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I'm studying a book of differential equations which says that if the Wronskian of two functions is zero then these functions are linearly dependent. the author doesn't prove it, he simply said as a easy consequence of basic properties of determinants, I tried to prove by myself without success.

I need help.

thanks a lot.

user42912
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    If that is how it is written, then it is rather misleading. See my answer here http://math.stackexchange.com/questions/429995/question-about-linear-dependence-and-independence-by-using-wronskian/430054#430054. When two functions a linearly dependent, the Wronskian is zero. When the Wronskian is zero, you cannot say anything about linear dependence of the functions. – Amzoti Jul 06 '13 at 21:36
  • The statement also occurs in Courant's Differential and Integral Calculus, volume II, and perhaps this authority accounts for the statement in some other books. But there is an error in the proof given there. – Bob Terrell Jul 07 '13 at 13:17
  • This can be generalised if one of the functions is not zero in the interval given. Otherwise, the statement is wrong. Consider, $x^2$ and $x|x|$. – sonu Apr 15 '18 at 04:34

3 Answers3

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The statement holds if $f_1$ and $f_2$ are solutions of a linear differential equation. The functions $x^3$ and $|x|^3$ are linearly independent in any open interval containing $0$, but their Wronskian is identically equal to $0$.

David C. Ullrich
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Julián Aguirre
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This appears to be false: that is, you wan have the Wronskian of two functions being identically zero on an interval without those two functions being linearly dependent. See for instance the Wikipedia article:

A common misconception is that $W = 0$ everywhere implies linear dependence, but Peano (1889) pointed out that the functions $x^2$ and $|x|x$ have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0.

However, it holds with additional assumptions on the functions (eg, one of them never takes value $0$ — cf. same link).

Clement C.
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  • If they take an isolated zero, the statement could still be recovered by a continuation at one point. This is pretty standard throughout calculus, because most of these ideas were developed before the modern version of a function. Back in the 18 and 19th centuries a function was thought of as some equation. This is why most textbooks sweep things like "divide by zero" under the rug. – Joel Jul 06 '13 at 22:03
  • In any case, I believe the context here is for functions as solutions to a differential equation. It was mentioned that this comes from the middle of a problem. So Julian's answer probably is appropriate to this point. – Joel Jul 06 '13 at 22:05
  • @Clement The link is not working anymore :/ – Sam Mar 10 '20 at 16:25
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    @Sabhrant Dang. Well, Wikipedia has a reference for this (Peano 1889) and a discussion. I edited my answer accordingly. – Clement C. Mar 10 '20 at 16:31
  • Can you please explain the case when we impose an additional condition, that one of the functions is non-zero. How does L.D follow from this? – Sam Mar 10 '20 at 16:38
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Let $f_1, f_2$ be two differentiable functions. The Wronskian of these functions is given by: $$f_1 f_2^\prime - f_2 f_1^\prime$$ and we are told that this difference is $0$.

This tells us that $$\frac{f_2^\prime}{f_2} = \frac{f_1^\prime}{f_1}$$ and so by integration we have $\ln(|f_1|) = \ln(|f_2|) + C$ for some constant $C$. This yields $$f_1 = \pm e^C \cdot f_2.$$

Hence these functions are linearly dependent.

Joel
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  • why the integration of $\frac{f_2^\prime}{f_2} = \frac{f_1^\prime}{f_1}$ give us $\ln(|f_1|) = \ln(|f_2|) + C$? – user42912 Jul 06 '13 at 21:25
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    @user42912 Differentiate the equality $\ln (|f_1|)=\ln (|f_2|)+C$. – Git Gud Jul 06 '13 at 21:27
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    What if there exists $x$ such that $f_1(x)=0$ or $f_2(x)=0$? In general, don't you have to argue why you can divide by $f_1$ and $f_2$? (since what you do is only valid if they're non-zero on the whole domain). – Clement C. Jul 06 '13 at 21:27
  • It's true I left this a little sketchy. We were not given much about the conditions on the functions involved here. For instance if $f_1(x) = 0$ for all $x \in [0,3/4]$ and $f_2(x) = 0$ for all $x \in [1/4,1]$, but both functions are nonzero on $(3/4,1]$ and $[0,1/4)$ repsectively, the Wronskian will still be zero on $[0,1]$. However these functions are clearly not linearly dependent. – Joel Jul 06 '13 at 21:43
  • "Hence these functions are linearly dependent." Only if $k=\pm e^C$ is the same for the entire interval under consideration. – user164587 Oct 18 '16 at 13:45
  • The problem being that although $C$ is constant the value of $\pm$ may vary from point to point. (Don't try to fix this; the example in the other answer shows that what you're proving is false.) – David C. Ullrich Jul 19 '19 at 12:10
  • @DavidC.Ullrich you are right, of course, and this is what user164587 was mentioning in his comment. I haven't bothered updating this answer in a while. What we would be looking for would be assumptions such as the functions being continuously differentiable and non-zero on a compact set. Since this was an intro ODE class, I just wanted to give the student a quick idea of how you would approach this problem. – Joel Jul 20 '19 at 16:51