📂Numerical Analysis

Error of the Euler Method When the Initial Value Is Slightly Different

Theorem

For $f$ defined on $[x_{0} , b] \times \mathbb{R}$, suppose that the solution $Y(x)$ of the initial value problem $\begin{cases} y ' = f(x,y) \\ y( x_{0} ) = Y_{0} \end{cases}$ satisfies $Y \in C^{3} [ x_{0} , b ]$ and that $\displaystyle f_{y} (x,y) = {{ \partial f (x,y) } \over { \partial y }}$ and $\displaystyle f_{yy} (x,y) = {{ \partial^{2} f (x,y) } \over { \partial y^{2} }}$ are continuous and bounded. Suppose the initial value $y_{h} (x_{0} )$ satisfies $Y_{0} - y_{h} (x_{0} ) = \delta_{0} h + O ( h^2 )$. Then the error produced by the Euler method satisfies the following for the solution $D(x)$ of the linear initial value problem $$ \begin{cases} \displaystyle D’ (x) = f_{y} (x, Y(x) ) D(x) + {{1} \over {2}} Y’’ (x) \\ D ( x_{0} ) = \delta_{0} \end{cases} $$ $$ Y(x_{n} ) - y_{h} (x_{n} ) = D(x_{n} ) h + O (h^2) $$

Explanation

Saying that the initial value $y_{h} (x_{0} )$ satisfies $Y_{0} - y_{h} (x_{0} ) = \delta_{0} h + O ( h^2 )$ means precisely that the initial value is allowed to differ slightly from the true value. Moreover, the error arising from that difference itself can be further reduced if $D$ can be computed.

Proof¹

Strategy: assume the strong Lipschitz condition and construct $g_{n}$ and $k_{n}$ in order to show $\epsilon_{n} = O ( h^2)$. It is a proof method with so much formula manipulation that it could even be called tenacious, but such proofs appear quite often in numerical analysis. Even if it is not a beautiful proof, it is good to get used to the method itself.

$x_{i} : = x_{0} + ih$, and in particular $x_{n+1} - x_{n} = h$. For convenience, write $$ Y(x_{n} ) := Y_{n} $$ $$ y(x_{n}) := y_{n} $$ and denote the $n$th error $\epsilon_{n}$ as follows. $$ \epsilon_{n} : = Y(x_{n}) - y (x_{n} ) $$

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Part 1.

Taking the $3$rd-order Taylor expansion of $Y(x_{n+1} )$ about $ x_{n}$, for some $x_{n} \le \xi_{n} \le x_{n+1}$, $$ Y_{n+1} = Y_{n} + h Y’_{n} + {{h^2} \over {2}} Y’’ ( x_{n} ) + {{h^3} \over {6}} Y’’’ ( \xi_{n} ) $$ Here, the Euler approximation yields the following equation. $$ \begin{align} \displaystyle \epsilon_{n+1} = \epsilon_{n} + h ( f( x_{n} , Y_{n} ) - f (x_{n} , y_{n}) ) + {{h^2} \over {2}} Y’’ ( x_{n} ) + {{h^3} \over {6}} Y’’’ ( \xi_{n} ) \end{align} $$ Regarding $f(x_{n} , y_{n})$ in $(1)$ as a function of $y_{n}$ and taking the Taylor expansion, for some $\zeta_{n} \in \mathscr{H} \left\{ y_{n} , Y_{n} \right\} $, $$ f(x_{n} , y_{n} ) = f(x_{n} , Y_{n} ) + (y_{n } - Y_{n}) f_{y} (x_{n} , Y_{n} ) + {{1} \over {2}} ( y_{n} - Y_{n} )^2 f_{yy} (x_{n} , \zeta_{n} ) $$ Therefore, $$ \epsilon_{n+1} = \epsilon_{n} + h \left( \epsilon_{ n } f_{y} (x_{n} , Y_{n} ) - {{1} \over {2}} \epsilon^2 f_{yy} (x_{n} , \zeta_{n} ) \right) + {{h^2} \over {2}} Y’’ ( x_{n} ) + {{h^3} \over {6}} Y’’’ ( \xi_{n} ) $$ Rearranging gives $$ \epsilon_{n+1} = [ 1+ f_{y} (x_{n} , Y_{n} ) ] \epsilon_{n} + {{h^2} \over {2}} Y’’ ( x_{n} ) + {{h^3} \over {6}} Y’’’ ( \xi_{n} ) - {{1} \over {2}} h f_{yy} (x_{n} , \zeta_{n} ) \epsilon_{n}^{2} $$

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Part 2.

Strong Lipschitz condition and the error of the Euler method: $$\max_{ x_{0 } \le x_{n} \le b } | Y_{x_{n}} - y_{h} (x_{n}) | \le \epsilon^{( b - x_{0} ) K} | \epsilon_{0} | + \left[ {{ \epsilon^{(b- x_{0}) K } - 1 } \over {K}} \right] \tau (h)$$

Since $\epsilon = O (h)$, $$ {{h^3} \over {6}} Y’’’ ( \xi_{n} ) - {{1} \over {2}} h f_{yy} (x_{n} , \zeta_{n} ) \epsilon_{n}^{2} = O (h^{3} ) $$ and since $O ( h^3)$ is assumed to be sufficiently small, we can define $g_{n+1}$, with $O ( h^3)$ removed, as follows. $$ g_{n+1} : = [ 1+ f_{y} (x_{n} , Y_{n} ) ] g_{n} + {{h^2} \over {2}} Y’’ ( x_{n} ) $$ Since the assumption allows us to take $g_{0} = h \delta_{0}$, setting $g_{n} = h \delta_{n}$ and canceling $h$ from both sides gives $$ \delta_{n+1} : = \delta_{n} + h \left[ f_{y} (x_{n} , Y_{n} ) \delta_{n} + {{1} \over {2}} Y’’ ( x_{n} ) \right] $$ Rearranging just a little, $$ {{\delta_{n+1} - \delta_{n} } \over {h}} = f_{y} (x_{n} , Y_{n} ) \delta_{n} + {{1} \over {2}} Y’’ ( x_{n} ) $$ and if we look closely at the form of this equation, it resembles the linear initial value problem $\begin{cases} \displaystyle D’ (x) = f_{y} (x, Y(x) ) D(x) + {{1} \over {2}} Y’’ (x) \\ D ( x_{0} ) = \delta_{0} \end{cases}$. In the end, $\delta_{n}$ is the Euler approximate solution of $D(x_{n} )$, that is, $D(x_{n} ) - \delta_{n} = O (h) $. Now, by the definition of $g_{n}$, multiplying both sides by $h$ gives $$ g_{n} = D ( x_{n} ) h + O ( h^2 ) $$

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Part 3.

Letting $k_{n} : = e_{n} - g_{n}$, $$ k_{n+1} = [ 1 + h f_{h} (x_{n} , y_{n} ) ] k_{n} + O(h^3) $$ and since $f_{y}$ is bounded, for some $K \ge 0$, $$ | k_{n+1} | \le ( 1 + h K ) | k_{n} | + O(h^3) $$ Solving recursively, $$ |k_{n}| = O(h^2) + O (h^3 ) = O(h^2) $$ Therefore, $$ \begin{align*} \epsilon_{n} =& g_{n} + k_{n} \\ =& [ h D (x_{n} ) + O(h^2) ] + O(h^2) \\ =& O(h^2) \end{align*} $$

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