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lecture24 - Dierential Equations Linear Algebra Lecture 24...

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Differential Equations & Linear Algebra Lecture 24 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010
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Contents Fourier series for even and odd functions
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Contents Fourier series for even and odd functions Separation of variables
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Fourier series expansion Recall that the Fourier convergence theorem says that if f and f are piecewise continuous on the interval - L x L , and that f is periodic of period 2 L , then f has a Fourier series expansion f ( x )= a 0 2 + m =1 a m cos mπx L + b m sin mπx L , (1) where the Fourier coefficients are given by the Euler-Fourier formulas a m = 1 L L - L f ( x ) cos mπx L dx, m = 0 , 1 , · · · , (2) b m = 1 L L - L f ( x ) sin mπx L dx, m = 1 , 2 , · · · . (3)
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Cosine series expansion for even functions Suppose that f satisfies the conditions in the Fourier convergence theorem, and further that f is an even function. Then it follows that the Fourier coefficients of f are a n = 2 L L 0 f ( x ) cos nπx L dx, n = 0 , 1 , · · · ; b n = 0 , n = 1 , 2 , · · · . (4) Thus f has the Fourier series f ( x )= a 0 2 + n =1 a n cos nπx L . (5) We call such a series a Fourier cosine series .
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Sine series expansion for even functions Similarly, suppose f and f are piecewise continuous on - L x L and that f is an odd periodic function of period 2 L . Then it follows that the Fourier coefficients of f satisfy a n = 0 , n = 0 , 1 , 2 , · · · ; b n = 2 L L 0 f ( x ) sin nπx L dx, n = 1 , 2 , · · · . (6) Thus f has the Fourier series f ( x )= n =1 b n sin nπx L . (7) We call such a series a Fourier sine series .
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Example Example Let f ( x ) = x, - L < x < L , and let f ( - L ) = f ( L ) = 0 . Let f be defined elsewhere so that it is periodic of period 2 L . Find the Fourier series for this function. Solution Since f is an odd function, its Fourier coefficients, by equation (7), are a n = 0 , n = 0 , 1 , · · · ; b n = 2 L L 0 x sin nπx L dx = 2 L L 2 sin nπx L - nπx L cos nπx L L 0 = 2 L ( - 1) n +1 , n = 1 , 2 , · · · .
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Solution Hence the Fourier series for f is f ( x ) = 2 L π n =1 ( - 1) n +1 n sin nπx L . (8) Observe that the periodic function f is discontinuous at the points ± L, ± 3 L, · · · . At these points the series (8) converges to the mean value of the left and right limits, namely zero. Interestingly, if we take the value of x = L/ 2 in equation (8), we obtain k =1 ( - 1) k +1 2 k - 1 = 1 - 1 3 + 1 5 - 1 7 + · · · = π 4 . (9)
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Exercise Exercise Let f ( x ) = - 1 , ( - π, 0) 1 , (0 , π ) . Find the Fourier series of f and determine the sum of the series. Then find the value of 1 - 1 3 + 1 5 - 1 7 + · · · . Solution a n = 0 , b n = 2 π π 0 sin nxdx = 2 (1 - cos ) , b 2 k = 0 , b 2 k - 1 = 4 (2 k - 1) π , k = 1 , 2 , · · · . f ( x )= 4 π k =1 sin(2 k - 1) x 2 k - 1 = 4 π sin x + sin 3 x 3 + · · · . The series converges to f for x ( - π, 0) (0 , π ) and to zero for x = 0 , ± π .
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Fourier series for functions defined in half of a period In solving partial differential equations it is often useful to expand a function f originally defined only on the interval [0 , L ] into a Fourier series of period 2 L . Then we may expand f by either sine or cosine, or both series through the following ways: 1 . Define a function g of period 2 L so that g ( x ) = f ( x ) , 0 x L, f ( - x ) , - L < x < 0 .
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