Modeling with First Order DE’s — Question 7

Question 7 – Solution

Strategy. Separate input, outflow and internal loss, then use the solution’s monotonicity to turn a whole-interval constraint into a justified endpoint test.

Step 1: Write the balances. If $M$ is pollutant mass in kg and $c=M/1000$, then $$M'=100u-100\frac{M}{1000}-0.10M,\qquad M(0)=80.$$ Each term is in kg/day. Dividing by the constant volume gives $$c'=0.10u-0.20c,\qquad c(0)=0.08.$$ The total loss coefficient combines flushing at $0.10$ day$^{-1}$ with internal disappearance at the same rate.

Step 2: Solve and identify monotonicity. An integrating factor $e^{0.20t}$ gives $$\boxed{c(t)=\frac u2+(0.08-u/2)e^{-0.20t}},\qquad t\ge 0.$$ Its initial value is $0.08$, and differentiating verifies $c'=0.10u-0.20c$. It is decreasing when $u<0.16$, constant when $u=0.16$, and increasing when $u>0.16$.

Step 3: Enforce the five-day bound. For $u\le 0.16$, the maximum is the acceptable initial concentration. For $u>0.16$, the maximum on $[0,5]$ is $c(5)$. Therefore the largest allowed input is $$\boxed{u_5=\frac{2(0.10-0.08e^{-1})}{1-e^{-1}} \approx 0.2233\text{ kg/m}^3}.$$ This value exceeds $0.16$, so its concentration rises monotonically from $0.08$ to exactly $0.10$. Any larger $u$ fails at day five, proving optimality rather than just feasibility.

Step 4: Compare a permanent operating requirement. The limiting pond concentration is $u/2$. A bound valid for all $t\ge 0$ requires and is guaranteed by $$\boxed{u\le 0.20\text{ kg/m}^3}.$$ For these inputs, $c(t)$ lies between its initial value and $u/2$. If $u>0.20$, the limit exceeds the bound, so it must eventually be violated. The larger finite-horizon input uses the initially cleaner pond as temporary dilution capacity; it cannot be continued indefinitely under the same requirement.