Higher Order Equations

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Using the same trick as in the case of equations with constant coefficients we can rewrite a higher order linear equation with nonconstant coefficients as a first order system of linear equations. This way we can also use the method of variation of parameters for the solution of higher order equations. In fact, consider a general linear differential equation of order $n$

$\begin{equation} \label{E:noncc} \frac{d^nx}{dt^n} + a_{n-1}(t)\frac{d^{n-1}x}{dt^{n-1}} +\cdots + a_1(t)\frac{dx}{dt}+a_0(t)x = g(t). \end{equation}$ As in Section ?? we define the functions $x_1(t),\ldots ,x_n(t)$ by $x_1(t)=x(t),\quad x_2(t)=\frac {dx}{dt}(t),\quad \ldots , \quad x_n(t)=\frac {d^{n-1}x}{dt^{n-1}}(t)$ and see that (??) is equivalent to $\begin {array}{rcl} \dps \frac {dx_j}{dt}&=&x_{j+1} \hspace {1.6in} (j=1,\ldots ,n-1)\\ \dps \frac {dx_n}{dt}&=& g(t) - a_0(t)x_1 - \cdots - a_{n-1}(t) x_n. \end {array}$ Introducing the vectors $X(t) = (x_1(t),\ldots ,x_n(t))^t \AND G(t) = (0,\ldots ,0,g(t))^t$ we conclude that (??) is equivalent to the first order system $\frac {dX}{dt} = A(t)X+G(t),$ where $A(t)$ is the matrix $A(t) = \left (\begin {array}{ccccc} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1\\ -a_0(t) & -a_1(t) & -a_2(t) & \cdots & -a_{n-1}(t) \end {array}\right ).$ In particular, $x_1(t)$ is a solution of (??) if $X(t)=(x_1(t),\ldots ,x_n(t))^t$ is a solution of the system $\dot X = A(t)X+G(t)$ which can in principle be solved by variation of parameters.

An Example of Second Order

As an example we find a solution to the initial value problem

$\begin{equation} \label{E:2ndorderex} \begin{array}{c} \dps \ddot x - \frac{2}{t} \dot x + \frac{2}{t^2} x = t,\\ \dps x(1) = 0,\quad \dot x(1) = 1. \end{array} \end{equation}$ This second order differential equation is equivalent to the system $\begin{equation} \label{E:2ndexam} \begin{array}{rcl} \dps \dot{X} & = & A(t) X + G(t) \\ \dps X(1) & = & (0,1)^t, \end{array} \end{equation}$ where $A(t) = \mattwo {0}{1}{-\frac {2}{t^2}}{\frac {2}{t}}\AND G(t) = \vectwo {0}{t}.$

A basis of solutions to the homogeneous equation $\dot X = A(t)X$ is $X_1(t) = \vectwo {t}{1} \AND X_2(t) = \vectwo {t^2}{2t}.$

The inverse of $Y(t)$ is given by $Y(t)^{-1} = \frac {1}{t^2} \mattwo {2t}{-t^2}{-1}{t}.$ Hence we obtain $D(t) = Y(t)^{-1} G(t) = \frac {1}{t^2} \mattwo {2t}{-t^2}{-1}{t}\vectwo {0}{t} = \vectwo {-t}{1}$

Note that $X(1) = \vectwo {0}{1} = c_1(1)\vectwo {1}{1} + c_2(1)\vectwo {1}{2}.$ So $c_1(1)=-1$ and $c_2(1)=1$ . Now solve the differential equations $\begin {array}{rclcl} \dot {c}_1 & = & d_1(t) & = & -t \\ \dot {c}_2 & = & d_2(t) & = & 1, \end {array}$ with these initial conditions obtaining

$\begin{eqnarray*} c_1(t) & = & -\frac{1}{2}(t^2+1) \\ c_2(t) & = & t. \end{eqnarray*}$

Theorem ?? implies that the solution to (??) is $X(t) = c_1(t)X_1(t) + c_2(t)X_2(t) = \left (\begin {array}{c} \frac {t}{2}(t^2-1)\\ \frac {1}{2}(3t^2-1) \end {array}\right ),$ and hence the solution of (??) is given by $x(t) = \frac {t}{2}(t^2-1).$

An Example of an Electrical Circuit

It is rare, however, that variation of parameters can actually be used to find a closed form solution to an inhomogeneous higher order linear differential equations with variable coefficients. In such instances it is best to resort to numerical integration of the first order system obtained from the higher order equation.

