This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 11/23/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.
 Winter '09
 NilesA.Pierce

Click to edit the document details