ps8_soln - The University of Chicago Department of Physics Page 1 of 8 Problem set 8 solutions[50 Michael A Fedderke Standard disclaimer while every

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The University of Chicago, Department of Physics Page 1 of 8 Problem set 8 solutions [50] Michael A. Fedderke Standard disclaimer: while every effort is made to ensure the correctness of these solutions, if you find anything that you suspect is an error, please email me at [email protected] . All problem references to Arfken and Weber, 7th edition. I. 10.1.4 [10] I assume that t (0 , π ) , else the Green’s Function is trivially identically zero on the whole domain. Consider the boundary value problem - y 00 - 1 4 y = δ ( x - t ); y (0) = 0; y ( π ) = 0 . (1) In the domains x t , the differential equation is homogeneous and second-order, so the solution space is spanned by two linearly independent solutions: - y 00 - 1 4 y = 0 y ( x ) = A cos( x/ 2) + B sin( x/ 2) , (2) so if we split the domain x [0 , π ] into sub-domains x [0 , t ) and x ( t, π ] , we have the solution G ( x ; t ) = A ( t ) cos( x/ 2) + B ( t ) sin( x/ 2) 0 x < t C ( t ) cos( x/ 2) + D ( t ) sin( x/ 2) t < x π , (3) where A, B, C, D depend in general on t but not on x . Imposing the BCs y (0) = 0 G ( x = 0 , t > 0) = 0 and y ( π ) = 0 G ( x = π, t < π ) = 0 sets A = D = 0 , while continuity at x = t demands that B ( t ) sin( t/ 2) = C ( t ) cos( t/ 2) C ( t ) = B ( t ) tan( t/ 2) . (4) The value of B ( t ) can be found by integrating both sides of the ode in the small domain ( t - , t + ) for some small positive and then sending 0 : Z t + t - dx - G 00 - 1 4 G = Z t + t - dx δ ( x - t ) (5) ⇒ - G 0 | t + t - - O ( ) = 1 (6) G 0 - - G 0 + = 1 (7) 1 2 B ( t ) cos( t/ 2) + 1 2 C ( t ) sin( t/ 2) = 1 . (8) Substituting (4) into (8), we have B ( t ) = 2 cos( t/ 2) (9) C ( t ) = 2 sin( t/ 2) (10) Therefore, we have the Green’s Function G ( x ; t ) = 2 cos( t/ 2) sin( x/ 2) 0 x t 2 sin( t/ 2) cos( x/ 2) t < x π . (11) Mathematical Methods of Physics v1.0 2 December 2015, 6:00pm Autumn 2015
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The University of Chicago, Department of Physics Page 2 of 8 II. 10.1.7 [10] I will initially assume that k 6 = 0 . Let’s find the Green’s function by solving ¨ ψ + k ˙ ψ = δ ( t - t 0 ); ψ (0) = ˙ ψ (0) = 0 . (12) I will also assume that t 0 > 0 otherwise the Green’s function is again trivially identically zero on t 0 . (I am implicitly assuming that t 0 is the only domain we are interested in since we have initial conditions imposed at t = 0 .) We split the domain up into t [0 , t 0 ) and t ( t 0 , ] . In either sub-domain, the ode is homogeneous: ¨ ψ + k ˙ ψ = 0 , and has a solution space spanned by two linearly independent solutions ( they are only linearly independent for k 6 = 0 ): ψ ( t ) = Ae - kt + B . So the solution is G ( t ; t 0 ) = A ( t 0 ) e - k ( t - t 0 ) + B ( t 0 ) 0 t < t 0 C ( t 0 ) e - k ( t - t 0 ) + D ( t 0 ) 0 < t 0 < t π , (13) where I have factored out e kt 0 from A ( t 0 ) and C ( t 0 ) for reasons to become clear shortly. The IC’s ψ (0) = ˙ ψ (0) = 0 demand that A = B = 0 , which can be thought of as a statement of causality: G ( t ; t 0 ) = 0 for t < t 0 . We demand continuity at t = t 0 so A + B = C + D C = - D , giving G ( t ; t 0 ) = 0 0 t < t 0 D ( t 0 ) 1 - e - k ( t - t 0 ) 0 < t 0 < t π , (14) To find D ( t 0 ) we again perform the trick of integrating the ode around t 0 in an -domain, to find that ˙ ψ | + - + | + - = 1 kD ( t 0 ) = 1 D = 1 /k, (15) so G ( t ; t 0 ) = 0 0 t < t 0 1 k 1 - e - k ( t - t 0 ) 0 < t 0 < t π . ( k 6 = 0) (16) I will not find the Green’s function for k = 0 because that case is much more easily handled by direct integration.
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