Wronskian determinant WronskianDeterminant 2002-02-19 01:41:25 2006-12-31 20:45:16 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Given functions $f_1, f_2, \dotsc, f_n$, then the \emph{Wronskian determinant} (or simply the Wronskian) $W(f_1, f_2, f_3, \dotsc, f_n)$ is the determinant of the square matrix \[ W(f_1, f_2, f_3, \dotsc, f_n) = \left\lvert\begin{array}{@{}ccccc@{}} f_1 & f_2 & f_3 & \cdots & f_n\\ f_1' & f_2' & f_3' & \cdots & f_n'\\ f_1'' & f_2'' & f_3'' & \cdots & f_n''\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ f_1^{(n-1)} & f_2^{(n-1)} & f_3^{(n-1)} & \cdots & f_n^{(n-1)}\\ \end{array}\right\rvert \] where $f^{(k)}$ indicates the $k$th derivative of $f$ (not exponentiation). The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent when there exists a non-trivial linear relation between them, a set of functions $\{f_1, f_2, f_3, \dotsc, f_n\}$ is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e., \[ a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t) = 0 \] for some $a_1, a_2, \dotsc, a_n$, not all zero, at any $t \in I$. Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian. \paragraph{Examples} Consider the functions $x^2$, $x$, and $1$. Take the Wronskian: \[ W = \left\lvert\begin{array}{@{}ccc@{}} x^2 & x & 1\\ 2x & 1 & 0\\ 2 & 0 & 0\\ \end{array}\right\rvert = -2 \] Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^2$ and $x$: \[ W = \left\lvert\begin{array}{@{}cc@{}} x^2 & x\\ 2x & 1\\ \end{array}\right\rvert = x^2 - 2x^2 = -x^2 \] Here $W = 0$ only when $x = 0$. Therefore $x^2$ and $x$ are independent except at $x = 0$. Consider $2x^2+3$, $x^2$, and $1$: \[ W = \left\lvert\begin{array}{@{}ccc@{}} 2x^2 + 3 & x^2 & 1\\ 4x & 2x & 0\\ 4 & 2 & 0\\ \end{array}\right\rvert = 8x - 8x = 0 \] Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since \[ 2x^2 + 3 = 2(x^2) + 3(1) \] Given $n$ linearly independant functions $f_1, f_2, \dotsc, f_n$, we can use the Wronskian to construct a linear differential equation whose solution space is exactly the span of these functions. Namely, if $g$ satisfies the equation \[ W(f_1, f_2, f_3, \dotsc, f_n, g) = 0, \] then $g = a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t)$ for some choice of $a_1, a_2, \dotsc, a_n$. As a simple illustration of this, let us consider polynomials of at most second order. Such a polynomial is a linear combination of $1$, $x$, and $x^2$. We have \[ W (1, x, x^2, g(x)) = \left| \begin{matrix} 1 & x & x^2 & g(x) \\ 0 & 1 & 2x & g'(x) \\ 0 & 0 & 2 & g''(x) \\ 0 & 0 & 0 & g'''(x) \end{matrix} \right| = 2 g''' (x) .\] Hence, the equation is $g''' (x) = 0$ which indeed has exactly polynomials of degree at most two as solutions.