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

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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:

$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$

if and only if

$p$ 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:

$q > 0$

$\Box$


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:

$\ds \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$

if and only if $m_1 < 0$.

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:

$q > 0$

$\Box$


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:

\(\ds p\) \(=\) \(\ds -\paren {a + i b + a - i b}\) Sum of Roots of Quadratic Equation
\(\ds \) \(=\) \(\ds -2 a\)
\(\ds \leadsto \ \ \) \(\ds p\) \(>\) \(\ds 0\)


Then:

\(\ds q\) \(=\) \(\ds \paren {a + i b} \paren {a - i b}\) Product of Roots of Quadratic Equation
\(\ds \) \(=\) \(\ds a^2 + b^2\)
\(\ds \leadsto \ \ \) \(\ds q\) \(>\) \(\ds 0\) whatever $a$ and $b$ are

$\Box$


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:

$\ds \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0$

$\Box$


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:

$\ds \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$

if and only if $m_1 < 0$.

It follows that:

$\ds \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 x e^{m_1 x} = 0$

$\Box$


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:

$\ds \lim_{x \mathop \to \infty} e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} = 0$

$\Box$


Thus in all cases:

$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$

$\blacksquare$


Sources

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