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According to Wikipedia, if the Wronskian of two functions is always zero, then they are not necessarily linearly dependent.

But it seems that if the two functions are solutions of the same homogeneous second-order linear differential equation, then the condition $W[y_1, y_2](t) = 0$ does indeed imply that they are linearly dependent.

Online, I found that if two functions are real analytic and their Wronskian is identically zero, then they are necessarily linearly dependent. But there is no reason that the solutions to a linear differential equation should be real analytic.

How can we prove that the condition $W[y_1, y_2](t) = 0$ implies the linear dependence of $y_1(t)$ and $y_2(t)$? More generally, how can we prove that the condition $W[y_1, \ldots, y_n](t) = 0$ implies the linear dependence of $y_1(t), \ldots, y_n(t)$?

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    Have you heard about Abel-Liouville's formula?2012-05-10
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    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'$.2012-05-10
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    http://en.wikipedia.org/wiki/Abel%27s_identity2012-05-10
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    This is treated in [Tenenbaum & Pollard's book](http://books.google.com/books?id=29utVed7QMIC&printsec=frontcover#v=onepage&q&f=false), 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 $I$.2012-05-11
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    @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 independent2012-05-11
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    @MichaelBoratko Thanks, it is exactly what I expected. In practice the key seems to be the uniqueness theorem.2012-05-11
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    @copper.hat I don't understand what the $\Phi$ is and what you intend for linear. Let's say I have a solution to $y''=-y$, for example $\cos(t)$. Now $\cos(0) = 1$ and $\cos(\pi/2)=0$ so it seems that $\Phi(0,\pi/2)=0$. However $\cos(\pi)=-1$ so $\Phi(0,\pi)=-1\ne 0 \ne 2\Phi(0,\pi/2)$.2012-05-11
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    @Cauchy: In your example, $\sigma(t) = (y(t), y'(t)) \in \mathbb{R}^2$, and $\Phi(t,t') = \begin{bmatrix} \cos (t-t') & \sin (t-t') \\ -\sin (t-t') & \cos (t-t') \end{bmatrix}$. The particular solution you mention corresponds to (after some algebra) an initial state of $\sigma_0 = \binom{1}{0}$.2012-05-11
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    @copper.hat Thanks, you are right. So, if I understood well, you basically say $\Phi(t,t')$ is a linear transformation $\sigma \rightarrow \sigma'$ for each $t'>t$. Basically you are transforming the initial conditions into the solution of the ODE at a particular $t'$. Then you say that the columns of the Wronskian matrix represent actually particular states and that if they are independent at a time $t$ they will be also at a time $t'$, but how does this relate to the independence of the functions? E.g. if for each $t$ $(f,f')$ and $(g,g')$ are independent, $k_1f+k_2g$ infers $k_1=k_2=0$?2012-05-11
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    If $\sigma_1,...,\sigma_n$ are solutions (well, you would call the solutions to be the $1$st element of each $\sigma_k$), then let $W(t) = \det \begin{bmatrix} \sigma_1(t) & \cdots & \sigma_n(t) \end{bmatrix}$. Then we have $W(t) = \det(\Phi(t,t')) W(t')$. In particular, if the Wronskian is $0$ anywhere, then it implies that the initial conditions are dependent, which implies that the solutions are dependent.2012-05-11
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    @copper.hat OK, I would be glad to accept your answer, if it were an answer and not a comment2012-05-12
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    No problem, glad to help.2012-05-12

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