Numerov's method (also called Cowell's method) is a numerical method to solve ordinary differential equations of second order in which the first-order term does not appear. It is a fourth-order linear multistep method. The method is implicit, but can be made explicit if the differential equation is linear.
Numerov's method was developed by the Russian astronomer Boris Vasil'evich Numerov.
The Numerov method can be used to solve differential equations of the form
                                                                        d                                  2                                            y                                      d                              x                                  2                                                                    =        −        g        (        x        )        y        (        x        )        +        s        (        x        )        .                In it, three values of                               y                      n            −            1                          ,                  y                      n                          ,                  y                      n            +            1                                   taken at three equidistant points                               x                      n            −            1                          ,                  x                      n                          ,                  x                      n            +            1                                   are related as follows:
                              y                      n            +            1                                    (          1          +                                                    h                                  2                                            12                                            g                          n              +              1                                )                =        2                  y                      n                                    (          1          −                                                    5                                  h                                      2                                                              12                                            g                          n                                )                −                  y                      n            −            1                                    (          1          +                                                    h                                  2                                            12                                            g                          n              −              1                                )                +                                            h                              2                                      12                          (                  s                      n            +            1                          +        10                  s                      n                          +                  s                      n            −            1                          )        +                              O                          (                  h                      6                          )        ,                where                               y                      n                          =        y        (                  x                      n                          )                ,                               g                      n                          =        g        (                  x                      n                          )                ,                               s                      n                          =        s        (                  x                      n                          )                , and                     h        =                  x                      n            +            1                          −                  x                      n                                  .
For nonlinear equations of the form
                                                                        d                                  2                                            y                                      d                              x                                  2                                                                    =        f        (        x        ,        y        )        ,                the method gives
                              y                      n            +            1                          −        2                  y                      n                          +                  y                      n            −            1                          =                                            h                              2                                      12                          (                  f                      n            +            1                          +        10                  f                      n                          +                  f                      n            −            1                          )        +                              O                          (                  h                      6                          )        .                This is an implicit linear multistep method, which reduces to the explicit method given above if                     f                 is linear in                     y                 by setting                     f        (        x        ,        y        )        =        −        g        (        x        )        y        (        x        )        +        s        (        x        )                . It achieves order-4 accuracy (Hairer, Nørsett & Wanner 1993, §III.10).
In numerical physics the method is used to find solutions of the unidimensional Schrödinger equation for arbitrary potentials. An example of which is solving the radial equation for a spherically symmetric potential. In this example, after separating the variables and analytically solving the angular equation, we are left with the following equation of the radial function                     R        (        r        )                :
                                          d                          d              r                                                (                      r                          2                                                                          d                R                                            d                r                                              )                −                                            2              m                              r                                  2                                                                    ℏ                              2                                                    (        V        (        r        )        −        E        )        R        (        r        )        =        l        (        l        +        1        )        R        (        r        )        .                This equation can be reduced to the form necessary for the application of Numerov's method with the following substitution:
                    u        (        r        )        =        r        R        (        r        )        ,                                    R        (        r        )        =                                            u              (              r              )                        r                          ,                                                                        d              R                                      d              r                                      =                              1            r                                                              d              u                                      d              r                                      −                                            u              (              r              )                                      r                              2                                                    =                              1                          r                              2                                                              (          r                                                    d                u                                            d                r                                              −          u          (          r          )          )                ,                                                          d                          d              r                                                (                      r                          2                                                                          d                R                                            d                r                                              )                =                                            d              u                                      d              r                                      +        r                                                            d                                  2                                            u                                      d                              r                                  2                                                                    −                                            d              u                                      d              r                                      =        r                                                            d                                  2                                            u                                      d                              r                                  2                                                                    .                And when we make the substitution, the radial equation becomes
                    r                                                            d                                  2                                            u                                      d                              r                                  2                                                                    −                                            2              m              r                                      ℏ                              2                                                    (        V        (        r        )        −        E        )        u        (        r        )        =                                            l              (              l              +              1              )                        r                          u        (        r        )        ,                or
                    −                                            ℏ                              2                                                    2              m                                                                                          d                                  2                                            u                                      d                              r                                  2                                                                    +                  (          V          (          r          )          +                                                    ℏ                                  2                                                            2                m                                                                                        l                (                l                +                1                )                                            r                                  2                                                              )                u        (        r        )        =        E        u        (        r        )        ,                which is equivalent to the one-dimensional Schrödinger equation, but with the modified effective potential
                              V                      eff                          (        r        )        =        V        (        r        )        +                                            ℏ                              2                                                    2              m                                                                          l              (              l              +              1              )                                      r                              2                                                    =        V        (        r        )        +                                            L                              2                                                    2              m                              r                                  2                                                                    ,                          L                      2                          =        l        (        l        +        1        )                  ℏ                      2                          .                This equation we can proceed to solve the same way we would have solved the one-dimensional Schrödinger equation. We can rewrite the equation a little bit differently and thus see the possible application of Numerov's method more clearly:
                                                                        d                                  2                                            u                                      d                              r                                  2                                                                    =        −                                            2              m                                      ℏ                              2                                                    (        E        −                  V                      eff                          (        r        )        )        u        (        r        )        ,                                    g        (        r        )        =                                            2              m                                      ℏ                              2                                                    (        E        −                  V                      eff                          (        r        )        )        ,                                    s        (        r        )        =        0.                We are given the differential equation
                              y          ″                (        x        )        =        −        g        (        x        )        y        (        x        )        +        s        (        x        )        .                To derive the Numerov's method for solving this equation, we begin with the Taylor expansion of the function we want to solve,                     y        (        x        )                , around the point                               x                      0                                  :
