Midpoint method

For the midpoint rule in numerical quadrature, see rectangle method.

Illustration of the midpoint method assuming that y_n equals the exact value y(t_n). The midpoint method computes y_{n + 1} so that the red chord is approximately parallel to the tangent line at the midpoint (the green line).

In numerical analysis, a branch of applied mathematics, the midpoint method is a one-step method for solving the differential equation

$y'(t) = f(t, y(t)), \quad y(t_0) = y_0$

numerically, and is given by the formula

$y_{n+1} = y_n + hf\left(t_n+\frac{h}{2},y_n+\frac{h}{2}f(t_n, y_n)\right), \qquad\qquad (1)$

for $n=0, 1, 2, \dots$ Here, h is the step size — a small positive number, t_n = t₀ + nh, and y_n is the computed approximate value of y(t_n).

The name of the method comes from the fact that in the formula above the function f is evaluated at t = t_n + h / 2, which is the midpoint between t_n at which the value of y(t) is known and t_{n + 1} at which the value of y(t) needs to be found.

The local error at each step of the midpoint method is of order $O\left(h^3\right)$, giving a global error of order $O\left(h^2\right)$. Thus, while more computationally intensive than Euler's method, the midpoint method generally gives more accurate results.

The method is an example of a class of higher-order methods known as Runge-Kutta methods.

Derivation of the midpoint method

Illustration of numerical integration for the equation y' = y,y(0) = 1. Blue: the Euler method, green: the midpoint method, red: the exact solution, y = e^t. The step size is h = 1.0.

The same illustration for h = 0.25. It is seen that the midpoint method converges faster than the Euler method.

The midpoint method is a refinement of the Euler's method

$y_{n+1} = y_n + hf(t_n,y_n),\,$

and is derived in a similar manner. The key to deriving Euler's method is the approximate equality

$y(t+h) \approx y(t) + hf(t,y(t)) \qquad\qquad (2)$

which is obtained from the slope formula

$y'(t) \approx \frac{y(t+h) - y(t)}{h} \qquad\qquad (3)$

and keeping in mind that y' = f(t,y).

For the midpoint method, one replaces (3) with the more accurate

$y'\left(t+\frac{h}{2}\right) \approx \frac{y(t+h) - y(t)}{h}$

when instead of (2) we find

$y(t+h) \approx y(t) + hf\left(t+\frac{h}{2},y\left(t+\frac{h}{2}\right)\right). \qquad\qquad (4)$

One cannot use this equation to find y(t + h) as one does not know y at t + h / 2. The solution is then to use a Taylor series expansion exactly as if using the Euler method to solve for y(t + h / 2):

$y\left(t + \frac{h}{2}\right) \approx y(t) + \frac{h}{2}y'(t)=y(t) + \frac{h}{2}f(t, y(t)),$

which, when plugged in (4), gives us

$y(t + h) \approx y(t) + hf\left(t + \frac{h}{2}, y(t) + \frac{h}{2}f(t, y(t))\right)$

and the midpoint method (1).

See also

• Rectangle method
• Heun's method
• Leapfrog integration and Verlet integration

References

• Griffiths,D. V.; Smith, I. M. (1991). Numerical methods for engineers: a programming approach. Boca Raton: CRC Press. p. 218. ISBN 0-8493-8610-1.