This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
 Spring '00
 Courdurier

Click to edit the document details