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Maths › Calculus › Differential ›

The D operator

Solving Differential Equations using the D operator

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Contents

Theory Of Differential Operator (differential Module)
The D Operator And The Fundamental Laws Of Algebra
The Use Of The D Operator To Find The Complementary Function For Linear Equations
Three Useful Formulae Based On The Operator D
Linear First Order D Equations With Constant Coefficients
Linear Second Order D Equations With Constant Coefficients
Physical Examples
Page Comments

Theory Of Differential Operator (differential Module)

Definition

A differential operator is an operator defined as a function of the differentiation operator.

It is helpful, as a matter of notation first, to consider differentiation as an abstract operation, accepting a function and returning another (in the style of a higher-order function in computer science).

The most commonly used differential operator is the action of taking the derivative itself. Common notations for this operator include:

$\displaystyle D\equiv\frac{d}{dx}$ and if generalize $\displaystyle D^n\equiv\frac{d^n}{dx^n}$

Note

is an operator and must therefore always be followed by some expression on which it operates.

Simple Equivalents

means $\displaystyle Du\equiv \frac{du}{dx}$ but $uD\equiv u\frac{d}{dx}$
$\displaystyle D^2y\equiv D\times Dy\equiv \frac{d}{dx}\left(\frac{dy}{dx} \right) = \frac{d^2y}{dx^2}$
Similarly $\displaystyle D^2\equiv \frac{d^2}{dx^2}$ and $D^3\equiv \frac{d^3}{dx^3}$

The D Operator And The Fundamental Laws Of Algebra

The following differential equation:

$2\,\frac{d^2y}{dx^2} + 5\,\frac{dy}{dx} + 2\,y = 0$

may be expressed as: $\left(2\,D^2+5\,D+2y \right) y=0$ or $2\,D^2\,y+5\,D\,y+2y=0$

This can clearly be factorised to give:

Examples

$D(e^{\alpha x})=\alpha e^{\alpha x}$

$D(ln(x))=\frac{1}{x}$

$D(\sqrt{x})=\frac{1}{2\sqrt{x}}$

But is it justifiable to treat D in this way?

Algebraic procedures depend upon three laws.

The Distributive Law: $\displaystyle m(a + b) = ma + mb$
The Commutative Law: $\displaystyle a b = b a$
The Index Law: $\displaystyle a^{m}\times a^{n} = a^{(m\,+\,n)}$

If D satisfies these Laws, then it can be used as an Algebraic operator(or a linear operator). However:

$D^m(D^n\,u)=D^{(m+n)}\;u$
only when u is a constant.

Thus we can see that D does satisfy the Laws of Algebra very nearly except that it is not interchangeable with variables.

In the following analysis we will write

$F(D)\;\equiv \;p_0D^n + p_1D^{n\,-\,1} + ....p_{n\,-\,1}D + p_n$

are constants and

is a positive integer. As has been seen, we can factorise this or perform any operation depending upon the fundamental laws of Algebra.

We can now apply this principle to a number of applications.

The Use Of The D Operator To Find The Complementary Function For Linear Equations

It is required to solve the following equations:

Example:

Example - Simple example

Problem

Solve the following equation:-

$\frac{d^2y}{dx^2} - 2\,\frac{dy}{dx} + y = 0$

Workings

Using the D operator this can be written as:-

$(D^2 - 2D + 1)\,y = 0$

$Or\;\;\;\;(D - 1)^2\,y = 0$

$Let\;\;\;\;(D - 1)\,y = u$

$Then\;\;\;\;(D - 1)\,u = 0$

$\therefore\;\;\;\;u = A\,e^{x}$

$\therefore\;\;\;\;(D - 1)\,y = A\,e^{x}$

$\;\;\;\;\frac{dy}{dx}\;-\,y = A\,e^{x}$

Solution

Integrating using $e^{-\,x}$ as the factor

$y\,e^{-x} = Ax + B$

$\mathbf{\therefore\;\;\;\;y = (Ax + B)\,e^{x}}$

Three Useful Formulae Based On The Operator D

Equation A

Let

represent a polynomial function

$\mathbf{F(D)\;e^{ax} = e^{ax}\;F\;(a)}$

Since

$D\;e^{ax} = a\;e^{ax}$

and

$D^2\;e^{ax} = a^2\;e^{ax}$

From which it can be seen that:

