# Why Wronskian works for DEs, but in general it doesn't show independence

For a set of functions $$f_1,f_2,...f_n$$, if their Wronskian determinant is identically zero $$W(f_1,...f_n)(x) = 0$$ for all $$x$$ in some interval $$I$$ we can't conlcude that these functions are linearly dependent.

But in case of differential equations, if $$f_1,f_2,...f_n$$ are solutions of some linear DE and the Wronskian is zero on some interval we say that the functions $$f_1,f_2,...f_n$$ are linearly dependent.

Is there some deep meaning why this is so?

• particular type means initial value problem? – Upstart Jul 29 '19 at 6:20
• – Kavi Rama Murthy Jul 29 '19 at 6:34
• What do you want to ask ? Pointout particularly where you stuck ? – nmasanta Jul 29 '19 at 7:43

Have a look at the functions $$f_1(x) = x^3$$ and $$f_2(x) = \vert x \vert^3$$. Obviously, $$f_1$$ and $$f_2$$ are linearly independent. Moreover, we have $$f_1'(x) = 3x^2$$ and a little bit more calculus reveals that $$f_2$$ is differentiable with $$f_2'(x) = 3 x \vert x \vert$$. Furthermore,
$$W(f_1, f_2)(x) \equiv 0.$$
So what we have shown is, that in general the statement "$$f_1, \dots, f_n$$ linearly independent $$\Longrightarrow$$ $$W(f_1, \dots, f_n) \neq 0$$" is false. Since this if false, the contraposition "$$W(f_1, \dots, f_n) \equiv 0 \Longrightarrow f_1, \dots, f_n$$ linearly dependent" is also wrong, which explains, why we can't conclude the linear dependence in general.
In the case that $$f_1, \dots, f_n$$ solve an linear ODE and $$W \equiv 0$$ one can conclude that they are linearly dependent. A prove based on this additional assumption can be found e. g. here.
From $$W(f_1,...f_n)(x)=0$$, only piecewise linear dependence follows. That means, the functions could be locally linear independent.