Page 310 - Physical Principles of Sedimentary Basin Analysis

292 Flexure of the lithosphere
∞

w(x)dx = 2w max α. (9.39)
−∞
The net area of the deflection is equal to two times the area of a rectangle with sides of
maximum deflection (w max ) and the length α.

Note 9.2 Equation (9.31) is solved for only a half-plate (x > 0) because the problem is
symmetric around x = 0. The half-plate formulation makes the load disappear from the
equation and it appears instead at the boundary x = 0. Equation (9.31) is first rewritten as

1 d
4
+ 1 w = 0 (9.40)
4 d ˆx 4
x
using the dimensionless x-coordinate ˆ = x/α. The parameter α has unit length and it is
the characteristic length scale of the deflection problem. The solution of equation (9.40)is
done by solving the corresponding characteristic equation

1 4
r + 1 = 0 (9.41)
4
which has four roots r k with k = 1, 2, 3, 4. The general solution of the fourth-order
equation (9.40) in terms of the roots r k is

4
x
x
w(ˆ) = c k e r k ˆ (9.42)
k=1
which is seen by inserting the solution into equation (9.40). The characteristic equation is
4
r iπ+i2πk

√ =−1 = e (9.43)
2
which gives
√
r k = 2 e i(1/4+k/2)π (9.44)

for k = 1,..., 4. The four roots of the characteristic equation are r 1 = e iπ/4 = 1 + i,
r 2 = e i3π/4 =−1 + i, r 3 = e i5π/4 =−1 − i and r 4 = e i7π/4 = 1 − i. We recall that
a
e a+ib = e (cos b +i sin b), and since we are solving the flexure problem for x > 0 we can
only use r 2 and r 3 , because r 1 and r 4 have positive real parts. Solution (9.42)is

x
w(ˆ) = e −ˆ x (c 2 + c 3 ) cos ˆx + i(c 2 − c 3 ) sin ˆx (9.45)
when r 2 and r 3 are inserted. We want a solution with only real numbers, which is achieved
by replacing the coefficients c 2 and c 3 by real coefficients d 1 = c 2 +c 3 and d 2 = i(c 2 −c 3 ).
The solution in terms of the real coefficients d 1 and d 2 is then

x
w(ˆ) = e −ˆ x d 1 cos ˆx + d 2 sin ˆx . (9.46)