Systems of Differential Equations — Question 10

Question 10 – Solution

Strategy. Compare Euclidean growth in scaled coordinates with decay in a suitable weighted norm.

Step 1: Derive the scaled cascade. With $N$ having ones just above the diagonal, $$\boxed{q'=(-I+4N)q,\qquad y=q_1,\quad y'=4q_2-q_1,\quad y''=16q_3-8q_2+q_1.}$$ The coordinate map is invertible, so this is a similarity transformation of the scalar companion system. Its triangular matrix has three eigenvalues $-1$.

Step 2: Compute an actual transient. Since $N^3=0$, $e^{(-I+4N)t}=e^{-t}(I+4tN+8t^2N^2)$. Therefore $$\boxed{q=e^{-t}(8t^2,4t,1)^T,\qquad \|q(t)\|_2=e^{-t}\sqrt{64t^4+16t^2+1}.}$$ The initial norm is one, but $\|q(1)\|_2=9/e>1$. Every state still tends to zero because its components are polynomials times $e^{-t}$.

Step 3: Prove a uniform weighted estimate. The new state satisfies $w'=(-I+N)w$, so $$\frac{d}{dt}\|w\|_2^2=-2\|w\|_2^2+2w_2(w_1+w_3) \leq-(2-\sqrt 2)\|w\|_2^2.$$ The stated inequality follows from Cauchy–Schwarz and $2\sqrt 2|w_2|\sqrt{w_1^2+w_3^2}\leq\sqrt 2\|w\|_2^2$. With $\alpha=1-1/\sqrt 2>0$, integration and $\|q\|_2\leq\|w\|_2\leq 16\|q\|_2$ give $$\boxed{\|q(t)\|_2\leq 16e^{-\alpha t}\|q(0)\|_2.}$$

Step 4: Interpret coordinate-dependent growth. Asymptotic stability allows a finite transient; it does not require every norm to decrease at every instant. For this trajectory, $\|w\|_2/16=e^{-t}\sqrt{t^4/4+t^2+1}$ decreases strictly, whereas $\|q\|_2$ has a pronounced transient. They measure the same state with different weights.