6 - MATH20411 1. Problem sheet 6: Separation of Variables...

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MATH20411 Problem sheet 6: Separation of Variables 1. a) u tt - u xx = 0 with u (0 ,t ) = u (ˆL ,t ) = 0, u ( x, 0) = f ( x ) and u t ( x, 0) = g ( x ): Assuming u = X ( x ) T ( t ) we can use the PDE with u xx = X 00 ( x ) T ( t ) and u tt = X ( x ) T 00 ( t ) to find u tt - u xx = XT 00 - TX 00 = 0. Dividing by XT (assumed not equal to zero) gives T 00 T - X 00 X = λ . Thus we have X 00 - λX = 0 and T 00 - λT = 0 with λ constant because X 00 X is independent of t and T 00 T is independent of x . The homogeneous boundary conditions give u (0 ,t ) = X (0) T ( t ) = 0, u ( L,t ) = X ( L ) T ( t ) = 0 . Since we seek XT 6 = 0 this gives X (0) = X ( L ) = 0 . Solving X 00 - λX = 0 with X (0) = X ( L ) = 0: λ = 0 : we have X 00 = 0 giving X = A + Bx . X (0) = X ( L ) = 0 give A = 0 and BL = 0 so that A = B = 0. Only the trivial solution arises for λ = 0. λ = ω 2 > 0 : we have X 00 - ω 2 X = 0 giving X = Ae ωx + Be - ωx . X (0) = X ( L ) = 0 gives A + B = 0 and Ae ωL + Be - ωL = 0, i.e. (substituting) A ( e ωL - e - ωL ) = 0 so that A = B = 0 since e ωL - e - ωL 6 = 0 . Only trivial solutions arise for λ > 0. λ = - ω 2 < 0 : we have X 00 + ω 2 X = 0 giving X = A cos( ωx ) + B sin( ωx ). X (0) = X ( L ) = 0 give A = 0 and B sin( ωL ) = 0. Thus we can have B 6 = 0 if and only if sin( ωL ) = 0 or ωL = for any n = 1, 2, 3, . ... Hence the only eigenvalues are λ = λ n = - ( L ) 2 for n N with corresponding eigenfunc- tions X = X n = sin( nπx L ). Solving T 00 - c 2 λT = 0 : For any λ = - ( L ) 2 we have T 00 + ( L ) 2 T = 0. Thus T = A cos ( nπt L ) + B sin ( nπt L ) . Since the solutions u = X n ( x ) T n ( t ) all satisfy a homogeneous PDE with homogeneous boundary conditions, the principle of superposition means that any linear combination of such solutions is also a solution. Thus a convergent sum: u = X n =1 A n X n T n = X n =1 ± A n cos ± nπt L + B n sin ± nπt L ¶¶ sin nπx L · is also a solution, for constants A n and B n . At t = 0 we have u ( x, 0) = f ( x ) and u t ( x, 0) = g ( x ) so that: f ( x ) = X n =1 A n X n T n (0) = X n =1 A n sin nπx L · g ( x ) = X n =1 ± - A n L · sin ± 0 L + B n L · cos ± 0 L ¶¶ sin nπx L · = X n =1 B n L · sin nπx L · Since the eigenfunctions X n = sin ( nπx L ) are orthogonal under the inner product ( f,g ) = R L 0 f ( x ) g ( x ) dx we have: A n R L 0 sin 2 ( πnx L ) dx = R L 0 f ( x ) sin ( πnx L ) dx . Integrating: R L 0 sin 2 ( πnx L ) dx = R L 0 1 2 (1 - cos ( 2 πnx L ) ) dx = L 2 - [ L 2 πn sin ( 2 πnx L ) ] L 0 = L 2 Thus A n = 2 L R L 0 f ( x )sin ( πnx L ) , n = 1 , 2 ,.... Similarly, for B n , we obtain the expression, B n = 2 L ( L )R L 0 g ( x )sin ( πnx L ) = 2 R L 0 g ( x )sin ( πnx L ) n = 1 , 2 ,....
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This note was uploaded on 09/26/2011 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.

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6 - MATH20411 1. Problem sheet 6: Separation of Variables...

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