HigherLin4

# 87 green functions the laplace transform can be used

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8.7: Green Functions. The Laplace transform can be used to efficiently compute Green functions. Recall that given the n th order differential operator L with constant coefficients given by L = D n + a 1 D n 1 + · · · + a n 1 D + a n , the Green function g ( t ) associated with L is the solution of the homogeneous initial-value problem L g = 0 , g (0) = 0 , g (0) = 0 , · · · g ( n 2) (0) = 0 , g ( n 1) (0) = 1 . The Laplace transform of this initial-value problem is L bracketleftbig D n g bracketrightbig ( s ) + a 1 L bracketleftbig D n 1 g bracketrightbig ( s ) + · · · + a n 1 L [D y ]( s ) + L [ g ]( s ) = 0 , where if G ( s ) = L [ g ]( s ) then L [D g ]( s ) = s G ( s ) g (0) = s G ( s ) , L bracketleftbig D 2 g bracketrightbig ( s ) = s 2 G ( s ) s g (0) g (0) = s 2 G ( s ) , . . . L bracketleftbig D n 1 g bracketrightbig ( s ) = s n 1 G ( s ) s n 2 g (0) s n 3 g (0) − · · · − g ( n 2) (0) = s n 1 G ( s ) , L bracketleftbig D n g bracketrightbig ( s ) = s n G ( s ) s n 1 g (0) s n 2 g (0) − · · · − g ( n 1) (0) = s n G ( s ) 1 . We thereby see that G ( s ) satisfies p ( s ) G ( s ) 1 = 0 , = G ( s ) = 1 p ( s ) , where p ( s ) is the characteristic polynomial of L, which is given by p ( s ) = s n + a 1 s n 1 + · · · + a n 1 s + a n . In other words, the Laplace transform of the Green function of L is the reciprocal of the characteristic polynomial of L.

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20 The problem of computing a Green function is thereby reduced to the problem of finding an inverse Laplace transform. This can often be done quickly. Example. Find the Green function g ( t ) for the operator L = D 2 + 6D + 13. Because p ( s ) = s 2 + 6 s + 13 = ( s + 3) 2 + 2 2 , the bottom right entry of table (8.13) with a = 3 and b = 2 shows that the Green function is given by g ( t ) = L 1 bracketleftbigg 1 p ( s ) bracketrightbigg = 1 2 L 1 bracketleftbigg 2 ( s + 3) 2 + 2 2 bracketrightbigg = 1 2 e 3 t sin(2 t ) . Example. Find the Green function g ( t ) for the operator L = D 2 + 2D 15. Because p ( s ) = s 2 + 2 s 15 = ( s 3)( s + 5), we use the partial fraction identity 1 ( s 3)( s + 5) = 1 8 s 3 1 8 s + 5 . The top right entry of table (8.13) with a = 3 and with a = 5 shows that the Green function is given by g ( t ) = L 1 bracketleftbigg 1 p ( s ) bracketrightbigg = 1 8 L 1 bracketleftbigg 1 s 3 bracketrightbigg 1 8 L 1 bracketleftbigg 1 s + 5 bracketrightbigg = e 3 t e 5 t 8 . Example. Find the Green function g ( t ) for the operator L = D 4 + 13D 2 + 36. Because p ( s ) = s 4 +13 s 2 +36 = ( s 2 +4)( s 2 +9) only depends on s 2 , we can use the partial fraction identity 1 ( z + 4)( z + 9) = 1 5 z + 4 1 5 z + 9 at z = s 2 . The bottom left entry of table (8.13) with b = 2 and with b = 3 shows that the Green function is given by g ( t ) = L 1 bracketleftbigg 1 p ( s ) bracketrightbigg = 1 10 L 1 bracketleftbigg 2 s 2 + 2 2 bracketrightbigg 1 15 L 1 bracketleftbigg 3 s 2 + 3 2 bracketrightbigg = sin(2 t ) 10 sin(3 t ) 15 . 8.8: Convolutions. Let f ( t ) and g ( t ) be any two functions defined over the interval [0 , ). Their convolution is a third function ( f g )( t ) that is defined by the formula ( f g )( t ) = integraldisplay t 0 f ( t τ ) g ( τ ) d τ , (8.15) whenever the above integral makes sense for every t 0. In particular, the convolution of f and g will be defined whenever both f and g are piecewise continuous over every [0 , T ].
21 The convolution can be thought of a some kind of product between two functions. It

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