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The D operator

Solving Differential Equations using the D operator
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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
D is an operator and must therefore always be followed by some expression on which it operates.

Simple Equivalents

  • Du 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+y \right) y=0 or 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 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(u + v)=Du+Dv
  • D^m(D^n\,u)=D^{(m+n)}\;u
  • D(uv) = u Dv 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

p_i are constants and 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 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 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
\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 V is any function of x

Applying Leibniz's theorem for the 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 {\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:-

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 2 + 5D

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

But D^2;=-4

\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 0 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 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 \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
(D^2 + 3D + 1)y = 0

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
or
\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 theta 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