Question on the Wronskian

  • Thread starter τheory
  • Start date
  • #1
43
0

Homework Statement


Given the two functions:
[itex]f(t) = t[/itex]
[itex]g(t) = |t|[/itex]

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:

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

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

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

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

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

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

Answers and Replies

  • #2
Deveno
Science Advisor
908
6

Homework Statement


Given the two functions:
[itex]f(t) = t[/itex]
[itex]g(t) = |t|[/itex]

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:

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

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

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

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

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

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

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
Dick
Science Advisor
Homework Helper
26,263
619
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:
  • #4
Deveno
Science Advisor
908
6
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
Dick
Science Advisor
Homework Helper
26,263
619
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|).

Right. That's why I was careful to say "Wronskian suggests" instead of "Wronskian shows". Good point to make explicit though.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
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|).

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.
 

Related Threads on Question on the Wronskian

  • Last Post
Replies
3
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
988
Replies
7
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
552
  • Last Post
Replies
2
Views
983
  • Last Post
Replies
4
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
929
Top