Wronskian

In this subsection, we discuss the linear independence or dependence of two solutions of Equation (8.2.1).

DEFINITION 8.2.2 (Wronskian)   Let $y_1$ and $y_2$ be two real valued continuously differentiable function on an interval $I \subset {\mathbb{R}}.$ For $x \in I,$ define

$$W(y_1, y_2) := \begin{vmatrix}y_1 & y_1^\prime \\ y_2 & y_2^\prime \end{vmatrix}$$

$$= y_1 y_2^\prime - y_1^\prime y_2.$$

$W$ is called the Wronskian of $y_1$ and $y_2.$

EXAMPLE 8.2.3  

1. Let $y_1 = \cos x$ and $y_2 = \sin x, \; x\in I \subset {\mathbb{R}}.$ Then

   $$W(y_1, y_2) = \begin{vmatrix}\sin x & \cos x \\ \cos x & - \sin x \end{vmatrix} \equiv -1 \;\;{\mbox{ for all }} \;\; x \in I.$$ (8.2.2)

   
   
   Hence $\{\cos x, \sin x \}$ is a linearly independent set.
2. Let $y_1 = x^2 \vert x\vert,$ and $y_2 = x^3$ for $x \in (-1, 1).$ Let us now compute $y_1^\prime$ and $y_2^\prime.$ From analysis, we know that $y_1$ is differentiable at $x = 0$ and
   $$y_1 (x ) = - 3 x^2 \; {\mbox{ if }} \; x < 0 \;\; {\mbox{ and }} y_1 (x ) = 3 x^2 \; {\mbox{ if }} \; x \ge 0.$$
   Therefore, for $x \geq 0,$
   $$W(y_1, y_2) = \begin{vmatrix}y_1 & y_1^\prime \\ y_2 & y_2^\prime \end{vmatrix} = \begin{vmatrix}x^3 & 3 x^2 \\ x^3 & 3 x^2 \end{vmatrix} = 0$$
   and for $x < 0,$
   $$W(y_1, y_2) = \begin{vmatrix}y_1 & y_1^\prime \\ y_2 & y_2^\pri... ...nd{vmatrix} = \begin{vmatrix}-x^3 & - 3 x^2 \\ x^3 & 3 x^2 \end{vmatrix} = 0.$$
   That is, for all $x \in (-1, 1),$ $\; \; W(y_1, y_2) = 0.$

   It is also easy to note that $y_1, y_2$ are linearly independent on $(-1, 1).$ In fact,they are linearly independent on any interval $(a, b)$ containing $0.$

Given two solutions $y_1$ and $y_2$ of Equation (8.2.1), we have a characterisation for $y_1$ and $y_2$ to be linearly independent.

THEOREM 8.2.4   Let $I \subset {\mathbb{R}}$ be an interval. Let $y_1$ and $y_2$ be two solutions of Equation (8.2.1). Fix a point $x_0 \in I.$ Then for any $x \in I,$

$$W(y_1, y_2) = W(y_1, y_2)(x_0) \exp(- \int_{x_0}^x q(s) d s).$$ (8.2.3)

Consequently,

$$W(y_1, y_2)(x_0) \neq 0 \Longleftrightarrow W(y_1, y_2) \neq 0 \;\; {\mbox{ for all }} \;\; x \in I.$$

Proof. First note that, for any $x \in I,$

$$W(y_1, y_2) = y_1 y_2^\prime - y_1^\prime y_2.$$

So

$$\frac{d}{dx} W(y_1, y_2) = y_1 y_2^{\prime\prime} - y_1^{\prime\prime} y_2$$ (8.2.4)

$$= y_1 \left( - q(x) y_2^\prime - r(x) y_2 \right) - \left( - q(x) y_1^\prime - r(x) y_1 \right) y_2$$ (8.2.5)

$$= q(x) \bigl( y_1^\prime y_2 - y_1 y_2^\prime \bigr)$$ (8.2.6)

$$= - q(x) W(y_1, y_2).$$ (8.2.7)

So, we have

$$W(y_1, y_2) = W(y_1, y_2)(x_0) \; \exp\bigl(- \int_{x_0}^x q(s) ds\bigr).$$

This completes the proof of the first part.

The second part follows the moment we note that the exponential function does not vanish. Alternatively, $W(y_1, y_2)$ satisfies a first order linear homogeneous equation and therefore

$$W(y_1, y_2) \equiv 0 \;\; {\mbox{ if and only if }} \; W(y_1, y_2)(x_0) = 0.$$

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Remark 8.2.5  

1. If the Wronskian $W(y_1, y_2)$ of two solutions $y_1, y_2$ of (8.2.1) vanish at a point $x_0 \in I,$ then $W(y_1, y_2)$ is identically zero on $I.$
2. If any two solutions $y_1, y_2$ of Equation (8.2.1) are linearly dependent (on $I$ ), then $W(y_1, y_2) \equiv 0$ on $I.$

THEOREM 8.2.6   Let $y_1$ and $y_2$ be any two solutions of Equation (8.2.1). Let $x_0 \in I$ be arbitrary. Then $y_1$ and $y_2$ are linearly independent on $I$ if and only if $W(y_1, y_2)(x_0) \neq 0.$

Proof. Let $y_1, y_2$ be linearly independent on $I.$
To show: $W(y_1, y_2)(x_0) \neq 0.$

