Obtaining stability criterion for 2nd order ODEs in coefficient form

[zenterix]

Homework Statement

A system is called stable if its long-term behavior does not depend significantly on the initial conditions.

Consider a system modeled by the following inhomogeneous 2nd order linear differential equation with constant coefficients

$$y''+by'+ay=r(t)\tag{1}$$

The general solution has the form

$$y=c_1y_1+c_2y_2+y_p\tag{2}$$

$y_p$ is a particular solution to (1) and $c_1y_1+c_2y_2$ is the general solution to the associated homogeneous equation.

If $c_1y_1+c_2y_2$ approaches zero asymptotically for every choice of $c_1$ and $c_2$ then the system is stable because the long-term behavior is due solely to the $y_p$ term (which does not depend on initial conditions).

Note that $c_1y_1+c_2y_2$ is called the transient and $y_p$ is called the steady-state solution.

Relevant Equations

My question is about the conditions for stability.

Below I go through the argument to show that the system in (1) is stable $\iff$ coefficients $a$ and $b$ are positive.

I am posting the argument here to try to make sure it is correct.

As we noted above, stability is all about the solution to the homogeneous equation.

For the equation

$$y''+by'+ay=0\tag{3}$$

we have discriminant

$$\Delta = b^2-4a\tag{4}$$

and the roots are

$$r=\frac{-b\pm\sqrt{b^2-4a}}{2}\tag{5}$$

We have three cases.

Case 1 (Distinct Real Non-Complex Roots) If we have a positive discriminant, then we have two distinct real roots. The solution is $c_1e^{r_1t}+c_2e^{r_2t}$ and the only way this tends to zero as $t\to\infty$ is if $r_1<0$ and $r_2<0$.

What does this mean for the coefficients of (3)?

$\Delta>0$ means $b^2>4a$.

$r_1<0$ implies

$$-b+\sqrt{\Delta}=-b+\sqrt{b^2-4a}<0$$

$$b>\sqrt{\Delta}>0$$

$$b^2>b^2-4a$$

$$4a>0$$

$$a>0$$

Therefore, we have shown that it must be that both $a$ and $b$ are positive.

Case 2 (Repeated Real Non-Complex Roots) If the discriminant is zero then $b^2=4a$ and we have a single root $r=\frac{-b}{2}$. Note that we already have $a>0$ from $0<b^2=4a$.

The solution to (3) is $e^{rt}(c_1+c_2t)$ and this approaches zero as $t\to\infty$ only if $r<0$. This happens if $b>0$.

Once again, stability requires both $a$ and $b$ to be positive.

Case 3 (Distinct Complex Non-Real Roots) If the discriminant is negative then $0<b^2<4a$. Thus $a>0$

The roots are

$$r_1=\frac{-b+ i\sqrt{4a-b^2}}{2}$$

$$r_2=\frac{-b- i\sqrt{4a-b^2}}{2}$$

and the solution is

$$e^{-bt/2}(c_1\cos{kt}+c_2\sin{kt})$$

where $k=\frac{1}{2}\sqrt{-\Delta}$.

For stability we need $-b<0$ and so $b>0$.

Note that $b/2$ is the real part of the roots. Therefore, we can express the condition for stability as the requirement that the real part of the complex roots is negative.

We can summarize all cases by saying that for stability the real part of the roots must all be negative.

Alternatively we can say that if $a$ and $b$ are positive then we will have stability.

 

[pasmith]

You can write the solutions of $\lambda^2 + b\lambda + a = 0$ as $$\lambda_{\pm} = -\frac{b}{2} \left( 1 \pm \sqrt{1 - \frac{4a}{b^2}}\right).$$ It follows that if the roots are a complex conjugate pair then the real part has the oppsite sign from $b$, whereas if the roots are real then there is always a root ($\lambda_{+}$) which has the opposite sign from $b$. Thus $b > 0$ is a necessary condition for stability. It is not however sufficient, since the other real root ($\lambda_{-}$) will have the same sign as $b$ if $a < 0$.