Let $f$ and $g$ be the solutions of the homogeneous linear equation:

$$y'' + p(x)y' + q(x)y = 0$$

and $p(x)$ and $q(x)$ are continuous in segment $I$.

Is it true, that if the wronskian of $f$ and $g$ is zero for every $x$ in $I$, than the two functions are linear dependent?

I know that if the wronskian isn't zero for one $x$ in segment $I$, than it's not zero for every $x$ (using Abel's identity), and they are linear independent as a result.

Please give me examples if I can't infer that they are linear dependent.


$$y_1y_2'-y_1'y_2=0$$ $$\frac{y_1y_2'-y_1'y_2}{y_1^2}=0$$ $$\frac{y_2}{y_1}=C$$

Unless I'm mistaken, this appears to imply if the Wronskian is $0$, the solutions are linearly dependent.

  • $\begingroup$ I think we have to use the fact that they are solutions of an equation and that p and q are continuous. take for example x and |x| in R. the wronskian is zero but they are linear independent. $\endgroup$ Jul 5 '15 at 9:06
  • $\begingroup$ @NoamMansur but $|x|$ isn't a solution to a differential equation $\endgroup$
    – krvolok
    Jul 5 '15 at 9:51
  • $\begingroup$ My point is I can do the same thing Mike did with x and |x| and the two functions are independent. $\endgroup$ Jul 5 '15 at 10:23

It is existence and uniqueness results for ODEs that give you what you want.

Theorem [Existence and Uniqueness]: Let $p$ and $q$ be continuous functions on $I=[a,b]$. Given $x \in [a,b]$ and constants $A$, $B$, there exists a unique twice continuously differentiable solution $y$ of $$ y''+py'+qy = 0,\\ y(x)=A,\;\; y'(x)=B. $$

Suppose $y_1$ and $y_2$ are solutions of the differential equation. The Wronskian is $$ w(y_1,y_2)(x)= \left|\begin{array}{cc}y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x)\end{array}\right| $$ The Wronskian vanishes at some $x \in [a,b]$ iff there is a non-zero vector with components $A$, $B$ such that $$ \left[\begin{array}{cc}y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x)\end{array}\right]\left[\begin{array}{c}A \\ B\end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right]. $$ Equivalently, $y=Ay_1 + by_2$ is a solution of the ODE with $y(x)=0,y'(x)=0$. Because solutions are unique, then $y\equiv 0$, because $0$ is another such solution. Hence, the Wronskian vanishes at some $x\in[a,b]$ iff there are constants $A$ and $B$ (not both $0$) such that $$ Ay_1 + By_2 \equiv 0. $$ You can see how this theorem generalizes to $n$-th order ODEs as well.

  • $\begingroup$ Thank you, very helpful. Now I really get it. $\endgroup$ Jul 8 '15 at 20:47
  • $\begingroup$ That is what I figured out along the way that made me finally get it, too. Glad I could help. $\endgroup$ Jul 8 '15 at 21:02

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