Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

In Wikipedia it says that if the Wronskian of two function is 0 everywhere it does not imply they are linearly dependent.

However, in books treating differential equations it seems that, if the two functions in question are solutions of a linear differential equation, then the condition $W=0$ does indeed imply they are linearly dependent.

In the web I found only that if two function are real analytic the condition $W=0$ guarantees their dependence, but there is no reason a solution to a linear ODE should be real analytic.

So I'd like to find a rigorous proof, possibly for the more general case, with $n$ functions.

Thank you in advance

share|cite|improve this question
Have you heard about Abel-Liouville's formula? – Josué Tonelli-Cueto May 10 '12 at 22:16
If $x$ is the solution to a 1 dim., $n$th order linear ode, the state $\sigma(t) = (x(t), x^{(1)}(t),...,x^{(n-1)}(t))$ encapsulates all relevant information about $x$ at the time $t$. Suppose $t'>t$, then the ode defines a linear map $\Phi(t',t)$ such that $\sigma(t') = \Phi(t',t)\sigma(t)$. Under reasonable smoothness conditions, the ode is reversible, ie, $\sigma(t)$ can be obtained by running the ode backwards starting at $\sigma(t')$, or in other words $\sigma(t) = \Phi(t,t')\sigma(t')$. The point is that linear independence at time $t$ is equivalent to linear independence at time $ t'$. – copper.hat May 10 '12 at 22:55
This is treated in Tenenbaum & Pollard's book, on page 779 and 780. The theorem starts on the very bottom of 779, maybe you can make do with just what is available in the preview on 780? The theorem states that if the functions are solutions of a homogeneous linear differential equation on some interval $I$ and if the Wronskian is identically zero on $I$, then the set of functions is linearly dependent on $I$. – process91 May 11 '12 at 4:02
@BillCook Thanks, I've read the article but, at least to my understanding, it simply implies that $W$ must be zero or nonzero everywhere. it says nowhere that if the Wronskian is 0 everywhere then the two solutions are independent – Cauchy May 11 '12 at 15:15

Your Answer


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

Browse other questions tagged or ask your own question.