ABSTRACT
Substituting C I and Cz into Eq. (8.27) yields the solution. No iteration is required for linear ODEs.
Example 8.2. The second-order shooting method using superposition
The heat transfer problem considered in Example 8.1 is governed by a linear boundary-value ODE. Consequently, the solution can be obtained by generating two solutions for two assumed values of U(O.O) = T'(O.O) and superimposing those two solutions. The results of the first two solutions by the implicit trapezoid method with Llx = 0.25 cm are presented in Table 8.2 and repeated in Table 8.6. From Table 8.6,
T~l) = 75.925926 and TJ2) = 126.543210. The specified value of 1"z is 100.0 C. Substituting these values into Eq. (8.30) gives
C = 100.000000 - 126.543210 = 0.524389 I 75.925926 - 126.543210
= 75.925926 - 100.000000 = 0.475611 Cz 75.925926 - 126.543210
Substituting these results into Eq. (8.27) yields
T(x) = 0.524389T(x)(I) + 0.475611 T(4Z)
(8.31)
(8.32)
(8.33)
Ax = 0.25cm
Solving Eq. (8.33) for y(x) gives the values presented in the final column of Table 8.6, which are identical to the values presented in Table 8.4.
