02 - Wronskian

Linear Independence

Let $y_1(x),y_2(x)$ be a pair of functions

$y_1(x),y_2(x)$ is linearly independent if

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0⟺c_1=c_2=0$$

(that is there is no non-trivial linear combination that vanishes identically)

Linear Dependence

Let $y_1(x),y_2(x)$ be a pair of functions then

If $c_i$ are not both zero then there exists combination such that

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0$$

Wronskian

Let $y_1(x),y_2(x)$ be a pair of functions then

$$W(x)=W[y_1,y_2]=\det \left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)=y_1(x)y_2^{\prime }(x)-y_2(x)y_1^{\prime }(x)$$

Proof - Logic for it $y_1,y_2$ are linearly dependent then there exists $c_1,c_2$ such that

If

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0$$

So also $c_1{y}_1^{\prime }(x)+c_2{y}_2^{\prime }(x)\equiv 0$ hence

$$\left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)\equiv 0$$

(assuming $y_1,y_2$ are differentiable)

Thus there exists non-trivial solutions $(c_1,c_2)$ if and only if determinant of matrix is zero
Which leads to the definition of the Wronskian

Wronskian of Linearly Dependent Functions

If two functions are linearly dependent then Wronskian vanishes

Wronskian is identically zero or never zero for Homogenous Second Order Linear ODEs

Let $y_1,y_2$ be two solutions to $(H)$

If $W=0$ at one point then $W\equiv 0$ everywhere
Conversely of $W\ne 0$ at one point then $W\ne 0$ everywhere

Proof

Let $y_1$ and $y_2$ be two solution to $(H)$ so

$$\begin{array}{rl}{P}_2{y}_1^{\prime \prime }+P_1{y}_1^{\prime }+P_0{y}_1& =0\\ P_2{y}_2^{\prime \prime }+P_1{y}_2^{\prime }+P_0{y}_2& =0\end{array}$$

Eliminating $P_0$ term then

$$P_2(y_1{y}_2^{\prime \prime }-y_2{y}_1^{\prime \prime })+P_1(y_1{y}_2^{\prime }-y_2{y}_1^{\prime })=0$$

So

$$P_2\frac{dW}{dx}+P_{1}W=0$$

Assuming that $P_2$ is not zero anywhere then

$$W(x)=\text{const}\cdot \exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)$$

Since the exponential can’t vanish then result follows

Relation between Wronskian and Linearly Dependence of Solution to a Homogenous Second-Order ODE

Let $(H)$ denote the Homogenous Equation ($Ly=0$)
Let $y_1(x),y_2(x)$ be two solutions of a given homogeneous second-order ODE $(H)$

$y_1$ and $y_2$ are linearly dependent if and only if their Wronskian is zero

Proof

Suppose $y_1$ and $y_2$ are two solutions of $(H)$

If they are linearly dependent then their Wronskian is zero

So instead suppose that their Wronskian is zero (everywhere)

If $y_1$ is the zero function then $y_1$ and $y_2$ are certainly linearly dependent
Otherwise assume there exists at least one value $x=a$ such that $y_1(a)\ne 0$

Let $u$ be such that $y_2(a)=\mu y_1(a)$ then

$$0=W(a)=y_1(a)y_2^{\prime }(a)-y_2(a)y_1^{\prime }(a)=y_1(a)(y_2^{\prime }(a)-\mu y_1^{\prime }(a))$$

Since $y_1(a)\ne 0$ by assumption then

$$y_2^{\prime }(a)=\mu y_1^{\prime }(a)$$

Define

$$y(x)=y_2(x)-\mu y_1(x)$$

By linearity then $y(x)$ is a solution of $(H)$
So $y(x)$ satisfies the initial conditions (by construction from above)

$$y(a)=0=y^{\prime }(a)$$

So by uniqueness of solution of $(H)$ (by Picard’s Theorem, assuming $P_2\ne 0$)
Thus

$$y(x)\equiv 0$$

as it satisfies the initial conditions and by uniqueness it must be the only solution

Hence $y_1$ and $y_2$ are linearly dependent