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 more generally $\displaystyle D^n\equiv\frac{d^n}{dx^n}$
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: $\left(2\,D^2+5\,D+2 \right) y=0$ or $2\,D^2+5\,D+2=0$
This can be factorised to give: $(2D+1)(D+2) = 0$
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 1
ProblemSolve the following equation:-
$\frac{d^2y}{dx^2} - 2\,\frac{dy}{dx} + y = 0$
WorkingsUsing the D operator this can be written as:-
$(D^2 - 2D + 1)\,y = 0$ $Or\;\;\;\;(D - 1)^2\,y = 0$ $Let\;\;\;\;(D - 1)\,y = u$ $Then\;\;\;\;(D - 1)\,u = 0$ $\therefore\;\;\;\;u = A\,e^{x}$ $\therefore\;\;\;\;(D - 1)\,y = A\,e^{x}$ $\;\;\;\;\frac{dy}{dx}\;-\,y = A\,e^{x}$
SolutionIntegrating using $e^{-\,x}$ as the factor $y\,e^{-x} = Ax + B$ $\mathbf{\therefore\;\;\;\;y = (Ax + B)\,e^{x}}$
Example 2
Problem$\mathbf{\frac{d2y}{dx^2}\;-\;2\,\frac{dy}{dx}\;-\;3y\;=\;0}$
Workings$D^2\;-\;2D\;-\;3\;=\;0$ $\therefore\;\;\;\;\;\;D\;=\;-\,3\;\;\;or\;\;\;+\,1$
Solution$\therefore\;\;\;\;\;\;y\;=\;Ae^{-3x}\;+\;Be^{x}$
Example 3
Problem$\mathbf{\frac{d^2y}{dx^2}\;+\;9\frac{dy}{dx}\;+\;8y\;=\;0}$
Workings$D^2\;+\;9D\;+\;8y\;=\;0$ $\therefore\;\;\;\;\;\;D\;=\;-8\;\;\;or\;\;\;-1$
Solution$y\;=\;Ae^{-8x}\;+\;Be^{-x}$
Example 4
ProblemSolve $\displaystyle \frac{d^2y}{dx^2} - 3\;\frac{dy}{ax} + 2\,y = 0$
WorkingsThis can be re-written in terms of the $D$ operator as:
$D^2-3D + 2y= 0\;=(D-1)(D-2)y$ Let $\displaystyle (D-2)y=u$
Then $\displaystyle (D-1)u=0$
Therefore $(D-2)\;y=A\,e^{x}\;or\;\displaystyle \frac{dy}{dx} - 2\,y = A\,e^{x}$
Integrating using $e^{-x}$ as the factor $y\,e^{-2x}=-\,A\,e^{-X}+B$
Solution$\mathbf{y=\B\,e^{2X}-A\,e^{X}}$
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 1
Problem$\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = e^{4x}$
WorkingsThis can be re-written as:
$(D^2 - 5D + 6)\,y = e^{4x}$
$\therefore\;\;\;\;y = e^{4x}\times\frac{1}{D^2 - 5D + 6}$
SolutionWe can put D = 4
$\therefore\;\;\;\;y = e^{4x}\times\frac{1}{4^2 - 5\times4 + 6} = \frac{1}{2}\,e^{4x}$
Example 2
Problem$\frac{d^2y}{dx^2}-2\frac{dy}{dx}-y=0}$
Workings$(D^2-2D-y)=0$ $\therefore\;\;\;\;\;\;D=\frac{+2\pm \sqrt{4+4}}{2}$ $=1 \pm \sqrt{2}$
Solution$y\;=\;Ae^{(\sqrt{2}-1)x}\;+\;Be^{(-\;\sqrt{2}-1)x}$
Example 3
Problem$3\frac{d^2y}{dx^2}-\frac{dy}{dx}-4y=0$
Workings$3D^2-D-4=0$ $\therefore\;\;\;\;\;\;D=\frac{1 \pm 7}{6}$ or $D=-1, \frac{4}{3}$
Solution$y=Ae^{\frac{4}{3}x}+Be^{-x}$
Example 4
Problem$\mathbf{\frac{d^2y}{dx^2}\;+\;2\frac{dy}{dx}\;+\;2y\;=\;0}$
Workings$D^2\;+\;2D\;+\;2\;=\;0$ $\therefore\;\;\;\;\;\;D\;=\;\frac{-2\;\pm \;\sqrt{4-8}}{2}\;=\;\frac{-2\;\pm \;2j}{2}$ $\therefore\;\;\;\;\;\;y\;=\;Ae^{(-1+j)x}\;+\;be^{(-1-j)x}$ $=A\,\left( e^{-x}\,\times\,e^{jx} \right)\;+\;B\left(e^{-x}\,\times\,e^{-jx} \right)$ $=\;e^{-x}\;\left(Ae^{jx}\;+\;Be^{-jx} \right)$ $=\;e^{-x}\;\{A(cos\,x\;+\;j\,sin\,x)\;+\;B(cos\,x\;-\;j\,sin\,x)\}$
Solution$\therefore\;\;\;\;\;\;y\;=\;e^{-x}\;(E\,cos\,x\;+\;F\,sin\,x)$
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 $\displaystyle D^{n-1}\left<e^{ax}V\right>= e^{ax}\;(D + a)^{n-1}\,V$ and so on $F(D)\left<e^{ax}V \right>\;=\:\left(p_0D^n + p_1D^{n-1} + .........+\;p_{n-1}D + p_n \right)\left<e^{ax}V \right>$ $= e^{ax}\left<p_0(D+a)^n + p_1(D+a)^{n-1} + .........+\;p_{n-1}(D+a) + p_n \right>V$
