Parabolic cylinder function

<p>In <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, the parabolic cylinder functions are <a href="/facts/Special_functions/JETSeucA">special functions</a> defined as solutions to the differential equation
</p>
<table><tbody><tr><td>                                                                        d                                  2                                            f                                      d                              z                                  2                                                                    +                  (                                                                      a                  ~                                                                    z                              2                                      +                                                            b                  ~                                                      z            +                                                            c                  ~                                                              )                f        =        0.              {\displaystyle {\frac {d^{2}f}{dz^{2}}}+\left({\tilde {a}}z^{2}+{\tilde {b}}z+{\tilde {c}}\right)f=0.}  </td> <td></td> <td>1</td></tr></tbody></table>
<p>This equation is found when the technique of <a href="/facts/Separation_of_variables/oJdkw5xq">separation of variables</a> is used on <a href="/facts/Laplace%27s_equation/8n1YkmFS">Laplace's equation</a> when expressed in <a href="/facts/Parabolic_cylindrical_coordinates/aXrXjqr1">parabolic cylindrical coordinates</a>.
</p><p>The above equation may be brought into two distinct forms (A) and (B) by  <a href="/facts/Completing_the_square/SN6xA8sc">completing the square</a> and rescaling z, called <a href="/facts/H._F._Weber/GT7ylX6X">H. F. Weber</a>'s equations:
</p>
<table><tbody><tr><td>                                                                        d                                  2                                            f                                      d                              z                                  2                                                                    −                  (                                                                      1                  4                                                                    z                              2                                      +            a                    )                f        =        0              {\displaystyle {\frac {d^{2}f}{dz^{2}}}-\left({\tfrac {1}{4}}z^{2}+a\right)f=0}  </td> <td></td> <td>A</td></tr></tbody></table>
<p>and
</p>
<table><tbody><tr><td>                                                                        d                                  2                                            f                                      d                              z                                  2                                                                    +                  (                                                                      1                  4                                                                    z                              2                                      −            a                    )                f        =        0.              {\displaystyle {\frac {d^{2}f}{dz^{2}}}+\left({\tfrac {1}{4}}z^{2}-a\right)f=0.}  </td> <td></td> <td>B</td></tr></tbody></table>
<p>If 
  
    
      
        f
        (
        a
        ,
        z
        )
      
    
    {\displaystyle f(a,z)}
  
 is a solution, then so are

f
        (
        a
        ,
        −
        z
        )
        ,
        f
        (
        −
        a
        ,
        i
        z
        )
        
           and 
        
        f
        (
        −
        a
        ,
        −
        i
        z
        )
        .
      
    
    {\displaystyle f(a,-z),f(-a,iz){\text{ and  }}f(-a,-iz).}

</p><p>If 
  
    
      
        f
        (
        a
        ,
        z
        )
        
      
    
    {\displaystyle f(a,z)\,}
  
 is a solution of equation (A), then 
  
    
      
        f
        (
        −
        i
        a
        ,
        z
        
          e
          
            (
            1
            
              /
            
            4
            )
            π
            i
          
        
        )
      
    
    {\displaystyle f(-ia,ze^{(1/4)\pi i})}
  
 is a solution of (B), and, by symmetry,

f
        (
        −
        i
        a
        ,
        −
        z
        
          e
          
            (
            1
            
              /
            
            4
            )
            π
            i
          
        
        )
        ,
        f
        (
        i
        a
        ,
        −
        z
        
          e
          
            −
            (
            1
            
              /
            
            4
            )
            π
            i
          
        
        )
        
           and 
        
        f
        (
        i
        a
        ,
        z
        
          e
          
            −
            (
            1
            
              /
            
            4
            )
            π
            i
          
        
        )
      
    
    {\displaystyle f(-ia,-ze^{(1/4)\pi i}),f(ia,-ze^{-(1/4)\pi i}){\text{ and }}f(ia,ze^{-(1/4)\pi i})}

are also solutions of (B).
</p>

Parabolic cylinder function open-in-new

Parabolic cylinder function