Repeated Eigenvalues — Question 8

Question 8 – Solution

Strategy. The nilpotent coupling does not change the eigenvalues, but it can change which quadratic size decreases at every state.

Step 1: Solve for every coupling strength. The repeated eigenvalue is $-1$. If $k=0$, the eigenspace is the whole plane; if $k>0$, it is $y=0$. For initial $(p,q)$, $$\boxed{X=e^{-t}(p+kqt,q)^T.}$$ The component equations directly verify completeness. Constant and linear factors times $e^{-t}$ both vanish forward, so every initial state decays.

Step 2: Locate the exact norm threshold. The squared-norm derivative is $$(x^2+y^2)'=-2x^2+2kxy-2y^2.$$ For $0\le k<2$, $2|xy|\le x^2+y^2$ makes this strictly negative off zero. At $k=2$ it equals $-2(x-y)^2$, with equality on $x=y$. For $k>2$, its value at $(1,1)$ is $2k-4>0$. Thus the threshold is $\boxed{k=2}$, despite identical eigenvalues throughout the family.

Step 3: Distinguish instantaneous equality from an interval. At $k=2$, a trajectory cannot stay on $x=y$ over an interval unless it is zero. On that line the velocity is $(x,-x)$, which is tangent to the line only when $x=0$. Integrating $-2(x-y)^2$ over any interval therefore gives strictly negative change for every nonzero solution. The norm strictly decreases between distinct times even if its derivative vanishes at an isolated time.

Step 4: Construct a uniformly valid quadratic choice. Choose $c=1+k^2/2>0$. Completing the square yields $$\boxed{H'=-2(x-ky/2)^2-(2+k^2/2)y^2<0\quad(X\ne 0).}$$ The quadratic $H=x^2+(1+k^2/2)y^2$ is positive definite for every $k$, including zero. Its decrease is compatible with temporary Euclidean growth when $k>2$; the two quantities measure different families of ellipses. The spectral decay is unchanged by this choice of measurement.