$ obtaining:

$\begin{equation} \label{e:elam} e^{\lambda t}\left(\lambda^n + a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda +a_0\right). \end{equation}$ Thus, $x(t)$ is a solution to (??) precisely when $\lambda$ is a root of $\begin{equation} \label{E:charpoly} p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda +a_0. \end{equation}$ We call $p(\lambda )$ the characteristic polynomial of the $n^{th}$ order equation (??). The roots of the characteristic polynomial are called eigenvalues. This terminology is explained in Section ??.

It follows immediately that there is one solution associated to each eigenvalue of the characteristic polynomial $p(\lambda )$ in (??). If the eigenvalues $\lambda =\sigma \pm i\tau$ are complex, then there are two solutions to (??), namely, $e^{\sigma t}\cos (\tau t)$ and $e^{\sigma t}\sin (\tau t)$ .

Let $\lambda _1,\ldots ,\lambda _k$ be the distinct eigenvalues of the characteristic polynomial $p(\lambda )$ of (??) with multiplicities $m_1,\ldots ,m_k$ . Then $\{t^je^{\lambda _\ell }: 1\leq \ell \leq k, 0\leq j<m_\ell \}$ is a basis of solutions to (??).

Proof: The previous calculations show that if $\lambda$ is an eigenvalue of $p(\lambda )$ , then $e^{\lambda t}$ is a solution to (??). The difficulty in proving this theorem is in showing that $t^je^{\lambda t}$ is also a solution to (??) for any $j<m$ , where $m$ is the multiplicity of the eigenvalue $\lambda$ . We prove this theorem in two steps. First, we show in Lemma ?? that the eigenvalues of $p$ are just the eigenvalues of the characteristic polynomial of the coefficient matrix $Q$ in (??). Second, we show in Lemma ?? that every eigenvalue of $Q$ is associated to precisely one Jordan block and that every eigenvector of $Q$ has a nonvanishing first component. Then the result follows from Theorem ??. $\blacksquare$

The characteristic polynomial of the matrix $Q$ in (??) is $p_Q(\lambda ) = (-1)^np(\lambda ).$

Proof: We prove the lemma by induction. For $n=1$ the matrix $Q$ is just $(-a_0)$ . Hence $p_Q(\lambda )=\det (Q-\lambda I_1) = -a_0-\lambda = -(\lambda +a_0)$ as desired.
Now suppose that the result is valid for all matrices of order $n-1$ . Then we may expand the determinant by cofactors along the $1^{st}$ column (see (??) in Chapter ??) and obtain $p_Q(\lambda ) = -\lambda \det \left (\begin {array}{cccc} -\lambda & 1 & \cdots & 0\\ 0 & -\lambda & \cdots & 0\\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & 1\\ -a_1 & -a_2 & \cdots & -\lambda -a_{n-1} \end {array}\right ) +(-1)^na_0\det \left (\begin {array}{ccccc} 1 & 0 & \cdots & 0 & 0\\ -\lambda & 1 & \cdots & 0 & 0\\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0\\ 0 & 0 & \cdots & -\lambda & 1\\ \end {array}\right ).$ By induction, the first determinant is $(-1)^{n-1}(\lambda ^{n-1}+a_{n-1}\lambda ^{n-2}+ \cdots +a_2\lambda +a_1),$ and by direct calculation the second determinant is $1$ . Therefore,
$\begin{eqnarray*} p_Q(\lambda) & = & -\lambda (-1)^{n-1}(\lambda^{n-1}+a_{n-1}\lambda^{n-2}+ \cdots+a_2\lambda+a_1) +(-1)^n a_0 \\ & = & (-1)^n(\lambda^n+a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda+a_0). \end{eqnarray*}$
This proves the lemma. $\blacksquare$

Let $\lambda$ be an eigenvalue of $Q$ . Then, up to scaling, there is a unique eigenvector $v$ corresponding to the eigenvalue $\lambda$ given by $v = \left (\begin {array}{c} 1\\ \lambda \\ \lambda ^2\\ \vdots \\ \lambda ^{n-1} \end {array}\right ).$ In particular, the geometric multiplicity of each eigenvalue of $Q$ is one, and there is exactly one Jordan block of $Q$ associated to the eigenvalue $\lambda$ .

