When studying Elementary Differential Equations by William, I found trouble understanding Theorem 5.1.5

It says the two solutions are linearly independent iff their Wronskian is never zero, but I think they can still be linearly independent even if the Wronskian is zero for some $x$.

In the proof, when $W(x_0)=0$, Theorem 5.1.4 is used to show $W\equiv 0$.

This theorem is the Abel's identity. It seems flawless, until I saw this answer. So $p(x)$ must be continuous on $(a,b)$, but it is not as long as $W(x_0)=0$, so we shouldn't use Abel's identity. This is because $$ y_1''+p(x)y_1'+q(x)y_1,\quad y_2''+p(x)y_2'+q(x)y_2 $$

$$ y_1''y_2+p(x)y_1'y_2+q(x)y_1y_2,\quad y_2''y_1+p(x)y_2'y_1+q(x)y_2y_1 $$

$$ p(x) = \frac{y_1''y_2-y_2''y_1}{y_2'y_1-y_1'y_2} = \frac{y_1''y_2-y_2''y_1}{W[y_1,y_2](x)}, $$$p(x)$ is undefined at $x_0\in(a,b)$.

Also, I noticed all this by considering an example: for two solutions $y_1=1+x$ and $y_2=1+x^2$, $W(x) = x^2+2x-1$, on interval $(0,\infty)$. So although $W(\sqrt{2}-1)=0$, I think the two functions are still linearly independent on the interval, at least according to the definitions here. But using Theorem 5.1.5 they should be linearly dependent, because $W$ is zero for a point in the interval?

Now it really confuses me despite thinking about it for hours. Which part did I miss? I am sorry if the question is too dumb, still not accustomed to the linearly independence of functions.


1 Answer 1


Two functions $f(x)$ and $g(x)$ are linearly independent if and only if neither is a constant multiple of the other. That's directly from the definition of linear independence: there do not exist constants $a_1$ and $a_2$, not both $0$, such that $a_1 f(x) + a_2 g(x) = 0$ for all $x$ in the domain of the functions. So, in your example, $1 + x$ and $1 + x^2$ are obviously linearly independent because their ratio is not constant.

What about Abel's identity? That gives a formula for the Wronskian of two solutions of the differential equation $y'' + p(x) y' + q(x) y = 0$ on an interval where $p(x)$ and $q(x)$ are defined and continuous, and in particular shows that in that interval the Wronskian is either always $0$ or never $0$. So if you find two linearly independent functions (such as your $1+x$ and $1+x^2$) whose Wronskian is $0$ at some point but not elsewhere, it tells you that something goes wrong with the coefficients of a second-order homogeneous linear differential equation that has these solutions at a point where the Wronskian is $0$.

And indeed, in your example you can show that the second-order homogeneous linear differential equation that has $1+x$ and $1+x^2$ as solutions is

$$ y'' - \frac{2(1+x)}{x^2 + 2 x - 1} y' + \frac{2}{x^2 + 2 x - 1} y = 0$$

where the coefficients $p(x)$ and $q(x)$ blow up at the roots of $x^2 + 2 x - 1$, and by no coincidence $x^2 + 2 x - 1$ is your Wronskian.

I really don't understand why DE textbooks make such a big deal about the Wronskian for second-order equations as a criterion for linear independence of solutions, when it's so easy to tell whether two functions are linearly independent.

  • $\begingroup$ I agree. So Theorem 5.1.5 is wrong (or replace iff by if)? We actually use it for exams, for continuous $p(t)$ though. So it doesn't change the answer but would change the way it should be justified. Thank you for your clarification and I will ask my teacher tomorrow. $\endgroup$
    – AntidusPig
    Feb 21 at 2:47
  • $\begingroup$ Theorem 5.1.5 is perfectly correct. In your example, the interval $(a,b)$ can't contain any roots of $x^2 + 2 x - 1$. $\endgroup$ Feb 21 at 15:24
  • $\begingroup$ Ah my bad! I got caught up in the logic, it is correct indeed. Thanks for the help! $\endgroup$
    – AntidusPig
    Feb 22 at 6:13

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