                    y        (        x        )        =        y        (                  x                      0                          )        +        (        x        −                  x                      0                          )                  y          ′                (                  x                      0                          )        +                                            (              x              −                              x                                  0                                                            )                                  2                                                                    2              !                                                y          ″                (                  x                      0                          )        +                                            (              x              −                              x                                  0                                                            )                                  3                                                                    3              !                                                y          ‴                (                  x                      0                          )        +                                            (              x              −                              x                                  0                                                            )                                  4                                                                    4              !                                                y          ⁗                (                  x                      0                          )        +                                            (              x              −                              x                                  0                                                            )                                  5                                                                    5              !                                                y          ′′′′′                (                  x                      0                          )        +                              O                          (                  h                      6                          )        .                Denoting the distance from                     x                 to                               x                      0                                   by                     h        =        x        −                  x                      0                                  , we can write the above equation as
                    y        (                  x                      0                          +        h        )        =        y        (                  x                      0                          )        +        h                  y          ′                (                  x                      0                          )        +                                            h                              2                                                    2              !                                                y          ″                (                  x                      0                          )        +                                            h                              3                                                    3              !                                                y          ‴                (                  x                      0                          )        +                                            h                              4                                                    4              !                                                y          ⁗                (                  x                      0                          )        +                                            h                              5                                                    5              !                                                y          ′′′′′                (                  x                      0                          )        +                              O                          (                  h                      6                          )        .                If we evenly discretize the space, we get a grid of                     x                 points, where                     h        =                  x                      n            +            1                          −                  x                      n                                  . By applying the above equations to this discreet space, we get a relation between the                               y                      n                                   and                               y                      n            +            1                                  :
                              y                      n            +            1                          =                  y                      n                          +        h                  y          ′                (                  x                      n                          )        +                                            h                              2                                                    2              !                                                y          ″                (                  x                      n                          )        +                                            h                              3                                                    3              !                                                y          ‴                (                  x                      n                          )        +                                            h                              4                                                    4              !                                                y          ⁗                (                  x                      n                          )        +                                            h                              5                                                    5              !                                                y          ′′′′′                (                  x                      n                          )        +                              O                          (                  h                      6                          )        .                Computationally, this amounts to taking a step forward by an amount                     h                . If we want to take a step backwards, we replace every                     h                 with                     −        h                 and get the expression for                               y                      n            −            1                                  :
                              y                      n            −            1                          =                  y                      n                          −        h                  y          ′                (                  x                      n                          )        +                                            h                              2                                                    2              !                                                y          ″                (                  x                      n                          )        −                                            h                              3                                                    3              !                                                y          ‴                (                  x                      n                          )        +                                            h                              4                                                    4              !                                                y          ⁗                (                  x                      n                          )        −                                            h                              5                                                    5              !                                                y          ′′′′′                (                  x                      n                          )        +                              O                          (                  h                      6                          )        .                Note that only the odd powers of                     h                 experienced a sign change. By summing the two equations, we derive that
                              y                      n            +            1                          −        2                  y                      n                          +                  y                      n            −            1                          =                  h                      2                                    y                      n                    ″                +                                            h                              4                                      12                                    y                      n                    ⁗                +                              O                          (                  h                      6                          )        .                We can solve this equation for                               y                      n            +            1                                   by substituting the expression given at the beginning, that is                               y                      n                    ″                =        −                  g                      n                                    y                      n                          +                  s                      n                                  . To get an expression for the                               y                      n                    ⁗                         factor, we simply have to differentiate                               y                      n                    ″                =        −                  g                      n                                    y                      n                          +                  s                      n                                   twice and approximate it again in the same way we did this above:
                              y                      n                    ⁗                =                                            d                              2                                                    d                              x                                  2                                                                    (        −                  g                      n                                    y                      n                          +                  s                      n                          )        ,                                              h                      2                                    y                      n                    ⁗                =        −                  g                      n            +            1                                    y                      n            +            1                          +                  s                      n            +            1                          +        2                  g                      n                                    y                      n                          −        2                  s                      n                          −                  g                      n            −            1                                    y                      n            −            1                          +                  s                      n            −            1                          +                              O                          (                  h                      4                          )        .                If we now substitute this to the preceding equation, we get
                              y                      n            +            1                          −        2                  y                      n                          +                  y                      n            −            1                          =                              h                          2                                      (        −                  g                      n                                    y                      n                          +                  s                      n                          )        +                                            h                              2                                      12                          (        −                  g                      n            +            1                                    y                      n            +            1                          +                  s                      n            +            1                          +        2                  g                      n                                    y                      n                          −        2                  s                      n                          −                  g                      n            −            1                                    y                      n            −            1                          +                  s                      n            −            1                          )        +                              O                          (                  h                      6                          )        ,                or
                              y                      n            +            1                                    (          1          +                                                    h                                  2                                            12                                            g                          n              +              1                                )                −        2                  y                      n                                    (          1          −                                                    5                                  h                                      2                                                              12                                            g                          n                                )                +                  y                      n            −            1                                    (          1          +                                                    h                                  2                                            12                                            g                          n              −              1                                )                =                                            h                              2                                      12                          (                  s                      n            +            1                          +        10                  s                      n                          +                  s                      n            −            1                          )        +                              O                          (                  h                      6                          )        .                This yields the Numerov's method if we ignore the term of order                               h                      6                                  . It follows that the order of convergence (assuming stability) is 4.