Undetermined Coefficients — Question 6

Question 6 – Solution

Strategy. Enforce the initial data before taking a resonant limit; separate divergent terms can cancel.

Step 1: Solve away from resonance. Set $K=[(1-\omega^2)(4-\omega^2)]^{-1}$. The trial $K\cos\omega x$ has the required residual. Odd initial derivatives vanish; adding $A\cos x+C\cos 2x$ and matching the value and second derivative gives $$\boxed{y_\omega=K\cos\omega x-\frac{\cos x}{3(1-\omega^2)} +\frac{\cos 2x}{3(4-\omega^2)},\quad \omega\ne 1,2.}$$ Indeed $A+C=-K$ and $A+4C=-K\omega^2$, which verify the remaining two data.

Step 2: Solve at the two simple resonances. For $L=(D^2+1)(D^2+4)$, direct differentiation gives $L(x\sin x)=6\cos x$ and $L(x\sin 2x)=-12\cos 2x$. Adding homogeneous cosine terms to enforce zero value and curvature yields $$\boxed{y_1=\frac{x\sin x}{6}+\frac{\cos 2x-\cos x}{9},\qquad y_2=-\frac{x\sin 2x}{12}+\frac{\cos x-\cos 2x}{9}.}$$ Both are even and vanish at zero. Their second derivatives at zero vanish by cancellation of $1/3$ with $-1/3$ in each case, so all four initial data hold.

Step 3: Take the limit of the combined response. Partial fractions give $$y_\omega=\frac 13\left(\frac{\cos\omega x-\cos x}{1-\omega^2} -\frac{\cos\omega x-\cos 2x}{4-\omega^2}\right).$$ For fixed $x$, the first ratio tends to $x\sin x/2$ by differentiating numerator and denominator with respect to $\omega$. The second tends to $(\cos x-\cos 2x)/3$. The resulting limit is exactly $y_1$.

Step 4: Distinguish the two limit questions. Every fixed nonresonant response is a bounded sum of cosines. At resonance, the linear sine term has unbounded values along its phase maxima, and the added cosines remain bounded. Thus boundedness holds exactly when $\omega\notin\{1,2\}$. A common bound for all $x\ge 0$ and all nonresonant $\omega$ sufficiently near $1$ would pass to the fixed-$x$ limit and bound $y_1$ everywhere, a contradiction.