Euler Method

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Proof Technique

Consider the first order ODE:

$(1): \quad y' = \map f {x, y}$ subject to the initial condition $\map y {x_0} = y_0$

where $\map f {x, y}$ is a continuous real function.

Let $\map y x$ be the particular solution of $(1)$.

For all $n \in \N_{>0}$, we define:

$x_n = x_{n - 1} + h$

where $h \in \R_{>0}$.

Then for all $n \in \N_{>0}$ such that $x_n$ is in the domain of $y$:

$y_{n + 1} = y_n + h \map f {x_n, y_n}$

is an approximation to $\map y {x_{n + 1} }$.


Let $(1)$ be integrated with respect to $x$ from $x_0$ to $x_1$.

\(\text {(2)}: \quad\) \(\ds \map y {x_1} - \map y {x_0}\) \(=\) \(\ds \int_{x_0}^{x_1} \map f {x, y} \rd x\)
\(\ds \leadsto \ \ \) \(\ds \map y {x_1}\) \(=\) \(\ds y_0 + \int_{x_0}^{x_1} \map f {x, y} \rd x\)
Assuming that the integrand in $(2)$ varies little over the interval $\closedint {x_0} {x_1}$:
\(\ds \leadsto \ \ \) \(\ds \map y {x_1}\) \(\approx\) \(\ds y_0 + \map f {x_0, y_0} \paren {x_1 - x_0}\)
\(\ds \) \(=\) \(\ds y_0 + h \, \map f {x_0, y_0}\)

Because $f$ is continuous, the assumption holds.

By making $h$ small, the difference:

$y_0 + h \, \map f {x_0, y_0} - \map f {x_1, y_1}$

can be made arbitrarily small.

$y_{n + 1}$ can be defined recursively:

\(\ds \map y {x_{n + 1} } - \map y {x_n}\) \(=\) \(\ds \int_{x_n}^{x_{n + 1} } \map f {x, y} \rd x\)
\(\ds \leadsto \ \ \) \(\ds \map y {x_{n + 1} }\) \(=\) \(\ds \map y {x_n} + \int_{x_n}^{x_{n + 1} } \map f {x, y} \rd x\)
\(\ds \) \(\approx\) \(\ds \map y {x_n} + \map f {x_n, y_n} \paren {x_{n + 1} - x_n}\)


The errors accumulate; with increasing $n$ the values of $y_{n + 1}$ are based on increasingly inaccurate values of $y_n$.

These can be reduced by making $h$ smaller, so the inaccuracies can be reduced by increasing the computation needed.


Also known as

Some sources give this (in the possessive form) as Euler's method.

Source of Name

This entry was named for Leonhard Paul Euler.