For which (constant) values $p$ and $q$, all the solutions of equation: $x''+px'+qx=0$ go to $0$ at $t \to +\infty$

Question

For which (constant) values $p$ and $q$, all the solutions of equation: $x''+px'+qx=0$ go to $0$ at $t \to +\infty$ ?

So I've found the solution of our 2nd order differential equation which is

$x(t) = C_1 e^{-\frac{1}{2} t \left(\sqrt{p^2 - 4 q} + p\right)} + C_2 e^{\frac12 t \left(\sqrt{p^2 - 4 q} - p\right)} $ .

Now I don't really know what to do with it. Do I have to calculate limits from $x(t)$ or something like that?

Any help will be highly appreciated.

P.S. Sorry for the title, I'm not sure how should I formulate it.

Answer 1

The corresponding characteristic equation is $$ r^2+pr+q=0. \tag{1}$$ Let roots of (1) be $r_{1,2}$.

Case 1: $p^2-4q>0$. Then $r_1,r_2$ are negative if $p>0, q>0$.

Case 2: $p^2-4q<0$. Then $r_1,r_2$ are complex and have negative real parts if $p>0$.

Case 3: $p^2-4q=0$ Then $r_1=r_2=-\frac{p}{2}<0$ if $p>0$. In this case, $q>0$ . Thus if $p>0, q>0$, (1) has roots with negative real part and hence $$ \lim_{t\to\infty}x(t)=0. $$

Answer 2

It asks for all solutions (which includes all possible $C_1$ and $C_2$ values, as these characterize the solutions) having $\lim_{t\to\infty} x(t) = 0$, so you choose those $p$ and $q$ which guarantee this.

E.g. what is needed to make the solution $C_1 = 1, C_2 = 0$ to behave this way? And what for the solution $C_1 = 0, C_2 = 1$?

For $\Delta = p^2 - 4 q \ge 0$ we have no imaginary components in the exponents and need $$ -\frac{1}{2} \left(\sqrt{p^2 - 4 q} + p\right) < 0 \quad \wedge \quad \frac{1}{2} \left(\sqrt{p^2 - 4 q} - p\right) < 0 \iff \\ \sqrt{p^2 - 4 q} + p > 0 \quad \wedge \quad \sqrt{p^2 - 4 q} - p < 0 $$ Now $$ \sqrt{p^2-4q} - p < 0 \iff \sqrt{p^2-4q} < p \iff q > 0 \wedge p > 0 $$ For $\Delta < 0$ we need $p>0$.

Answer 3

Hint/Sketch:
I will assume that $p,q \in \mathbb{R}$. So based on your work, the general solution (almost every time) is $$x(t) = C_1 e^{\frac{-1}{2} t (\sqrt{p^2 - 4 q} + p)} + C_2 e^{\frac12 t (\sqrt{p^2 - 4 q} - p)}$$ 1. If $p^2-4q>0$, then everything is real. So we need both $e^{\text{something}}$ to go to $0$. That means we need the something to be negative. So we want: $\sqrt{p^2 - 4 q} + p>0$ and $\sqrt{p^2 - 4 q} - p<0$ at the same time (because $t$ is positive).

2.If $p^2-4q<0$, then we are caring about the real part of the something only because the imaginary part will mean oscillation only (As user539887 pointed it out in the comments). In that case, the $\sqrt{p^2-4q}$ is pure imaginary, and the $p$ is real. so we need the $\frac{-tp}{2}$ to be negative, so $p>0$.

3.If $p^2 - 4 q=0$, the general solution is different: $$x(t)=C_1e^{-\frac{tp}{2}}+C_2te^{-\frac{tp}{2}}$$ In this case, you will need the something to be negative, so $p>0$.

Answer 4

The general solution of the homogeneous equation have form $x(t) = c_1e^{r_1t} + c_2e^{r_2t}$ with roots of its characteristic $r_1$ and $r_2$. The limit $$\lim_{t\to \infty} x(t) = c_1\lim_{t\to\infty}e^{r_1t} + c_2\lim_{t\to\infty}e^{r_2t}$$ is $0$, if the limt of terms $e^{r_1t}$ and $e^{r_2t}$ are $0$. Therefore the roots $r_1$ and $r_2$ are negaitive real number that it is $$r = \frac{-p\pm\sqrt{p^2-4q}}{2} < 0.$$ It is two inequalities $$p > \sqrt{p^2 - 4q}\quad \mathrm{and}\quad p^2 - 4q \geq 0$$ that they become $$q > 0\quad \mathrm{and}\quad p^2 \geq 4q.$$ Therefore, if two inequalities $q > 0$ and $p^2 \geq 4q$ are hold, then the general solution of the homogeneous equation $\ddot{x} + p\dot{x} + qx = 0$ have limit $$\lim_{t\to \infty} x(t) = 0.$$ If $D = p^2 - 4q < 0$, what will happen ? The roots have form $r_1 = \alpha + i\beta$ and $r_2 = \alpha - i\beta$ with $$\alpha = -\frac{p}{2}\quad \mathrm{and}\quad \beta = \frac{\sqrt{p^2 - 4q}}{2}.$$ And the solution become $$\begin{aligned}x(t) &= e^{\alpha t}(c_1e^{i\beta t} + c_2e^{-i\beta t}) \\ &= e^{\alpha t}[(c_1+c_2)\cos (\beta t)+ i(c_1-c_2)\sin (\beta t)] \\ &= ae^{\alpha t}(e^{i(\beta t - \gamma)}) \\ &= ae^{\alpha t+ i(\beta t - \gamma)}\end{aligned}$$ with $$a = \sqrt{(c_1+c_2)^2 + (c_1-c_2)^2}\quad\mathrm{and}\quad e^{i\gamma} = \frac{c_1+c_2}{a}+ i\frac{c_1-c_2}{a}.$$ Therefore if $\alpha < 0$ and $\beta < 0$ which is $$p > 0\quad \mathrm{and}\quad p^2 < 4q,$$ the general solution of the homogeneous equation have also limit $$\lim_{t\to \infty} x(t) = 0.$$ Therefore if the coefficient $p > 0$ in the homogeneous, its general solution have the limit $0$.