$F(D)\;e^{ax}\;= \;\left( p_0D^n + p_1D^{n\,-\,1} + ....p_{n\,-\,1}D + p_n \right)e^{ax}$

$= \;\left( p_0a^n + p_1a^{n\,-\,1} + ....p_{n\,-\,1}a + p_n \right)e^{ax}$

$e^{ax}\,F\;(a)$

Example:

Example - Equation A example

Problem

$\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = e^{4x}$

Workings

This can be re-written as:

$(D^2 - 5D + 6)\,y = e^{4x}$

$\therefore\;\;\;\;y = e^{4x}\times\frac{1}{D^2 - 5D + 6}$

Solution

We can put D = 4

$\therefore\;\;\;\;y = e^{4x}\times\frac{1}{4^2 - 5\times4 + 6} = \frac{1}{2}\,e^{4x}$

Equation B

$\mathbf{F(D)\left<e^{ax}V \right> = e^{ax}F(D + a)V}$

Where

is any function of x

Applying Leibniz's theorem for the

differential coefficient of a product.

$D^n\left<e^{ax}V \right> = (D^ne^{ax})V + n(D^{n-1}e^{ax})(DV) + \frac{1}{2}n(n-1)(D^{n-2}e^{ax})(D^2V) + .....e^{ax}(D^nV)$

$= a^ne^{ax}V + na^{n-1}e^{ax}DV + \frac{1}{2}n(n-1)a^{n-2}e^{ax}D^2V + .....e^{ax}D^nV$

$= e^{ax}(a^n + na^{n-1}D + \frac{1}{2}n(n-1)a^{n-2}D^2 + .....+\:D^n)V$

$= e^{ax}\;(D + a)^n\,V$

Similarly ${\displaystyle D^{n-1}\left<e^{ax}V\right>= e^{ax}\;(D + )^{n-1}\,V$ and so on

$F(D)\left<e^{ax}V \right>\;=\:\left(p_0D^n + p_1D^{n-1} + .........+\;p_{n-1}D + D \right)\left<e^{ax}V \right>$

$= e^{ax}\left<p_0(D+a)^n + p_1(D+a)^{n-1} + .........+\;p_{n-1}(D+a) + p_n \right>V$

therefore

$F(D)\left<e^{ax}V \right> = e^{ax}\;F\;(D+a)\;V$

Example:

Example - Equation B example

Problem

Find the Particular Integral of:

$(D^2 - 5D + 6)\,y = x^2$

Workings

$y = \frac{x^2}{(D^2 - 5D + 6)}=\left(\frac{1}{(2 - D)} - \frac{1}{(3 - D)} \right)\,x^2$

$= \frac{1}{2}(1+\frac{1}{2}D+\frac{1}{4}D^2+\frac{1}{8}D^3+.....)x^2 - \frac{1}{3}(1+\frac{1}{3}D+\frac{1}{9}D^2+\frac{1}{27}D^3+...)x^2$

We have used D as if it were an algebraic constant but it is in fact an operator where $D\,(x^2) = 2x\;and\;D^2\,(x^2) = 2.$

Solution

$y= \frac{1}{6}x^2 + \frac{5}{18}x + \frac{19}{108}}$

Equation C - Trigonometrical Functions

$\mathbf{F(D^2)\;cos \,ax= F(-\,a^2)\;cos\,ax}$

$D^2\;cos \,ax= -\,a^2\;cos\,ax}$

$D^4\;cos \,ax= (-\,a^2)^2\;cos\,ax}$

And so on

$F(D^2)\;cos\,ax = \left(p_0D^n + p_1D^{n-1} + .......+p_{n-1}D + p_n \right)cos\;ax$

$= \left<p_0(-\,a^2)^n + p_1(-\,a^2)^{n-1} + ........+p_{n-1}(\,-\,a^2) + p_n \right>cos\;ax$

Therefore

$F(D^2)\;cos\;ax = F(-\,a^2)\;cos\;ax$

similarly

$\mathbf{\therefore\;\;\;\;\;\;F(D^2)\;sin\;ax = F(-\,a^2)\;sin\;ax}$

Example:

Example - Trigonometric example

Problem

Find the Particular Integral of:-

$\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = sin\,2x$

Workings

This can be re-written as:-

$y = \frac{1}{D^2 - 5D + 6}\;sin\,2x$

(1)

