Convolution Integrals — Question 3

Question 3 – Solution

Strategy. Keep the Gaussian force inside an integral and separate the permanent oscillation from the remaining tail integral.

Step 1: Transform and invert. The initial terms give $(s^2+1)Y-2s+1=G$, hence $$Y=\frac{2s-1+G}{s^2+1},\qquad \boxed{y=2\cos t-\sin t+z(t),\quad z(t)=\int_0^t\sin(t-u)e^{-u^2}\,du.}$$ These transform operations are valid for real $s>0$.

Step 2: Differentiate with the endpoint terms. The sine kernel vanishes at its endpoint, so $z'=\int_0^t\cos(t-u)e^{-u^2}\,du$. Differentiating again gives $z''=e^{-t^2}-z$ because $\cos 0=1$. Also $z(0)=z'(0)=0$. Thus the formula satisfies the full IVP.

Step 3: Isolate the asymptotic oscillation. Set $A=\int_0^\infty e^{-u^2}\cos u\,du$ and $B=\int_0^\infty e^{-u^2}\sin u\,du$, both absolutely convergent. Expanding $\sin(t-u)$ gives $$y_\infty=(2-B)\cos t+(A-1)\sin t,\qquad y-y_\infty=-\int_t^\infty\sin(t-u)e^{-u^2}\,du.$$ For $t>0$, use $1\le u/t$ on the tail: $$|y-y_\infty|\le\int_t^\infty e^{-u^2}\,du \le\frac 1t\int_t^\infty u e^{-u^2}\,du =\boxed{\frac{e^{-t^2}}{2t}}.$$

Step 4: Establish persistent motion. Since $|A|\le\sqrt\pi/2<1$, the coefficient $A-1$ is nonzero. The limiting sinusoid therefore has positive amplitude and has recurring positive and negative extrema. Along their times the error tends to zero, so $y$ cannot tend to zero (or to any single constant). The homogeneous modes $\cos t,\sin t$ have no damping. The figure compares the exact integral solution with $y_\infty$; the two quickly become indistinguishable.