Systems of Differential Equations — Question 8

Question 8 – Solution

Strategy. Normalize a scalar-derived fundamental matrix at the actual base time before transporting states or forcing.

Step 1: Lift the scalar basis. The derivative columns give $$F(t)=\begin{pmatrix}1&t&e^t\\0&1&e^t\\0&0&e^t\end{pmatrix},\qquad \det F=e^t\ne 0.$$ Differentiating the rows gives $F'=AF$, including the last row because $(e^t)'=e^t$. Thus $F$ is fundamental on the entire real line.

Step 2: Normalize and compose. With $r=t-s$, direct multiplication gives $$\boxed{\Phi(t,s)=\begin{pmatrix}1&r&e^r-1-r\\0&1&e^r-1\\0&0&e^r\end{pmatrix}.}$$ It equals $I$ at $t=s$, and $\det\Phi=e^{t-s}$. The identity $F(t)F(s)^{-1}F(s)F(a)^{-1}=F(t)F(a)^{-1}$ proves $\Phi(t,s)\Phi(s,a)=\Phi(t,a)$ for all real base times.

Step 3: Transport the forcing in both directions. The zero initial vector at time one gives $$X(t)=\int_1^t\Phi(t,v)e_3\,dv,\qquad \boxed{y(t)=e^{t-1}-1-(t-1)-\tfrac 12(t-1)^2.}$$ Indeed its first component integrates $e^{t-v}-1-(t-v)$. For $t<1$ the integral is oriented: it is minus the integral from $t$ to $1$. Writing $r=t-1$, we obtain $y'=e^r-1-r$, $y''=e^r-1$, $y'''=e^r$. These give the zero state at $t=1$ and residual $y'''-y''=1$ on all of $\mathbb R$.

Step 4: Separate basis from transition. Here $F(0)\ne I$. For example, $F(t)e_3=(e^t,e^t,e^t)^T$ starts at $(1,1,1)^T$, not at $e_3$. The correct expression is $F(t)F(0)^{-1}e_3$. For any constant invertible $C$, $$(F(t)C)(F(s)C)^{-1}=F(t)CC^{-1}F(s)^{-1}=\Phi(t,s).$$ A basis change alters coordinates of solution columns, but not the transport of a specified physical state.