# Preface

This tutorial was made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica and programming before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel. The Mathematica commands in this tutorial are all written in bold black font, while Mathematica output is in normal font.

Finally, you can copy and paste all commands into your Mathematica notebook, change the parameters, and run them because the tutorial is under the terms of the GNU General Public License (GPL). You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately. The tutorial accompanies the textbook Applied Differential Equations. The Primary Course by Vladimir Dobrushkin, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043

# Annihilator operators

The annihilator of a function is a differential operator which, when operated on it, obliterates it. There is nothing left. We say that the differential operator $$L\left[ \texttt{D} \right] ,$$ where $$\texttt{D}$$ is the derivative operator, annihilates a function f(x) if $$L\left[ \texttt{D} \right] f(x) \equiv 0 .$$ For example, the differential operator $$\texttt{D}^2$$ annihilates any linear function. In other words, if an operator annihilates a function f, then f belongs to the kernel of the operator.

Since we consider only linear differential operators, any such operator is a polynomial in $$\texttt{D}$$

$L\left[ \texttt{D} \right] = a_n \texttt{D}^n + a_{n-1} \texttt{D}^{n-1} + \cdots a_1 \texttt{D} + a_0 \qquad \mbox{or, when it operates on a function y,} \qquad L\left[ \texttt{D} \right] y = a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots a_1 y' + a_0 y .$
As usual, we drop the identical operator $${\bf I} = \texttt{D}^0 .$$ If all coefficients $$a_n , \ a_{n-1}, \ \ldots , a_1 , \ a_0$$ are constants, we get a constant coefficient linear differential operator that corresponds to a linear differential equation with constant coefficients $$L\left[ \texttt{D} \right] y =0 .$$ Otherwise, we have a variable coefficient linear differential operator. Note that constant coefficient linear differential operators commute, but variable coefficient not.

It is known, see Applied Differential Equations. The Primary Course by Vladimir Dobrushkin, CRC Press, 2015, that for any set of k linearly independent functions y1, y2, ... , yk, there exists a unique (up to an arbitrary nonzero multiple) linear differential operator of order k that annihilates the given set of functions. This differential operator is defined by the Wronskian

\begin{eqnarray} \label{Ebd14.wronskian} L[f] &=& W[ y_1 , y_2 , \ldots , y_k , f] = \det \begin{bmatrix} y_1 & y_2 & \cdots & y_k & f \\ y'_1 & y'_2 & \cdots & y'_k & f' \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ y_1^{(k)} & y_2^{(k)} & \cdots & y_k^{(k)} & f^{(k)} \end{bmatrix} \\ &=& \left( W[y_1 , \ldots , y_k ] \,\texttt{D}^k + \cdots + W[y'_1 , \ldots , y'_k ] \,\texttt{I} \right) f . \notag \end{eqnarray}
Usually, the latter differential operator is divided by the Wronskian $$W[y_1 , y_2 , \ldots , y_k ] \ne 0$$ of the given set of functions to isolate the differential operator $$\texttt{D}^k$$ in the expression for L. With such division, the annihilating operator L of lowest order is uniquely defined by the set of given functions $$\{ y_1 , y_2 , \ldots , y_k \} .$$ Obviously, if L is the annihilating operator for the set of functions $$\{ y_1 , y_2 , \ldots , y_k \} ,$$ its product L M with any other operator $$M\left[ \texttt{D} \right]$$ is also an annihilating operator for the given set of functions. Therefore, we usually are looking for an annihilating operator of least possible order.

Any two linearly independent functions y1 and y2 span the kernel of the linear differential operator, which is referred to as the annihilator operator:

$L\left[ x, \texttt{D} \right] = \texttt{D}^2 + p(x)\, \texttt{D} + q(x) , \quad \mbox{where} \quad p(x) = - \frac{y_1 y''_2 - y''_1 y_2}{y_1 y'_2 - y'_1 y_2} = - \frac{W' (x)}{W(x)} , \quad q(x) = \frac{y'_1 y''_2 - y''_1 y'_2}{y_1 y'_2 - y'_1 y_2} .$
In the above formula, W(x) denotes the Wronskian of two functions: $$W(x) = y_1 y'_2 - y'_1 y_2 .$$

Example: Let $$y_1 (x) = x \quad\mbox{and} \quad y_2 = 1/x$$ be two linearly independent functions on any interval not containing zero. Then the differential operator that annihilates these two functions becomes

$L\left[ x, \texttt{D} \right] = \texttt{D}^2 + \frac{1}{x}\, \texttt{D} + \frac{1}{x^2} . \qquad ■$

