# Numerov's method

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Numerov'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 method

The Numerov method can be used to solve differential equations of the form

$\left({\frac {d^{2}}{dx^{2}}}+a(x)\right)y(x)=0$ The function $a(x)$ is sampled in the interval [a..b] at equidistant positions $x_{n}$ . Starting from function values at two consecutive samples $x_{n-1}$ and $x_{n}$ the remaining function values can be calculated as

$y_{n+1}={\frac {\left(2-{\frac {5h^{2}}{6}}a_{n}\right)y_{n}-\left(1+{\frac {h^{2}}{12}}a_{n-1}\right)y_{n-1}}{1+{\frac {h^{2}}{12}}a_{n+1}}}$ ### Nonlinear equations

For nonlinear equations of the form

${\frac {d^{2}}{dx^{2}}}y=f(x,y)$ the method is given by

$y_{n+1}=2y_{n}-y_{n-1}+{\tfrac {1}{12}}h^{2}(f_{n+1}+10f_{n}+f_{n-1}).$ This is an implicit linear multistep method, which reduces to the explicit method given above if f is linear in y by setting $f(x,y)=-a(x)y(x)$ . It achieves order 4 Template:Harv.

## Application

In numerical physics the method is used to find solutions of the radial Schrödinger equation for arbitrary potentials.

$\left[-{\hbar ^{2} \over 2\mu }\left({\frac {1}{r}}{\partial ^{2} \over \partial r^{2}}r-{l(l+1) \over r^{2}}\right)+V(r)\right]R(r)=ER(r)$ The above equation can be rewritten in the form

$\left[{\partial ^{2} \over \partial r^{2}}-{l(l+1) \over r^{2}}+{2\mu \over \hbar ^{2}}\left(E-V(r)\right)\right]u(r)=0$ with $u(r)=rR(r)$ . If we compare this equation with the defining equation of the Numerov method we see

$a(x)={\frac {2\mu }{\hbar ^{2}}}\left(E-V(x)\right)-{\frac {l(l+1)}{x^{2}}}$ and thus can numerically solve the radial Schrödinger equation.

## Derivation

$y(x)=y(x_{0})+(x-x_{0})y'(x_{0})+{\frac {(x-x_{0})^{2}}{2!}}y''(x_{0})+{\frac {(x-x_{0})^{3}}{3!}}y'''(x_{0})+{\frac {(x-x_{0})^{4}}{4!}}y''''(x_{0})+{\frac {(x-x_{0})^{5}}{5!}}y'''''(x_{0})+{\mathcal {O}}(h^{6})$ $y(x_{0}+h)=y(x_{0})+hy'(x_{0})+{\frac {h^{2}}{2!}}y''(x_{0})+{\frac {h^{3}}{3!}}y'''(x_{0})+{\frac {h^{4}}{4!}}y''''(x_{0})+{\frac {h^{5}}{5!}}y'''''(x_{0})+{\mathcal {O}}(h^{6})$ Computationally, this amounts taking a step forward by an amount h. If we want to take a step backwards, replace every h with -h for the equation of $y(x_{0}-h)$ :

$y(x_{0}-h)=y(x_{0})-hy'(x_{0})+{\frac {h^{2}}{2!}}y''(x_{0})-{\frac {h^{3}}{3!}}y'''(x_{0})+{\frac {h^{4}}{4!}}y''''(x_{0})-{\frac {h^{5}}{5!}}y'''''(x_{0})+{\mathcal {O}}(h^{6})$ Note that only the odd powers of h experienced a sign change. On an evenly spaced grid, the nth site on a computational grid corresponds to position $x_{n}$ if the step-size between grid points are of length $h$ (hence h should be small for the computation to be accurate). This means we have sampling points $(x_{n-1},y_{n-1})$ and $(x_{n+1},y_{n+1})$ . Taking the equations for $y_{n-1}$ and $y_{n+1}$ from continuous space to discrete space, we see that

$y_{n+1}=y(x_{n}+h)=y(x_{n})+hy'(x_{n})+{\frac {h^{2}}{2!}}y''(x_{n})+{\frac {h^{3}}{3!}}y'''(x_{n})+{\frac {h^{4}}{4!}}y''''(x_{n})+{\frac {h^{5}}{5!}}y'''''(x_{n})+{\mathcal {O}}(h^{6})$ $y_{n-1}=y(x_{n}-h)=y(x_{n})-hy'(x_{n})+{\frac {h^{2}}{2!}}y''(x_{n})-{\frac {h^{3}}{3!}}y'''(x_{n})+{\frac {h^{4}}{4!}}y''''(x_{n})-{\frac {h^{5}}{5!}}y'''''(x_{n})+{\mathcal {O}}(h^{6})$ The sum of those two equations gives

$y_{n-1}+y_{n+1}=2y_{n}+{h^{2}}y''_{n}+{\frac {h^{4}}{12}}y''''_{n}+{\mathcal {O}}(h^{6})$ We solve this equation for $y''_{n}$ and replace it by the expression $y''_{n}=-a_{n}y_{n}$ which we get from the defining differential equation.

$h^{2}a_{n}y_{n}=2y_{n}-y_{n-1}-y_{n+1}+{\frac {h^{4}}{12}}y''''_{n}+{\mathcal {O}}(h^{6})$ We take the second derivative of our defining differential equation and get

$y''''(x)=-{\frac {d^{2}}{dx^{2}}}\left[a(x)y(x)\right]$ We replace the second derivative ${\frac {d^{2}}{dx^{2}}}$ with the second order difference quotient and insert this into our equation for $a_{n}y_{n}$ (note that we take the mixed forward and backward finite difference, not the double forward difference or the double backward difference)

$h^{2}a_{n}y_{n}=2y_{n}-y_{n-1}-y_{n+1}-{\frac {h^{4}}{12}}{\frac {a_{n-1}y_{n-1}-2a_{n}y_{n}+a_{n+1}y_{n+1}}{h^{2}}}+{\mathcal {O}}(h^{6})$ $y_{n+1}={\frac {\left(2-{\frac {5h^{2}}{6}}a_{n}\right)y_{n}-\left(1+{\frac {h^{2}}{12}}a_{n-1}\right)y_{n-1}}{1+{\frac {h^{2}}{12}}a_{n+1}}}+{\mathcal {O}}(h^{6}).$ This yields Numerov's method if we ignore the term of order $h^{6}$ . It follows that the order of convergence (assuming stability) is 4.