Variation of Parameters — Question 7

Question 7 – Solution

Strategy. Cancel the growing mode using a convergent integral, then prove the resulting tail expression is bounded.

Step 1: Construct the family. Variation of parameters with $e^t,e^{-t}$ gives $$y_v=v\sinh t+\int_0^t\sinh(t-s)f(s)\,ds.$$ Since $0<f\le 1$, $0<I\le\int_0^\infty e^{-s}\,ds=1$, with strict inequality $I<1$.

Step 2: Identify the necessary slope. Expanding the hyperbolic sine yields $$e^{-t}y_v=\frac v2(1-e^{-2t})+\frac 12\int_0^t e^{-s}f(s)\,ds -\frac 12e^{-2t}\int_0^t e^sf(s)\,ds.$$ The last term has magnitude at most $(e^{-t}-e^{-2t})/2$. Thus $e^{-t}y_v\to(v+I)/2$. Boundedness requires $\boxed{v_*=-I}$.

Step 3: Prove sufficiency and sign. Splitting the integral at $s=t$ and substituting $v=-I$ gives $$y_*(t)=-\int_0^\infty K(t,s)f(s)\,ds, \quad K(t,s)=\begin{cases}e^{-t}\sinh s,&s\le t,\\ e^{-s}\sinh t,&s\ge t.\end{cases}$$ This identity follows by grouping the $e^{t}$ and $e^{-t}$ terms above. Moreover, $$\int_0^\infty K(t,s)\,ds =e^{-t}(\cosh t-1)+e^{-t}\sinh t=1-e^{-t}.$$ Therefore $|y_*|\le 1-e^{-t}<1$. For $t>0$, the kernel and forcing are positive for $s>0$, so $y_*(t)<0$.

Step 4: Prove decay and sensitivity. Given $\varepsilon>0$, choose $A$ with $f(s)\le\varepsilon$ for $s\ge A$. For $t>A$, the integral over $[0,A]$ has magnitude at most $e^{-t}\int_0^A\sinh s\,ds\to 0$; the remaining magnitude is at most $\varepsilon\int_A^\infty K\le\varepsilon$. Hence $y_*\to 0$. If $v=v_*+\delta$, then $y_v=y_*+\delta\sinh t$, so $y_v/e^t\to\delta/2$. It tends to $+\infty$ for $\delta>0$ and $-\infty$ for $\delta<0$.