Some Theorems about Zeros and Oscillation
To motivate this section, let’s go back and consider the simple second-order ODE
\begin{align}
\frac{d^2y}{dx^2} + y = 0 \tag{1}
\end{align}
It is obvious that the two linearly independent solutions are \( y_1 = \cos x \) and \( y_2 = \sin x \), which have some curious characteristics. First, from elementary trigonometry, we know that they are just sinusoidal waves that have a (constant) wavelength/period of \( 2\pi \). This means that they will oscillate about the \( x \)-axis indefinitely, i.e. they have infinitely many zeros (the points on the \(x\)-axis where the function equals \(0\)). What is even more special is that these zeros of \( \cos x \) and \( \sin x \) appear alternately, i.e. there is exactly one zero for \( \cos x \) (\(\pi/2 + n\pi\), for all integers \(n\)) between two consecutive zeros for \( \sin x \) (\( n\pi \), also for all integers \(n\)) and vice versa. Below, we will show that this is not a coincidence: the nature of the corresponding second-order linear ODEs (such as (1)) will enforce these properties upon their two linearly independent solutions.
The arguments below are mainly taken from the reference textbook Differential Equations with Applications and Historical Notes by Simmons.
A Wronskian Lemma
We introduced the Wronskian a while ago. It has been explained that linearly independent functions will have a non-zero Wronskian at some point \( x_0 \). We further strengthen this statement by showing that if \( y_1 \) and \( y_2 \) are some two solutions of any standard second-order linear ODE:
\begin{align}
\frac{d^2y}{dx^2} + P(x) \frac{dy}{dx} + Q(x) y = 0 \tag{2}
\end{align}
then their Wronskian \( W[y_1,y_2](x) \) being zero/non-zero at a point \( x_0 \) implies that it will always be zero/non-zero throughout all \( x \), so the location of \( x_0 \) does not matter. Therefore, if \( y_1 \) and \( y_2 \) are two linearly independent solutions to (2), since their Wronskian will not be equal to zero at one location at least, it will be non-zero for all \( x \).
Lemma 1. If \(y_1(x)\) and \(y_2(x)\) are two solutions to (2), then their Wronskian \( W[y_1,y_2](x) \) will either be identically zero or never vanish.
To prove that, we first find the derivative of the Wronskian:
\begin{align}
W &= y_1y_2^\prime -y_1’y_2 \\
W’ &= y_1’y_2^\prime + y_1y_2^{\prime\prime} -y_1^{\prime\prime}y_2 -y_1’y_2^\prime \\
&= y_1y_2^{\prime\prime} -y_1^{\prime\prime}y_2 \tag{3}
\end{align}
As \(y_1(x)\) and \(y_2(x)\) are solutions to (2), we have
\begin{align}
\begin{aligned}
y_1^{\prime\prime} + Py_1^{\prime} + Qy_1 &= 0 \\
y_2^{\prime\prime} + Py_2^{\prime} + Qy_2 &= 0
\end{aligned} \tag{4}
\end{align}
Multiplying the first equation by \( y_2 \) and the second equation by \( y_1 \), then doing subtraction between them, will give
\begin{align}
(y_1y_2^{\prime\prime} + Py_1y_2^{\prime} + Qy_1y_2) -(y_1^{\prime\prime}y_2 + Py_1^{\prime}y_2 + Qy_1y_2) &= 0 \\
(y_1y_2^{\prime\prime} -y_1^{\prime\prime}y_2) + P(y_1y_2^{\prime} -y_1^{\prime}y_2) &= 0 \\
\frac{dW}{dx} + PW &= 0 \tag{5}
\end{align}
where we have recalled (3). This relation is known as Abel’s Identity. The general solution to (5), a first-order linear ODE, is just
\begin{align}
W(x) = Ce^{-\int P(x) dx} \tag{6}
\end{align}
Since the exponential factor is always positive and never zero, \( W \) will either be identically zero (if \( C = 0 \)) or never zero (if \( C \neq 0\)).
Main Theorems
Now we shall begin proving the desired results for oscillation. The first one is
Theorem 2. If \( y_1(x) \) and \(y_2(x)\) are two linearly independent solutions to (2), then the two respective sets of zeros for them are distinct and occur alternately, i.e. there is exactly one zero of \(y_1\) between any two successive zeros of \(y_2\) and vice versa.
