This preview shows pages 1–2. 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: AM 3815 - Solutions to Assignment 4 1. We separate the variables in the usual way to get- T T =- X 00 X = . We first consider the BVP X 00 =- X , X (- ) = X ( ), X (- ) = X ( ). We know the eigenvalues of this problem are real, and we consider three possible cases. Case 1: If > 0, then X = A cos x + B sin x . The X (- ) = X ( ) BCs imply A cos(- ) + B sin(- ) = A cos + B sin A cos - B sin = A cos + B sin 2 B sin = which holds for either B = 0 or = n 2 , where n 1 is an integer (recall that we have assumed > 0). The X (- ) = X ( ) BCs imply- A sin(- ) + B cos(- ) =- A sin + B cos A sin + B cos =- A sin + B cos 2 A sin = which holds for either A = 0 or = n 2 , where n 1 is an integer. Note that if 6 = n 2 , then A = B = 0 and we would have only a trivial solution to the eigenvalue problem. For given n 1, then, the eigenfunction corresponding to n = n 2 is X n ( x ) = A n cos nx + B n sin nx ....
View Full Document
- Spring '11