Condition for Solutions to Constant Coefficient Homogeneous LSOODE to tend to Zero

== Theorem ==
Let:
:$(1): \quad y + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let the general solution to $(1)$ be $\map y {x, C_1, C_2}$.
Then:
:$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
:$p$ and $q$ are both strictly positive.
== Proof ==
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
:$y = \begin{cases}
C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
 & \\
C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
 & \\
e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
\end{cases}$
where:
:$m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + p m + q = 0$
:$a + i b = m_1$
:$a - i b = m_2$
=== Sufficient Condition ===
Let:
:$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
Let $y$ be of the form:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
Then it follows that $m_1 < 0$ and $m_2 < 0$.
From Sum of Roots of Quadratic Equation:
:$p = -\paren {m_1 + m_2}$
from which it follows that:
:$p > 0$
From Product of Roots of Quadratic Equation:
:$q = m_1 m_2$
from which it follows that:
:$q > 0$
Let $y$ be of the form:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
From Limit at Infinity of Polynomial over Complex Exponential:
:$\ds \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$
 $m_1 < 0$.
Thus it follows that $m_1 < 0$.
Again from Sum of Roots of Quadratic Equation:
:$p = -\paren {2 m_1}$
from which it follows that:
:$p > 0$
From Product of Roots of Quadratic Equation:
:$q = m_1^2$
from which it follows that:
:$q > 0$
Let $y$ be of the form:
:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
Thus it follows that $a < 0$.
From Sum of Roots of Quadratic Equation:
Then:
Thus it is seen that in all three cases:
:$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
implies that $p$ and $q$ are both strictly positive.
=== Necessary Condition ===
Let $p$ and $q$ both be strictly positive.
Let $p^2 > 4 q$.
Then both $m_1$ and $m_2$ are real and unequal.
Thus from Solution of Constant Coefficient Homogeneous LSOODE:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
From Product of Roots of Quadratic Equation:
:$m_1 m_2 = q > 0$
and so either $m_1$ and $m_2$ are either both strictly positive or strictly negative.
It follows that $m_1 + m_2$ is also either strictly positive or strictly negative.
From Sum of Roots of Quadratic Equation:
:$-\paren {m_1 + m_2} = p > 0$
and so $m_1 + m_2 < 0$
Hence both $m_1$ and $m_2$ are strictly negative.
It follows that: 
:$\ds \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0$
Let $p^2 = 4 q$.
Then from Solution to Quadratic Equation with Real Coefficients:
:$m_1 = m_2 = -\frac p 2$
and so:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
From Limit at Infinity of Polynomial over Complex Exponential:
:$\ds \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$
 $m_1 < 0$.
It follows that: 
:$\ds \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 x e^{m_1 x} = 0$
Let $p^2 < 4 q$.
Then both $m_1$ and $m_2$ are complex and unequal:
:$m_1 = a + i b = -\dfrac p 2 + i \dfrac {\sqrt {4 q - p^2} } 2$
:$m_2 = a + i b = -\dfrac p 2 - i \dfrac {\sqrt {4 q - p^2} } 2$
Then from Solution of Constant Coefficient Homogeneous LSOODE:
:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
As $p > 0$ it follows that $a < 0$.
Thus it follows that: 
:$\ds \lim_{x \mathop \to \infty} e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} = 0$
Thus in all cases:
:$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
== Sources ==
* : $\S 3.17$: Problem $2$
Category:Constant Coefficient Homogeneous LSOODEs