6.2 Introduction-to-Partial-Differential-Equations-Third-Edition-by-K-Sankara-Rao

ELLIPTIC DIFFERENTIAL EQUATIONS 157 Thus the required temperature distribution is given by ¦u (x, y) f cosech §  nπ b · 4Ta2 [(1)n 1] sin § nπ x · sinh ª nπ (y  b)»¼º n1 ©¨ a ¹¸ n3π 3 ¨© a ¹¸ «¬ a EXAMPLE 2.12 Solve 0 d x d a, 0 d y d b ’2u 0, u (x, 0) 0, u (x, b) 0 satisfying the BCs: u (0, y) 0, w u (a, y) T sin3 π y wx a Solution Using the variables separable method, one of the acceptable general solutions is given by Eq. (2.38). Hence u (x, y) (c1e px  c2e px ) (c3 cos py  c4 sin py) Using the BC: u (x, 0) 0, we get implying c3 = 0. Therefore, 0 c3 (c1e px  c2e px ) u (x, y) c4 sin py (c1e px  c2e px ) Now, using the BC: u (x, b) 0, we obtain 0 c4 sin pb (c1e px  c2e px ) c4 z 0 (why?) implying sin pb 0 which gives n 1, 2, 3, } pb nπ or p nπ , b Thus, u (x, y) c4 sin § nπ y · (c1e px  c2e px ) ¨© b ¹¸ Renaming the constants, we have u (x, y) sin § nπ y · ª A exp § nπ x ·  B exp §  nπ x ·º , n 1, 2,} ©¨ b ¹¸ ¬« ©¨ b ¸¹ ¨© b ¸¹»¼ If we use the BC: u (0, y) 0, we get 0 sin § nπ y · ( A  B) ¨© b ¹¸ www.MathSchoolinternational.com