Real Eigenvalues — Question 4

Question 4 – Solution

Strategy. Euclidean distance mixes nonorthogonal modes; a quadratic quantity in modal coordinates separates their decay.

Step 1: Decouple the eigenmodes. The eigenpairs are $-1,(1,0)^T$ and $-2,(-6,1)^T$. Since $(0,1)^T=6(1,0)^T+(-6,1)^T$, $$\boxed{x=6(e^{-t}-e^{-2t}),\qquad y=e^{-2t}.}$$ Every initial vector has this two-mode form with suitable coefficients, so all solutions tend to zero.

Step 2: Exhibit transient amplification. The derivative $x'=6(-e^{-t}+2e^{-2t})$ changes from positive to negative at $t=\ln 2$. Thus $x_{\max}=3/2$, while $y=1/4$ there. Consequently$$\boxed{\|X(\ln 2)\|=\sqrt{37}/4>1=\|X(0)\|}.$$ This is the maximum of $x$, not a claim about the time of maximum norm.

Step 3: Test instantaneous norm growth. Direct differentiation gives $$(\|X\|^2)'=-2x^2+12xy-4y^2.$$ At $(1,1)$ it equals $6>0$, so the norm can increase. At the given initial state $(0,1)$ it equals $-4$: this trajectory first moves closer, but later exceeds its initial distance. Negative eigenvalues do not make this quadratic expression negative at every state.

Step 4: Use a norm adapted to the modes. The invertible coordinates satisfy $u'=-u$, $v'=-2v$. Thus $E=(x+6y)^2+y^2=u^2+v^2$ is positive except at zero and $\boxed{E'=-2u^2-4v^2<0}$ for every nonzero state. The Euclidean circles and the level ellipses of $E$ measure different sizes. Decrease of the latter is compatible with temporary Euclidean growth, and the explicit modes still establish eventual decay.