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Unit 12: Sturm Comparison and Separation Theorems




          where P(x), Q(x) and R(x) are finite polynomials that contain no common factor. This equation is  Notes
          inhomogeneous and has variable coefficients. After dividing through P(x), we obtain the more
          concurrent, equivalent form,
                                        2
                                       d y      dy
                                            a 1 ( )  a 0 ( )y  f  ( )              ...(1)
                                                      x
                                                            x
                                              x
                                       dx  2    dx
          Provided p   0. If p(x) = 0 at some point x = x , we call x = x  a singular point of the equation. If P(x)
                                             0         0
            0, x  is a regular or ordinary point of the equation. If  P(x)   0 for all points x in the interval
               0
          where we want to solve the equation, we say the equation is non-singular or regular in  the
          interval.
          If a (x), a (x) and f(x) are continuous on some open interval a < x < b that contains the initial point,
             1   0
          then a unique solution of the form
                                      y = Au (x) + Bu (x) + G(x)
                                           1      2
          where A,  B are constants and are fixed by initial conditions. Before  we try  to construct  the
          general solution of equation (1), we will outline a series of sub-problems that are more tractable.

          12.2 The Method of Reduction of Order

          As a first simplification we discuss the solution of the homogeneous differential equation

                                        2
                                       d y      dy
                                              x
                                            a 1 ( )  a 0 ( )y  0                   ...(2)
                                                      x
                                       dx  2    dx
          on the assumption that we know one solution, say y(x) = u (x), and only need to find the second
                                                         1
          solution. We will look for a solution of the form y(x) = U(x)u (x). Differentiating y(x) using the
                                                           1
          product rule gives
                                         dy  dU      du 1
                                               u 1  U   ,
                                         dx  dx      dx
                                         2
                                                          2
                                     2
                                    d y  d U  u  2  dU du 1  U  d u 1
                                    dx 2  dx 2  1  dx dx  dx 2
          If we substitute these expressions into (2) we obtain
                                          2
                         2
                        d U  u  2  dU du  1  U  d u  1  a  ( )  dU  u  U  du  1  a  ( )Uu  0
                                                                  x
                                                x
                        dx 2  1  dx dx   dx  2  1  dx  1   dx    0    1
          We can now collect terms to get
                                                    2
                            2
                           d u      du             d U  dU   du
                                   x
                        U     1  a 1 ( )  1  a 0 ( )u 1  u 1  2  1  a u  0
                                           x
                                                                  1 1
                            dx 2     dx            dx  2  dx  dx
          Now, since u (x) is a solution of (2), the term multiplying U is zero. We have therefore obtained
                     1
          a differential equation for dU/dx, and, by defining Z = dU/dx, we have
                                        dZ     du
                                      u     Z  2  1  a u  0
                                       1             1 1
                                        dx      dx
          Dividing through by Zu  we have
                              1
                                        1 dZ  2 du 1  a  0,
                                        Z dx  u dx   1
                                              1

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