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 numerically solving the differential equation,

$y'(t)=f(t,y(t)),\quad y(t_{0})=y_{0}$.

The explicit midpoint method is given by the formula

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

the implicit midpoint method by

$y_{n+1}=y_{n}+hf\left(t_{n}+{\frac {h}{2}},{\frac {1}{2}}(y_{n}+y_{n+1})\right),\qquad \qquad (1i)$

for $n=0,1,2,\dots$ Here, $h$ is the step size — a small positive number, $t_{n}=t_{0}+nh,$ and $y_{n}$ is the computed approximate value of $y(t_{n}).$ The explicit midpoint method is also known as the modified Euler method, the implicit method is the most simple collocation method, and, applied to Hamiltionian dynamics, a symplectic integrator.

The name of the method comes from the fact that in the formula above the function $f$ giving the slope of the solution is evaluated at $t=t_{n}+h/2,$ (note : $t$ is ${\frac {t_{n}+t_{n+1}}{2}}$) 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's error generally decreases faster as $h\to 0$.

The methods are examples 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 methods, 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 explicit midpoint method (1e).

The implicit method (1i) is obtained by approximating the value at the half step $t+h/2$ by the midpoint of the line segment from $y(t)$ to $y(t+h)$

$y\left(t+{\frac {h}{2}}\right)\approx {\frac {1}{2}}{\bigl (}y(t)+y(t+h){\bigr )}$

and thus

${\frac {y(t+h)-y(t)}{h}}\approx y'\left(t+{\frac {h}{2}}\right)\approx k=f\left(t+{\frac {h}{2}},{\frac {1}{2}}{\bigl (}y(t)+y(t+h){\bigr )}\right)$

Inserting the approximation $y_{n}+h\,k$ for $y(t_{n}+h)$ results in the implicit Runge-Kutta method

${\begin{aligned}k&=f\left(t_{n}+{\frac {h}{2}},y_{n}+{\frac {h}{2}}k\right)\\y_{n+1}&=y_{n}+h\,k\end{aligned}}$

which contains the implicit Euler method with step size $h/2$ as its first part.

Because of the time symmetry of the implicit method, all terms of even degree in $h$ of the local error cancel, so that the local error is automatically of order ${\mathcal {O}}(h^{3})$. Replacing the implicit with the explicit Euler method in the determination of $k$ results again in the explicit midpoint method.

See also

• Rectangle method
• Heun's method
• Leapfrog integration and Verlet integration
• Direct Nonlinear Analysis using DNA-Alpha method

The DNA-Alpha method will find exact solutions of any nonlinear equation without iteration. The equation must be in the following form:

   y = f(x), where f(x) is a high order nonlinear function in x variable

and

   y'(x) = f'(x), where f'(x) is the known first derivative of the function.

Instead of using the mid-point method, it uses two previously solved values and the Mean Value Theorem to find three estimations of next step. The three pair of values will be used to form a quadratic equation and a root of the quadratic equation will become the next step. The required initial conditions are : y(0) and y'(0).

It can be used for finding all exact roots of a N-degree polynomial equation and traces the curve easily. The DNA-Alpha method can be used to solve a nonlinear equation exactly without any iteration. The time needed depends only on the evaluation time of the function to be solved. It can also be used to trace any function's curve easily.

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.
• Süli, Endre; Mayers, David (2003), An Introduction to Numerical Analysis, Cambridge University Press, ISBN 0-521-00794-1.