# Question on the Wronskian

1. Feb 16, 2012

### τheory

1. The problem statement, all variables and given/known data
Given the two functions:
$f(t) = t$
$g(t) = |t|$

Use the Wronskian to determine if the two functions are dependent or independent.

2. The attempt at a solution
I have already found the correct answer to this, which is that it is independent but I have some questions as to how this is. When I first tried to solve this I found that it was dependent, based on the following reasoning:

$W[f(t),g(t)] = \left| {\begin{array}{cc} t & |t| \\ 1 & \pm 1 \\ \end{array} } \right|$

Calculating the determinant:
$W[f(t),g(t)] = (t)(\pm 1) - (|t|)(1)$
$W[f(t),g(t)] = (t)(\pm 1) - |t|$

Given that $t<0$ and $0<t$ will determine the sign of $g'(t) = \frac{d}{dt} |t|$, then utilizing the following conditions:

If $t = +1$, then $g'(t) = +1$.
If $g'(t) = +1$, then:
$W[f(t),g(t)] = (t)(\pm 1) - |t|$
$W[f(1),g(1)] = (1)(+1) - |1| = 0$

Likewise, if $t = -1$, then $g'(t) = -1$.
If $g'(t) = -1$, then:
$W[f(t),g(t)] = (t)(\pm 1) - |t|$
$W[f(-1),g(-1)] = (-1)(-1) - |-1| = 1 - 1 = 0$

Thus, based on the above reasoning, I thought the answer was dependent, but in fact, it is independent; can someone point out my mistake?

2. Feb 16, 2012

### Deveno

the Wronskian is undefined at t = 0, and is therefore not identically 0 everywhere.

indeed suppose that there existed a,b in R, with af(t) + bg(t) identically 0, for all t.

for t > 0, we get that at + bt = 0, and since t ≠ 0, a + b = 0.

for t < 0, we have that at - bt = 0, so that a - b = 0.

therefore, 2a = 0, so a = 0, and thus b = 0, so f and g are linearly independent over R.

3. Feb 16, 2012

### Dick

I would say you can't really apply the Wronskian if the interval you are testing on contains 0. |t| isn't differentiable at t=0. But they are linearly independent since there are no constants such that a*t+b*|t|=0 for all t except for a=b=0. On the other hand as your Wronskian suggests, if you are working on an interval of nonnegative numbers, 1*t+(-1)*|t|=0 or an interval of nonpositive numbers, 1*t+1*|t|=0. So they are linearly dependent over some intervals.

Last edited: Feb 16, 2012
4. Feb 16, 2012

### Deveno

i think it fair to point out that the Wronskian test is a negative test:

Wronskian not identically 0 => independent functions.

Wronskian identically 0: might be independent, might not.

(counter-example: f(t) = t2, g(t) = t|t|).

5. Feb 16, 2012

### Dick

Right. That's why I was careful to say "Wronskian suggests" instead of "Wronskian shows". Good point to make explicit though.

6. Feb 16, 2012

### LCKurtz

To add a little more, W(f,g) identically 0 does imply dependence if f and g are solutions of a 2nd order linear DE. To get away from the derivative doesn't exist example, consider {t,t2} on [-1,1]. Here their Wronskian is W = t2. Since that doesn't satisfy that it is either never 0 or identically 0 on the interval, what it shows, besides that they are independent, is that they aren't solutions to a second order linear DE.