ilovephysics.com :: free online physics help, tutorials, forums, classes, misconceptions in / Derivation of the Wronskian

M@Man · 2005-09-08 15:58:31

I was doing my Applied Math homework the other day and ended up accidentally deriving the Wronskian, so I thought I'd throw it out there as a challenge to others.

The Wronskian is a tool used to tell if a set of functions is linearly independent. Say you have $n$ functions $f_{n}$ . The Wronskian is a matrix in which the (i,j) element is the $(i-1)^{th}$ derivative of $f_{j}$ . In other words, you construct a matrix in which the first row is your list of functions, the second row is the first derivative of the functions in the first row, the third row is the second derivative of the functions, and so on.

Here's the challenge: prove that your set of functions is linearly independent ~~if and only if~~ the determinant of the Wronskian matrix is nonzero.

---> corrected by kylekatarn: "prove that your set of functions is linearly independent if the determinant of the Wronskian matrix is nonzero. (There are cases in which the determinant of the Wronskian matrix is zero, but the functions are still linearly independent.)

Tools you will need:
-the definition of linear independence
-some skills at linear algebra and matrix factorization
-a little bit of cleverness to generate your system of equations

Good luck!

By the way, the Wronskian is used primarily in the study of Differential Equations, because it can be shown using linear algebra that a linear differential equation of order $n$ has $n$ linearly independent solutions. So, if you find 2 solutions to a 2nd order differential equation, you'll need to test to see if they are linearly independent before you can conclude that you have all the independent solutions. Once you've done that, you can take a linear combination of your independent solutions to form the general solution to the linear differential equation.

Last edited by M@Man (2005-09-08 23:56:58)

kylekatarn · 2005-09-08 20:34:24

Aren't there some special cases where the Wronskian is zero but the functions are linearly independent?

M@Man · 2005-09-08 23:48:40

Oops, you're absolutely right, kylekatarn. I erred when I said "if and only if." I should have simply said if. A good case that proves your point is the example of $f(x)=x^3$ vs. $| x^3 |$ . These functions are linearly independent, but their Wronskian is identically zero. I'll fix my original post to say "if" instead of "if and only if."

Last edited by M@Man (2005-09-08 23:54:26)

kylekatarn · 2005-09-09 03:44:14

ok! no problem:)

daphysicist · 2006-02-28 01:18:41

$x=y$
$x^2=xy$
$y+x=y$
$y+y=y$
$2=1$

Last edited by daphysicist (2006-02-28 01:24:08)

M@Man · 2006-02-28 18:41:38

daphysicist,

1.) Why are you posting that problem here and not in its own thread?

2.) It looks like you skipped some steps between your second and third lines. I believe the traditional argument goes something like:

i.) Suppose $x=y$ .
ii.) Then $x^2=xy$
iii.) $x^2 - y^2=xy - y^2$
iv.) $(x+y)(x-y)=y(x-y)$
v.) $x+y = y$
vi.) $y+y = y$ by hypothesis. Thus
vii.) $2y = y$
viii.) $2=1$ .

The misstep, of course - which can be seen by supposing $x = y = 1$ , for instance, occurs between steps (iv.) and (v.), in which $(x-y)$ is divided from both sides, and, since $x=y$ , $x-y = 0$ , so you are dividing by zero, which is illegal. Still, it's a fun "proof."