Suppose not. Then $W(y_1, y_2)(x_0) = 0.$ So, by Theorem 2.5.1 the equations

$$c_1 y_1(x_0) + c_2 y_2(x_0) = 0 \;\; {\mbox{ and }} \;\; c_1 y_1^\prime(x_0) + c_2 y_2^\prime(x_0) = 0$$ (8.2.8)

admits a non-zero solution $d_1, d_2.$ (as $0 = W(y_1, y_2)(x_0) = y_1(x_0) y_2^\prime(x_0) - y_1^\prime(x_0) y_2(x_0).$ )

Let $y = d_1 y_1 + d_2 y_2.$ Note that Equation (8.2.8) now implies

$$y(x_0) = 0 \; {\mbox{ and }} \; y^\prime(x_0) = 0.$$

Therefore, by Picard's Theorem on existence and uniqueness of solutions (see Theorem 8.1.9), the solution $y \equiv 0$ on $I.$ That is, $d_1 y_1 + d_2 y_2 \equiv 0$ for all $\; x\in I$ with $\vert d_1\vert + \vert d_2\vert \neq 0.$ That is, $y_1, y_2$ is linearly dependent on $I.$ A contradiction. Therefore, $W(y_1, y_2)(x_0) \neq 0.$ This proves the first part.

Suppose that $W(y_1, y_2)(x_0) \neq 0$ for some $x_0 \in I.$ Therefore, by Theorem 8.2.4, $W(y_1, y_2) \neq 0$ for all $x \in I.$ Suppose that $c_1 y_1(x) + c_2 y_2(x) = 0$ for all $x \in I$ . Therefore, $c_1 y_1^\prime(x) + c_2 y_2^\prime(x) = 0$ for all $x \in I$ . Since $x_0 \in I,$ in particular, we consider the linear system of equations

$$c_1 y_1(x_0) + c_2 y_2(x_0) = 0 \; {\mbox{ and }} \; c_1 y_1^\prime(x_0) + c_2 y_2^\prime(x_0) = 0.$$ (8.2.9)

But then by using Theorem 2.5.1 and the condition $W(y_1, y_2)(x_0) \neq 0,$ the only solution of the linear system (8.2.9) is $c_1 = c_2 = 0.$ So, by Definition 8.1.8, $y_1, y_2$ are linearly independent. height6pt width 6pt depth 0pt

Remark 8.2.7   Recall the following from Example 2:

1. The interval $I = (-1, 1).$
2. $y_1 = x^2 \vert x\vert, \; y_2 = x^3$ and $W(y_1, y_2) \equiv 0$ for all $x \in I.$
3. The functions $y_1$ and $y_2$ are linearly independent.

This example tells us that Theorem 8.2.6 may not hold if $y_1$ and $y_2$ are not solutions of Equation (8.2.1) but are just some arbitrary functions on $(-1, 1).$

The following corollary is a consequence of Theorem 8.2.6.

COROLLARY 8.2.8   Let $y_1, y_2$ be two linearly independent solutions of Equation (8.2.1). Let $y$ be any solution of Equation (8.2.1). Then there exist unique real numbers $d_1, d_2$ such that

$$y = d_1 y_1 + d_2 y_2 \; {\mbox{ on }} \; I.$$

Proof. Let $x_0 \in I.$ Let $y(x_0) = a, \; y^\prime(x_0) = b.$ Here $a$ and $b$ are known since the solution $y$ is given. Also for any $x_0 \in I,$ by Theorem 8.2.6, $\; W(y_1, y_2)(x_0) \neq 0$ as $y_1, y_2$ are linearly independent solutions of Equation (8.2.1). Therefore by Theorem 2.5.1, the system of linear equations

$$c_1 y_1(x_0) + c_2 y_2(x_0) = a \; {\mbox{ and }} \; c_1 y_1^\prime(x_0) + c_2 y_2^\prime(x_0) = b$$ (8.2.10)

has a unique solution $d_1, d_2.$
Define $\zeta(x) = d_1 y_1 + d_2 y_2$ for $x \in I.$ Note that $\zeta$ is a solution of Equation (8.2.1) with $\zeta(x_0) = a$ and $\zeta^\prime(x_0) = b.$ Hence, by Picard's Theorem on existence and uniqueness (see Theorem 8.1.9), $\zeta = y$ for all $x \in I.$ That is, $y = d_1 y_1 + d_2 y_2.$ height6pt width 6pt depth 0pt

EXERCISE 8.2.9  

1. Let $y_1$ and $y_2$ be any two linearly independent solutions of $y^{\prime\prime} + a(x) y = 0.$ Find $W(y_1, y_2).$
2. Let $y_1$ and $y_2$ be any two linearly independent solutions of
   $$y^{\prime\prime} + a(x) y^\prime+ b(x) y = 0, \; x \in I.$$
   Show that $y_1$ and $y_2$ cannot vanish at any $x = x_0 \in I.$
3. Show that there is no equation of the type
   $$y^{\prime\prime} + a(x) y^\prime+ b(x) y = 0, \; x \in [0, 2 \pi]$$
   admiting $y_1 = \sin x$ and $y_2 = x - \pi$ as its solutions; where $a(x)$ and $b(x)$ are any continuous functions on $[0, 2\pi].$ [Hint: Use Exercise 8.2.9.2.]