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15.1 Eigenfunction Expansions   513


                                                                               x
                                                            –0.5     0    0.5     1     1.5    2
                                                                   0


                                                                  –1



                                                                  –2


                                                                  –3



                                                                  –4
                                                            FIGURE 15.1 Eigenfunction expansion in Example 15.3.
                                           Figure 15.1 shows a graph of f compared with the fifteenth partial sum of this series. The
                                        graph is drawn on a slightly larger interval than [0,π/2]. This eigenfunction expansion appears to
                                        converge rapidly to f (x) on [0,π/2], but the graphs diverge from each other outside this interval.
                                        The eigenfunction expansion is unrelated to f (x) outside of the interval of the Sturm-Liouville
                                        problem.

                                           The following example illustrates an eigenfunction expansion in which the eigenfunctions
                                        are not just sines and cosines.


                                 EXAMPLE 15.4
                                        We will expand f (x)= x in a series of the eigenfunctions of the regular Sturm-Liouville problem


                                                              y + 4y + (3 + λ)y = 0; y(0) = y(1) = 0
                                        on [0,1]. First we must generate the eigenvalues and eigenfunctions. Put y = e rx  to obtain the
                                        characteristic equation of the differential equation:
                                                                      2
                                                                     r + 4r + (3 + λ) = 0,
                                        with roots
                                                                               √
                                                                       r =−2 ±  1 − λ.
                                        Take cases on λ.
                                        Case 1: λ = 1
                                        Then
                                                                     y(x) = c 1 e −2x  + c 2 xe −2x .
                                        Now
                                                                y(0) = c 1 = 0 and y(1) = c 2 e −2  = 0
                                        so c 2 = 0 and y is the trivial solution. Therefore, 1 is not an eigenvalue of this problem.

                                        Case 2: 1 − λ> 0
                                                     2
                                        Write 1 − λ = α with α> 0 to obtain
                                                                   y(x) = c 1 e (−2+α)x  + c 2 e (−2−α)x .






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                                   October 14, 2010  15:20  THM/NEIL   Page-513        27410_15_ch15_p505-562
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