Information About

Wronskian

Given a set of n functions f₁, ..., f_n, the Wronskian W(f₁, ..., f_n) is given by:

:

W(f_1, \ldots, f_n) =
\begin{vmatrix} 
f_1 & f_2 & \cdots & f_n \
f_1' & f_2' & \cdots & f_n' \
dots & dots & \ddots & dots \
f_1^{(n-1)} & f_2^{(n-1)} & \cdots & f_n^{(n-1)}
\end{vmatrix}

That is, it is the Determinant of the Matrix constructed by placing the functions in the first row, the first derivative of each function in the second row, and so on through the n-1 derivative, thus forming a Square Matrix sometimes called a fundamental matrix.

In a second-order linear differential equation, the Wronskian can be computed more easily by Abel's Identity .

THE WRONSKIAN AND LINEAR INDEPENDENCE

The Wronskian can be used to determine whether a set of Differentiable functions is Linearly Independent on a given Interval :

If the Wronskian is non-zero at some point in an interval, then the associated functions are ''linearly independent'' on the interval.

Differential Equation

W = 0

If a set of functions is ''linearly dependent'' on an interval, then the corresponding Wronskian is ''uniformly zero'' on the interval.

In fact the two bulleted statements are Logically Equivalent (by Transposition ); they are simply alternative statements of the same truth. A proof of the theorem is given below.

EXAMPLES

Consider the functions $x^2,$ $x,$ and $1,$ defined for ''x'' a Real Number . Take the Wronskian:

::

W = 
\begin{vmatrix}
x^2 & x & 1 \
2x & 1 & 0 \
2 & 0 & 0
\end{vmatrix}
= -2.

:We see that

W

is not uniformly zero, so these functions must be linearly independent.

Consider the functions $2x^2+3$ , $x^2$ , and $1$ . These functions are clearly dependent, since $2x^2 + 3 = 2(x^2) + 3(1).$ Thus, the Wronskian must be zero, which follows by a quick calculation:

::

W = 
\begin{vmatrix}
2x^2 + 3 & x^2 & 1 \
4x & 2x & 0 \
4 & 2 & 0
\end{vmatrix}
= 8x-8x = 0.

x^3

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