We consider here the temporal behavior of an RLC-circuit introduced in Section ??. This circuit is described by the differential equation $\ddot x + \frac {R}{L}\dot x + \frac {1}{CL}x = \frac {1}{CL}V(t),$ where $x(t)$ is the voltage drop at the capacitor, $R$ is the resistance, $L$ is the inductance, $C$ is the capacitance, and a voltage source is producing a time dependent voltage $V(t)$ , see (??) in Chapter ??. Now we assume that the circuit additionally contains a microphone that has a time dependent resistance $R_{mic}(t)$ of the form $R_{mic}(t) = R_0 + R_1\cos (\omega t).$ This corresponds to the fact that a periodic signal of frequency $\frac {2\pi }{\omega }$ is entering the microphone. Then the differential equation for the electrical circuit becomes

$\begin{equation} \label{E:RCLMcir} \ddot x + \frac{R+R_{mic}(t)}{L}\dot x + \frac{1}{CL}x = \frac{1}{CL}V(t). \end{equation}$ For simplicity we set $R=0 \AND C=L=R_0 = R_1 = \omega = 1$ and assume that the voltage source is producing the constant voltage $V(t) = 1.$ Thus we obtain the following linear differential equation of second order $\ddot x + (1+\cos t)\dot x + x = 1,$ which is equivalent to the system $\dot X= A(t)X + G(t)$ , where $\begin{equation} \label{E:RCLM} A(t) = \left( \begin{array}{cc}0 & 1\\-1 & -(1+\cos t)\end{array}\right) \AND G(t) = \vectwo{0}{1}. \end{matlabEquation} It is not obvious how to find the general solution\index{general solution} of this equation by hand. In fact, it is not even clear how to construct solutions of the homogeneous equation. Hence we have to rely on numerical methods and compute solutions using {\tt ode45}\index{\computer!ode45} in \Matlabp. The right hand side in \eqref{E:RCLM} is stored in the function m-file {\tt f17\_4\_5.m}. For completeness, \begin{verbatim} function f = f17_4_5(t,x) A = [0 1; -1 -(1+cos(t))]; f = A*x + [0; 1]; \end{verbatim} We compute the solution of \eqref{E:RCLM} with initial condition $X(0)=(1,1)^t$ for $t\in[0,30]$ by typing \begin{verbatim} [t,x] = ode45('f17_4_5',[0 30],[1,1]'); \end{verbatim} The result of this computation is shown in Figure~\ref{Fig:micro1}. \begin{figure}[htb] \centerline{% \psfig{file=../figures/microx1.eps,width=3.0in} \psfig{file=../figures/microx2.eps,width=3.0in}} \caption{The two components of the solution of \protect\eqref{E:RCLM} for $t\in[0,30]$ with initial condition $X(0)=(1,1)^t$.} \label{Fig:micro1} \end{figure} It can be observed that $x_1(t)$ tends to $1$ and $x_2(t)$ tends to $0$ for increasing time $t$. In other words, the solution approaches the time independent solution $X(t)=(1,0)^t$ of \eqref{E:RCLM}. \subsection*{Reduction of Order} \index{reduction of order} In order to find solutions to a higher order linear differential equation we have rewritten the higher order equation as a first order system of linear equations, and then we applied variation of parameters to that system. There is another method known as {\em reduction of order\/} that uses the same idea for the construction principle of the solution --- but this idea is applied directly to the higher order equation. We remark that {\em reduction of order\/} can sometimes work even when a method like undetermined coefficients fails, since its applicability does not depend on the specific type of the inhomogeneity. However, as in the case of variation of parameters, the practical use of this technique is very limited. The reason is that the integrations that are involved in the computations can rarely be performed by hand. We illustrate this method by the second order linear differential equation \begin{equation} \label{e:inhom2} \ddot{x} + a_1(t)\dot{x} + a_0(t)x = g(t). \end{equation}$ Roughly speaking, reduction of order is a way to reduce the second order inhomogeneous equation to an inhomogeneous first order equation.

