The D operator
Examples
- $D(x^2+2x)=2x+2$
- $D(e^{\alpha x})=\alpha e^{\alpha x}$
- $D(ln(x))=\frac{1}{x}$
- $D(\sqrt{x})=\frac{1}{2\sqrt{x}}$
- $D(sin(x))=cos(x)$
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 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:
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: $F(D)\;e^{ax}\;= \;\left( p_0D^n + p_1D^{n\,-\,1} + ....p_{n\,-\,1}D + p_n \right)e^{ax}$ $= \;\left( p_0a^n + p_1a^{n\,-\,1} + ....p_{n\,-\,1}a + p_n \right)e^{ax}$ $= \;e^{ax}\,F\;(a)$
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 $F(D)\left<e^{ax}V \right> = e^{ax}\;F\;(D+a)\;V$
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
$\mathbf{\therefore\;\;\;\;\;\;F(D^2)\;sin\;ax = F(-\,a^2)\;sin\;ax}$
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
$2\,\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 2y = 0$ $Or\;\;\;\;(2D^2 + 5D + 2)\,y = 0$ $\therefore\;\;\;\;(2D + 1)(D + 2)\,y = 0$ $y = A\,e^{-\frac{1}{2}x} + B\,e^{-2x}$
$(D^2 + 3D + 1)y = 0$
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:- $\mathbf{y = A\;e^{(p + jq)x} + B\;e^{(p - jq)x}}$
$\frac{d^2y}{dx^2} + 4\,\frac{dy}{dx} + 8\,y = 0$ $(D^2 + 4D + 8)\,y = 0$
The roots of this equation are :- $-\frac{-\,4\;\pm \sqrt{16 - 20}}{2}=2\;\pm \sqrt{-\,1}$ $\therefore\;\;\;\;y = A\;e^{(-2\,+\;j)x}+B\,e^{(-2 - j)x}$