Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at $\mathrm{\Omega }=\sqrt{k\mathrm{/}m}$

Non-resonant case. [assume-case 1] Suppose $D\mathrm{\ne }0$ and put $z(t):=f_0\exp (i\mathrm{\Omega }t)\mathrm{/}D$. By [F2], $z^{\mathrm{\prime }}=i\mathrm{\Omega }z$ and $z^{\mathrm{\prime }\mathrm{\prime }}=-{\mathrm{\Omega }}^{2}z$, hence $m{z}^{\mathrm{\prime }\mathrm{\prime }}+c{z}^{\mathrm{\prime }}+kz=(-m{\mathrm{\Omega }}^2+ic\mathrm{\Omega }+k)z=Dz=f_0\exp (i\mathrm{\Omega }t)\mathrm{.}$ The differential expression has real coefficients and the real and imaginary parts of a complex-valued $C^2$ function may be differentiated separately, so $x_p=\mathrm{Re}z$ satisfies $m{x}_p^{\mathrm{\prime }\mathrm{\prime }}+c{x}_p^{\mathrm{\prime }}+k{x}_p=\mathrm{Re}(f_0\exp (i\mathrm{\Omega }t))=f_0\cos (\mathrm{\Omega }t)$ on $\mathbb{R}$. The displayed real form follows from [F2] and $1\mathrm{/}D=\bar{D}\mathrm{/}\mathrm{\mid }D{\mathrm{\mid }}^2$, $\bar{D}=k-m{\mathrm{\Omega }}^2-ic\mathrm{\Omega }$, $\mathrm{\mid }D{\mathrm{\mid }}^2=(k-m{\mathrm{\Omega }}^2{)}^2+c^2{\mathrm{\Omega }}^2$. For the amplitude, put $a=k-m{\mathrm{\Omega }}^2$, $b=c\mathrm{\Omega }$ and $r=\mathrm{\mid }D\mathrm{\mid }=\sqrt{a^2+b^2}>0$. By [F5], $(a\cos u+b\sin u{)}^2+(a\sin u-b\cos u{)}^2=r^2$, so $\mathrm{\mid }{x}_p(t)\mathrm{\mid }\le \mathrm{\mid }{f}_0\mathrm{\mid }\mathrm{/}r$. If $\mathrm{\Omega }=0$, then $D=k$ and $x_p\equiv f_0\mathrm{/}k$, attaining that bound. If $\mathrm{\Omega }>0$, the cosine range in [F5] supplies a real $u$ with $\cos u=a\mathrm{/}r$; the Pythagorean identity gives $\mathrm{\mid }\sin u\mathrm{\mid }=b\mathrm{/}r$ since $b\ge 0$, and replacing $u$ by $-u$ if necessary makes $\sin u=b\mathrm{/}r$ while preserving its cosine. At $t=u\mathrm{/}\mathrm{\Omega }$ the numerator $a\cos u+b\sin u$ equals $r$, attaining $\mathrm{\mid }{x}_p(t)\mathrm{\mid }=\mathrm{\mid }{f}_0\mathrm{\mid }\mathrm{/}r$. This also covers $f_0=0$, proving the stated amplitude. When $\mathrm{\Omega }>0$ and $f_0\mathrm{\ne }0$, take $\varphi =u$ from that construction. For this angle, and for any other angle with the stated sine and cosine, the difference formula in [F5] gives $\cos (\mathrm{\Omega }t-\varphi )=(a\cos (\mathrm{\Omega }t)+b\sin (\mathrm{\Omega }t))\mathrm{/}r$; therefore $x_p(t)=(f_0\mathrm{/}r)\cos (\mathrm{\Omega }t-\varphi )$ for every $t$, proving the phase relation with the declared signed convention.

Resonance. [assume-case 2] Suppose $c=0$ and $\mathrm{\Omega }={\omega }_0$, and put $x_p(t)=f_{0}t\sin ({\omega }_{0}t)\mathrm{/}(2m{\omega }_0)$, so that $f(t)=f_0\cos ({\omega }_{0}t)$ and $k=m{\omega }_0^2$. By [F3], $x_p^{\mathrm{\prime }}(t)=\frac{f_0(\sin ({\omega }_{0}t)+{\omega }_{0}t\cos ({\omega }_{0}t))}{2m{\omega }_0},x_p^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{f_0(2{\omega }_0\cos ({\omega }_{0}t)-{\omega }_0^{2}t\sin ({\omega }_{0}t))}{2m{\omega }_0},$ so $m{x}_p^{\mathrm{\prime }\mathrm{\prime }}+k{x}_p=f_0\cos ({\omega }_{0}t)-\frac{1}{2}{f}_0{\omega }_{0}t\sin ({\omega }_{0}t)+\frac{1}{2}{f}_0{\omega }_{0}t\sin ({\omega }_{0}t)=f_0\cos ({\omega }_{0}t)$: it is a solution on $\mathbb{R}$. For each $j\in \mathbb{N}$, put $t_j=(\pi \mathrm{/}2+2\pi j)\mathrm{/}{\omega }_0$. These times are nonnegative and tend to $+\mathrm{\infty }$ because ${\omega }_0>0$ and $\pi >0$. Integer periodicity and the quarter-turn value in [F3] give $\sin ({\omega }_0{t}_j)=1$, so $\mathrm{\mid }{x}_p(t_j)\mathrm{\mid }=\mathrm{\mid }{f}_0\mathrm{\mid }{t}_j\mathrm{/}(2m{\omega }_0)\to +\mathrm{\infty }$ whenever $f_0\mathrm{\ne }0$; hence $x_p$ is unbounded on $[0,\mathrm{\infty })$.

Transient decay. [assume-case 3, cases-exhaustive: $D\mathrm{\ne }0$; $c=0$, $\mathrm{\Omega }={\omega }_0$; $c>0$] Suppose $c>0$; then $D=0$ would force $c\mathrm{\Omega }=0$, hence $\mathrm{\Omega }=0$ and $D=k\mathrm{\ne }0$, a contradiction, so $D\mathrm{\ne }0$ for every $\mathrm{\Omega }$ and step 1.1 supplies the solution $x_p$. Let $x:[0,\mathrm{\infty })\to \mathbb{R}$ be any solution of the forced model. Then $x_h:=x-x_p$ satisfies the homogeneous model by linearity [F1]; by [F4] it is one of the three families of the classification theorem, and since $c>0$, Strictly damped homogeneous oscillator with $k>0$: every solution tends to zero gives $x_h(t)\to 0$ as $t\to +\mathrm{\infty }$ and boundedness of $x_h$ on $[0,\mathrm{\infty })$; $x_p$ is bounded by [F5] because $\mathrm{\mid }\cos \mathrm{\mid }\le 1$, $\mathrm{\mid }\sin \mathrm{\mid }\le 1$ and $D\mathrm{\ne }0$ in the real form of step 1.1. Hence $x(t)-x_p(t)=x_h(t)\to 0$ and $x=x_p+x_h$ is bounded.