Suppose that $x_h(t)$ is a solution of the homogeneous second order equation $\ddot {x} + a_1(t)\dot {x} + a_0(t)x = 0$ . In analogy to the method of variation of parameters, we try to find a solution $x(t)$ of the inhomogeneous equation (??) of the form $x(t) = c(t) x_h(t),$ where $c(t)$ is a smooth function. To determine $c(t)$ we substitute $x(t)$ into the inhomogeneous equation (??). Before proceeding, we compute the derivatives of $x(t)$ :

$\begin{eqnarray*} \dot x & = & \dot c x_h + c \dot{x}_h,\\ \ddot x & = & \ddot c x_h + 2 \dot c \dot{x}_h + c \ddot{x}_h. \end{eqnarray*}$

Next, we substitute $\dot {x}$ and $\ddot {x}$ into (??) and find

$\begin{eqnarray*} && \ddot c x_h +2\dot c \dot{x}_h + c \ddot{x}_h + a_1(t)\left(\dot c x_h + c \dot{x}_h \right)+ a_0(t) c x_h =\\ && x_h \left(\ddot c + \dot c(2 \frac{\dot{x}_h}{x_h} + a_1(t))\right) + c(\ddot{x}_h +a_1(t) \dot{x}_h +a_0(t) x_h) = g(t). \end{eqnarray*}$

Since $x_h$ is a solution of the homogeneous equation we have that $\ddot {x}_h +a_1(t) \dot {x}_h +a_0(t)x_h=0$ . After dividing by $x_h$ the function $c(t)$ satisfies $\ddot c + \dot c\left ( 2\frac {\dot {x}_h}{x_h} +a_1(t)\right ) = \frac {g(t)}{x_h}.$ Introducing $y(t)=\dot c(t)$ , we arrive at the differential equation $\frac {dy}{dt} = -y\left ( 2\frac {\dot {x}_h}{x_h} +a_1(t)\right ) + \frac {g(t)}{x_h}$ which is linear and of first order. If we can find a solution $y(t)$ of this equation then we can compute $c(t)$ by integration and we have constructed a particular solution $x(t)=c(t)x_h(t)$ of the inhomogeneous equation (??). Thus

Reduction of Order Consider the inhomogeneous linear ODE of second order (??).

Let $x_h(t)$ be a nonzero solution of the homogeneous equation $\ddot {x} + a_1(t)\dot {x} + a_0(t)x = 0.$
Let $c(t)$ be a function such that its derivative $\dot c(t)$ is a solution of the linear differential equation $\begin{equation} \label{eq:ro} \frac{dy}{dt} = -\left(\frac{2\dot x_h(t)+a_1(t) x_h(t)}{x_h(t)}\right)y +\frac{g(t)}{x_h(t)}. \end{equation}$

Then $x_p(t)=c(t) x_h(t)$ is a particular solution of the second order inhomogeneous equation (??).

Observe that (??) is again a scalar first order differential equation that can in principle be solved by variation of parameters (see Theorem ??). Indeed, to see this one has to substitute $-\left (\frac {2\dot x_h(t)+a_1(t) x_h(t)}{x_h(t)}\right ) \AND \frac {g(t)}{x_h(t)}.$ for $a(t) \AND g(t)$ in (??).

A Specific Case: Constant Coefficients and Real Eigenvalues

As a special case of Theorem ??, we suppose that the coefficients $a_0$ and $a_1$ do not depend on $t$ and that $\lambda$ is a real root of the characteristic polynomial of the homogeneous equation $\ddot {x} + a_1\dot {x} + a_0x = 0$ . Then we can choose $x_h(t)=e^{\lambda t}$ and (??) becomes

$\begin{equation} \label{eq:redeqreal} \frac{dy}{dt} = -(2\lambda +a_1) y + e^{-\lambda t}g(t). \end{equation}$ If $\dot c(t)$ is a solution to (??), then $x(t)=c(t) e^{\lambda t}$ is a particular solution of the second order inhomogeneous equation (??).