Using equation 1 we can put $D^2 = -\,4$

$\therefore\;\;\;\;y = \frac{1}{-\,4\;-\,5D + 6}sin\,2x$

$=\frac{1}{2 - 5D}\;sin\,2x$

If we multiply the top and bottom of this equation by

$=\frac{2 + 5D}{4 - 25D^2}\;sin\,2x$

But

$\therefore\;\;\;\;y\;=\frac{2 + 5D}{104}\;sin\,2x = \frac{1}{104}\left(2\;sin\,2x + 5D\;sin\,2x \right)$

Solution

But since $D\;sin\,2x = 2\;cos\,2x$

$\mathbf{y = \frac{1}{104}(2\;sin\,2x + 10\;cos\,2x)}$

Linear First Order D Equations With Constant Coefficients

These equations have

on the right hand side

$(D - \alpha )y = 0$

This equation is

$\frac{dy}{dx} - \alpha \,y = 0$

Using an Integrating Factor of $\displaystyle e^{-\alpha x}$ the equation becomes:-

$\frac{d}{dx}\left(y\,e^{-\alpha x} \right) = 0$

$\therefore\;\;\;\;y\,e^{-\alpha x} = C$

$\mathbf{Thus\;\;\;\;y = C\,e^{\alpha x}}$

Which is the General Solution.

Linear Second Order D Equations With Constant Coefficients

$p_0\,\frac{d^2y}{dx^2} + p_1\,\frac{dy}{dx} + p_2\,y = 0\;\;\;\;\;where\;\;\;\;p_0\neq 0$

$or\;\;\;\;\left(p_0\,D^2 + p_1\,D + p_2 \right)\,y = 0$

$i.e.\;\;\;\;\;p_0(D - \alpha )(D - \beta )\,y = 0$

Where $\apha\;and\;\beta$ are the roots of the quadratic equation. i.e. the auxiliary equation.

$p_0\,m^2 + p_1\,m + p_2 = 0$

$(D - \alpha)[(D - \beta)]\,y = 0$

$\therefore\;\;\;\;(D - \beta)\,y = C\,e^{\alpha x}$

Where

is an arbitrary Constant

$\therefore\;\;\;\;\frac{dy}{dx} - \beta\,y = C\,e^{\alpha x}$

This equation can be re-written as:-

$\frac{d}{dx}(y\,e^{-\beta x}) = C\,e^{\alpha\,x}\times e^{-\beta\,x} = C\,e^{(\alpha - \beta)}$

Integrating

$y\,e^{-\beta\,x} = \frac{C\,e^{(\alpha\,-\,\beta)}}{\alpha\,-\,\beta} + K$

$\therefore\;\;\;\;y = \frac{C}{\alpha - \beta}\;e^{\alpha\,x} + K\,e^{\beta\,x}$

Thus when $\displaystyle \alpha\neq \beta$ we can write the General Solution as:-

$\mathbf{y = A\,e^{\alpha\,x} + B\,e^{\beta\,x}}$

Where A and B are arbitrary Constants.

Example:

Example - Linear second order example

Problem

$2\,\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 2y = 0$

$Or\;\;\;\;(2D^2 + 5D + 2)\,y = 0$

$\therefore\;\;\;\;(2D + 1)(D + 2)\,y = 0$

$y = A\,e^{-\frac{1}{2}x} + B\,e^{-2x}$

Workings

The roots of this equation are:-

$\frac{-\,3\;\pm \sqrt{9 - 4}}{2} = \frac{-\,3\;\pm \sqrt{5}}{2}$

Therefore the General Solution is

$\;y + A\;e^{\frac{-3 + \sqrt{5}}{2}x} + B\;e^{\frac{-3\,- \sqrt{5}}{2}x$

The Special Case where $\displaystyle \alpha = \beta$

From Equation (41)

$\frac{d}{dx}(y\,e^{-\alpha\,x}) = C$

$\therefore\;\;\;\;y\,e^{- \alpha\,x} = Cx + K$

$\mathbf{y = (Cx + K)\,e^{\alpha\,x}}$

$9\;\frac{d^2y}{dx^2} - 6\,\frac{dy}{dx} + 1 = 0$

$Or\;\;\;(9D^2 - 6D + 1)\;y = 0$

$\therefore\;\;\;\;(3D - 1)(3D - 1) = 0$

$\therefore\;\;\;\;y = (A\,x + B)\;e^{\frac{1}{3}x}$

The roots of the Auxiliary Equation are complex.

