{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# lecture24 - Dierential Equations Linear Algebra Lecture 24...

This preview shows pages 1–11. Sign up to view the full content.

Differential Equations & Linear Algebra Lecture 24 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Contents Fourier series for even and odd functions
Contents Fourier series for even and odd functions Separation of variables

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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)
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
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 , · · · .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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)
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}