Real Eigenvalues — Question 7

Question 7 – Solution

Strategy. Express the initial state in a basis of three real modes, then read boundedness from individual modal coordinates.

Step 1: Resolve all three components. The triangular matrix has distinct eigenvalues $-2,-1,1$. Corresponding vectors are $v_1=(1,0,0)^T$, $v_2=(1,1,0)^T$, $v_3=(1,1,1)^T$, forming an invertible triangular basis. Solving for coefficients yields $$\boxed{X=(p-q)e^{-2t}v_1+(q-r)e^{-t}v_2+r e^t v_3.}$$ Each term satisfies the ODE; their sum at zero is $(p,q,r)^T$. Equivalently the independent coordinates $x-y,y-z,z$ solve scalar equations with rates $-2,-1,1$, proving completeness.

Step 2: Find the stable plane and its exception. Forward boundedness requires and is implied by $r=0$; this is the plane $z=0$ of initial states. If $q\ne 0$, then $X/e^{-t}\to qv_2$, so the slow limiting line is $\operatorname{span}(v_2)$. If $q=0$ and $p\ne 0$, only the faster mode $p e^{-2t}v_1$ remains. The zero state is separate. All states in this plane tend to zero.

Step 3: Find the backward-bounded line. As $t\to-\infty$, the negative-rate exponentials grow. Their coefficients must vanish: $p-q=0$ and $q-r=0$. Thus the exact set is the line $\boxed{p=q=r}$. On it $X=r e^t v_3$, tending to zero backward and, for $r\ne 0$, growing without bound forward. At $r=0$ it is the equilibrium.

Step 4: Intersect the two conditions. A solution bounded for all real time lies both in $r=0$ and in $p=q=r$, so $\boxed{X\equiv 0}$ is the only one. This is not merely an inference from the norm of a sum. The inverse basis map is fixed and linear; bounded $X$ forces all its modal coordinates $x-y,y-z,z$ to be bounded. Each is one scalar exponential, so a forbidden growing coefficient cannot be hidden by cancellation with another mode.