If the roots of the are complex then the General Solution will be of the form $\displaystyle p\;\pm j\,q$ , and the solution will be given by:-

$\mathbf{y = A\;e^{(p + jq)x} + B\;e^{(p - jq)x}}$

$\frac{d^2y}{dx^2} + 4\,\frac{dy}{dx} + 8\,y = 0$

$(D^2 + 4D + 8)\,y = 0$

Solution

The roots of this equation are :-

$-\frac{-\,4\;\pm \sqrt{16 - 20}}{2}=2\;\pm \sqrt{-\,1}$

$\therefore\;\;\;\;y = A\;e^{(-2\,+\;j)x}+B\,e^{(-2 - j)x}$

Physical Examples

Example:

Example - Small oscilations

Problem

Show that if

satisfies the differential equation $\displaystyle \frac{ d^2\theta}{dt^2}\;+\;2k\;\frac{d\theta }{dt}\;+\;n^2\,\theta \;=\;0$ with k < n and if when $\displaystyle t\;+\;0\;:\;\theta \;+\;\alpha \;and\;\frac{d\theta }{dt}\;=\;0$

$Then\;\;\;\;\theta \;=\;e^{-kt}\left(\alpha \,cos\,pt\;+\;\frac{k\,\alpha }{p} \;sin\,pt\right)$

$where\;\;\;\;p^2\;=\;n^2\;-\;k^2$

The complete period of small oscillations of a simple pendulum is 2 secs. and the angular retardation due to air resistance is 0.04 X the angular velocity of the pendulum. The bob is held at rest so the the string makes a small angle $\alpha\;=\;1^0$ with the downwards vertical and then let go. Show that after 10 complete oscillations the string will make an angle of about 40' with the vertical.(LU)

Workings

$\frac{d^2\theta }{dt^2}\;+\;2k\;\frac{d\theta }{dt}\;+\;n^2\,\theta \;=\;0$

$\therefore\;\;\;\;\frac{-\;2k\;\pm \;\sqrt{4k^2\;-\;4n^2}}{2}$

$\therefore\;\;\;\;D\;=\;-k\;\pm \sqrt{k^2\;-\;n^2}$

Using the "D" operator we can write

$D^2\;+\;2kD\;+\;n^2\;=\;0$

$=\;-k\;\pm jp\;\;\;\;\;where\;\;\;\;p^2\;=\;k^2\;-\;n^2$

$\theta \;=\;e^{-kt}\left(A\;cos\,pt\;+\;B\;sin\,pt \right)$

$\dot{\theta }\;=\;-\,k\,e^{-kt}\,A\,cos\,pt\;-\;e^{-kt}\;A\;p\,sin\,pt\;-\;ke^{-kt}\;B\,sin\,pt\;+\;e^{-kt}\;B\,pcos\,pt$

When t = 0 $\alpha$ = 0 and $\dot{\theta}$ = 0

$\therefore\;\;\;\;A\;=\;\alpha$

and

$0=-\;k\;\alpha \;+\;Bp\;\;\;\;\therefore\;\;\;\;B\;=\;\frac{k\alpha }{p}$

$\therefore\;\;\;\;\theta \;=\;e^{-kt}\left(\alpha \;cos\,pt\;+\;\frac{k\alpha }{p}\;sin\,pt \right)$

$Periodic\;Time\;=\;2\;secs.\;=\;\frac{2\,\pi }{p}\;\;\;\;\;\therefore\;\;\;\;p\;=\;\pi$

Solution

At t = 0

$\theta \;=\;1^0\;=\;\frac{\pi }{180}\;rds.$

We have been given that k = 0.02 and the time for ten oscillations is 20 secs.

$\therefore\;\;\;\;\theta _{10 cycles}\;=\;e^{-0.02\times20}\left(cos\,\pi \times20\;+\;\frac{0.02}{20}\;sin\,\pi \times20} \right)\frac{\pi }{180}\f] \f[=\;e^{-0.4}\times1\times\frac{\pi}{ 180}\;=\;\frac{1}{1.49182}\;\times\;\frac{\pi }{180}times\;60\approx 40\,seconds$

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