Repeated Roots — Question 9

Question 9 – Solution

Strategy. The exponential transformation turns the repeated-root endpoint problem into linear interpolation, but its continuation beyond the interval is extrapolation.

Step 1: Solve the endpoint equations. Write $y=(A+Bt)e^{-t}$. The first datum sets $A=a$, and the second gives $a+BL=be^L$. Thus $$\boxed{y(t)=[a+\frac{be^L-a}{L}t]e^{-t}}.$$ The two constants are uniquely determined because $L\ne 0$. This also proves existence by substitution into the repeated-root family.

Step 2: Prove positivity between the data points. For $0\le t\le L$, $$u(t)=e^ty(t)=(1-t/L)a+(t/L)be^L.$$ The weights are nonnegative and sum to 1, while both endpoint values are positive. Hence $u>0$ and therefore $y>0$ throughout the closed interval.

Step 3: Determine when positivity persists. The affine function $u=a+Bt$ starts positive. It remains nonnegative for all $t\ge 0$ exactly when $B\ge 0$. Equivalently, $$\boxed{b\ge ae^{-L}}.$$ At equality, $u$ is constant and $y=ae^{-t}$. If $b<ae^{-L}$, the slope of $u$ is negative and its eventual zero is unavoidable despite both observed values being positive.

Step 4: Check the concrete continuation. Here $B=e^{-1}-1<0$, so $$y(t)=[1-(1-e^{-1})t]e^{-t}.$$ Its unique zero is $$\boxed{t_z=\frac 1{1-e^{-1}}=\frac e{e-1}>1}.$$ The response is negative for every $t>t_z$ and tends to zero from below. The given endpoint at 1 is indeed $[e^{-1}]e^{-1}=e^{-2}$. Linear interpolation of positive transformed data stays positive between them; extending that same negative-slope line beyond the interval eventually crosses zero. The endpoint data determine the continuation uniquely, but do not force it to preserve the observed sign forever.