Systems of Differential Equations — Question 9

Question 9 – Solution

Strategy. Each internal transfer leaves one compartment and enters another at the same rate; this exposes conserved quantities before any solution is sought.

Step 1: Translate the transfers. The balances are $$\boxed{x'=-xy,\qquad y'=xy-2y,\qquad z'=2y, \qquad (x,y,z)(0)=(1,1,0).}$$ The system is autonomous, but nonlinear because of $xy$. Writing $x'=-y x$ does not make $-y$ a prescribed coefficient: $y$ is another unknown state coordinate, not a known function of time.

Step 2: Prove conservation and nonnegativity. Adding the rows gives $(x+y+z)'=0$, hence $x+y+z=2$. Along any solution, $$x(t)=x(0)e^{-\int_0^t y(s)\,ds},\qquad y(t)=y(0)e^{\int_0^t(x(s)-2)\,ds},\quad z(t)=z(0)+2\int_0^t y(s)\,ds.$$ These formulas prove nonnegativity for nonnegative data on every forward existence interval. Thus $0\le x,y,z\le 2$. The polynomial field is locally Lipschitz; boundedness prevents finite-time escape, so the solution continues for all $t\ge 0$.

Step 3: Obtain a second conserved quantity. With $x(0)=1$ and $0\le y\le 2$, the exponential formula gives $x(t)\ge e^{-2t}>0$ at finite times. Therefore $\ln x$ is legitimate, and $$(x+y-2\ln x)'=-2y-2(-xy)/x=0.$$ Its initial value is $2$, so $\boxed{x+y-2\ln x=2}$.

Step 4: Reduce and recover the state. The two conservation laws give $$\boxed{y=2-x+2\ln x,\quad z=-2\ln x,\quad x'=-x(2-x+2\ln x),\quad x(0)=1.}$$ The physical branch has $x>0$, $y\ge 0$, $z\ge 0$, in particular $x\le 1$. Conversely, on this branch differentiation gives $y'=(x-2)y$ and $z'=2y$, proving recovery of the original system. Taking logarithms at $x=0$ would be invalid; that boundary is excluded from this reduction.