Linear Differential Operators

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In Section ?? we describe a powerful method for solving certain second order linear differential equations. To describe this method, it is convenient to introduce the notion of linear differential operators.

We denote by $D$ the simplest differential operator, that is, $D=\frac {d}{dt}.$ From differential calculus we know that $D$ acts linearly on (differentiable) functions, that is,

$\begin{eqnarray*} D(x(t)+y(t)) & = & Dx(t) + Dy(t) \\ D(cx(t)) & = & cDx(t), \end{eqnarray*}$

where $c\in \R$ . Thus we say that $D$ is a linear differential operator.

Higher order derivatives can be written in terms of $D$ , that is, $\frac {d^2x}{dt^2} = \frac {d}{dt}\left (\frac {dx}{dt}\right ) = D(Dx) = D^2x,$ where $D^2$ is just the composition of $D$ with itself. Similarly, $\frac {d^nx}{dt^n} = D^nx.$ It follows that $D^2,\ldots ,D^n$ are all compositions of linear operators and therefore each is linear. We can even form a polynomial in $D$ by taking linear combinations of the $D^k$ . For example,

$\begin{equation} \label{E:pdo} D^4 - 3D^3 + D^2 -5D + 10 \end{equation}$ is a differential operator. We use the following polynomial notation to denote these operators. Let $q(\lambda )$ be the polynomial $q(\lambda ) = \lambda ^4 -3\lambda ^3 + \lambda ^2 - 5\lambda +10.$ Then we denote the linear operator in (??) by $q(D)$ . For example,

$\begin{eqnarray*} q(D)(\sin(2t)) & = & 16\sin(2t) + 24\cos(2t) - 4\sin(2t) - 10\cos(2t) + 10\sin(2t) \\ & = & 22\sin(2t) + 14\cos(2t). \end{eqnarray*}$

With this notation in mind, we can reformulate much of the discussion of higher order equations in terms of linear differential operators. Begin by rewriting the homogeneous equation (??) $\frac {d^nx}{dt^n}+a_{n-1}\frac {d^{n-1}x}{dt^{n-1}}+\cdots + a_1\frac {dx}{dt}+a_0x = (D^n+a_{n-1}D^{n-1}+\cdots +a_1D+a_0)x = 0.$ as

$\begin{equation} \label{E:opern2} p(D)x = 0, \end{equation}$ where $p(\lambda )$ is the characteristic polynomial of (??).

We think of the differential operator $p(D)$ as operating on functions (that are sufficiently differentiable).

The differential operator $p(D)$ is linear, that is,

$\begin{eqnarray*} p(D)(x+y) & = & p(D)x + p(D)y \\ p(D)(cx) & = & cp(D)x, \end{eqnarray*}$

for all sufficiently differentiable functions $x$ and $y$ and all scalars $c$ .

The proof is left as an exercise. See Exercise ??.

Using the linearity of these differential operators allows us to reformulate certain aspects of Section ?? in this new language.

(a): Solutions to the homogeneous equation (??) are just functions in the null space of $p(D)$ .
(b): Using operator notation we can simplify (??) as $\begin{equation} \label{e:elam2} p(D)\left(e^{\lambda t}\right) = p(\lambda)e^{\lambda t}. \end{equation}$ It follows from (??) that the functions $e^{\lambda t}$ are eigenvectors of the operator $p(D)$ with eigenvalue $p(\lambda )$ . Usually the functions $e^{\lambda t}$ are called eigenfunctions. Perversely, we follow convention and reserve the term eigenvalue just for those $\lambda$ that are roots of the characteristic polynomial, that is, those values of $\lambda$ for which $p(\lambda )=0$ .
(c): We can rewrite the inhomogeneous equation as: $p(D)x = g.$ Showing that the inhomogeneous equation is solvable is equivalent to showing that the function $g(t)$ is in the range of the operator $p(D)$ .

Superposition and the Inhomogeneous Equation

Let $p(\lambda )$ be a polynomial, let $g_1(t),g_2(t)$ be continuous functions, and let $\alpha _1,\alpha _2$ be scalars. We can find a particular solution $x_p(t)$ to the inhomogeneous differential equation $p(D)x = \alpha _1g_1(t) + \alpha _2g_2(t)$ by first finding solutions $x_j(t)$ to $p(D)x_j = g_j$ and then setting $x_p(t) = \alpha _1x_1(t) + \alpha _2x_2(t).$

Proof: The proof follows directly from the linearity of $p(D)$ . Just compute $p(D)(\alpha _1x_1 + \alpha _2x_2) = \alpha _1p(D)x_1 + \alpha _2p(D)x_2 = \alpha _1g_1 + \alpha _2g_2.$ Thus, the particular solution is a superposition of the solutions $x_1$ and $x_2$ . $\blacksquare$

The Method of Elimination

In Section ?? we showed how to solve an $n^{th}$ order constant coefficient linear differential equation by converting that equation to a constant coefficient first order system of differential equations. We now show that the process is reversible — we can solve a first order system of $n$ equations by finding solutions to an associated $n^{th}$ order equation. This procedure is called the method of elimination. We first discuss this method abstractly using the language of differential operators and then discuss the pragmatic implementation of the method.

