Tag: Wronskian

Use the Wronskian to find all solutions of a given second-order differential equation

by RoRi

The Wronskian $W(x)$ is defined by

$W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x)$

for given functions $u_1$ and $u_2$ .

Let $v_1$ and $v_2$ be two solutions of the second-order linear differential equation

$y'' + ay' + by = 0,$

such that $\frac{v_2}{v_1}$ is not constant.

Let $y = f(x)$ be any solution of the given differential equation. Use the properties of the Wronskian proved in the previous two exercises (here and here) to prove that there exist constants $c_1$ and $c_2$ such that
$c_1 v_1(0) + c_2 v_2(0) = f(0), \qquad c_1 v_1'(0)+ c_2 v_2'(0) = f'(0).$
Prove that every solution of the differential equation has the form $y = c_1 v_1 + c_2 v_2$ .

by RoRi

The Wronskian $W(x)$ is defined by

$W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x)$

for given functions $u_1$ and $u_2$ .

Let $W$ be the Wronskian of two solutions $u_1$ and $u_2$ of the differential equations

$y'' + ay' + by = 0,$

where $a$ and $b$ are constants.

Prove that $W$ satisfies the first-order linear differential equation
$W' + aW = 0$

and hence,

$W(x) = W(0) e^{-ax}.$

By this formula we can see that if $W(0) \neq 0$ then $W(x) \neq 0$ for all $x$ .
Assume $u_1$ is not identically zero and prove that $W(0) = 0$ if and only if $\frac{u_2}{u_1}$ is constant.

First, we evaluate $W'(x) + aW(x)$ where $W(x) = u_1(x) u_2'(x) - u_1'(x) u_2(x)$ is the Wronskian of the two functions $u_1(x)$ and $u_2(x)$ .
$\begin{align*} W'(x) + aW(x) &= u_1 (x) u_2''(x) - u_1''(x) u_2(x) + au_1(x) u_2'(x) - au_1'(x) u_2(x) \\ &= u_1 (x) \left( u_2''(x) + au_2'(x) \right) - u_2(x) \left(u_1''(x) + au_1'(x) \right) \\ &= u_1(x) \left( -bu_2(x)\right) - u_2(x) \left( -bu_1(x) \right) \\ &= 0. \qquad \blacksquare \end{align*}$

Furthermore, by Theorem 8.3 (page 310 of Apostol), since $W(x)$ is a solution of $W'(x) + aW(x) = 0$ we know

$W(x) = ce^{-ax} = W(0)e^{-ax}$

since $W(0) = c$ . Hence, $W(x) \neq 0$ if $W(0) \neq 0$ .
Assume $W(0) = 0$ . Then $W(x) = 0$ for all $x$ . By part (a) of the previous exercise (Section 8.14, Exercise #21) we know $\frac{u_2}{u_1}$ is constant.
Conversely, assume $\frac{u_2}{u_1}$ is constant. Then, again by the previous exericse, we have $W(x) = 0$ for all $x$ . Hence, $W(0) = 0. \qquad \blacksquare$

by RoRi

The Wronskian $W(x)$ is defined by

$W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x)$

for given functions $u_1$ and $u_2$ .

If the Wronskian $W(x)$ of two functions $u_1(x)$ and $u_2(x)$ is zero for all $x$ in an open interval $I$ , prove that $\frac{u_2(x)}{u_1(x)}$ is constant for all $x \in I$ . Equivalently, if $\frac{u_2}{u_1}$ is not constant on $I$ then there is some $c \in I$ such that $W(c) \neq 0$ .
Prove that the derivative of the Wronskian is given by
$W'(x) = u_1(x) u_2''(x) - u_2(x)u_1''(x).$

Proof. (Note: I think we need the additional assumption that $u_1(x) \neq 0$ for any $x \in I$ .) With our additional assumption we have,
$d \left( \frac{u_2}{u_1} \right) = \frac{u_1 u_2' - u_1' u_2}{(u_1)^2} = 0$

since $W = u_1 u_2' - u_1' u_2 = 0$ by assumption. Thus, $\frac{u_2}{u_1}$ is constant by the zero derivative theorem $. \qquad \blacksquare$
Proof. This is a direct computation of the derivative of the Wronskian,
$\begin{align*} &&W &= u_1(x) u_2'(x) - u_1'(x) u_2(x) \\ \implies && W' &= u_1(x)u_2''(x) + u_1'(x) u_2'(x) - u_1''(x) u_2(x) - u_1'(x) u_2'(x) \\ \implies && W' &= u_1(x)u_2''(x) - u_1''(x) u_2(x). \qquad \blacksquare \end{align*}$