Real Eigenvalues — Question 1

Question 1 – Solution

Strategy. Decompose into independent eigendirections, then distinguish the dominant slow mode from initial data that cancel it.

Step 1: Resolve the modes. The characteristic polynomial is $(\lambda+1)(\lambda+3)$. Eigenvectors are $v_1=(1,1)^T$ for $-1$ and $v_2=(1,-1)^T$ for $-3$. They form a basis, so modal coordinates satisfy $u'=-u$, $v'=-3v$. Initially $u=v=1$, giving $$\boxed{X=e^{-t}\binom 11+e^{-3t}\binom 1{-1}.}$$ This decoupling also proves completeness for arbitrary initial data.

Step 2: Retain the orbit domain. Here $u=e^{-t}>0$ and $v=e^{-3t}=u^3$ for all real $t$. Thus the orbit is $$\frac{x-y}{2}=\left(\frac{x+y}{2}\right)^3,\qquad x+y>0.$$ The origin is a limiting point, not a point attained at finite time.

Step 3: Read the approach direction. As time increases, $u$ decreases toward zero. Moreover $y/x=(1-e^{-2t})/(1+e^{-2t})\to 1$. Both the state direction and the velocity direction approach the line $y=x$; indeed $dy/dx=(1-3e^{-2t})/(1+3e^{-2t})\to 1$. At $(2,0)$ the velocity is $(-4,2)$, consistent with the arrows.

Step 4: Keep the exceptional line. For initial $(p,q)$, the slow and fast coefficients are $(p+q)/2$ and $(p-q)/2$. Every solution tends to zero. A nonzero solution approaches along the fast line $y=-x$ exactly when $p+q=0$; otherwise its limiting line is $y=x$. The zero state remains fixed and has no approach direction. Both eigenvalues are negative, so the origin is a stable node.