Let $A$ be an $n\times n$ matrix and let $p_A(\lambda )$ be the characteristic polynomial of $A$ . Let $X(t)=(x_1(t),\ldots ,x_n(t))^t$ be a solution to the system of ODEs $\frac {dX}{dt} = AX.$ Then each coordinate function $x_j(t)$ satisfies the $n^{th}$ order differential equation $\begin{equation} \label{E:Elimination} p_A(D)x_j = 0. \end{equation}$

Proof: The proof of this theorem follows from the Cayley-Hamilton theorem, as follows. Rewrite the differential equation using operator notation as $DX = AX,$ where $DX$ indicates differentiation of the vector $X(t)$ by $\frac {d}{dt}$ and $AX$ indicates multiplication of the vector $X(t)$ by the matrix $A$ . Since the coefficients of the matrix $A$ are constants (independent of $t$ ), it follows that $DAX=ADX$ . Hence $D^2X = D(AX) = A(DX) = A^2X.$ Hence $D^kX = A^kX$ by induction, and $p(D)X = p(A)X$ for any polynomial $p(\lambda )$ by linearity. The Cayley-Hamilton theorem (Theorem ?? of Chapter ??) states that $p_A(A)=0$ . Hence $p_A(D)X=0$ . So, in coordinates, $p_A(D)x_j=0$ . $\blacksquare$

Implementation of the Method of Elimination

Consider the first order system of differential equations

$\begin{equation} \label{E:Elim1} \begin{array}{rcl} \dot{x} & = & 2x-3y\\ \dot{y} & = & -5x+4y. \end{array} \end{equation}$ We can eliminate $y$ from the second equation in (??) by solving for $y$ in the first equation, differentiating, and substituting, as follows: $\begin{equation} \label{E:Elim1a} \begin{array}{rcl} y & = & \frac{1}{3}(2x-\dot{x})\\ \dot{y} & = & \frac{1}{3}(2\dot{x}-\ddot{x}) \end{array} \end{equation}$ On substituting (??) into the second equation in (??), we find $\frac {1}{3}(2\dot {x}-\ddot {x}) = -5x + \frac {4}{3}(2x-\dot {x}).$ Simplification leads to the differential equation $\ddot {x} - 6\dot {x} - 7x = 0.$ Since the characteristic polynomial of the coefficient matrix of (??) is $\lambda ^2-6\lambda -7=(\lambda -7)(\lambda +1)$ , this equation is the one predicted by Theorem ??.

Since the roots of the characteristic polynomial are $7$ and $-1$ , it follows that $x(t)$ has the form $x(t) = \alpha e^{7t} + \beta e^{-t}.$ We can solve for $y(t)$ using (??) and obtain $y(t) = \frac {1}{3}(2x-\dot {x}) = -\frac {5}{3} \alpha e^{7t} + \beta e^{-t}.$ Thus the general solution to the first order system is: $\vectwo {x(t)}{y(t)} = \frac {1}{3}\alpha e^{7t}\vectwo {3}{-5} + \beta e^{-t}\vectwo {1}{1}.$ Note that the vectors $(3,5)^t$ and $(1,1)^t$ are eigenvectors of the coefficient matrix of the system (??).

Note that the second half of the method of elimination — in the previous example where we back substituted for the function $y(t)$ — does not always work. For example, if the system of differential equations decouples, then the second half of the method will fail. See Exercise ?? for an example.

Thus the method of elimination provides another alternative to computing solutions to first order systems of differential equations; however, this procedure is not carried out easily for systems of more than two or three equations.

Exercises

In Exercises ?? – ?? apply the given linear differential operator $p(D)$ to the functions:
(a) $\sin (3t)$ and (b) $t^2 e^{-t}$ .

$p(D) = D+1$ .

$p(D) = D^2-D$ .

$p(D) = D^4-2D^2+5$ .

$p(D) = 2D^3-D^2+10D-1$ .

In Exercises ?? – ?? write the homogeneous differential equation $p(D)x=0$ using $d/dt$ notation, factor the characteristic polynomial, and then find the general solution.

$p(D) = D + 1$ .

$p(D) = -D^2 - 5$ .

$p(D) = -4D^5 +4D^4 -8D^2$ .

Attempt to use the method of elimination to solve the following system $\begin {array}{rcl} \dot {x} & = & 2x + z\\ \dot {y} & = & 3y + z\\ \dot {z} & = & z. \end {array}$ Use the characteristic equation (??) to solve for $x(t)$ and then attempt to use the system of differential equations to solve for $y(t)$ and $z(t)$ . What goes wrong?

Prove Lemma ??.

In Exercises ?? – ?? apply the method of elimination to find the general solution of the given system of differential equations.

$\begin {array}{rcl} \dot {x} & = & y\\ \dot {y} & = & 2x+y \end {array}$

$\begin {array}{rcl} \dot {x} & = & x-y\\ \dot {y} & = & x+y \end {array}$

$\begin {array}{rcl} \dot {x} & = & x+y\\ \dot {y} & = & 4x-2y \end {array}$

$\begin {array}{rcl} \dot {x} & = & 2x-y\\ \dot {y} & = & y \end {array}$

In Exercises ?? – ?? write the homogeneous differential equation $p(D)x=0$ using $d/dt$ notation, use MATLAB to find the roots of the characteristic polynomial, and then find the general solution.

$p(D) = 8D^2 - 4D +8$ .

$p(D) = D^4 + 4D^2 - 6D$ .

$p(D) = D^3 -6$ .