Proof: Let $\lambda$ be an eigenvalue of $Q$ and let $v=(v_1,\ldots ,v_n)^t$ be a corresponding eigenvector. Writing $Qv = \lambda v$ in coordinates we obtain $\left (\begin {array}{c} v_2\\ v_3\\ \vdots \\ v_n\\ -\sum _{j=0}^{n-1} a_j v_{j+1} \end {array}\right )= \left (\begin {array}{c} \lambda v_1\\ \lambda v_2\\ \vdots \\ \lambda v_{n-1}\\ \lambda v_n \end {array}\right ).$ Note that if $v_1=0$ then $v_2=v_3=\cdots =v_n=0$ , which contradicts the fact that $v$ is an eigenvector. Hence $v_1\neq 0$ and we may rescale $v$ so that $v_1=1$ . It follows that $\begin {array}{rclrl} v_2 & = & \lambda v_1 & = & \lambda \\ v_3 & = & \lambda v_2 & = & \lambda ^2 \\ & & \vdots & & \\ v_n & = & \lambda v_{n-1} & = & \lambda ^{n-1}. \end {array}$ as desired. $\blacksquare$

We restate the results in Theorem ?? in real form.

Let $\lambda$ be an eigenvalue of (??) with multiplicity $m$ .

If $\lambda$ is real, then $t^je^{\lambda t}$ is a solution of (??) for $0\leq j < m$ .
If $\lambda =\sigma +i\tau$ is complex, then $t^je^{\sigma t}\cos (\tau t)$ and $t^je^{\sigma t}\sin (\tau t)$ are solutions of (??) for $0\leq j < m$ .
Every solution of (??) is a linear combination of the functions constructed in (a) and (b).

Examples of Fourth Order

Consider the linear homogeneous ordinary differential equation

$\begin{equation} \label{eq:soex3} \frac{d^4x}{dt^4} + 2\frac{d^2x}{dt^2} + x= 0. \end{equation}$ This type of equation arises if one considers small deformations of an elastic beam under pressure where the pressure is acting at both ends of the beam in the direction of the beam. In this case $x(t)$ represents the resulting deformation of the beam at the spatial point $t$ .

First, we find all eigenvalues of the characteristic polynomial by solving $\lambda ^4 + 2\lambda ^2 + 1 = (\lambda ^2+1)^2 = 0.$ This fourth order polynomial has two roots namely $\pm i$ , each of algebraic multiplicity two. By Theorem ??, the general solution of (??) has the form $x(t) = c_1 \cos (t)+ c_2 \sin (t) +c_3 t\cos (t)+ c_4 t\sin (t).$ For instance, if we consider a beam which is hinged at both of its ends — say at $t=0$ and $t=\pi$ — then the corresponding solution has to satisfy $x(0)=x(\pi )=0 \AND \frac {d^2x}{dt^2}(0)=\frac {d^2x}{dt^2}(\pi )=0.$ These conditions on the solution $x(t)$ lead to a homogeneous system of four linear equations in the four unknowns $c_1,c_2,c_3,c_4$ . This system can easily be solved by hand, and it follows that $c_1$ , $c_3$ and $c_4$ have to vanish, whereas $c_2$ is arbitrary. Thus, under these conditions a small deformation of the beam is given by $x(t)=c_2 \sin (t)$ .

In general, however, we can not expect to find the roots of a fourth order characteristic polynomial by inspection. For example, suppose we add a first order term to (??) obtaining the differential equation

$\begin{equation} \label{eq:soex3a} \frac{d^4x}{dt^4} + 2\frac{d^2x}{dt^2} - \frac{dx}{dt} + x= 0. \end{equation}$ The characteristic polynomial of (??) is $p(\lambda ) = \lambda ^4 + 2\lambda ^2 - \lambda + 1 .$ We can use MATLAB to find the roots of $p(\lambda )$ , as follows. Polynomials are entered into MATLAB by entering the coefficients in an array from highest order to lowest. So we enter $p(\lambda )$ into MATLAB by typing

p = [1 0 2 -1 1];

To find the roots of $p(\lambda )$ just type the command roots(p). MATLAB responds with

ans =
  -0.3438 + 1.3584i
  -0.3438 - 1.3584i
   0.3438 + 0.6254i
   0.3438 - 0.6254i

By Theorem ??, the general solution of (??) has the form $\begin {array}{rcl} x(t) & = & (c_1 \cos (1.3584t)+ c_2 \sin (1.3584t))e^{-0.3438t} \\ & & + (c_3 \cos (0.6254t)+ c_4 \sin (0.6254t))e^{0.3438t}.\end {array}$ Note that the typical solution to the fourth order differential equation (??) contains sine and cosine terms with two different frequencies. This is not surprising since the characteristic polynomial has two distinct complex conjugate pairs of roots. See Section ??.

