# Wronskian property

1. Dec 5, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
If two differentiable functions are linearly dependent on the interval I, then their Wronskian is identically zero on I.

2. Relevant equations

3. The attempt at a solution

$y_1 = Cy_2$

$\displaystyle \frac{y_1}{y_2} = C$

$\displaystyle \left( \frac{y_1}{y_2} \right)' = 0$

$\displaystyle \left( \frac{y_1}{y_2} \right)' = \frac{y_1' y_2 - y_2' y_1}{y_2^2}= 0$

$\displaystyle y_1 y_2' - y_2 y_1'= 0$

This seems fine, but how can I justify my second step? How can I divide by $y_2$ without knowing if $y_2$ is zero somewhere on the interval I?

2. Dec 5, 2016

### LCKurtz

Why don't you just substitute your equations $y_1 = Cy_2$ directly into the Wronskian and see what happens? No division necessary.

3. Dec 5, 2016

### Staff: Mentor

$y_1 = Cy_2$ means, if $y_2(x_0) = 0$ then $y_1(x_0) = 0$ as well and $(y_1 y_2' - y_2 y_1')(x_0) = 0$ without any intermediate steps.
If $y_2(x_0) \neq 0$ then I suppose we may assume, that $y_2(x) \neq 0$ in an open neighborhood of $x_0$ so that your argument can be applied.

4. Dec 5, 2016

### Mr Davis 97

Those answer my question. Also, I have one more quick question about the Wronskian, from the linearly independent side of things.

I have seen two theorems relating the Wronskian and linear independence.

1. Two differentiable functions y1 and y2 are linearly independent on I if and only if their Wronskian is never zero on I.

2. If the Wronksian of two functions is not zero for some point on I, then they are linearly independent.

These seem like that their saying different things. One says that the Wronskian is never zero on the interval I, and other says that it is not zero for some point on I. Also, one has an iff and the other is just if. Are these the same or different theorems?

5. Dec 5, 2016

### Staff: Mentor

In the following I used $\left( A \Rightarrow B \right) \Leftrightarrow \left( \lnot B \Rightarrow \lnot A \right)$

The first statement consists of two parts:
1.a.) $y_1 \neq Cy_2 \Longrightarrow \;\forall \; x\in I \;:\; W(x) \neq 0$
1.b.) $\;\forall \; x\in I \;:\; W(x) \neq 0 \Longrightarrow y_1 \neq Cy_2$
This (1.b.)) is equivalent to
1.c.) $y_1 = Cy_2 \Longrightarrow \; \exists \; x\in I \;:\; W(x) = 0$

The second statement says
2.) $\; \exists \; x\in I \;:\; W(x) \neq 0 \Longrightarrow y_1 \neq Cy_2$
This is equivalent to
3.) $y_1 = Cy_2 \Longrightarrow \;\forall \; x\in I \;:\; W(x) = 0$
which you have proven above. But 3.) implies 1.c.) which is equivalent to 1.b.)

So it remains open 1.a.) or the equivalent statement
4.) $\; \exists \; x\in I \;:\; W(x) = 0 \Longrightarrow y_1=Cy_2$

Are you sure 4.) or 1.a.) is true? What about $y_1=x\, , \,y_2=x^2$ on $[-1,1]$ at $x_0=0$?

Last edited: Dec 5, 2016
6. Dec 6, 2016

### vela

Staff Emeritus

I remember being taught the same thing when I took calculus. It seemed wrong to me then, and it still does now. But I have seen that statement repeated elsewhere, and it makes me wonder why. If I'm not missing some subtlety, then why does this misconception seem to persist?

Fresh_42 suggested one counterexample. Here's another related to Fourier analysis. The functions $y_1 = \sin x$ and $y_2 = \sin 2x$ on $[0,2\pi]$ are linearly independent, but their Wronskian equals $2\sin^3 x$, which clearly vanishes at three points in the interval. I think the supposed theorem is just wrong.

7. Dec 6, 2016

### LCKurtz

The relevant theorem is this: The Wronskian of two solutions $y_1(x)$ and $y_2(x)$ of the linear DE $y'' + p(x)y' + q(x)y = 0$ is non-zero or vanishes identically. Here $p(x)$ and $q(x)$ are assumed to be continuous. Consequently, in this setting it is true that linear independence implies non-zero Wronskian and conversely.

What this implies is that the preceding counterexamples are not solutions of such a second order DE. For example, consider $y_1(x) = x$ and $y_2(x) = x^2$. These are linearly independent and their Wronskian is zero at $x = 0$. And they are solutions of $x^2y'' -2xy' +2x^2y = 0$. But you can't put that DE into the above form because you must divide by $x^2$ to get a leading coefficient $1$. So the theorem doesn't apply on any interval containing $x=0$, and the examples aren't counterexamples at all.

8. Dec 6, 2016

### vela

Staff Emeritus
Ah, that makes sense. Thanks for clearing up something I've wondered about for literally decades (but couldn't be bothered to look into more carefully).