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 $y_1(t), \ldots, y_n(t)$?

  • $\begingroup$ Have you heard about Abel-Liouville's formula? $\endgroup$ May 10, 2012 at 22:16
  • $\begingroup$ 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'$. $\endgroup$
    – copper.hat
    May 10, 2012 at 22:55
  • $\begingroup$ en.wikipedia.org/wiki/Abel%27s_identity $\endgroup$
    – Bill Cook
    May 10, 2012 at 23:44
  • 1
    $\begingroup$ @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 $\endgroup$
    – Cauchy
    May 11, 2012 at 15:15
  • 2
    $\begingroup$ Possible duplicate of Intuition of Wronskian determinant and linear independence $\endgroup$
    – L F
    Jul 19, 2019 at 19:24

1 Answer 1


Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e. \begin{equation*} \left| \begin{array}{cccc} y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\ y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0) \end{array} \right| = 0. \end{equation*} Then the corresponding matrix is not invertible, and the system of equations \begin{array}{c} c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\ c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\ \end{array} has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.

Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.

These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.

We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.

Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations \begin{array}{c} c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\ c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\ \end{array} has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .