= Solution
The cross-sectional <area> $A(s,t)$ multiplied by the mean downstream <speed> $u(s,t)$ is the <volume flux> $Q=Au$. The term $A_t$ is local storage change per unit river <length>, $(Au)_s$ is net downstream export per unit <length>, and $M>0$ is lateral <volume> input per unit <length> per unit <time>, for example distributed tributary or runoff supply.
For a fixed reach $[s_1,s_2]$, <incompressibility> and constant <density> give
$$
\frac d{dt}\int_{s_1}^{s_2}A\,ds=Q(s_1,t)-Q(s_2,t)+\int_{s_1}^{s_2}M\,ds.
$$
Apply the <fundamental theorem of calculus> and localize the resulting <integral> identity to obtain $A_t+Q_s=M$. The <density> cancels because every term transports the same <water>; no sediment source is included in this balance.
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