Differential and Polynomial Operators

Differential and Polynomial Operators

Definition of an Operator. Linear Property of Polynomial Operators

An operator is a mathematical device which converts one function into another. For example, differention and integration convert one function into another.

By convention, the letter $D$ is used to denote the operation of differentiation and is called a differential operator.

Hence, if $y$ is an $n$th order differentiable function, then

\[\tag{24.11} D^0 y = y,\quad Dy=y',\quad D^2 y = y'', \cdots,D^n y = y^{(n)}\]

By forming a linear combination of differential operators of orders 0 to $n$, we obtain the expression:

\[\tag{24.12} P(D) = a_0 + a_1 D + a_2 D^2 + \cdots + a_n D^n, a_n \neq 0,\]

where $a_0, a_1, \cdots, a_n$ are constants.

Because of the resemblance of $P(D)$ to a polynomial, we shall refer to it as a polynomial operator of order n.

Definition 24.13 Let $P(D)$ be the polynomial operator (24.12) of order $n$ and let $y$ be an $n$th order differntiable function. Then we define P(D)y to mean:

\[\tag{24.14} P(D)y = (a_n D^n + \cdots + a_1 D + a_0)y = a_n D^n y + \cdots + a_1 D y + a_0 y\]

By $(24.11)$ we can also write $(24.14)$ as:

\[\tag{24.15} P(D)y = a_n y^{(n)} + \cdots + a_1 y' + a_0 y,\quad a_n \neq 0\]

Hence by $(24.15)$ we can write the linear equation with constant coefficients:

\[\tag{24.16} a_n y^{(n)} + \cdots + a_1 y' + a_0 y = Q(x),\quad a_n \neq 0\]

as

\[\tag{24.17} P(D)y = Q(x)\]

where $P(D)$ is the polynomial operator $(24.12)$.

Theorem 24.2 If $P(D)$ is the polynomial operator $(24.12)$ and $y_1,~y_2$ are two nth order differentiable functions, then

\[\tag{24.21} P(D)(c_1 y_1 + c_2 y_2) = c_1 P(D) y_1 + c_2 P(D) y_2,\]

where $c_1$ and $c_2$ are constants.

(jmh: See the book for proof, but it’s just distributive property and the sum rule for derivatives.)

An operator which has the property $(24.21)$ is called a linear operator. Hence the polynomial operator $(24.12)$ is linear.

Principle of Superposition

In place of the linear differential equation:

\[\tag{a} P(D)y = Q_1 + Q_2 + \cdots + Q_n,\]

let us write the $n$ equations:

\[\tag{b} P(D)y = Q_1, P(D)y = Q_2, \cdots, P(D)y = Q_n.\]

Let $y_1 p, y_2 p, \cdots, y_{np}$ be respective particular solutions of the $n$ equations of $(b)$. Therefore:

\[\tag{c} P(D)y_{1p} = Q_1, P(D)y_{2p} = Q_2,\cdots, P(D)y_{np} = Q_n.\]

Adding all the equations in $(c)$ and making use of the distributive property of polynomial operators, there results:

\[\tag{d} P(D)(y_{1p} + y_{2p} + \cdots + y_{np}) = Q_1 + Q_2 + \cdots + Q_n,\]

which implies that

\[\tag{e} y_p = y_{1p} + y_{2p} + \cdots + y_{np}\]

is a solution of $(a)$.

(jmh: This means when finding particular solutions, we can split $Q(x)$ into small parts that may be more convenient to manipulate, find the particular solution for each of those parts, and then sum them to get a particular solution for the whole. In practice it’s not clear how much advantage there is to this, and this seems mostly to be establishing that this is a valid thing to do so we can use that fact later.)

Solution of a Linear Differential Equation with Constant Coefficients by Means of Polynomial Operators

(jmh: You can use this method to find general solutions, but it’s easier to find the complementary function by finding roots of the characteristic equation of the associated homogenous equation and then apply this method only for finding a partiuclar solution. To find the general solution with this method, don’t ignore the arbitrary constants of integration like I’ve done here.)

(jmh: my notes - see the book for details)

We will find the general solution of the nonhomogenous linear differential equation:

\[\tag{a} y''' - 3y' + 2y = e^{-x}\]

by finding the complentary function and a particular solution.

We can solve the related non-homogenous differential equation to find the complementary function:

\[\tag{b} y_c (x) = (c_1 + c_2 x)e^x + c_3 e^{-2x}\]

We can then rewrite $(a)$ in operator notation as:

\[\tag{c} (D-1)(D-1)(D+2)y_p = e^{-x}\]

Now let:

\[\tag{d} u = (D-1)(D+2)y_p\]

and substitute $u$ into $(c)$ to get:

\[\tag{e} (D-1)u = u' - u = e^{-x}\]

This is a linear differential equation that can be solved for $u = \frac{-e^{-x}}{2}$ (ignoring the constant of integration since we’re finding a particular solution). Substituting that into $(d)$ gives:

\[\tag{f} (D-1)(D+2)y_p = \frac{-e^{-x}}{2}\]

Now we repeat the process. Let:

\[\tag{g} v = (D+2)y_p\]

and substitute $v$ into $(f)$ to get:

\[\tag{h} (D-1)v = v' - v = \frac{-e^{-x}}{2}\]

This is linear differential equation with a solution of $v = \frac{e^{-x}}{4}$ (again ignoring the arbitrary constant of integration). Substituting that into $(g)$ gives:

\[\tag{i} (D+2)y_p = y_p ' + 2y_p = \frac{e^{-x}}{4}\]

This is a linear differential equation with a solution of $y_p = \frac{e^{-x}}{4}$ (again ignoring the arbitrary constant of integration).

We now add $y_c$ and $y_p$ to get the general solution:

\[\tag{j} y = y_c + y_p = (c_1 + c_2 x)e^x + c_3 e^{-2x} + \frac{e^{-x}}{4}\]