Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 2 / 8A (Differential Equations)

For the damped harmonic oscillator, the characteristic roots are $-k\pm i\omega$. For $\omega \ne 0$, the real general solution isIf the frequency parameter is zero, the characteristic root is repeated and the complete limiting basis is $e^{-kt},t{e}^{-kt}$, so $y_1=e^{-kt}(A+Bt)$. No sign restriction on $k$ is needed for these formulas, although $k>0$ corresponds to decay.

Before the impulse the zero initial data force $y_2=0$. An impulse modeled by the Dirac delta function requires continuity of $y_2$: a jump in $y_2$ would introduce an unwanted derivative of a delta in $y_2^{\prime \prime }$. Integrating the equation across $t=a$ then gives the velocity jump $y_2^{\prime }(a+)-y_2^{\prime }(a-)=1$.

The causal Green function of a damped oscillator isThe jump conditions give the causal impulse responseHere $H$ is the Heaviside step function. For $\omega =0$, interpret the quotient continuously as $t-a$, giving $H(t-a)(t-a)e^{-k(t-a)}$. Both initial values are zero because $a>0$.

The step response of a damped oscillator is the time integral of its causal Green function of a damped oscillator. Put $s=t-b$ and define $K(s)={\int }_0^s{e}^{-ku}\sin (\omega u)/\omega du$ for $s\ge 0$. Then $K(0)=K^{\prime }(0)=0$ and its oscillator equation has constant right side $1$. Integrating explicitly givesThis formula applies when $\omega \ne 0$. It is a convolution of the causal impulse response with the shifted Heaviside step function, so the response is zero before the forcing starts and has the stated zero initial data.

The integral definition also handles the degenerate parameters without division ambiguities. For $\omega =0,k\ne 0$,for $\omega =k=0$, $y_3=H(s)s^2/2$. These are the continuous zero-frequency limits of the step response.

Use the shift derivative of a step response, writing $y_3(t,b)=H(s)K(s)$ with $s=t-b$. Differentiate with respect to the forcing location:The possible delta term vanishes because $K(0)=0$, while $K^{\prime }(s)=g(s)$. Therefore, as a distribution and also as the ordinary piecewise response,Moving the start of a unit step differentiates its forcing into a negative impulse. The zero initial conditions and linearity ensure that the corresponding responses obey the same identity, including the zero-frequency cases.