As an example, apply reduction of order to find a solution to the inhomogeneous ODE

$\begin{equation} \label{e:inhomex1} \ddot{x} - 2\dot{x} + x = \frac{e^t}{t^3}. \end{equation}$ Observe that this equation cannot be solved by the method of undetermined coefficients since the right hand side in (??) is not a solution of a homogeneous linear differential equation with constant coefficients.

The characteristic polynomial of (??) has the double eigenvalue $\lambda = 1$ . Since $a_1=-2$ and $g(t) = e^t/t^3$ , (??) takes the form $\frac {dy}{dt} = -(2\lambda -2) y + e^{-\lambda t}\left ( \frac {e^t}{t^3}\right ) = \frac {1}{t^3}.$ Hence $y(t) = -\frac {1}{2t^2}$ and integrating $y(t)$ leads to $c(t) = \int y(t) dt = \frac {1}{2t}.$ Now Theorem ?? guarantees that $x(t) = c(t)e^t = \frac {e^t}{2t}$ is a solution to (??). The other solutions to the inhomogeneous equation are found by adding the general solution to the homogeneous equation.

Exercises

In Exercises ?? – ?? transform the given second order ODE into a first order system and then solve the initial value problem by variation of parameters.

$\begin {array}{c} \dps \ddot x - \frac {6}{t^2} x = 14t^3,\\ \dps x(-1) = -1,\quad \dot x(-1) = 5. \end {array}$ Hint: the linearly independent solutions of the homogeneous equation are: $x^1_h(t) = t^3,\quad x^2_h(t)=\frac {1}{t^2}.$

$\begin {array}{c} \dps \ddot x + \frac {2}{t} \dot x - \frac {2}{t^2} x = 4,\\ \dps x(1) = 1,\quad \dot x(1) = 2. \end {array}$ Hint: the linearly independent solutions of the homogeneous equation are: $x^1_h(t) = t,\quad x^2_h(t)=\frac {1}{t^2}.$

Use reduction of order to find a solution to the equation $\begin{equation} \label{e:inhomex2} \ddot{x} + 3\dot{x}+2x = t. \end{equation}$ Compare the level of effort with the solution obtained using the method of undetermined coefficients.

Consider the second order differential equation $\begin{equation} \label{ex:at1} \frac{d^2x}{dt^2} + p(t)\frac{dx}{dt} + q(t)x = g(t). \end{equation}$ Suppose that $x_1(t) = 1, \quad x_2(t) = 1+t, \AND x_3(t) = 1+t+t^2$ are all solutions to (??). Then determine $p(t)$ , $q(t)$ , and $g(t)$ .

Set $R=0$ and $C=L=1$ in the electrical circuit equation (??). Show that this equation may be rewritten as the first order system $\begin{equation} \label{E:ECsy} \begin{array}{rcl} \dot{x} & = & y \\ \dot{y} & = & -x -R_{mic}(t)y + V(t). \end{array} \end{equation}$

In Exercises ?? – ?? use MATLAB to find solutions for the electrical circuit (??). In each exercise, set $R=0$ and $C=L=1$ in addition to the specified information, and use the first order system (??).

Parameters for the circuit: $R_{mic}(t) = 1+\cos (5t)$ and $V(t) = 1$ ;
initial conditions and time interval: $x(0) = 1$ , $\dot {x}(0) = 0.9$ and $t\in [0,20]$ .

Parameters for the circuit: $R_{mic}(t) = 1+\sin t$ and $V(t) = \sin (2t)$ ;
initial conditions and time interval: $x(0) = 2$ , $\dot {x}(0) = 1$ and $t\in [0,40]$ .

Parameters for the circuit: $R_{mic}(t) = 1+\cos t$ and $V(t) = 0.4+e^{-t}\sin (3t)$ ;
initial conditions and time interval: $x(0) = 1$ , $\dot {x}(0) = -1.5$ and $t\in [0,30]$ .

Parameters for the circuit: $R_{mic}(t) = 0.02+\sin t$ and $V(t) = 1$ ;
initial conditions and time interval: $x(0) = 1.2$ , $\dot {x}(0) = 1.1$ and $t\in [0,60]$ .