hw10_solns - Homework #10 - Solutions Math 2930 Fall 2008...

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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 + y = 0 ; y (0) = 0 , y (1) = y prime (1) (a) Show that = 0 is an eigenvalue with associated eigenfunction y ( x ) = x . Solution: If = 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 = 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 n x , where n is the n th positive root of the equation tan z = z . Solution: If = 2 > 0 and y ( x ) = A cos x + B sin x then y (0) = A = 0 so y ( x ) = B sin x . Then y (1) = y prime (1) implies that B sin = B cos . If B negationslash = 0 then must be a positive root of the equation tan = . (c) Draw a sketch showing these roots. Solution: We see from figure 1 below that the point of intersection, n , of the lines y = tan z and y = z lies just to the left of the vertical line z = (2 n + 1) / 2. Figure 1: Plot of y = tan z and y = z (d) Deduce from this sketch that n (2 n + 1) / 2 when n is large. Solution: From figure above, n lies closer and closer to the line z = (2 n + 1) / 2 as n gets larger and larger. 3.8 #12 Consider the eigenvalue problem y primeprime + y = 0 ; y (- ) = y ( ) , y prime (- ) = y prime ( ) (a) Show that = 0 is an eigenvalue with associated eigenfunction y ( x ) = 1. 1 Solution: If = 0 and y ( x ) = A + Bx then y (- ) = y ( ) means that A- B = A + B , so B = 0 and y ( x ) = A . But then y prime (- ) = y prime ( ) implies nothing about A . Hence = 0 is an eigenvalue with associated eigenfuction y ( x ) = 1. (b) Show that there are no negative eigenvalues. Solution: If =- 2 < 0 and y ( x ) = Ae x + Be- x then the conditions y (- ) = y ( ) and y prime (- ) = y prime ( ) yield the equations Ae + Be- = Ae- + Be (1) Ae - Be- = Ae- - Be (2) Addition of these equations yields 2 Ae = 2 Ae- . Since e negationslash = e- because 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 = 2 > 0 and y ( x ) = A cos x + B sin x then the endpoint conditions yield the equations A cos + B sin = A cos - B sin (3)- A sin + B cos = A sin + B cos (4) The first equation implies that B sin = 0, the second that A sin = 0. If A and B are not both zero, then it follows that sin = 0, so = 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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hw10_solns - Homework #10 - Solutions Math 2930 Fall 2008...

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