Explanation It is given that ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 , 0 < x < π , 0 < y < π ... ( 1 ) With boundary conditions, u ( 0 , y ) = u ( π , y ) = 0 , 0 ≤ y ≤ π ... ( 2 ) and initial conditions, u ( x , 0 ) = f ( x ) , 0 ≤ x ≤ π , ... ( 3 ) u ( x , π ) = 0 , 0 ≤ x ≤ π ... ( 4 ) Consider the solution of the equation be u ( x , y ) = X ( x ) Y ( y ) . Above equation satisfies equation (1), so it becomes, X ″ ( x ) Y ( y ) + X ( x ) Y ″ ( y ) = 0 X ″ ( x ) X ( x ) = − Y ″ ( y ) Y ( y ) = − λ X ″ ( x ) + λ X ( x ) = 0 ... ( 5 ) and Y ″ ( y ) − λ Y ( y ) = 0 ... ( 6 ) Boundary condition in equation (2) becomes, X ( 0 ) = X ( π ) = 0 So, equation (5) becomes, X ″ ( x ) + λ X ( x ) = 0 with boundary conditions X ( 0 ) = X ( π ) = 0 . The auxiliary equation becomes r 2 + λ = 0 . If λ = 0 , then r 2 = 0 r = 0 Thus r = 0 is the repeated root. So, the general solution becomes X ( x ) = C 1 x + C 2 . By boundary condition, X ( 0 ) = C 2 C 2 = 0 and X ( π ) = C 1 π + C 2 C 1 π = 0 C 1 = 0 As, C 1 = 0 and C 2 = 0 , then X ( x ) = 0 . So, there is a trivial solution. Now, if λ > 0 , then the roots becomes, r 2 + λ = 0 r 2 = − λ r = ± i λ The general solution becomes, X ( x ) = c 1 cos λ x + c 2 sin λ x By boundary condition, X ( 0 ) = c 1 cos 0 + c 2 sin 0 0 = c 1 and X ( π ) = c 1 sin λ π + c 2 cos λ π 0 = c 2 sin λ π Here c 2 ≠ 0 , then sin λ π = 0 sin λ π = sin n π λ = n λ = n 2 which is given by

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 10.7, Problem 3E

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The formal solution to the given initial value problem u(0,y)=u(π,y)=0,0≤y≤π, u(x,0)=f(x),0≤x≤π, u(x,π)=0,0≤x≤π.

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Chapter 10.7, Problem 2E

Chapter 10.7, Problem 5E

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Chapter 10 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 6E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...

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The formal solution to the given initial value problem u ( 0 , y ) = u ( π , y ) = 0 , 0 ≤ y ≤ π , u ( x , 0 ) = f ( x ) , 0 ≤ x ≤ π , u ( x , π ) = 0 , 0 ≤ x ≤ π .

Solution Summary: The author explains that the formal solution to the given initial value problem is u(x,y)=displaystyle undersetn=1

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The formal solution to the given initial value problem u(0,y)=u(π,y)=0,0≤y≤π, u(x,0)=f(x),0≤x≤π, u(x,π)=0,0≤x≤π.

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Chapter 10 Solutions