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Unformatted text preview: ACM 95b/100b Problem Set 5 Solutions Prepared by: Lei Zhang February 25, 2007 Total: 146 points Include grading section: 2 points Problem 1 (3 × 8 points) Solve each of the following boundary value problems a) y 00 + y = 0 , y (0) = 0 , y ( π ) = 1 b) y 00 + y = 0 , y (0) = 1 , y ( L ) = 0 c) y 00 + 4 y = sin x, y (0) = 0 , y ( π ) = 0 Solution 1 a) The equation has general solution: y ( x ) = c 1 sin( x ) + c 2 cos( x ) . Applying the boundary condition at x = 0, y (0) = c 2 = 0 ⇒ c 2 = 0 . The boundary condition at x = π requires that, y ( π ) = c 1 = 1 ⇒ c 1 = 1 . Thus, the solution is, y ( x ) = sin( x ) . b) As in part (a), the general solution is, y ( x ) = c 1 sin( x ) + c 2 cos( x ) . Applying the boundary condition at x = 0, y (0) = c 1 = 1 ⇒ c 1 = 1 . The boundary condition at x = L requires that, y ( L ) = sin( L ) + c 2 cos( L ) = 0 . If L = ( n + 1 2 ) π , n ∈ Z , then the condition becomes sin( L ) = 0, and the boundary value problem has no solution. Otherwise, c 2 = tan( L ), so y ( x ) = sin( x ) tan( L ) cos( x ) . c) The solution to the homogeneous problem is y ( x ) = c 1 sin(2 x ) + c 2 cos(2 x ) . 1 The Wronskian is W = 2. Thus, y p ( x ) = 1 2 sin 2 x Z sin( t ) cos(2 t ) dt 1 2 cos(2 x ) Z sin( t ) sin(2 t ) dt = 1 2 sin(2 x ) cos( x ) 2 cos(3 x ) 6 1 2 cos(2 x ) sin( x ) 2 sin(3 x ) 6 = 1 2 (sin(2 x ) cos( x ) cos(2 x ) sin( x )) + 1 12 (cos(2 x ) sin(3 x ) sin(2 x ) cos(3 x )) = 1 4 sin( x ) + 1 12 sin( x ) = sin( x ) 3 . The lefthand boundary condition implies that c 2 = 0. Further, the righthand boundary condition implies that, y ( π ) = c 1 sin(2 π ) + sin( π ) 3 = 0 , which is satisfied for any choice of c 1 ; the solution is not unique. y ( x ) = c 1 sin(2 x ) + sin( x ) 3 . Problem 2 (15 points) Consider the eigenvalue problem y 00 + λy = 0 , y (0) = 0 , y ( L ) = 0 . (1) Show that there are no complex eigenvalues. Solution 2 The equation has solution y ( x ) = c 1 e i √ λx + c 2 e i √ λx . Applying the boundary conditions leads to the equations, c 1 + c 2 = 0 c 1 e i √ λL + c 2 e i √ λL = 0 . This has a nontrivial solution only if the matrix of coefficients is singular, or e i √ λL e i √ λL = 0 . Now, assume that √ λ = ν + iσ . Then the condition is e σL e iνL e σL e iνL = 0 . This can be rewritten e σL (cos( νL ) + i sin( νL )) e σL (cos( νL ) i sin( νL )) = 0 , or, equating real and imaginary parts, ( e σL e σL ) cos( νL ) = 0 ( e σL + e σL ) sin( νL ) = 0 ....
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 Winter '09
 NilesA.Pierce
 Fourier Series, Periodic function, Boundary value problem, Partial differential equation

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