Variation of Parameters — Question 8

Question 8 – Solution

Strategy. After the forcing ends, its two weighted integrals become the coefficients of the remaining homogeneous oscillation.

Step 1: Solve while the first forcing acts. Variation of parameters gives $y(t)=\int_0^t\sin(t-s)g(s)\,ds$. For $g_2$ on $[0,\pi]$, evaluation gives $$y_2(t)=\frac{2\sin t-\sin 2t}{3}.$$ Its value and slope at zero vanish, and $y_2''+y_2=\sin 2t$, independently verifying the integral evaluation.

Step 2: Match at the shutoff time. At $\pi$, $y_2=0$ and $y_2'=-4/3$. The subsequent homogeneous response is $$\boxed{y_2(t)=-\tfrac 43\sin(t-\pi)=\tfrac 43\sin t,\quad t\ge\pi.}$$ It matches both data. Since $g_2(\pi)=0$, $y''=g_2-y$ also matches. The resulting function is a classical $C^2$ solution, but it does not return to rest.

Step 3: Derive the cancellation conditions. For any such forcing and $t\ge\pi$, $$y(t)=\sin t\int_0^\pi\cos s\,g(s)\,ds -\cos t\int_0^\pi\sin s\,g(s)\,ds.$$ Independence of sine and cosine shows that the response vanishes identically after $\pi$ exactly when both displayed integrals are zero. Cancellation of just one moment is insufficient.

Step 4: Construct a return to rest. For $g_3$, product-to-sum identities give $\int_0^\pi\cos s\sin 3s\,ds=0$ and $\int_0^\pi\sin s\sin 3s\,ds=0$. Explicitly, $$\boxed{y_3(t)=\begin{cases}(3\sin t-\sin 3t)/8,&0\le t\le\pi,\\0,&t\ge\pi.\end{cases}}$$ On the first interval it has residual $\sin 3t$ and zero initial data; at $\pi$ its value and slope are zero. The forcing’s signed contributions cancel both final-state components, despite a nonzero response during the forcing interval.