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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

A formal solution of the given initial-boundary value problem. It has been determined that the required formal solution of the given initial-boundary value problem is, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t . Given: The initial-boundary value problem, ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , 0 < x < L , t > 0 , u 0 , t = u L , t = 0 , t > 0 , u x , 0 = f x , 0 < x < L , ∂ u ∂ t x , 0 = g x , 0 < x < L . The initial conditions, f x = h 0 x a , 0 < x ≤ a h 0 L − x L − a , a < x < L , g x = 0 . Concept used: The general formal solution of the given wave equation is, u x , t = ∑ n = 1 ∞ A n cos n π α L t + B n sin n π α L t sin n π L x , where: A n = 2 L ∫ 0 L f x sin n π L x d x & B n = 2 n π α ∫ 0 L g x sin n π L x d x . Calculation: It is given that g x = 0 . Put g x = 0 in B n = 2 n π α ∫ 0 L g x sin n π L x d x to get, B n = 0 . It is also given that f x = h 0 x a , 0 < x ≤ a h 0 L − x L − a , a < x < L . Put this value in A n = 2 L ∫ 0 L f x sin n π L x d x to get, A n = 2 L ∫ 0 a h 0 x a sin n π L x d x + 2 L ∫ a L h 0 L − x L − a sin n π L x d x . Simplifying, A n = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a ∫ a L L − x sin n π L x d x = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a ∫ a L L sin n π L x d x − ∫ a L x sin n π L x d x = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a L ∫ a L sin n π L x d x − ∫ a L x sin n π L x d x So, A n = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a L ∫ a L sin n π L x d x − ∫ a L x sin n π L x d x Now, consider the indefinite integral ∫ x sin n π L x d x . Solving using integration by parts, ∫ x sin n π L x d x = x ∫ sin n π L x d x − ∫ ∫ sin n π L x d x d x = − x cos n π L x n π L − ∫ − cos n π L x n π L d x = − x cos n π L x n π L + 1 n π L ∫ cos n π L x d x = − x cos n π L x n π L + 1 n π L sin n π L x n π L On further simplifying, ∫ x sin n π L x d x = L 2 n 2 π 2 sin n π L x − L n π x cos n π L x Put this value in the obtained expression for A n to get, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L x − L n π x cos n π L x 0 a + 2 h 0 L L − a − L cos n π L x n π L a L − L 2 n 2 π 2 sin n π L x − L n π x cos n π L x a L Solving, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L cos n π L L n π L − cos n π L a n π L − L 2 n 2 π 2 sin n π L L − L n π L cos n π L L − L 2 n 2 π 2 sin n π L a + L n π a cos n π L a Simplifying, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L L n π cos n π − L n π cos n π L a − L 2 n 2 π 2 sin n π − L 2 n π cos n π − L 2 n 2 π 2 sin n π L a + L n π a cos n π L a On further solving, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L 2 n π cos n π + L 2 n π cos n π L a − L 2 n 2 π 2 sin n π + L 2 n π cos n π + L 2 n 2 π 2 sin n π L a − L n π a cos n π L a Simplifying, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a L 2 n π cos n π L a + L 2 n 2 π 2 sin n π L a − L n π a cos n π L a = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a L n π L − a cos n π L a + L 2 n 2 π 2 sin n π L a Finally, A n = 2 h 0 L a n 2 π 2 sin n π L a − 2 h 0 n π cos n π L a + 2 h 0 n π cos n π L a + 2 h 0 L L − a n 2 π 2 sin n π L a = 2 h 0 L n 2 π 2 1 a + 1 L − a sin n π L a = 2 h 0 L n 2 π 2 L a L − a sin n π L a = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a So, A n = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a . Put A n = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a and B n = 0 in the general formal solution to get, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t . This is the required formal solution of the given initial-boundary value problem. Conclusion: It has been determined that the required formal solution of the given initial-boundary value problem is, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t .

