Undetermined Coefficients — Question 10

Question 10 – Solution

Strategy. Convert the pulse into a polynomial third derivative and enforce triple zeros at both ends.

Step 1: Translate the initial and terminal states. The shift gives $v'''=p$ on $(0,1)$ and $v'''=0$ on $(1,\infty)$. Multiplication by the smooth, nonzero $e^{-x}$ preserves matching of the first three state components. Thus $v(0)=v'(0)=v''(0)=0$, and complete cancellation requires $v(1)=v'(1)=v''(1)=0$. These terminal conditions also suffice: the post-pulse quadratic solution is then zero.

Step 2: Determine the polynomial exactly. The zero-data polynomial solution of $v'''=p$ has degree six and leading coefficient $1$, since $D^3x^6=120x^3$. The three zero conditions at each endpoint require divisibility by $x^3(x-1)^3$. Equal degrees and leading coefficients therefore force $$v=x^3(x-1)^3=x^6-3x^5+3x^4-x^3.$$ Differentiating three times gives $p=120x^3-180x^2+72x-6$. Hence $$\boxed{(a,b,c)=(-180,72,-6),\qquad y(x)=\begin{cases}e^{-x}x^3(x-1)^3,&0\le x\le 1,\\0,&x\ge 1.\end{cases}}$$ The formulas agree at the shared endpoint and uniquely determine the pulse.

Step 3: Prove the degree is necessary. For a polynomial $p$ of degree at most two, the zero-data solution $v$ has degree at most five. The terminal requirements would still make it divisible by the degree-six polynomial $x^3(x-1)^3$. It must therefore be zero, and so must $p=v'''$. No nonzero lower-degree polynomial pulse works.

Step 4: Check the join and the scaled shape. The triple endpoint zeros verify all six state conditions. The computed third derivative verifies the forcing before the switch; the zero branch verifies it afterward. At $1$, the left third derivative of $y$ is $e^{-1}v'''(1)=6/e$, while the right third derivative is zero. Thus $y$ is $C^2$ but not $C^3$ across the switch, exactly as allowed.

On $[0,1]$, $e^xy=-[x(1-x)]^3$. Since $x(1-x)$ has maximum $1/4$ at $x=1/2$, the scaled response has its unique minimum $\boxed{-1/64}$ there.