Variation of Parameters — Question 6

Question 6 – Solution

Strategy. Use the repeated-root basis to produce a positive kernel whose integral is explicit.

Step 1: Solve the shifted parameter system. Write $y=e^{-x}v$. The equation becomes $v^{(4)}=e^xf(x)$. The original basis becomes $1,x,x^2/2,x^3/6$ for $v$, with parameter derivatives $$(u_0',u_1',u_2',u_3')=e^xf(x)(-x^3/6,x^2/2,-x,1).$$ The homogeneous basis is fundamental: its Wronskian is $e^{-4x}\ne 0$. Integrating from zero gives $$\boxed{y(x)=\int_0^xG(x-t)f(t)\,dt,\qquad G(s)=\frac{s^3e^{-s}}6.}$$

Step 2: Verify the response kernel. The function $G$ satisfies $(D+1)^4G=0$ for $s>0$, and $G(0)=G'(0)=G''(0)=0$, $G'''(0)=1$. Repeated Leibniz differentiation therefore adds no endpoint terms until the fourth derivative, where it adds exactly $f(x)$. Combining derivatives in $(D+1)^4y$ cancels the integral terms. All four initial data vanish because the relevant integrals have zero length.

Step 3: Integrate the positive kernel sharply. For $s\ge 0$, $G(s)\ge 0$. Repeated integration by parts yields $$\int_0^xG(s)\,ds=1-e^{-x}\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)=H(x).$$ Consequently, $$\boxed{|y(x)|\le M H(x)\le M\quad(x\ge 0).}$$ The forcing $f\equiv M$ gives $y=MH$ and attains the pointwise bound. Also $H(x)\to 1$, so no smaller all-time constant than $M$ works for all admissible forcings when $M>0$. The case $M=0$ gives the zero solution.

Step 4: Bound input error without claiming decay. For identical initial data, the response difference has zero initial vector and forcing difference bounded by $\varepsilon$. Thus $|y_1-y_2|\le\varepsilon H(x)\le\varepsilon$. But $f\equiv 1$ gives $y=H(x)\to 1$, not zero. A positive integrable kernel controls size; it does not force the output to vanish for a persistent input.