Wronskian zero and linear dependence

Question

Let $y''(x)+p(x)y'(x)+q(x)y(x)=0$ such that $p(x),q(x)$ are continuous.

Let $y_1(x),y_2(x)$ be two solutions of the equation.

Suppose that $y_1(x),y_2(x)$ both attain a maximum at a point $x_0$.

Prove that $y_1,y_2$ are linearly dependent.

Is it possible to say that $y_1'(x_0)=y_2'(x_0)=0$ and then, using Abel's formula, we get that $W[y_1,y_2,x]\equiv 0$ and so $y_1,y_2$ are linearly dependent?

Moreover, in general can we say that if $p,q$ are continuous and $W[y_1,y_2,x]\equiv 0$ then the solutions $y_1,y_2$ are dependent?

Edit:

$W[y_1,y_2,x]=y_1(x)y_2'(x)-y_2(x)y_1'(x)\equiv 0$, so define $f(x)=\frac{y_2(x)}{y_1(x)}$.

Then $f'(x)=\frac{y_1(x)y_2'(x)-y_2(x)y_1'(x)}{(y_1(x))^2}\equiv 0$.

So $\exists c\in\mathbb{R}: f(x)\equiv c$ and finally $\exists c\in\mathbb{R}: y_2(x)\equiv c\cdot y_1(x)$.

But, for this we need $y_1(x)\neq 0$.

Answer 1

Well, let's see . . .

The Wronskian of two solutions $y_1$, $y_2$ to

$y''(x) + p(x) y'(x) + q(x) y(x) = 0 \tag 1$

is

$W[y_1, y_2, x] = \det \left ( \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \right ) = y_1 y_2' - y_2 y_1'; \tag 2$

thus,

$W'[y_1, y_2, x] = y_1' y_2' + y_1 y_2'' - y_2' y_1' - y_2 y_1'' = y_1 y_2'' - y_2 y_1''; \tag 3$

from (1) we have

$y_i'' = - p y_i' - qy_i, \; i = 1, 2; \tag 4$

inserting (4) into (3) yields

$W'[y_1, y_2, x] = y_1 (-py_2'- qy_2) - y_2 (-py_1' - qy_1)$ $= -py_1y_2' - q y_1 y_2 + py_2 y_1' + qy_1 y_2 = -p(y_1 y_2' - y_2 y_1') = -p W[y_1, y_2, x], \tag 5$

that is,

$W'[y_1, y_2, x] = -p W[y_1, y_2, x], \tag 6$

a first-order linear ordinary differential equation for $W[y_1, y_2, x]$; the solution is easily seen to be

$W[y_1, y_2, x] = W[y_1, y_2, x_0] \exp \left ( \displaystyle -\int_{x_0}^x p(s) \; ds \right ), \tag 7$

which is Abel's formula or identity; it may indeed be used to address the inquiries made in the question.

We start with the OPs final subject: suppose

$W[y_1, y_2, x] \equiv 0; \tag 8$

then

$\det \left ( \begin{bmatrix} y_1(x_0) & y_2(x_0) \\ y_1'(x_0) & y_2'(x_0) \end{bmatrix} \right ) = W[y_1, y_2, x_0] = 0, \tag 9$

which implies that the columns of the matrix on the left of this equation are linearly dependent; therefore there exist $a$, $b$, not both zero, with

$ay_1(x_0) + by_2(x_0) = ay_1'(x_0) + by_2'(x_0) = 0; \tag{10}$

now consider the function

$z(x) = ay_1(x) + by_2(x); \tag{11}$

by linearity, $z(x)$ is a solution to (1) and furthermore,

$z(x_0) = z'(z_0) = 0; \tag{12}$

it then follows from uniqueness of solutions to (1) that

$z(x) = 0 \tag{13}$

everywhere; therefore, by (11),

$ay_1(x) + by_2(x) = 0 \tag{14}$

for all $x$, $y_1(x)$ and $y_2(x)$ are linearly dependent; the OP's final query is thus answered in the affirmative.

Now suppose $y_1(x)$ and $y_2(x)$ both attain a maximum at $x_0$; then

$y_1'(x_0) = y_2'(x_0) = 0, \tag{15}$

whence

$W[y_1, y_2, x_0] = 0; \tag{16}$

we may now exploit Abel's formula (7) to infer that

$W[y_1, y_2, x] = W[y_1, y_2, x_0] \exp \left ( \displaystyle -\int_{x_0}^x p(s) \; ds \right ) = 0 \cdot \exp \left ( \displaystyle -\int_{x_0}^x p(s) \; ds \right ) = 0; \tag {17}$

it now follows by our preceding result that $y_1(x)$, $y_2(x)$ are linearly dependent for all $x$.

So it is indeed possible to say that

$y_1'(x_0) = y_2'(x_0) = 0 \Longrightarrow W[y_1, y_2, x_0] = 0 \Longrightarrow W[y_1, y_2, x] = 0, \tag{18}$

and hence that $y_1$ and $y_2$ are linearly dependent.

Answer 2

Consider $$z(x)=y_1(x)y_2(x_0)-y_1(x_0)y_2(x).$$ Then $$ z(x_0)=0\\ z'(x_0)=y_1'(x_0)y_2(x_0)-y_1(x_0)y_2'(x_0)=0 $$ so that by the existence and uniqueness theorem $z(x)=0$ everywhere, which means that $y_1$ and $y_2$ are linearly dependent.