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, $t_{n}=t_{0}+nh,$ 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 error at each step of the midpoint method is of order $O\left(h^{3}\right).$ 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

<math>y(t + h) \approx y(t) + hf\left(t + \frac{h}{2}, y(t) + \frac{h}{2}f(t, y(t))\right)</smath>

and the midpoint method (1).

Sample Fortran Program

      Program Calc

      Double Precision f,y,a,b,J,mult,sum,c1,c2,ci
      Sum=0
      c2=0
      ci=0


      Print*,'Enter the start and end of the interval'
      Read*,a,b
      If (b.gt.a) then
         goto 1
      Else
         goto2
      End If
      
  1   Do J=a,b,.00000001
         c1=J
         Y=F(((c1+c2)/2))
         Mult=Y*.00000001
         Sum=sum+mult
         c2=c1
      End Do

  2   Do J=a,b,-.00000001
         c1=J
         Y=F(((c1+c2)/2))
         Mult=Y*.00000001
         Sum=sum+mult
         c2=c1
      End Do

      Print*,Sum
  3   Format (F20.5)
      End

      Double Precision Function f(x)
      Double Precision x

      F=(4)/((x**2)+1)

      Return
      End

See also

• Rectangle method
• Heun's method

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.{{cite book}}: CS1 maint: multiple names: authors list (link)