Chapter-4 
              
                                                                                    Legendre’s Polynomials  
                                                                                                                                                              
             4.1 Introduction       
                    The following second order linear differential equation with variable 
             coefficients is known as Legendre’s differential equation, named after 
             Adrien Marie Legendre (1752-1833), a French mathematician, who is 
             best known for his work in the field of elliptic integrals and theory of 
             numbers :                          
                                 
          
                                                            	 
 
 
                                       …(1) 
             where n is a non-negative integer.       
                    Legendre’s  differential  equation  occurs  in  many  physical  and 
             engineering problems involving spherical geometry and gravitation.      
             4.2 Legendre’s Differential Equation        
                    We know that the differential equation of the form  
                                               
                                 
          
                                                            	 
 
 
                                       …(1)     
             is called Legendre’s differential equation (or simply Legendre’s equation), 
             where n is a non-negative integer.        
                    This equation can  also be put in the following form: 
                                                                    
          
                                          
  
 
 
                                                            
                                                         
                    Clearly, the only singular points of (1) are x = 1, x = − 1 and x = , 
             which  are  regular.  Therefore,  the  Legendre’s  differential  equation  is  a 
             Fuchsian differential equation. 
                                                                                                                                                                   92 
               
                    The points other than singular points e.g., x = 0, x = 2, etc. behave like 
              ordinary points of (1).        
                      Let the series solution of (1) be of the form 
                                                              
                                                            …(2) 
                                                                               
                                                                                      "
!
                                                              
"
!                                                                                 …(3) 
                                                          !
 !
                                                                    
                                     $
              and                       #                       "
! 
"
!                                                                            …(4)       
                                                          
                      Putting the above values of y, y and y# in (1), we have 
                                                                        "
!	             	                                                         "
!	
                                
            
                                                            
            
                    
                     !
! "
! "
!                                                 !
! "
! "
!                                                         
                                                                                 
             $%                                              
                                                        	                    "
!                            


                                       
 
                                                                                                                                    
                                            
          
                   $
              or                         "
! "
!                                       
                                                                                                                                                    
                                                                           &
           
                         
                   
         '
                                                                             "
! "
! 
	 "
!  
                                                          
 
                                                                 "
!	
                                       
            
              or           !
! "
! 
"
!                                              &                                                     ' 
                                                                                              
                   
            
                                                                                              "
! 
 "
!                                                   
 
                                                                                
                                                                                  "
!	
                                          
            
                    
              or             !
! "
! "
!                                                 
                                                                         &                                                                         ' 
                                                                           
                   
                           
                   
                                                                            "
!
 "
! 
 "
!                                                              
 
                                                                           "
!	
                                        
            
              or            !
! "
! 
"
!                                                      
                                                                          
                   
                             
                                                                            "
! "
!

                                               
           …(5)             
                                                               
               which is an identity in x and, therefore, the coefficients of various powers 
              of x in it should be zero. 
                      Thus,  equating  to  zero,  the  coefficient  of  the  smallest  power  of  , 
              namely $ in (5), we get the following  indicial equation 
                                           
                                                    
                                     (                 *
                                 
" "                  
             or           " "                     
                 ) 
  
  
           93 
            
           which gives two indicial roots k = k  = 1 and k = k  = 0.  
                                                                       1                       2
                  Note that the roots of indicial equation are unequal and differ by an 
           integer.       
                  Now, to get the recurrence relation, we equate to zero, the coefficient 
           of $in (5). Thus, we have   
                          
          
                            
                      
                             
                       ! " 
 !         "
! ! "
!	 "
!	

 
 
                                          
            
           or                      !     "
!	 
"
!
!	                                                          …(6)       
                                               
"
!
"
!                             $%
                  Next, equating to zero, the coefficient of                                 in (5), we obtain 
                                                
        
                                                  " 
  "  
                                                                     …(7)       
                  For k = 0, we note from (7)  that C1 is indeterminate.    
                                                                                  
         
                Thus, putting k = 0 in (6), we get  !  !	 
!
!	                  …(8)                                                        
                                                                                        !
!
                  We now express C , C , C …. in terms of C  and C , C  C …. in terms 
                                                 2     4     6                             0           3     5     7
           of C1 by assuming that C1 is finite.        
                        
                  Putting m = 2 in (8), we have  	   


  


                  …(9)       
                                                                                 	                   	+
                   Putting m = 4 in (8) and using (9), we obtain 
                                       
                     
     
      
                                  ,    	 
-
	  	  
 

-
                                             …(10) 
                                             ,-                        ,+
           and so on.        
                  Next, putting m = 3 in (8), we obtain 
                                              
                        
     
                                          -    
	
    

	                                          …(11)       
                                                   -	                       -+
                  Again, putting m = 5 in (8) and using (11), we have 
                                                                                                                                        94 
                                               
                     
     
     
     
                                          .   - 
,
-  -  
	 

,                               …(12) 
                                                    .,                            .+
            and so on.        
                  Now, the solution (2) can be re-written as:  
                                
 
   
   
  / 
  0 
  1 
 2, where k = 0 
                                                %                    /          0           1
                                  
                        0              
               /           1        
            or                    
  
  
2 
  
  
  
2                  …(13)        
                                                       0                     %         /          1
                  Using the values of C , C , C , C ,…... in the above equation, we get  
                                                       2      3     4      5
                                                               
      
     
                                3  4
4% 
 4$ 4 4$% 
4/0  25 
                                                +        
            0+       
     
     
     
                                                 
 3  4$% 
4 / 
 4$/ 4$% 4 
401 
 25       …(14) 
                                                 %               /+                            1+
            which is the required general series solution, C0 and C1 being arbitrary 
            constants. 
            4.3.       Solution  of  Legendre’s  Differential  Equation  in 
            Descending Powers         
                  Consider Legendre’s differential equation of the type 
                                           
                             
        
                                                      	 
 
 
                                       …(1) 
            where n is a non-negative integer.        
                  It  is  possible  to  obtain  the  solution  of  (1)  in  terms  of  descending 
            powers of x. Due to its applications to physical problems, this form of 
            solution of Legendre’s differential equation is more important.        
                  For  such  a  solution,  let  us  assume  that  the  Legendre’s  differential 
            equation (1) has a series solution of the form 
                                                   $ 
                                                             …(2)                                                           
                                                                  
                  Then,  by  Frobenius  method,  we  can  find  two  linearly  independent 
            solutions of (1) in descending powers of x as: