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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)

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:

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

Note
$\inline D$ is an operator andmust therefore always be followed by some expression on which it operates.

Simple Equivalents

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

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: $\inline \left(2\,D^2+5\,D+y \right) y=0$ or $\inline 2\,D^2+5\,D+2=0$

This can be factorised to give:

$(2D+1)(D+2) = 0$

Examples

$D(x^2+2x)=2x+2$
$D(e^{\alpha x})=\alpha e^{\alpha x}$
$D(ln(x))=\frac{1}{x}$
$D(\sqrt{x})=\frac{1}{2\sqrt{x}}$
$D(sin(x))=cos(x)$

But is it justifiable to treatD in this way?

Algebraic procedures depend upon three laws.

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

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

$\inline D(u + v)=Du+Dv$
$\inline D^m(D^n\,u)=D^{(m+n)}\;u$
$\inline D(uv) = u Dv$ only whenu is a constant.

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

In the following analysis we will write

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

$\inline p_i$ are constants and $\inline n$ 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 isrequired to solve the following equations:

Example:

Example - Simple example

Problem

Solve the following equation:-

Workings

Using the D operator this can be written as:-

Solution

Integrating using

as the factor

Three Useful Formulae Based On The Operator D

Equation A

Let $\inline F(D)$ 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

Workings

This can be re-written as:

Solution

We can put D = 4

Equation B

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

Where $\inline V$ is any function ofx

Applying Leibniz's theorem for the $\inline n{th}$ 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 $\inline {\displaystyle D^{n-1}\left<e^{ax}V\right>= e^{ax}\;(D + )^{n-1}\,V$ 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:

Workings

We have used D as if it were an algebraic constant but it is in fact an operator where

Solution

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:-

Workings

This can be re-written as:-

(1)

Using equation 1 we can put

If we multiply the top and bottom of this equation by

But

Solution

But since

Linear First Order D Equations With Constant Coefficients

These equations have $\inline 0$ on the right hand side

$(D - \alpha )y = 0$

This equation is

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

Using an Integrating Factor of $\inline \displaystyle e^{-\alpha x}$ 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 $\inline \apha\;and\;\beta$ 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 $\inline C$ 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 $\inline \displaystyle \alpha\neq \beta$ we can write theGeneral Solution as:-

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

WhereA andB are arbitrary Constants.

Example:

Example - Linear second order example

Problem

Workings

The roots of this equation are:-

Therefore the General Solution is

The Special Case where

From Equation (41)

The roots of the Auxiliary Equation are complex.

If the roots of the are complex then the General Solution will be of the form

, and the solution will be given by:-

Solution

The roots of this equation are :-

Physical Examples

Example:

Example - Small oscilations

Problem

Show that if

satisfies the differential equation

with k < n and if when

The complete period of small oscillations of a simple pendulum is 2 secs. and the angularretardation due to air resistance is 0.04 X the angular velocity of the pendulum. The bob is heldat rest so the the string makes a small angle