The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data

Algebraic reduction: for all $\lambda$, $p(\lambda )=m\left[\right(\lambda +\frac{c}{2m}{)}^2-\frac{\mathrm{\Delta }}{4m^2}]=m[(\lambda +\gamma {)}^2-\frac{\mathrm{\Delta }}{4m^2}]\mathrm{.}$ Hence: if $\mathrm{\Delta }>0$ the real roots of $p$ are exactly ${\lambda }_{\pm }=(-c\pm \sqrt{\mathrm{\Delta }})\mathrm{/}(2m)$ with ${\lambda }_{+}>{\lambda }_{-}$; if $\mathrm{\Delta }=0$ then $\lambda =-c\mathrm{/}(2m)$ is the only root, a double one; and if $\mathrm{\Delta }<0$ then $p$ has no real root, $\mathrm{\Delta }=-4m^2{\omega }^2$ for the positive number $\omega =\sqrt{4mk-c^2}\mathrm{/}(2m)$, and the complex roots of $p$ are $-\gamma \pm i\omega$ because $(\pm i\omega {)}^2=\mathrm{\Delta }\mathrm{/}(4m^2)$. In particular $p({\lambda }_{\pm })=0$ in case 1 and $p(\lambda )=0$ in case 2.

Verification of the real-root families. [assume-case 1, assume-case 2] For a real number $\mu$ with $p(\mu )=0$ the function $t\mapsto e^{\mu t}$ satisfies $m(e^{\mu t}{)}^{\mathrm{\prime }\mathrm{\prime }}+c(e^{\mu t}{)}^{\mathrm{\prime }}+k{e}^{\mu t}=(m{\mu }^2+c\mu +k)e^{\mu t}=0$ by [F3]; and $t\mapsto t{e}^{\mu t}$ satisfies $m(t{e}^{\mu t}{)}^{\mathrm{\prime }\mathrm{\prime }}+c(t{e}^{\mu t}{)}^{\mathrm{\prime }}+kt{e}^{\mu t}=e^{\mu t}[(2m\mu +c)+tp(\mu )],$ since $(t{e}^{\mu t}{)}^{\mathrm{\prime }}=e^{\mu t}(1+\mu t)$ and $(t{e}^{\mu t}{)}^{\mathrm{\prime }\mathrm{\prime }}=e^{\mu t}(2\mu +{\mu }^{2}t)$ by [F3]; this vanishes exactly when $2m\mu +c=0$, that is $\mu =-c\mathrm{/}(2m)=\lambda$, and then also $p(\mu )=0$. By step 1.1 the roots ${\lambda }_{+},{\lambda }_{-}$ in case 1 and the double root $\lambda$ in case 2 do satisfy $p=0$, so by linearity of the differential expression every $A{e}^{{\lambda }_{+}t}+B{e}^{{\lambda }_{-}t}$ and every $(A+Bt)e^{\lambda t}$ is a solution on every interval.

Verification of the oscillatory family. [assume-case 3] Let $\mu :=-\gamma +i\omega$, so that $p(\mu )=0$ by step 1.1; by [F3] the complex-valued curve $z(t):=\exp (\mu t)$ satisfies $m{z}^{\mathrm{\prime }\mathrm{\prime }}+c{z}^{\mathrm{\prime }}+kz=p(\mu )z=0$. Real and imaginary parts of a complex-valued solution of a real linear constant-coefficient equation solve that equation, by componentwise differentiation; by [F4], $z(t)=e^{-\gamma t}(\cos \omega t+i\sin \omega t)$, so $e^{-\gamma t}\cos \omega t$ and $e^{-\gamma t}\sin \omega t$ are real solutions, and every $e^{-\gamma t}(A\cos \omega t+B\sin \omega t)$ is one by linearity. This family too is defined on all of $\mathbb{R}$.

Matching initial data: in case 1, the two linear conditions $A+B=x_0$ and ${\lambda }_{+}A+{\lambda }_{-}B=v_0$ are solved uniquely by the coefficients displayed in clause 4 because ${\lambda }_{+}-{\lambda }_{-}\mathrm{\ne }0$, and substitution verifies $x(0)=x_0$, $x^{\mathrm{\prime }}(0)=v_0$. In case 2, $x(t)=(A+Bt)e^{\lambda t}$ has $x(0)=A$ and $x^{\mathrm{\prime }}(0)=\lambda A+B$ by [F3], giving $A=x_0$, $B=v_0-\lambda x_0$. In case 3, $x(t)=e^{-\gamma t}(A\cos \omega t+B\sin \omega t)$ has $x(0)=A$ and, by [F3], $x^{\mathrm{\prime }}(0)=-\gamma A+\omega B$, giving $A=x_0$ and $B=(v_0+\gamma x_0)\mathrm{/}\omega$ since $\omega \mathrm{\ne }0$. Thus in each case there is a global solution with the prescribed data.

Exhaustiveness and uniqueness. [cases-exhaustive: $\mathrm{\Delta }>0$, $\mathrm{\Delta }=0$, $\mathrm{\Delta }<0$] Let $x$ solve the model on an interval $J$ with nonempty interior and choose an arbitrary $t_{\ast }\in J^{\circ }$. For the ordered fundamental pairs $(e^{{\lambda }_{+}t},e^{{\lambda }_{-}t})$, $(e^{\lambda t},t{e}^{\lambda t})$ and $(e^{-\gamma t}\cos \omega t,e^{-\gamma t}\sin \omega t)$, the determinant of their value/derivative matrix is respectively $({\lambda }_{-}-{\lambda }_{+})e^{({\lambda }_{+}+{\lambda }_{-})t}$, $e^{2\lambda t}$ and $\omega e^{-2\gamma t}$, by [F3]. Each is nonzero, so there is a unique combination $y$ from the corresponding global family with $y(t_{\ast })=x(t_{\ast })$ and $y^{\mathrm{\prime }}(t_{\ast })=x^{\mathrm{\prime }}(t_{\ast })$. Put $S=\{t\in J^{\circ }:(x(t),x^{\mathrm{\prime }}(t))=(y(t),y^{\mathrm{\prime }}(t))\}$. This set is nonempty and relatively closed by phase-curve continuity. It is relatively open: at $t_1\in S$ choose a closed time neighborhood in $J^{\circ }$ and a state ball centered at the common phase value, containing both curves on a smaller time neighborhood. Choose $h>0$ within that neighborhood with $h{\sup }_{\bar{B}(a_0,R)}\mathrm{\parallel }{G}_0{\mathrm{\parallel }}_2\le R\mathrm{/}2$ and $Lh<1$. Since $G_0$ is autonomous, translate $t_1$ to zero and apply [F2] with $\rho =R\mathrm{/}2$ to the two phase curves; uniqueness gives equality on this neighborhood. The interval $J^{\circ }$ is connected, so $S=J^{\circ }$. Continuity extends the equality to any endpoints belonging to $J$, proving $x=y{\mathrm{\mid }}_J$. Thus every local solution is a restriction of exactly one global member of the appropriate family. In particular the global members matched in step 3.1 are the unique solutions with the prescribed data at zero, and all four clauses follow. A proper interval solution therefore has a strictly larger all-$\mathbb{R}$ extension, whereas an all-$\mathbb{R}$ solution has no larger real time interval; this proves the maximality assertion. The same classification applies to a restricted homogeneous model on $I$ by restricting these family members to $J\subseteq I$.