Differential Annihilators

Differential Annihilators

We are about to look at a new technique to solving higher order linear non homogenous differential equations known as The Method of Annihilators. This method provides an alternate way to solve many higher order linear non homogenous differential equations that could have otherwise been solved using the method of undetermined coefficients. Before look at this method, we will have to look at an important definition that the reader should already be somewhat familiar with.

Definition: The Differential Operator denoted $D$ maps each differentiable function $f(x)$ to its derivative $f'(x)$, that is $D(f(x)) = f'(x)$.

For example, $D(e^x) = e^x$, $D(\sin x) = \cos x$, and $D(a_0x^n + a_1x^{n-1} + … + a_{n-1}x + a_n) = na_0x^{n-1} + (n-1)a_1x^{n-2} + … + a_{n-1}$.

We will also need to be familiar with multiples, addition of constants, and squares of the differential operator to produce other differential operators. In particular, for $a \in \mathbb{F}$, the differential operator $(aD)(f(x)) = af'(x)$, the differential operator $(D + 1)(f(x)) = f'(x) + f(x)$, and the differential operator $D^2(f) = D(D(f)) = f''(x)$.

For example, $(2D + 1)(\sin x) = 2(\sin x)' + \sin x = 2 \cos x + \sin x$. Furthermore, $(D - 2)^2(e^x) = (D - 2)((D - 2)(e^x)) = (D - 2)(e^x - 2e^x) = e^x - 2e^x - 2(e^x - 2e^x) = e^x$

We will focus our interest on differential operators that map a specific function to zero.

Definition: If $(aD + b)^n (f(x)) = 0$ then the differential operator $(aD + b)^n$ is said to be an Annihilator of $f(x)$.

For example, $(D - 1)$ is an annihilator of $y = e^x$ because $(D - 1)(e^x) = e^x - e^x = 0$. Furthermore, $D^{n+1}$ is an annihilator of polynomials $y = a_0x^n + a_1x^{n-1} + … + a_{n-1}x + a_n$ of degree less than or equal to $n$, because when we differentiate such a polynomial $n+1$ times we the zero polynomial. Furthermore, if $\beta \in \mathbb{R}$ then $(D^2 + \beta^2)$ is an annihilator of $y = \sin \beta x$ and $y = \cos \beta x$ since:

\begin{align} \quad (D^2 + \beta^2)(\sin \beta x) = D^2 (\sin \beta x) + \beta^2 (\sin \beta x) = -\beta^2 \sin \beta x + \beta^2 \sin \beta x = 0 \end{align}
\begin{align} \quad (D^2 + \beta^2)(\cos \beta x) = D^2 (\cos \beta x) + \beta^2 (\cos \beta x) = -\beta^2 \cos \beta x + \beta^2 \cos \beta x = 0 \end{align}

There are many nice properties of these differential operators. For example, for $a, b \in \mathbb{R}$ we have that $(aD + b)^2(f(x)) = (a^2D^2 + 2abD + b^2)(f(x))$ since:

\begin{align} \quad (aD + b)^2 (f(x)) = (aD + b)[(aD + b)(f(x))] = (aD + b)[af'(x) + bf(x)) = aD(af'(x) + bf(x)) + b(af'(x) + bf(x)) \\ = a^2f''(x) + abf'(x) + abf'(x) + b^2f(x) = (a^2D^2 + 2abD + b^2)(f(x)) \end{align}

We should thus note that the combination that forms the differential operator can thus be factored/expanded like regular polynomials for simplification purposes.

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