The proof is a so-called convexity argument. By Lemma 1, their Wronskian \(W'[y_1,y_2] = y_1y_2^{\prime\prime} -y_1^{\prime\prime}y_2 \) is never zero and has a constant sign (either always positive or negative). For the first part, the zeros of \(y_1\) and \(y_2\) cannot be situated at a common point \( x_0 \), for otherwise \(W'[y_1,y_2](x_0) = (0)y_2^{\prime\prime} -y_1^{\prime\prime}(0) = 0 \) and this violates the premise. For the second part, suppose that \(x_1\) and \(x_2\) are two consecutive zeros of \( y_2 \), and we are going to show that \( y_1\) will vanish exactly once between them. At these two zeros, \(y_2(x_1) = y_2(x_2) = 0\) and so the Wronskians there are reduced to \(W(x_1) = y_1(x_1)y_2^\prime(x_1) \) and \(W(x_2) = y_1(x_2)y_2^\prime(x_2) \). Since both Wronskians cannot be zero, the two respective factors \(y_1(x_1) \neq 0 \neq y_1(x_2)\) and \(y_2^\prime(x_1) \neq 0 \neq y_2^\prime(x_2) \) must also be non-zero. Now, \( y_2^\prime(x_1) \) and \( y_2^\prime(x_2) \) must have opposite signs because if \(y_2\) are increasing at \(x_1\), it must be decreasing at \( x_2 \) (otherwise, \(y_2\) will have a zero in-between and this contradicts the assumption) and vice versa. Since the Wronskian has a constant sign by Lemma 1, \(y_1(x_1)\) and \(y_1(x_2)\) will need to have opposite signs, and by the Intermediate Value Theorem, \(y_1(x) = 0\) will vanish somewhere between \((x_1, x_2)\). However, this can only occur once and there will only be one zero of \(y_1\) between the two successive zeros of \(y_2\): Otherwise, the exact same argument can be applied to the more-than-one zeros of \(y_1\) and hence there will be another zero of \(y_2\) within, which contradicts the hypothesis that the original zeros of \(y_2\) are consecutive.
Next,
Theorem 3. Given that \( v(x) \) is a non-trivial solution to the canonical second-order linear ODE
\begin{equation}
v^{\prime\prime} + g(x)v = 0 \tag{7}
\end{equation}
if \( g(x) < 0 \) over some interval, then \( v(x) \) has at most one zero over the same interval.
Any second-order linear ODE can be cast into the canonical form as explained in this tutorial so this result can be extended freely. The proof hinges on showing that if there is a zero \( x_0 \), this will be the only zero. Now \(v(x_0) = 0\), and because \( v(x) \) is taken to be a non-trivial solution, by the uniqueness of ODE solution with a given initial condition (see this post), \(v'(x_0) \neq 0\). (Otherwise, under \(v(x_0) = v'(x_0) = 0\), the trivial solution of identically zero clearly satisfies (7), so due to uniqueness the only possibility is \( v(x) = 0 \) and this contradicts the non-trivial assumption.) Without the loss of generality, take \( v'(x_0) > 0 \), so over an adequately short enough interval to the right of \(x_0\), \( v(x) \) is positive. As \( g(x) < 0 \), (7) shows that \( v^{\prime\prime}(x) = -g(x)v(x) > 0 \) is also positive over that region. The second derivative being positive implies that the slope \( v'(x) \) will also be increasing. Repeating this argument, to the right of \( x_0 \), \( v'(x) \) and hence \( v(x) \) will always be positive and \( v'(x) \) will keep increasing too, so there will not be any zero there. Similarly, there will also be no zero to the left of \( x_0 \), and the same logic can be said if \(v'(x_0) < 0\). Hence we conclude that there can be at most one zero for \(v(x)\) under such a scenario.
Therefore, if we want to have an oscillating solution, \( g(x) \) in (7) must stay positive. However, it is only a necessary requirement. If indeed \( g(x) > 0 \) is positive but decays at a fast enough rate, then the magnitude of \( v^{\prime\prime} = -gv < 0 \) may be too small to cause a reversal in the sign of \( v’ \) and \( v \) will not traverse back to cut the \(x\)-axis again. The theorem below shows a sufficient condition for the oscillation to occur surely if \( g(x) \) does not decrease too rapidly.
Theorem 4. For a non-trivial solution \( v(x) \) to the canonical second-order linear ODE (7) where now \( g(x) > 0 \) for all \( x > 0 \), if
\begin{equation}
\int_1^{\infty} g(x) dx = \infty \tag{8}
\end{equation}
then \( v(x) \) will have infinitely many zeros on the positive \(x\)-axis.
