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# hw4S - Partial Dierential Equations Math 442 C13/C14 Fall...

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Partial Differential Equations – Math 442 C13/C14 Fall 2009 Homework 4 — due October 9 1. Consider the boundary value problem A ′′ + λA = 0 , A (0) + aA (0) = 0 , A ( L ) = 0 . (a) Show that if a < 0, then there is no negative eigenvalue. (b) Under which conditions is there a zero eigenvalue? (c) Show there are infinitely many positive eigenvalues for any value of a . Bonus: We showed in (a) that if a < 0 then there is no negative eigenvalue. If turns out that for some positive a , this problem has a negative eigenvalue (and for some others it does not). Write down a condition on a which determines whether such an eigenvalue exists. Solution. (a) Let’s say we have a negative eigenvalue λ < 0. Then if we say A ( x ) = e rx , we have r 2 + λ = 0 , r = ± - λ. Since λ < 0, this means the roots are real, let us write them as r = ± b where b > 0. Then we have A ( x ) = C 1 e bx + C 2 e bx . Plugging in the initial conditions gives C 1 b - C 2 b + a ( C 1 + C 2 ) = 0 , C 1 e bL + C 2 e bL = 0 . The second equation can be written as C 2 = - C 1 e 2 bL , and plugging this into the first gives C 1 ( a + b ) + C 1 e 2 bL ( b - a ) = C 1 ( b (1 + e 2 bL ) + a (1 - e 2 bL )) . This must be zero, but C 1 cannot be zero (since then so would C 2 be), so we need to solve b (1 + e 2 bL ) + a (1 - e 2 bL ) = 0 . (1) However, this equation has no solution! Notice that since bL > 0, we know e 2 bL > 1, and thus 1 - e 2 bL < 0. Since a is also less than zero, we know that a (1 - e 2 bL ) > 0 . But since b (1 + e 2 bL ) > 0 as well, it is not possible to solve (1). (b) If we have a zero eigenvalue, this means we have a solution to A ′′ ( x ) = 0 with those boundary conditions. However, this means that A ( x ) = αx + β , and plugging this into the boundary conditions gives α + = 0 , αL + β = 0 .

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