Systems of Differential Equations — Question 3

Question 3 – Solution

Strategy. Multiply each flow rate by the concentration in its source tank, then check the total balance.

Step 1: Balance volumes and salt separately. Tank $1$ receives $3+2=5$ L/min and sends $5$; tank $2$ receives $5$ and sends $2+3=5$. Their volumes are constant. The return concentration is $y/200$ and the forward concentration is $x/100$, giving $$\boxed{x'=-x/20+y/100,\qquad y'=x/20-y/40.}$$ Fresh water supplies no salt. Each coefficient has units min$^{-1}$.

Step 2: Change from amounts to concentrations. For $X=(x,y)^T$, $$X'=\begin{pmatrix}-1/20&1/100\\1/20&-1/40\end{pmatrix}X, \qquad X(0)=(100,0)^T.$$ Dividing each balance by its own tank volume gives $$\boxed{c'=\begin{pmatrix}-1/20&1/50\\1/40&-1/40\end{pmatrix}c, \qquad c(0)=(1,0)^T.}$$ This is a constant invertible change of state; the two matrices need not match.

Step 3: Check the external balance and equilibria. Adding the amount equations gives $$\boxed{(x+y)'=-3y/200},$$ exactly the outside drain rate times its source concentration. At an equilibrium this forces $y=0$, and then $y'=x/20=0$ forces $x=0$. The only equilibrium is the salt-free state.

Step 4: Prove positivity and bounds. The field is linear and therefore locally Lipschitz. On $x=0$, $y\ge 0$, we have $x'=y/100\ge 0$; on $y=0$, $x\ge 0$, we have $y'=x/20\ge 0$. The given inward-boundary criterion therefore proves that the nonnegative quadrant is forward invariant. The constant linear system exists for all forward times, so the conclusion holds for every $t\ge 0$. Now the total balance implies $0\le x+y\le 100$. Hence $0\le x,y\le 100$ g, $0\le c_1\le 1$ and $0\le c_2\le 1/2$ g/L. These are guaranteed bounds, not claims that every bound is attained.