The situation becomes more transparent when we switch to constant coefficient linear differential operators. Any constant coefficient linear differential operator is a polynomial (with constant coefficients) with respect to the derivative operator $$\texttt{D} .$$ Therefore, a constant coefficient linear differential operator is generated by the characteristic polynomial $$L\left( \lambda \right) = a_n \lambda^n + \cdots + a_1 \lambda + a_0 .$$ It is well known from algebra that any polynomial with real coefficients of order n can be factors into simple terms

$L\left[ \lambda \right] = a_n L_1 [\lambda ] \, L_2 [\lambda ] \cdots L_s [\lambda ] ,$
where each multiple is either $$L_k \left( \lambda \right) = \left( \lambda - \alpha_k \right)^{m_k}$$ or $$L_k \left( \lambda \right) = \left[ \left( \lambda - \alpha_k \right)^{2} + \beta_k^2 \right]^{m_k} ,$$ with mk being the multiplicity of the root, either real $$\lambda = \alpha_k$$ or complex pair $$\lambda = \alpha_k \pm {\bf j} \beta_k .$$ So a constant coefficient linear differential operator is uniquely identified by its leading term and the roots of its characteristic polynomial. Of course, the complex case of second order differential operator $$L_k \left( \lambda \right) = \left( \lambda - \alpha_k \right)^{2} + \beta_k^2 = \left( \lambda - \alpha_k + {\bf j} \beta_k \right) \left( \lambda - \alpha_k - {\bf j} \beta_k \right)$$ can be reduced to two first order differential operators, but on expense of complex coefficients.

Unfortunately, most functions cannot be annihilated by a constant coefficients linear differential operator. For instance, the reciprocal of a linear function such as 1/x cannot be annihilated by a linear constant coefficient differential operator. But some could be; the corresponding set of functions for which we can determine an annihilator includes polynomials, exponentials times polynomials, and previous functions times either sine or cosine. To each of these function we assign a control number, summarized in the table below.

function  formula  control number
polynomial  $$p_n t^n + \cdots + p_1 t + p_0$$   0
polynomial times exponential  $$\left( p_n t^n + \cdots + p_1 t + p_0 \right) e^{at}$$  a
polynomial times exponential & sine  $$\left( p_n t^n + \cdots + p_1 t + p_0 \right) e^{at} \, \sin bt$$  a + b j
polynomial times exponential & cosine  $$\left( p_n t^n + \cdots + p_1 t + p_0 \right) e^{at}\, \cos bt$$  a + b j

A control number is just a root of characteristic polynomial that corresponds to the annihilating operator. For instance, if a control number is known to be α, we know that the annihilating polynomial for such function must be $$\left( \texttt{D} - \alpha \right)^m ,$$ for some positive integer m (called the multiplicity). Since the characteristic polynomial for any constant coefficient differential operator can be factors into simple terms, it is natural to start analyzing with some such simple multiple.

Let us start with a simple function---polynomial of degree n. It is known from calculus that such functions is annihilated by the (n+1)-th power of the derivative operator: $$\texttt{D}^{n+1} \left( p_n t^n + \cdots + p_1 t + p_0 \right) \equiv 0 .$$

Our next move is to show that the annihilator of the product of the polynomial and an exponential function can be reduced to an elementary case of just polynomials, discussed previously. Therefore, we consider a first order differential operator

