soln4 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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Unformatted text preview: MATH 300 Fall 2007 Advanced Boundary Value Problems I Solutions to Problem Set 4 To Be Completed by: Friday November 9, 2007 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 194, #5.6.2] Consider the eigenvalue problem d 2 dx 2 + ( - x 2 ) = 0 d dx (0) = 0 d dx (1) = 0 Show that all eigenvalues satisfy > 0 (be sure to show that 6 = 0). Solution: The eigenvalue problem is a regular Sturm Liouville problem with p ( x ) = 1 , q ( x ) =- x 2 , ( x ) = 1 for 0 x 1 . An eigenvalue and the corresponding eigenfunction are related by the Rayleigh quotient: = R ( ) =- p ( x ) ( x ) ( x ) 1 + Z 1 p ( x ) ( x ) 2- q ( x ) ( x ) 2 dx Z 1 ( x ) 2 ( x ) dx . From the boundary conditions, we have- p ( x ) ( x ) ( x ) 1 = (0) (0)- (1) (1) = 0 , and the Rayleigh quotient becomes = R ( ) = Z 1 ( x ) 2 + x 2 ( x ) 2 dx Z 1 ( x ) 2 dx , and all the eigenvalues of the boundary value problem are nonnegative. Now note that is a solution to the differential equation 00 ( x ) + ( - x 2 ) ( x ) = 0 , x 1 so that and are both continuous on the interval [0 , 1] . Now, if = 0 is an eigenvalue of this boundary value problem, and is the corresponding eigenfunction, then we have 0 = R ( ) = Z 1 ( x ) 2 + x 2 ( x ) 2 dx Z 1 ( x ) 2 dx , so that Z 1 ( x ) 2 + x 2 ( x ) 2 dx = 0 , and this implies that ( x ) 2 = 0 and x 2 ( x ) 2 = 0 for 0 x 1 . Therefore, ( x ) = 0 for all x [0 , 1] , so that is constant on [0 , 1] . However, since x 2 ( x ) 2 = 0 for all x [0 , 1] , then ( x ) = 0 for all x [0 , 1] , which contradicts the fact that is an eigenfunction. Thus = 0 is not an eigenvalue, and all eigenvalues satisfy > . Question 2. [p 210, #5.8.6] Consider the boundary value initial value problem (with h > 0) given by 2 u t 2 = c 2 2 u x 2 u x (0 , t )- hu (0 , t ) = 0 u ( x, 0) = f ( x ) u x ( L, t ) = 0 u t ( x, 0) = g ( x ) . (a) Show that there are an infinite number of different frequencies of oscillation. (b) Estimate the large frequencies of oscillation (c) Solve the initial value problem. Solution: Since the partial differential equation is linear and homogeneous and the boundary conditions are linear and homogeneous, we can use separation of variables. Assuming a solution of the form u ( x, t ) = ( x ) G ( t ) , x L, t and separating variables, we have two ordinary differential equations: 00 ( x ) + ( x ) = 0 , x L, G 00 ( t ) + c 2 G ( t ) = 0 , t > , (0)- h (0) = 0 ( L ) = 0 (a) We use the Rayleigh quotient to show that > 0 for all eigenvalues ....
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soln4 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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