Consider the simple groundwater flow problem shown in Figure 2.25(a). The equation of flow for steady-state saturated flow in the plane is

(A3.1)

The mathematical statement of the boundary conditions is

on and (A3.2)

on (A3.3)

on (A3.4)

We will solve for using the separation-of-variables technique.

Assume that the solution is a product solution of the form

(A3.5)

Equation (A3.1) then becomes

(A3.6)

Dividing through by yields

(A3.7)

The left-hand side is independent of . Therefore the right-hand side, despite its appearance, must also be independent of , since it is identically equal to the left-hand side. Similarly, the right-hand side is independent of , and so too is the left-hand side. If both sides are independent of and , each side must equal a constant. Therefore,

and (A3.8)

The constant may be positive, negative, or zero. All three cases lead to product solutions, but only the case leads to a solution that is physically meaningful for this problem. Equations (A3.8) become

and (A3.9)

These are ordinary differential equations whose solutions are well known:

and (A3.10)

The product solution Eq. (A3.5) becomes

(A3.11)

We can evaluate the coefficients , , , and by invoking the boundary conditions. Differentiating Eq. (A3.11) with respect to yields

(A3.12)

and invocation of Eq. (A3.2) implies that . From Eq. (A3.11), we are left with

(A3.13)

Invoking the boundary conditions of Eqs. (A33) and (A3.4) yields and . The solution is, therefore,

(A3.14)

This equation is identical to Eq. (2.81), presented in Section 2.11 without proof.

It is immediately clear that Eq. (A3.14) satisfies the boundary conditions of Eqs. (A3.3) and (A3.4). Differentiation with respect to yields zero in satisfaction of Eq. (A3.2). Differentiation with respect to twice also yields zero, so the solution Eq. (A3.14) satisfies the equation of flow [Eq. (A3.1)].