$L\left[ \texttt{D} \right] = \texttt{D} - \alpha ,$
where α is a real number. We apply this operator to a product of a smooth function f(t) and the exponential function:
$L\left[ \texttt{D} \right] f(t)\, e^{\alpha \,t} = \texttt{D}\, f(t)\, e^{\alpha \,t} - \alpha \, f(t)\, e^{\alpha \,t} = f' (t)\, e^{\alpha \,t} + \alpha \, f(t)\, e^{\alpha \,t} - \alpha \, f(t)\, e^{\alpha \,t} .$
Therefore,
$\left( \texttt{D} - \alpha \right) f(t)\, e^{\alpha \,t} = e^{\alpha \,t} \,\texttt{D}\, f(t) = e^{\alpha \,t} \, f' (t) = f' (t)\, e^{\alpha \,t} .$
Applying the above formula to the constant function f(t) = 1, we get
$\left( \texttt{D} - \alpha \right) e^{\alpha \,t} = e^{\alpha \,t} \,\texttt{D}\, 1 = e^{\alpha \,t} \, 0 \equiv 0.$
So the function $$e^{\alpha \,t}$$ belongs to the kernel of the differential operator $$\left( \texttt{D} - \alpha \right) .$$ We apply the same formula to a monomial of degree n, $$f(t) = t^n .$$ This yields
$\left( \texttt{D} - \alpha \right) t^n \, e^{\alpha \,t} = e^{\alpha \,t} \,\texttt{D}\, t^n = e^{\alpha \,t} \, n\, t^{n-1} ,$
which is a monomial of degree n-1. Next application of the same operator gives
$\left( \texttt{D} - \alpha \right)^2 t^n \, e^{\alpha \,t} = \left( \texttt{D} - \alpha \right) e^{\alpha \,t} \, n\, t^{n-1} = e^{\alpha \,t} \, n(n-1)\, t^{n-2} .$
So to annihilate the product $$t^n \, e^{\alpha \,t} ,$$ where n is a positive integer, we need to apply $$L\left[ \texttt{D} \right] = \texttt{D} - \alpha$$ n+1 times to obtain
$\left( \texttt{D} - \alpha \right)^{n+1} t^n \, e^{\alpha \,t} = e^{\alpha \,t} \,\texttt{D}^{n+1}\, t^n = 0 .$
In particular,
$\left( \texttt{D} - \alpha \right)^{2} t \, e^{\alpha \,t} = 0 \qquad \mbox{and} \qquad \left( \texttt{D} - \alpha \right)^{2} \, e^{\alpha \,t} = 0 .$

Lemma: If f(t) is a smooth function and $$\gamma \in \mathbb{C}$$ is a complex number, then for any constant coefficient linear differential operator $$L[\texttt{D}] = a_n \texttt{D}^n + a_{n-1} \texttt{D}^{n-1} + \cdots + a_1 \texttt{D} + a_0$$ of degree n

$e^{-\gamma \,t} \, L \left[ \texttt{D} \right] f(t) \,e^{\gamma \,t} = e^{-\gamma \,t} \,L\left[ \frac{\text d}{{\text d}t} \right] f(t)\, e^{\gamma t} = L \left[ \texttt{D} + \gamma \right] f(t) .$

Lemma: If f(t) is a smooth function and $$\gamma \in \mathbb{C}$$ is a complex number, then for any constant coefficient linear differential operator $$L[\texttt{D}]$$ of degree n

$L\left[ \frac{\text d}{{\text d}t} \right] f(t)\, e^{\gamma t} = \left[ \frac{1}{n!} \,L^{(n)} (\gamma )\, f^{(n)} (t) + \frac{1}{(n-1)!} \,L^{(n-1)} (\gamma )\, f^{(n-1)} (t) + \cdots + P' (\gamma )\,f' (t) + P(\gamma )\, f(t) \right] e^{\gamma t} ,$
where $$L[\lambda ] = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_1 \lambda + a_0 . \qquad$$

Now we turn our attention to the second order differential operator

$L \left[ \texttt{D} \right] = \left( \texttt{D} - \alpha \right)^{2} + \beta^2 = \left( \lambda - \alpha + {\bf j} \beta \right) \left( \lambda - \alpha - {\bf j} \beta \right) .$
Using our approach, we seek a solution of $$L\left[ \texttt{D} \right] y =0$$ in the form $$y(t) = v(t)\, e^{\alpha\,t} ,$$ then function v will be a solution of the reduced equation
$L_0 \left[ \texttt{D} \right] v =0 \qquad\mbox{or} \qquad \left[ \texttt{D}^{2} + \beta^2 \right] v =0 .$
The latter equation is not hard to solve finding two linear independent solutions:
$v(t) =\cos \left( \beta t \right) \qquad\mbox{and} \qquad v(t) = \sin \left( \beta t \right) .$
This leads to two linearly independent solutions of the equation $$y'' - 2\alpha \, y' + \left( \alpha^2 + \beta^2 \right) y =0$$ to be
$y(t) = e^{\alpha\,t} \, \cos \left( \beta t \right) \qquad\mbox{and} \qquad y(t) = e^{\alpha\,t} \,\sin \left( \beta t \right) .$
Now we are ready to find the kernel of the differential operator of order 2n:
$L_n \left[ \texttt{D} \right] = \left[ \left( \texttt{D} - \alpha \right)^{2} + \beta^2 \right]^n ,$
where n is a positive integer. It is a linear combination of 2n functions
$y (t) = e^{\alpha\,t} \left( c_0 + c_1 t + \cdots + c_{n-1} t^{n-1} \right) \cos \left( \beta t \right) + e^{\alpha\,t} \left( C_0 + C_1 t + \cdots + C_{n-1} t^{n-1} \right) \sin \left( \beta t \right) ,$
with some constants c's and C's.