The Initial Value Problem for Higher Order Equations

For motivation, recall the introductory mechanical examples. The motion of mass points are described by second order ODEs. To determine a specific motion we need to specify the position of the point together with its velocity at a certain time $t_0$ . In other words, to obtain existence and uniqueness of solutions to second order equations, both $x(t_0)\AND \frac {dx}{dt}(t_0)$ have to be specified.

The initial value problem $\begin{equation} \begin{array}{c} \dps \frac{d^nx}{dt^n} + a_{n-1}\frac{d^{n-1}x}{dt^{n-1}} + \cdots + a_1\frac{dx}{dt}+a_0x = 0\\ \dps \frac{d^{n-1}x}{dt^{n-1}}(t_0)=x_{n-1},\ldots,\frac{dx}{dt}(t_0)=x_1,\quad x(t_0)=x_0. \end{array} \end{equation}$ has a unique solution $x(t)$ .

Proof: The result is an immediate consequence of Proposition ??, since the corresponding solution of the system of linear differential equations exists and is unique (see Theorem ?? in Chapter ??). $\blacksquare$

Inhomogeneous Higher Order Equations

Linearity implies that to find the general solution the inhomogeneous equation (??) we need to find one solution to that equation and all solutions to the corresponding homogeneous equation (??). Sections ?? and ?? are devoted to finding specific solutions to inhomogeneous equations of various types using the method of undetermined coefficients. Our study of inhomogeneous equations will continue in Chapter ?? where we use Laplace transforms to study discontinuous forcing and in Section ?? where we discuss the method of reduction of order.

Exercises

In Exercises ?? – ?? rewrite the given higher order equation as a first order system.

$\ddot {x}+5\dot {x}+x = 0$

$\ddot {x}-10x = 0$

$\frac {d^3x}{dt^3}-4\frac {d^2x}{dt^2}-10x = 0$

$\frac {d^4x}{dt^4}+2\frac {d^2x}{dt^2}-3\frac {dx}{dt}-2x = 0$

In Exercises ?? – ?? find the general solution to the given homogeneous linear differential equation.

$\ddot {x}-3\dot {x}-4x = 0$

$\ddot {x}+2\dot {x}+2x = 0$

$\ddot {x}-6\dot {x}+9x = 0$

$\frac {d^3x}{dt^3} -3\frac {d^2x}{dt^2}+3\frac {dx}{dt}-x=0$

$\frac {d^3x}{dt^3}+4\frac {dx}{dt} = 0$

Find a solution to the differential equation $\ddot {x} -3\dot {x} + 2x = 0$ satisfying $x(0)=0$ and $x(\ln 2) = 4$ .

In Exercises ?? – ??, use MATLAB to find the roots of the characteristic polynomial for the given homogeneous linear differential equation, and then find the general solution to that equation.

$\frac {d^3x}{dt^3} -6\frac {d^2x}{dt^2}+11\frac {dx}{dt}-6x=0$

$\frac {d^3x}{dt^3} + \frac {d^2x}{dt^2}+ 10x=0$

$\frac {d^3x}{dt^3} +\frac {d^2x}{dt^2}-4\frac {dx}{dt}-4x=0$

$\frac {d^4x}{dt^4}-2\frac {d^3x}{dt^3}+\frac {d^2x}{dt^2}+8\frac {dx}{dt}-20x=0$

$\frac {d^4x}{dt^4}-3\frac {d^2x}{dt^2}-4x=0$

$\frac {d^4x}{dt^4}-4\frac {d^3x}{dt^3}+14\frac {d^2x}{dt^2}-4\frac {dx}{dt}+13x=0$