Variation of Parameters — Question 8

Question 8 – Solution

Strategy. Distinguish a nonnegative response kernel from the sign of its derivative.

Step 1: Compute the response kernel. The derivative matrix of $1,\cos x,\sin x$ has Wronskian $1$. Its parameter equations give $$u_0'=g,\qquad u_c'=-g\cos x,\qquad u_s'=-g\sin x.$$ Integrating from zero and combining by the cosine difference identity yields $$\boxed{y(x)=\int_0^x[1-\cos(x-t)]g(t)\,dt.}$$ The kernel and its first derivative vanish at zero, its second derivative is $1$ there, and its third plus first derivative vanishes. These facts verify the equation and zero data.

Step 2: Prove positivity without a monotonicity assumption. Both factors in the integrand are nonnegative. If $g(t_0)>0$ for some $0\le t_0<x$, continuity gives an interval of positive forcing inside $[0,x)$. The zeros of $1-\cos(x-t)$ are isolated, so the integral is strictly positive. In particular, a continuous nonnegative forcing that is not identically zero on $[0,x)$ suffices.

Step 3: Evaluate the post-pulse moments. For the stated pulse, $$\int_0^\pi\sin^2t\,dt=\frac\pi 2,\qquad \int_0^\pi\sin^2t\cos t\,dt=0,\qquad \int_0^\pi\sin^3t\,dt=\frac 43.$$ The first follows from the double-angle identity, the second from the derivative of $\sin^3t/3$, and the third by substituting $u=\cos t$. Thus $$\boxed{y(x)=\frac\pi 2-\frac 43\sin x\quad(x\ge\pi).}$$ For plotting the initial interval, evaluating the same kernel gives $y=x/2-2\sin x/3+\sin 2x/12$ on $[0,\pi]$; its state matches the displayed branch at $\pi$.

Step 4: Separate positivity from monotonicity. The post-pulse range is $$\boxed{\frac\pi 2-\frac 43\le y(x)\le\frac\pi 2+\frac 43.}$$ Both bounds are attained infinitely often; the lower bound is positive. Yet $y'=-4\cos x/3$ changes sign, and $y'(2\pi)=-4/3<0$. Thus the nonnegative input gives a nonnegative response that continues oscillating after the pulse. Positivity of $1-\cos s$ does not imply positivity of its derivative $\sin s$.