Page 28 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

18 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and

b
2
2
Y ηη =∓ λ Y (2.29)
a
X(ξ)Y (0) = X(ξ)Y (1) = 0
Y (0) = Y (1) = 0 (2.30)

The solution of the differential equation in the η direction is

Y (η) = A cosh(bλη/a) + B sinh(bλη/a) (2.31)

Applying the first boundary condition (at η = 0) we find that A = 0. When we apply the
boundary condition at η = 1 however, we find that it requires that

0 = B sinh(bλ/a) (2.32)

so that either B = 0or λ = 0. Neither of these is acceptable since either would require that
Y (η) = 0 for all values of η.
We next try the positive sign. In this case

2
X ξξ = λ X (2.33)
b
2
2
Y ηη =− λ Y (2.34)
a
with the same boundary conditions given above. The solution for Y (η)isnow
Y (η) = A cos(bλη/a) + B sin(bλη/a) (2.35)

The boundary condition at η = 0 requires that

0 = A cos(0) + B sin(0) (2.36)

so that again A = 0. The boundary condition at η = 1 requires that

0 = B sin(bλ/a) (2.37)

Since we don’t want B to be zero, we can satisfy this condition if

λ n = anπ/b, n = 0, 1, 2, 3,... (2.38)
Thus

Y (η) = B sin(nπη) (2.39)