Variation of Parameters — Question 10

Question 10 – Solution

Strategy. Keep the variation-of-parameters integrals exact, then bound their tails to identify the surviving oscillation.

Step 1: Write and check the exact response. With $W=1$, the parameter derivatives are $u_1'=-e^{-t^2}\sin t$ and $u_2'=e^{-t^2}\cos t$. Hence $$\boxed{y(t)=\sin t\int_0^t e^{-s^2}\cos s\,ds -\cos t\int_0^t e^{-s^2}\sin s\,ds.}$$ Equivalently, $y=\int_0^t\sin(t-s)e^{-s^2}\,ds$. Differentiating gives $y'=\int_0^t\cos(t-s)e^{-s^2}\,ds$ and $y''=e^{-t^2}-y$; both initial data vanish.

Step 2: Isolate the tails. Absolute convergence follows from the integrability of $e^{-s^2}$ (for $s\ge 1$, it is at most $e^{-s}$). Subtracting the full integrals gives $$y(t)-P(t)=-\int_t^\infty\sin(t-s)e^{-s^2}\,ds.$$ Thus $|y-P|\le\int_t^\infty e^{-s^2}\,ds\to 0$.

Step 3: Obtain an explicit error bound. For $s\ge t>0$, $1\le s/t$. Consequently $$|y-P|\le\frac 1t\int_t^\infty s e^{-s^2}\,ds =\boxed{\frac{e^{-t^2}}{2t}}.$$ Combining the tails before estimating avoids an unnecessary sum of two bounds.

Step 4: Prove a nonzero surviving oscillation. Absolute convergence permits pairing intervals. Their $k$th pair contributes $$\int_0^\pi\sin u[e^{-(2k\pi+u)^2} -e^{-((2k+1)\pi+u)^2}],du>0,\qquad k\ge 0.$$ Therefore $S>0$. Along $t=2n\pi$, $y(t)\to-S$; along $t=(2n+1)\pi$, $y(t)\to S$. The solution has no limit and does not decay to zero. A decaying forcing can leave a persistent homogeneous oscillation.