Theis solution

Remarks

The idea of the proof: the problem has no length or time scale except through $r^2 S / (T t)$, so one looks for a similarity solution $s = F(u)$. The equation becomes $u F'' + (1 + u) F' = 0$, so $F'(u) = -c \, e^{-u} / u$, and $F(\infty) = 0$ gives $F = c \, E_1(u)$. Since $r \, \partial_r s = 2 u F'(u) = -2 c e^{-u}$, the condition at the well fixes $c = Q / (4 \pi T)$, and $s(r, 0) = 0$ holds because $u \to \infty$ as $t \to 0$.

Example: for small $u$, $W(u) = -\gamma - \ln u + O(u)$, with $\gamma \approx 0.5772$ the Euler–Mascheroni constant. So at late times $s \approx \frac{Q}{4 \pi T} \ln \frac{2.25 \, T t}{r^2 S}$, since $e^{-\gamma} \cdot 4 \approx 2.25$; this is the Cooper–Jacob approximation. A plot of drawdown against $\ln t$ is then a straight line of slope $Q / (4 \pi T)$, which gives $T$ from a pumping test.

The infinite extent is needed: in a bounded aquifer with a fixed-head boundary, the drawdown tends to a steady state instead of growing like $\ln t$.

Going deeper

For a student

The drawdown depends on $r$ and $t$ only through $r^2 / t$, so curves at different distances collapse onto one type curve $W(u)$. Matching measured data to this curve gives $T$ and $S$. The same function solves the problem of a line source of heat in an infinite plane, which is the classical source of the analogy. Extensions handle leaky aquifers (Hantush–Jacob), unconfined aquifers, and wells of finite radius.