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Unformatted text preview: Homework #10 - Solutions Math 2930 Fall 2008 Due November 4, 2008 3.8 #8 Consider the eigenvalue problem y primeprime + λy = 0 ; y (0) = 0 , y (1) = y prime (1) (a) Show that λ = 0 is an eigenvalue with associated eigenfunction y ( x ) = x . Solution: If λ = 0 then y primeprime = 0 and y ( x ) = A + Bx . Using y (0) = 0 gives A = 0 and so y ( x ) = Bx . Since y (1) = y prime (1) implies that B = B then λ = 0 is an eigenvalue with associated eigenfuction y ( x ) = x (here we let B=1 as explained on the last paragraph of page 234 -before example 3). (b) Show that the remaining eigenfuctions are given by y n ( x ) = sin β n x , where β n is the n th positive root of the equation tan z = z . Solution: If λ = β 2 > 0 and y ( x ) = A cos βx + B sin βx then y (0) = A = 0 so y ( x ) = B sin βx . Then y (1) = y prime (1) implies that B sin β = Bβ cos β . If B negationslash = 0 then β must be a positive root of the equation tan β = β . (c) Draw a sketch showing these roots. Solution: We see from figure 1 below that the point of intersection, β n , of the lines y = tan z and y = z lies just to the left of the vertical line z = (2 n + 1) π/ 2. Figure 1: Plot of y = tan z and y = z (d) Deduce from this sketch that β n ≈ (2 n + 1) π/ 2 when n is large. Solution: From figure above, β n lies closer and closer to the line z = (2 n + 1) π/ 2 as n gets larger and larger. 3.8 #12 Consider the eigenvalue problem y primeprime + λy = 0 ; y (- π ) = y ( π ) , y prime (- π ) = y prime ( π ) (a) Show that λ = 0 is an eigenvalue with associated eigenfunction y ( x ) = 1. 1 Solution: If λ = 0 and y ( x ) = A + Bx then y (- π ) = y ( π ) means that A- Bπ = A + Bπ , so B = 0 and y ( x ) = A . But then y prime (- π ) = y prime ( π ) implies nothing about A . Hence λ = 0 is an eigenvalue with associated eigenfuction y ( x ) = 1. (b) Show that there are no negative eigenvalues. Solution: If λ =- α 2 < 0 and y ( x ) = Ae αx + Be- αx then the conditions y (- π ) = y ( π ) and y prime (- π ) = y prime ( π ) yield the equations Ae απ + Be- απ = Ae- απ + Be απ (1) Ae απ- Be- απ = Ae- απ- Be απ (2) Addition of these equations yields 2 Ae απ = 2 Ae- απ . Since e απ negationslash = e- απ because α negationslash = 0 then it follows that A = 0. Similarly B = 0 (by subtracting). Thus there are no negative eigenvalues. (c) Show that the n th positive eigenvalue is n 2 and that it has two linearly independent associated eigenfunctions, cos nx and sin nx . Solution: If λ = α 2 > 0 and y ( x ) = A cos αx + B sin αx then the endpoint conditions yield the equations A cos απ + B sin απ = A cos απ- B sin απ (3)- A sin απ + B cos απ = A sin απ + B cos απ (4) The first equation implies that B sin απ = 0, the second that A sin απ = 0. If A and B are not both zero, then it follows that sin απ = 0, so α = n = an integer. In this case A and B are both arbitrary. Thus cos nx and sin nx are two different eigenfunctions associated with the single eigenvalue...
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This note was uploaded on 11/09/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell.
- Fall '07