Wronskian and related

Some definitions and theorems

A second order linear ODE can be written as $$y'' + p(x)y' + q(x)y = r(x) \tag{1},\ \ \ \ \ x \in \mathcal{I} $$where $I$ is an interval. If $r (x) = 0, \forall x \in I$ , then it is a homogeneous equation.
The existence and uniqueness for $(1)$ is given by

Existence and Uniqueness Theorem

Let $p, q, r$ be continuous in $I$ . If $x_{0} \in I$ and $k_{0}, k_{1}$ are two arbitrary real numbers, then $(1)$ has a unique solution $y (x)$ on $I$ such that $y (x_{0}) = k_{0}$ and $y^{'} (x_{0}) = k_{1}$

Definition

Two functions $f$ and $g$ are defined in $I$ . If there exists constant $a, b$ not both zero such that $$af(x) + bg(x) = 0,\ \ \ \forall x \in \mathcal{I}$$then $f$ and $g$ are linearly dependent in $I$ , otherwise LI.

Wronskian

$W (f, g) = | \begin{matrix} f (x) & g (x) \\ f^{'} (x) & g^{'} (x) \end{matrix} | $ $ i s c a l l e d t h e w r o n s k i a n o f $ f $ a n d $ g $ . > [! n o t e] N o t e > L e t $ f, g $ b e d i f f e r e n t i a b l e . I f t h e y a r e L D i n $ I $, t h e n $ W (f, g) = 0, \forall x \in I $ . I f $ W (f, g) \neq 0 $ a t a n y p o i n t $ x_{0} $ i n $ I $, t h e n $ f $ a n d $ g $ a r e L I > T h e c o n v e r s e i s n o t t r u e . I f $ W (f, g) = 0 $ f o r a l l $ x $ i n $ I $, i t d o e s n^{'} t m e a n $ f, g $ a r e L D$

Theorem of Wronskians

If $W (y_{1}, y_{2}) = 0$ at any point in the interval $I$ of the solution set of a linear homogeneous ODE, then it must imply that $W = 0$ everywhere on $I$ , and thus the solutions $y_{1}, y_{2}$ are Linearly Dependent
If $W (y_{1}, y_{2}) \neq 0$ at any point in the interval $I$ , then it must imply that $W \neq 0$ for all of $I$ , meaning that solutions $y_{1}, y_{2}$ are LI.

Proof

1.

Consider the linear combination $y (x) = a y_{1} + b y_{2}$ where $a, b$ are arbitrary constants not equal to 0. If $W (x_{0}) = 0$ , meaning that the system of solutions $$\begin{align}ay_{1} + by_{2} &= 0 \ ay_{1}' + by_{2}' &= 0 \end{align}$$must have a non-trivial solutions (infinite, actually) for $(a, b)$ .
We can clearly see that from the above equations, $y (x_{0}) = 0$ and $y^{'} (x_{0}) = 0$ .
By uniqueness theorem, the only possible solution for $y$ is $y (x) \equiv 0 \forall x \in I$
This means $$ay_{1} + by_{2} = 0, \ \ \ \forall x \in \mathcal{I}$$meaning $y_{1}, y_{2}$ are LD

2.

Showing that they are LI is trivial, from the above proof
Instead we shall prove that $y_{1}, y_{2}$ are indeed the generating set for the solution set of ODE.
I am genuinely tripping balls in this part. Will cover some other day

Relating to $W (y_{1}, y_{2})$ in $I$

When we try to compute $W^{'}$ we find that $$W' + p(x)W = 0$$The solutions of this linear homogeneous ODE are of the form $$W = C\int e^{-\int pdx}$$and since the $\exp$ term cannot be equivalent $0$ . Which must imply that if $W (x_{0}) = 0$ then $C = 0$ , therefore $W = 0$ everywhere in $I$ . Thus $y_{1}, y_{2}$ are LD.
Similarly if $W (x_{0}) \neq 0$ , $C \neq 0$ meaning that $W$ can never be zero anywhere in $I$ , thus $y_{1}, y_{2}$ are LI.

Some definitions and theorems

Proof

1.

2.

Relating to W(y1,y2) in I

Relating to $W (y_{1}, y_{2})$ in $I$