Wronskian determinant


Given functionsMathworldPlanetmath f1,f2,,fn, then the Wronskian determinant (or simply the WronskianDlmfMathworld) W(f1,f2,f3,,fn) is the determinantDlmfMathworldPlanetmath of the square matrixMathworldPlanetmath

W(f1,f2,f3,,fn)=|f1f2f3fnf1f2f3fnf1′′f2′′f3′′fn′′f1(n-1)f2(n-1)f3(n-1)fn(n-1)|

where f(k) indicates the kth 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 relationMathworldPlanetmath between them, a set of functions {f1,f2,f3,,fn} is also said to be dependent over an interval I when there exists a non-trivial linear relation between them, i.e.,

a1f1(t)+a2f2(t)++anfn(t)=0

for some a1,a2,,an, not all zero, at any tI.

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 equationMathworldPlanetmath are independent, we may use the Wronskian.

Examples

Consider the functions x2, x, and 1. Take the Wronskian:

W=|x2x12x10200|=-2

Note that W is always non-zero, so these functions are independent everywhere. Consider, however, x2 and x:

W=|x2x2x1|=x2-2x2=-x2

Here W=0 only when x=0. Therefore x2 and x are independent except at x=0.

Consider 2x2+3, x2, and 1:

W=|2x2+3x214x2x0420|=8x-8x=0

Here W is always zero, so these functions are always dependent. This is intuitively obvious, of course, since

2x2+3=2(x2)+3(1)

Given n linearly independant functions f1,f2,,fn, 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(f1,f2,f3,,fn,g)=0,

then g=a1f1(t)+a2f2(t)++anfn(t) for some choice of a1,a2,,an.

As a simple illustration of this, let us consider polynomials of at most second order. Such a polynomial is a linear combinationMathworldPlanetmath of 1, x, and x2. We have

W(1,x,x2,g(x))=|1xx2g(x)012xg(x)002g′′(x)000g′′′(x)|=2g′′′(x).

Hence, the equation is g′′′(x)=0 which indeed has exactly polynomials of degree at most two as solutions.

Title Wronskian determinant
Canonical name WronskianDeterminant
Date of creation 2013-03-22 12:22:59
Last modified on 2013-03-22 12:22:59
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 13
Author rspuzio (6075)
Entry type Definition
Classification msc 34-00
Synonym Wronskian
Related topic GrammianDeterminant