At steady state, $q_x=R$ and $q(L)=0$, so
$$
q=R(x-L),
\qquad
Khh_x=R(L-x).
$$
Using $h(0)=0$ gives
$$
\boxed{h(x)=\left[\frac RK(2Lx-x^2)\right]^{1/2},
\qquad q(0)=-RL.}
$$
Near the river,
$$
\boxed{h(x)\sim\left(\frac{2RLx}{K}\right)^{1/2}\quad(x\ll L).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 332 / 1 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-332/1/a/ii/solution