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{{For|the midpoint rule in numerical [[quadrature]]|rectangle method}}


[[File:Midpoint method illustration.png|right|thumb|Illustration of the midpoint method assuming that <math>y_n</math> equals the exact value <math>y(t_n).</math> The midpoint method computes <math>y_{n+1}</math> 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 [[Numerical ordinary differential equations|numerically]] solving the [[ordinary differential equation|differential equation]],
:<math> y'(t) = f(t, y(t)), \quad y(t_0) = y_0 </math>
and is given by the formula
:<math> 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)</math>
for <math>n=0, 1, 2, \dots</math> Here, <math>h</math> is the ''step size'' &mdash; a small positive number, <math>t_n=t_0 + n h,</math> and <math>y_n</math> is the computed approximate value of <math>y(t_n).</math> The midpoint method is also known as the '''modified Euler method'''.<ref>{{harvnb|Süli|Mayers|2003|p=328}}</ref>


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The name of the method comes from the fact that in the formula above the function <math>f</math> is evaluated at <math>t=t_n+h/2,</math> which is the midpoint between <math>t_n</math> at which the value of ''y''(''t'') is known and <math>t_{n+1}</math> at which the value of ''y''(''t'') needs to be found.
 
The local error at each step of the midpoint method is of order <math>O\left(h^3\right)</math>, giving a global error of order <math>O\left(h^2\right)</math>.  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==
[[File:Numerical integration illustration, h=1.png|right|thumb|Illustration of numerical integration for the equation <math>y'=y, y(0)=1.</math> Blue: the [[Euler method]], green: the midpoint method, red: the exact solution, <math>y=e^t.</math> The step size is <math>h=1.0.</math>]]
[[File:Numerical integration illustration, h=0.25.png|right|thumb|The same illustration for <math>h=0.25.</math> It is seen that the midpoint method converges faster than the Euler method.]]
 
The midpoint method is a refinement of the Euler's method
:<math> y_{n+1} = y_n + hf(t_n,y_n),\, </math>
and is derived in a similar manner.
The key to deriving Euler's method is the approximate equality
:<math> y(t+h) \approx y(t) + hf(t,y(t)) \qquad\qquad (2)</math>
which is obtained from the slope formula
:<math> y'(t) \approx \frac{y(t+h) - y(t)}{h} \qquad\qquad (3)</math>
and keeping in mind that <math> y' = f(t, y).</math>
 
For the midpoint method, one replaces (3) with the more accurate
:<math> y'\left(t+\frac{h}{2}\right) \approx \frac{y(t+h) - y(t)}{h} </math>
when instead of (2) we find
:<math> y(t+h) \approx y(t) + hf\left(t+\frac{h}{2},y\left(t+\frac{h}{2}\right)\right). \qquad\qquad (4)</math>
 
One cannot use this equation to find <math> y(t+h)</math> as one does not know <math>y</math> at <math>t+h/2.</math> The solution is then to use a [[Taylor series]] expansion exactly as if using the [[Euler method]] to solve for <math>y(t+h/2)</math>:
:<math>y\left(t + \frac{h}{2}\right) \approx y(t) + \frac{h}{2}y'(t)=y(t) + \frac{h}{2}f(t, y(t)),</math>
which, when plugged in (4), gives us
:<math>y(t + h) \approx y(t) + hf\left(t + \frac{h}{2}, y(t) + \frac{h}{2}f(t, y(t))\right)</math>
and the midpoint method (1).
 
==See also==
* [[Rectangle method]]
* [[Heun's method]]
* [[Leapfrog integration]] and [[Verlet integration]]
 
==Notes==
{{reflist}}
 
==References==
 
* {{cite book
|author=Griffiths,D. V.; Smith, I. M.
|title=Numerical methods for engineers: a programming approach
|publisher=CRC Press
|location=Boca Raton
|year=1991
|isbn=0-8493-8610-1
|oclc=
|doi=
|page=218
}}
* {{Citation | last1=Süli | first1=Endre | last2=Mayers | first2=David | title=An Introduction to Numerical Analysis | publisher=[[Cambridge University Press]] | isbn=0-521-00794-1 | year=2003}}.
 
{{Numerical integrators}}
[[Category:Numerical differential equations]]
[[Category:Runge–Kutta methods]]

Latest revision as of 22:28, 13 March 2013

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Illustration of the midpoint method assuming that yn equals the exact value y(tn). The midpoint method computes yn+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)),y(t0)=y0

and is given by the formula

yn+1=yn+hf(tn+h2,yn+h2f(tn,yn)),(1)

for n=0,1,2, Here, h is the step size — a small positive number, tn=t0+nh, and yn is the computed approximate value of y(tn). The midpoint method is also known as the modified Euler method.[1]

The name of the method comes from the fact that in the formula above the function f is evaluated at t=tn+h/2, which is the midpoint between tn at which the value of y(t) is known and tn+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(h3), giving a global error of order O(h2). 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=et. 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

yn+1=yn+hf(tn,yn),

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

y(t+h)y(t)+hf(t,y(t))(2)

which is obtained from the slope formula

y(t)y(t+h)y(t)h(3)

and keeping in mind that y=f(t,y).

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

y(t+h2)y(t+h)y(t)h

when instead of (2) we find

y(t+h)y(t)+hf(t+h2,y(t+h2)).(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(t+h2)y(t)+h2y(t)=y(t)+h2f(t,y(t)),

which, when plugged in (4), gives us

y(t+h)y(t)+hf(t+h2,y(t)+h2f(t,y(t)))

and the midpoint method (1).

See also

Notes

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References

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