Systems of Differential Equations — Question 7

Question 7 – Solution

Strategy. Separate invertibility of the derivative matrix from the extra consistency conditions imposed when its rank drops.

Step 1: Solve when the derivative matrix is invertible. Here $$M_\kappa=\begin{pmatrix}1&1\\1&\kappa\end{pmatrix},\quad B=\begin{pmatrix}-1&0\\0&1\end{pmatrix},\quad \det M_\kappa=\kappa-1.$$ For $\kappa\ne 1$, inversion yields $$\boxed{X'=\frac 1{\kappa-1} \begin{pmatrix}-\kappa&-1\\1&1\end{pmatrix}X.}$$ This constant linear system has a unique solution for every initial state.

Step 2: Test instantaneous algebraic consistency. At $\kappa=1$, the left sides coincide, so the right sides must satisfy $y=-x$. At a state $(a,-a)$, the derivative pairs $(p,q)$ satisfy $p+q=-a$, an entire line of possibilities. Outside $y=-x$, no derivative can satisfy both equations even at that instant.

Step 3: Enforce the constraint along a solution. A classical solution must have $y(t)=-x(t)$ at every time. Differentiating this identity gives $x'+y'=0$. The first equation then forces $x=0$, and the constraint forces $y=0$. Therefore $$\boxed{X(t)\equiv(0,0)^T\text{ is the only classical solution at }\kappa=1.}$$ It indeed satisfies both equations. Nonzero points on the consistency line admit instantaneous derivatives but no solution interval through them. Even at the origin, most algebraic derivative choices cannot persist.

Step 4: Check the limiting initial velocity. Substituting $(1,-1)$ into the explicit system gives $x'=(-\kappa+1)/(\kappa-1)=-1$ and $y'=0$ for every $\kappa\ne 1$. Thus $(x+y)'=-1$, although initially $x+y=0$. This velocity does not keep the state on the constraint line required at $\kappa=1$. Its finiteness does not imply a limiting solution through $(1,-1)$; Step 3 proves that no such solution exists at that parameter.