Systems of Differential Equations — Question 9

Question 9 – Solution

Strategy. Expose an internal oscillator that resonantly forces the displacement oscillator.

Step 1: Reveal the oscillator cascade. The original equation gives $$\boxed{W'=\begin{pmatrix}0&1&0&0\\-1&0&1&0\\0&0&0&1\\0&0&-1&0\end{pmatrix}W.}$$ The coordinate map is triangular with determinant one; its inverse is $(y,y',y'',y''')=(W_1,W_2,W_3-W_1,W_4-W_2)$. The last pair satisfies $W_3'=W_4$, $W_4'=-W_3$ independently.

Step 2: Solve the resonant forcing. We have $W_3=A\cos t+B\sin t$ and $y''+y=W_3$. A resonant particular solution is $(A/2)t\sin t-(B/2)t\cos t$. Enforcing the first two data gives $$\boxed{y=a\cos t+(b+B/2)\sin t+(A/2)t\sin t-(B/2)t\cos t.}$$ It has $y''(0)=A-a$ and $y'''(0)=B-b$, checking all four data. Applying $D^2+1$ twice verifies the fourth-order equation.

Step 3: Classify bounded data. If $A=B=0$, only bounded sine and cosine remain. If $(A,B)\ne(0,0)$, choose a phase $\theta$ with $A\sin\theta-B\cos\theta=\sqrt{A^2+B^2}$. Along $t_n=\theta+2\pi n\geq 0$, the secular part grows as $t_n\sqrt{A^2+B^2}/2$, while the other terms remain bounded. Consequently $$\boxed{y\text{ bounded}\ \Longleftrightarrow\ y''(0)=-y(0),\quad y'''(0)=-y'(0).}$$ The whole state is bounded exactly on this same two-dimensional subspace.

Step 4: Distinguish two kinds of conserved energy. The internal oscillator has $$E_{\mathrm{int}}=\tfrac 12(W_3^2+W_4^2)=\tfrac 12(A^2+B^2), \qquad E_{\mathrm{int}}'=W_3W_4-W_4W_3=0.$$ This is not positive definite on the full state space. If a conserved positive- definite form $W^TPW$ existed, its smallest eigenvalue would bound $\|W(t)\|^2$ by a constant for every initial state. The unbounded solutions just constructed contradict that conclusion. Purely imaginary characteristic roots with repeated factors therefore do not imply boundedness of all trajectories.