# Concept of the wronksian

1. Nov 5, 2006

### CPL.Luke

hmm I don't think I have the concept of the wronksian down all the way yet.

mainly I'm having trouble seeing how the the wronksian not equaling zero implies that all the solutions are linearly independant of eachother.

2. Nov 6, 2006

### Alpha Russ Omega

Wronskian

Remember when back in Algebra you had to decide wether two lines were parallel, or the same line? When the distance between the two lines you had to work with was zero, you decided that this could only be when the two lines are one and the same, point for point. Think of that analogy here.

First of all, getting zero for a Wronskian does not imply linear dependence - that's only when the Wronskian is identically zero - which means zero for all X. [Pretty strict condition]. When you use the Wronskian to check for linear independence, what you are actually doing is checking if the two solutions are multiples of each other. If you calculate the Wronksian and get identically zero, that is, zero for every point on the interval, that means that that the functions are linearly dependant. If the Wronskian yields anything else, the solutions are linearly independant and they are not multiples of each other. I hope this helps!

3. Nov 6, 2006

### HallsofIvy

Staff Emeritus
If you have n functions, fi(x), and n numbers, ai, such that a1f1(x)+ ...+ anfn(x)= 0 for all x, then you can make that into n equations for the ais by taking n derivatives (to get n equations) and taking x= 0 (or any other fixed number). The Wronskian, in that case, evaluated at x= 0 or whatever number, is simply the coefficient matrix. As long as that matrix does not have 0 determinant, you could find the unique solution multiplying by its inverse matrix. Obviously, ai= 0 for all i is a solution. If the determinant of the Wronskian is not 0, it is the only solution and so the functions are independent. If the determinant is 0, then there is no unique solution. (In general, if the determinant of the coefficient matrix for n equations is 0, either there is no solution or there are an infinite number of solutions. In this case, ai= 0 is an obvious solution so there must be an infinite number of solutions.)

4. Nov 7, 2006

### CPL.Luke

but then why do we differentiate the function to fill up the matrix?

5. Nov 7, 2006

### Galileo

Then you can use it to show a homogeneous linear DE of order N cannot have more than N linear independent solutions by just taking a linear combination of rows. If y is a solution to the DE, then the D.E. for y is just a coefficient of the vector from a linear combination of the rows of the Wronskian

Last edited: Nov 7, 2006
6. Nov 8, 2006

### HallsofIvy

Staff Emeritus
All I can do is repeat what I said before:
If you have n functions, fi(x), and n numbers, ai, such that a1f1(x)+ ...+ anfn(x)= 0 for all x, then you can make that into n equations for the ais by taking n derivatives (to get n equations) and taking x= 0 (or any other fixed number).