therefore $F(D)\left<e^{ax}V \right> = e^{ax}\;F\;(D+a)\;V$
Example 1
ProblemFind 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$y= \frac{1}{6}x^2 + \frac{5}{18}x + \frac{19}{108}}$
Example 2
Problem$\mathbf{4\,\frac{d^2y}{dx^2}\;+\;16\,\frac{dy}{dx}\;+\;25y\;=\;0}$
Workings$i.e.\;\;\;\;\;\;\;(4D^2\;+\;16D\;+\;25)\;=\;0$ $\therefore\;\;\;\;\;\;D\;=\;\frac{-\;16\;\pm \sqrt{16^2\;-\;4\times4\times25}}{8}$ $=\;-\;2\;\pm \;\frac{1}{2}\sqrt{16\;-\;25}\;=\;-\;2\;\pm j\,\frac{3}{2}$ Thus $y\;=\;Ae^{(-2+j\frac{3}{2})x}\;+\;Be^{(-2-j\frac{3}{2})x}$ $=\;e^{-2x}\left(Ae^{jx}\;+\;Be^{-jx} \right)$
Solution$\therefore\;\;\;\;\;\;y\;=\;e^{-2x}\left(E\,cos\,\frac{3}{2}\;+\;F\a,sin\,\frac{3}{2}
\right)$
Example 3
Problem$\mathbf{\frac{d^2x}{dt^2}\;-\;16\,x\;=\;0}$
Workings$(D\;+\;4)(D\;-\;4)\;=\;0$ $\therefore\;\;\;\;\;\;D\;=\;\pm\; 4$
Solution$x\;=\;Ae^{4t}\;+\;Be^{-4t}$
Example 4
Problem$\mathbf{\frac{d^2\lambda} {dx^2}\;+\;2\,\lambda \;=\;0}$
Workings$D^2\;+\;2\;=\;0$ $\therefore\;\;\;\;\;\;\;D\;=\;\pm j\sqrt{2}$ $thus\;\;\;\;\;\;\;\lambda \;=\;Ae^{j\sqrt{2}}\;+\;Be^{-j\sqrt{2}}$
Solution$\therefore\;\;\;\;\;\;\;\lambda \;=\;F\,cos\,\sqrt{2}\,x\;+\;E\,sin\,\sqrt{2}\,x$
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 1
ProblemFind the Particular Integral of:- $\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = sin\,2x$
WorkingsThis 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 $D^2;=-4$
$\therefore\;\;\;\;y\;=\frac{2 + 5D}{104}\;sin\,2x = \frac{1}{104}\left(2\;sin\,2x + 5D\;sin\,2x \right)$
SolutionBut since $D\;sin\,2x = 2\;cos\,2x$
$\mathbf{y = \frac{1}{104}(2\;sin\,2x + 10\;cos\,2x)}$
Example 2
Problem$\mathbf{\frac{d^2x}{dt^2}\;+\;5\frac{dx}{dt}\,+\,4(x\,-\,1)\;=\;0}$
Workings$\frac{d^2x}{dt^2}\;+\;5\frac{dx}{dt}\;+\;4x\;=\;4$ $Let\;\;\;\;\;D^2\;+\;5D\;+\;4\;=\;0$ $\therefore\;\;\;\;\;x\;=\;-\;1\;\;or\;\;-\;4$
Solution$x\;=\;Ae^{-t}\;+\;Be^{-4t}\;+\;1$
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 $\alpha\;and\;\beta$ 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:- $\frac{d}{dx}(y\,e^{-\beta x}) = C\,e^{\alpha\,x}\times e^{-\beta\,x} = C\,e^{(\alpha - \beta)x}$
Integrating $y\,e^{-\beta\,x} = \frac{C\,e^{(\alpha\,-\,\beta)x}}{\alpha\,-\,\beta} + K$ $\therefore\;\;\;\;y = \frac{C}{\alpha - \beta}\;e^{\alpha\,x} + K\,e^{\beta\,x}$
- 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 1
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:- $\mathbf{y = A\;e^{(p + jq)x} + B\;e^{(p - jq)x}}$
$\frac{d^2y}{dx^2} + 4\,\frac{dy}{dx} + 8\,y = 0$ $(D^2 + 4D + 8)\,y = 0$
SolutionThe 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 1
ProblemShow 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 \f$\alpha\f$ = 0 and $\dot{\theta}$ = 0 $\therefore\;\;\;\;A\;=\;\alpha$ and $0=-\;k\;\alpha \;+\;Bp\;\;\;\;\therefore\;\;\;\;B\;=\;\frac{k\alpha }{p}$ $\therefore\;\;\;\;\theta \;=\;e^{-kt}\left(\alpha \;cos\,pt\;+\;\frac{k\alpha }{p}\;sin\,pt
\right)$ $Periodic\;Time\;=\;2\;secs.\;=\;\frac{2\,\pi }{p}\;\;\;\;\;\therefore\;\;\;\;p\;=\;\pi$
SolutionAt 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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