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
(
t
0
)
=
y
0
.
The explicit midpoint method is given by the formula
y
n
+
1
=
y
n
+
h
f
(
t
n
+
h
2
,
y
n
+
h
2
f
(
t
n
,
y
n
)
)
,
(
1
e
)
the implicit midpoint method by
y
n
+
1
=
y
n
+
h
f
(
t
n
+
h
2
,
1
2
(
y
n
+
y
n
+
1
)
)
,
(
1
i
)
for
n
=
0
,
1
,
2
,
…
Here,
h
is the step size — a small positive number,
t
n
=
t
0
+
n
h
,
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
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
(
h
3
)
, giving a global error of order
O
(
h
2
)
. Thus, while more computationally intensive than Euler's method, the midpoint method's error generally decreases faster as
h
→
0
.
The methods are examples of a class of higher-order methods known as Runge-Kutta methods.
The midpoint method is a refinement of the Euler's method
y
n
+
1
=
y
n
+
h
f
(
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
)
≈
y
(
t
)
+
h
f
(
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 methods, one replaces (3) with the more accurate
y
′
(
t
+
h
2
)
≈
y
(
t
+
h
)
−
y
(
t
)
h
when instead of (2) we find
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
+
h
2
,
y
(
t
+
h
2
)
)
.
(
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
+
h
2
)
≈
y
(
t
)
+
h
2
y
′
(
t
)
=
y
(
t
)
+
h
2
f
(
t
,
y
(
t
)
)
,
which, when plugged in (4), gives us
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
+
h
2
,
y
(
t
)
+
h
2
f
(
t
,
y
(
t
)
)
)
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
(
t
+
h
2
)
≈
1
2
(
y
(
t
)
+
y
(
t
+
h
)
)
and thus
y
(
t
+
h
)
−
y
(
t
)
h
≈
y
′
(
t
+
h
2
)
≈
k
=
f
(
t
+
h
2
,
1
2
(
y
(
t
)
+
y
(
t
+
h
)
)
)
Inserting the approximation
y
n
+
h
k
for
y
(
t
n
+
h
)
results in the implicit Runge-Kutta method
k
=
f
(
t
n
+
h
2
,
y
n
+
h
2
k
)
y
n
+
1
=
y
n
+
h
k
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
O
(
h
3
)
. Replacing the implicit with the explicit Euler method in the determination of
k
results again in the explicit midpoint method.