A formal solution of the given initial-boundary value problem. It has been determined that the required formal solution of the given initial-boundary value problem is, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t . Given: The initial-boundary value problem, ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , 0 < x < L , t > 0 , u 0 , t = u L , t = 0 , t > 0 , u x , 0 = f x , 0 < x < L , ∂ u ∂ t x , 0 = g x , 0 < x < L . The initial conditions, f x = h 0 x a , 0 < x ≤ a h 0 L − x L − a , a < x < L , g x = 0 . Concept used: The general formal solution of the given wave equation is, u x , t = ∑ n = 1 ∞ A n cos n π α L t + B n sin n π α L t sin n π L x , where: A n = 2 L ∫ 0 L f x sin n π L x d x & B n = 2 n π α ∫ 0 L g x sin n π L x d x . Calculation: It is given that g x = 0 . Put g x = 0 in B n = 2 n π α ∫ 0 L g x sin n π L x d x to get, B n = 0 . It is also given that f x = h 0 x a , 0 < x ≤ a h 0 L − x L − a , a < x < L . Put this value in A n = 2 L ∫ 0 L f x sin n π L x d x to get, A n = 2 L ∫ 0 a h 0 x a sin n π L x d x + 2 L ∫ a L h 0 L − x L − a sin n π L x d x . Simplifying, A n = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a ∫ a L L − x sin n π L x d x = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a ∫ a L L sin n π L x d x − ∫ a L x sin n π L x d x = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a L ∫ a L sin n π L x d x − ∫ a L x sin n π L x d x So, A n = 2 h 0 a L ∫ 0 a x sin n π L x d x + 2 h 0 L L − a L ∫ a L sin n π L x d x − ∫ a L x sin n π L x d x Now, consider the indefinite integral ∫ x sin n π L x d x . Solving using integration by parts, ∫ x sin n π L x d x = x ∫ sin n π L x d x − ∫ ∫ sin n π L x d x d x = − x cos n π L x n π L − ∫ − cos n π L x n π L d x = − x cos n π L x n π L + 1 n π L ∫ cos n π L x d x = − x cos n π L x n π L + 1 n π L sin n π L x n π L On further simplifying, ∫ x sin n π L x d x = L 2 n 2 π 2 sin n π L x − L n π x cos n π L x Put this value in the obtained expression for A n to get, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L x − L n π x cos n π L x 0 a + 2 h 0 L L − a − L cos n π L x n π L a L − L 2 n 2 π 2 sin n π L x − L n π x cos n π L x a L Solving, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L cos n π L L n π L − cos n π L a n π L − L 2 n 2 π 2 sin n π L L − L n π L cos n π L L − L 2 n 2 π 2 sin n π L a + L n π a cos n π L a Simplifying, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L L n π cos n π − L n π cos n π L a − L 2 n 2 π 2 sin n π − L 2 n π cos n π − L 2 n 2 π 2 sin n π L a + L n π a cos n π L a On further solving, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a − L 2 n π cos n π + L 2 n π cos n π L a − L 2 n 2 π 2 sin n π + L 2 n π cos n π + L 2 n 2 π 2 sin n π L a − L n π a cos n π L a Simplifying, A n = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a L 2 n π cos n π L a + L 2 n 2 π 2 sin n π L a − L n π a cos n π L a = 2 h 0 a L L 2 n 2 π 2 sin n π L a − L n π a cos n π L a + 2 h 0 L L − a L n π L − a cos n π L a + L 2 n 2 π 2 sin n π L a Finally, A n = 2 h 0 L a n 2 π 2 sin n π L a − 2 h 0 n π cos n π L a + 2 h 0 n π cos n π L a + 2 h 0 L L − a n 2 π 2 sin n π L a = 2 h 0 L n 2 π 2 1 a + 1 L − a sin n π L a = 2 h 0 L n 2 π 2 L a L − a sin n π L a = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a So, A n = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a . Put A n = 2 h 0 L 2 n 2 π 2 a L − a sin n π L a and B n = 0 in the general formal solution to get, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t . This is the required formal solution of the given initial-boundary value problem. Conclusion: It has been determined that the required formal solution of the given initial-boundary value problem is, u x , t = 2 h 0 L 2 π 2 a L − a ∑ n = 1 ∞ 1 n 2 sin n π L a sin n π L x cos n π α L t .

Solution Summary: The author determines that the required formal solution of the given initial-boundary value problem is u(x,t)=2h_0L2,pi

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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