The lower limit of \(1\) in the integral may be replaced by any point \(> 0\) and this theorem can be adapted for the negative \(x\)-axis as well. To prove this, we assume the opposite where \( v(x) \) only has finitely many zeros over the positive \(x\)-axis. Then there will be a point \(x_0 > 1\) so that \( v(x) \neq 0 \) for \( x \geq x_0 \) beyond. Without loss of generality, we assume it is the case that \( v(x) > 0 \) remains positive for \( x \geq x_0 \). Consider
\begin{align}
s(x) = -\frac{v'(x)}{v(x)} \tag{9}
\end{align}
and by Quotient Rule and (7) (plus (9) itself):
\begin{align}
s'(x) &= \frac{v'(x)^2-v(x)v^{\prime\prime}(x)}{v(x)^2} \\
&= s(x)^2 + g(x) \tag{10}
\end{align}
Integrating (10) reveals that
\begin{align}
[s]_{x_0}^x = s(x) -s(x_0) = \int_{x_0}^x s(x)^2 dx + \int_{x_0}^x g(x) dx \tag{11}
\end{align}
If \( x \) is large enough, (11) shows that \( s \) will be positive due to (8) and by (9) \( v’ \) will be negative. As \( v^{\prime\prime} = -gv < 0 \), \( v’ \) will keep decreasing beyond that point, and eventually \(v\) will vanish again, contradicting the assumption that there is no more zero further away.
Airy Equation
The Airy Equation is a second-order ODE in the form of
\begin{align}
\frac{d^2y}{dx^2} -xy = 0 \tag{12}
\end{align}
Compared with (7), we identify \( g(x) = -x \). When \( x < 0 \), \( g(x) = -x > 0 \) and clearly \( \int_{-\infty}^{-1} -x dx = \infty \), the version of Theorem 4 adapted for the negative \(x\)-axis then implies that (12) is oscillatory to the left. On the other hand, when \( x > 0 \), \( g(x) < 0 \), Theorem 3 then illustrates that (12) will become exponential to the right. It can then be seen that the Airy Equation demonstrates, in its simple form, a transition from oscillatory to exponential behavior. In the next post, we will derive the solution to the Airy Equation and explain its properties in greater depth.
Exercise
Consider the second-order Euler Form ODE
\begin{equation}
4x^2\frac{d^2y}{dx^2} + y = 0 \tag{13}
\end{equation}
Show that it satisfies the first requirement of Theorem 4 (\(g(x) > 0\)) but fails the second one (8), and by solving it directly, further show that it only has at most one zero on the positive \(x\)-axis, hence cannot achieve the conclusion of Theorem 4.
Answer
Rewriting the equation as
\begin{align}
\frac{d^2y}{dx^2} + \frac{y}{4x^2} = 0
\end{align}
\( g(x) = 1/(4x^2) > 0 \) obviously, but
\begin{align}
\int_1^{\infty} \frac{1}{4x^2} dx &= [-\frac{1}{4x}]_1^{\infty} \\
&= 0 -(-\frac{1}{4(1)}) = \frac{1}{4}
\end{align}
does not tend to infinity. The solution of the ODE can be found by doing the substitution \( z = \ln x \) as suggested in this post and the new auxiliary equation is \( r^2 -r + \frac{1}{4} = 0 \) which has a repeated root of \( r = 1/2 \). So its general solution will be
\begin{align}
y = c_1 \sqrt{x} + c_2 \sqrt{x} \ln x
\end{align}
Assume there is a zero \( x_0 \) such that \( c_1 \sqrt{x_0} + c_2 \sqrt{x_0}\ln {x_0} = 0 \), since \( \sqrt{x}\ln {x} \) grows strictly faster than \( \sqrt{x} \) when \( x > 1 \), beyond \( x_0 \) the \(c_2\) part of the solution will be greater than the \(c_1\) part in magnitude:
\begin{align}
\left|\frac{c_2\sqrt{x_0}\ln{x_0}}{c_1\sqrt{x_0}}\right| = \left|\frac{c_2\ln{x_0}}{c_1}\right| = 1
\end{align}
and
\begin{align}
\left|\frac{c_2\sqrt{x}\ln{x}}{c_1\sqrt{x}}\right| = \left|\frac{c_2\ln{x}}{c_1}\right| > 1
\end{align}
for \( x > 1 \), so there will be no more zeros.
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