Appendix III

Appendix III
Example of an Analytical Solution to a Boundary-Value Problem

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

\frac{\delta^2 h}{\delta x^2} + \frac{\delta^2 h}{\delta y^2} = 0 (A3.1)

The mathematical statement of the boundary conditions is

\frac{\delta h}{\delta y} = 0 on y = 0 and y = y_L (A3.2)

h = h_0 on x = 0 (A3.3)

h = h_1 on x = x_L (A3.4)

We will solve for h(x, y) using the separation-of-variables technique.

Assume that the solution is a product solution of the form

h(x, y) = X(x) \cdot Y(y) (A3.5)

Equation (A3.1) then becomes

Y\frac{\delta^2x}{\delta x^2} + X\frac{\delta^2y}{\delta y^2} = 0 (A3.6)

Dividing through by XY yields

\frac{1}{X}\frac{\delta^2X}{\delta x^2} = \frac{1}{Y}\frac{\delta^2Y}{\delta y^2} (A3.7)

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

\frac{1}{X}\frac{\delta^2X}{\delta x^2} = G and \frac{1}{Y}\frac{\delta^2Y}{\delta y^2} = G (A3.8)

The constant G may be positive, negative, or zero. All three cases lead to product solutions, but only the case G = 0 leads to a solution that is physically meaningful for this problem. Equations (A3.8) become

\frac{1}{X}\frac{\delta^2X}{\delta x^2} = 0 and \frac{1}{Y}\frac{\delta^2Y}{\delta y^2} = 0 (A3.9)

These are ordinary differential equations whose solutions are well known:

X = Ax + B and Y = Cy + D (A3.10)

The product solution Eq. (A3.5) becomes

h(x, y) = (Ax + B)(Cy +D) (A3.11)

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

\frac{\delta h}{\delta y} = (Ax + B)C (A3.12)

and invocation of Eq. (A3.2) implies that C = 0. From Eq. (A3.11), we are left with

h(x, y) = (Ax + B)D = Ex + F (A3.13)

Invoking the boundary conditions of Eqs. (A33) and (A3.4) yields F = h_0 and E = -(h_0 - h_1)/x_L. The solution is, therefore,

h(x, y) = h_0 - (h_0 - h_1)\frac{x}{x_L} (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 y yields zero in satisfaction of Eq. (A3.2). Differentiation with respect to x twice also yields zero, so the solution Eq. (A3.14) satisfies the equation of flow [Eq. (A3.1)].