b (Rainfall similarity)

Use the dimensionless equation from part (a). For the similarity ansatz

h = t^{a} f (η)

η = x / t^{b}

, its three terms have time factors

h_{t} = t^{a - 1} (a f - b η f^{'}), (h h_{x})_{x} = t^{2 a - 2 b} (f f^{'})^{'}, 1 = t^{0} .

(1)

A nontrivial similarity solution retaining storage, lateral drainage and nonzero rainfall requires the exponents to agree:

a - 1 = 2 a - 2 b = 0, a = b = 1.

(2)

Unlike a source-free diffusion equation, the nonzero prescribed rainfall fixes an amplitude-versus-time scale directly, so no additional conserved quantity or boundary condition is required to determine these exponents. Degenerate spatially uniform solutions are an exception to this all-terms balance. The rain-fed aquifer similarity solution satisfies

(f f^{'})^{'} + η f^{'} - f + 1 = 0, h = t f (x / t) .

(3)

If the dimensional equation is used directly without scaling out coefficients, its similarity profile instead obeys

ϕ (f - η f^{'}) = K (f f^{'})^{'} + R

. The normalization above is the one in which the subsequent condition

f (\infty) = 1

applies.

Solution

 0  0

The condition

f (0) = 0

means the river maintains the water table at the impermeable base level:

h (0, t) = 0

. The liquid is exposed to atmospheric pressure there. It does not imply zero drainage, because

h_{x}

may be singular at the outlet.

Far from the river, lateral drainage has not significantly affected storage. Rainfall simply fills pore volume, so the dimensional height is

h \sim Rt / ϕ

. In the dimensionless variables this is

h \sim t

, giving

f (\infty) = 1

Although the physical aquifer is finite, these boundary conditions describe its early similarity solution before the drainage-affected region reaches

x = L

. In dimensional variables its horizontal extent is of order

K R t / ϕ

, so the semi-infinite approximation requires

t ≪ T = ϕ L / K R

Solution

 0  0

Introduce the positive outlet constant

J = lim_{η ↓ 0} f (η) f^{'} (η) .

(1)

For a draining similarity profile,

J

is finite and positive. A regular linear profile would have

f f^{'} \to 0

and

(f f^{'})^{'} \geq 0

at the origin, incompatible with the rain-fed aquifer similarity solution

(f f^{'})^{'} (0) = - 1

. Instead there is a square-root drainage boundary. At leading order,

(f^{2})^{'} = 2 f f^{'} \to 2 J

, giving

f (η) \sim 2 J η, f^{'} (η) \sim \frac{J}{2 η} .

(2)

More explicitly, the rain-fed aquifer similarity solution implies

(f f^{'})^{'} = - 1 + O (η)

, hence

f f^{'} = J - η + O (η^{3/2})

and

f^{2} = 2 J η - η^{2} + O (η^{5/2})

. The height vanishes but its derivative diverges; their product remains finite.

Use

t_{d}, x_{d}, h_{d}

temporarily for dimensional quantities. The nondimensionalization in part (a) gives

h_{d} = \frac{R}{ϕ} t_{d} f (η), η = \frac{ϕ x _{d}}{K R t _{d}}, \partial_{x_{d}} h_{d} = \frac{R}{K} f^{'} (η) .

(3)

Taking the drainage flux to be positive out of the left-hand end,

Q_{0} = - q (0, t_{d})

, gives

Q_{0} (t_{d}) = K x_{d} ↓ 0 lim h_{d} \partial_{x_{d}} h_{d} = \frac{K R ^{3/2}}{ϕ} t_{d} η ↓ 0 lim f (η) f^{'} (η) = \frac{K R ^{3/2}}{ϕ} J t_{d} .

(4)

The printed expression in terms of

f (0)

and

f^{'} (0)

must be understood as this limit; a literal product of their separate endpoint values is undefined. This is discharge per unit transverse width. Very close to the river, the divergent slope marks the limit of the Dupuit approximation, while the outer solution still supplies the leading discharge.