Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was studying the book of ODE of Simmons, and he says that the Wronskian ($W$) is $0$ iff the solutions are LI. Even he present a proof, a counterexample that $W = 0$ does not implies LD is $$ \left| x \right|, \ x $$ and he used that to prove the theorem, but since it is not well tested I can not follow, in addition to not really understand what he did.

Anyway, I want to know some proof of this fact, the fact that the solutions form a vector space (easy) of dimension $n$ (difficult to me T_T).


If someone knows some book to study ODE (for a beginner), I'll appreciate it.

share|cite|improve this question
You have 8 more questions with no accepted answers. This fact may discourage people to answer you ( – Ilya Sep 25 '11 at 21:48
The pair $|x|,x$ is not a counterexample. Either your interval has $0$ in which case the wronskian doesn't exist there, or it doesn't include $0$ and then they are actually linearly dependent. However, the pair $x^2,x|x|$ both have continuous derivatives, their wronskian vanishes everywhere, and they are not linearly independent on intervals that include $0$. Generally, $W=0$ can only guarantee linear dependence if the functions are analytic. (I don't think the other direction requires analyticity, though.) – anon Sep 25 '11 at 23:01
What does "v.e" in the title mean? I don't see any pair of neighboring words in the question that even start with those letters. I suppose "LI" means linearly independent, but what is "LD"? – Henning Makholm Sep 25 '11 at 23:03
@Henning: Obviously LD means linearly *de*$\text{}$pendent. But I have no idea what v.e means either. It's possible OP mistyped "VS" for vector space (as 'e' and 's' are right next to each other on the keyboard). And I think OP's asking (1) how the mentioned proof could be valid given the stated 'counterexample' (which I addressed), (2) for help understanding the proof, and (3) for other ODE books for a beginner. – anon Sep 25 '11 at 23:15
@anon, of course "linearly dependent". It was not as easy to figure out because it is not clearly a predicate where it appears. – Henning Makholm Sep 25 '11 at 23:20

You might take a look at Birkhoff and Rota, "Ordinary Differential Equations". It treats the Wronskian in section II.3.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.