Repeated Roots — Question 6

Question 6 – Solution

Strategy. Divide by the proposed exponential bound and maximize the resulting ratio, rather than judging only the root’s sign.

Step 1: Find the transient peak. We have $y'=(1-t)e^{-t}$, so the response rises until $t=1$ and falls thereafter. Hence $$\boxed{\max_{t\ge 0} y(t)=1/e}.$$ Also $t/e^t\to 0$, for example by l’Hôpital’s rule, proving eventual decay.

Step 2: Test the exact root exponent. The proposed bound is equivalent to $t\le C$ for all $t\ge 0$, which no finite $C$ can satisfy. The exponential decay is modified by an unbounded polynomial prefactor.

Step 3: Find the sharp slightly slower bound. Dividing by $e^{-(1-\varepsilon)t}$ gives the ratio $t e^{-\varepsilon t}$. Its derivative is $e^{-\varepsilon t}(1-\varepsilon t)$, so its unique maximum occurs at $t=1/\varepsilon$ and equals $1/(e\varepsilon)$. Thus $$\boxed{C_\varepsilon=\frac 1{e\varepsilon}}.$$ This constant is sufficient by the maximum calculation and necessary because equality holds at that time.

For $\varepsilon=1/2$, the bound is $(2/e)e^{-t/2}$ and contact occurs at $t=2$, where both values are $2e^{-2}$. Both derivatives equal $-e^{-2}$, confirming tangency rather than a crossing.

Step 4: Bound every fixed repeated-root solution. The triangle inequality gives $$|w(t)|e^{(1-\varepsilon)t} \le |A|e^{-\varepsilon t}+|B|t e^{-\varepsilon t} \le |A|+\frac{|B|}{e\varepsilon}.$$ Therefore $\boxed{|w(t)|\le (|A|+|B|/(e\varepsilon))e^{-(1-\varepsilon)t}}$. This displayed constant need not be optimal for arbitrary $A,B$. A bound with the exact exponent 1 exists precisely when $B=0$; for $B\ne 0$, the ratio $|A+Bt|$ is unbounded even though $w$ itself decays.