Systems of Differential Equations — Question 6

Question 6 – Solution

Strategy. Turn the old independent variable into a dependent variable whose rate is fixed and whose initial value records the time origin.

Step 1: Identify the original structure. With $X=(x,y)^T$, $$X'=\begin{pmatrix}t&0\\0&-1\end{pmatrix}X+ \begin{pmatrix}0\\\sin t\end{pmatrix},\qquad X(2)=(3,-1)^T.$$ It is linear, nonhomogeneous and nonautonomous. Time-dependent coefficients do not violate linearity in the dependent variables.

Step 2: Add and initialize the clock. Write $U(\tau)=(s(\tau),u(\tau),v(\tau))^T$. The autonomous system is $$\boxed{s'=1,\qquad u'=su,\qquad v'=-v+\sin s, \qquad U(0)=(2,3,-1)^T.}$$ Here every prime denotes differentiation with respect to $\tau$. There is no explicit $\tau$ on the right-hand side.

Step 3: Prove the change-of-clock correspondence. The first row forces $s(\tau)=2+\tau$. Given an original solution, define $u(\tau)=x(2+\tau)$ and $v(\tau)=y(2+\tau)$. The chain rule verifies the last two rows and the initial data. Conversely, define $x(t)=u(t-2)$ and $y(t)=v(t-2)$ from an augmented solution with $s(0)=2$; substitution recovers the original system. Intervals shift by the same amount. Choosing $s(0)=0$ would instead sample the coefficients at $t=\tau$, changing the stated starting time.

Step 4: Distinguish autonomy, linearity and equilibrium. The augmented system is nonlinear: $su$ multiplies two dependent variables, and $\sin s$ is a nonlinear function of a dependent variable. It cannot be classified as a homogeneous linear system merely because the original equations were linear. An equilibrium would require all three right-hand sides to vanish, but $s'=1$ never vanishes. Thus the augmented autonomous system has no equilibrium. A clock keeps moving even if some physical state components happen to be constant.