Variation of Parameters — Question 6

Question 6 – Solution

Strategy. The sign of the integral kernel gives comparison and error bounds without evaluating each forcing integral.

Step 1: Derive the response kernel. Here $W=-2$, so $u_1'=e^{-t}g/2$ and $u_2'=-e^tg/2$. Integrating from zero and combining gives $$\boxed{y(t)=\int_0^t\sinh(t-s)g(s)\,ds.}$$ Differentiation gives $y'=\int_0^t\cosh(t-s)g(s)\,ds$ and $y''=g(t)+y(t)$. Both initial data vanish.

Step 2: Prove sharpness. For $0\le s\le t$, $\sinh(t-s)\ge 0$. Consequently $$|y(t)|\le M\int_0^t\sinh(t-s)\,ds =\boxed{M(\cosh t-1)}.$$ Constant forcing $g=M$ gives equality for every $t$, and $g=-M$ gives equality in magnitude. The bound is therefore sharp for this forcing class; it also bounds response error when $g$ denotes a forcing error.

Step 3: Prove positivity carefully. If $g\ge 0$, the integral is nonnegative. At fixed $t>0$, it is strictly positive exactly when $g$ is not identically zero on $[0,t]$. Indeed, continuity then gives a positive subinterval lying before $t$, where the kernel is strictly positive. Forcing that is positive only at later times cannot make the response positive at this time.

Step 4: Evaluate a comparison example. For $g=(1+\cos 2t)/2$, an elementary particular solution is $-1/2-\cos(2t)/10$. Fitting zero data gives $$\boxed{y=\tfrac 35\cosh t-\tfrac 12-\tfrac 1{10}\cos 2t.}$$ Direct differentiation yields $y''-y=(1+\cos 2t)/2$ and zero data. Since $0\le g\le 1$, the kernel proves $0<y(t)<\cosh t-1$ for $t>0$: both $g$ and $1-g$ are nonzero on every such interval.