A method of a [[Iterative Numerical Integration]] for solving a given [[Differential Equations|Differential Equation]] $y(t)$. Note that this method (like many others) is technically part of the family of [[Runge-Kutta Methods]] $\huge \begin{cases} y'(t) = f(t,y)\\ y(t_{0})=y_{0} \end{cases} $ With a given step size $h$, each iteration of Euler's method is computed as such: $\huge \begin{align} z_{i+1} &= z_{i} +h f(t_{i},z_{i}) \end{align} $ For smaller and smaller step sizes of $h$, this method will diverge slower. ### [[Iterative Numerical Integration#Error|Error]] #### Local Error $ \large \begin{align} y(t) & = y(t_{i})(t-t_{i})+\frac{1}{2}y''(\xi) (t-t_{i})^{2} \\ & = y_{i} + f(t_{i},y_{i})(t-t_{i})+\frac{1}{2}y''(\xi)(t-t_{i})^{2}\\ y_{i+1} & =y_{i}+f(t_{i},y_{i})(t_{i+1}-t_{i})+\frac{1}{2}y''(\xi)(t_{i+1}-t_{i})^{2} \\ y_{i+1} & = y_{i} + hf(t_{i},y_{i}) + \frac{h^{2}}{2}y''(\xi) \\ z_{i+1} &= z_{i} + hf(t_{i},z_{i}) \\ & y_{i} + hf(t_{i},y_{i}) \end{align} $ $ \huge |z_{i+1}-y_{i+1} | = \left| \frac{h^{2}}{2}y''(\xi) \right| \leq \underbrace{ {\frac{h^{2}}{2}|y''_\text{ext}|} }_{ \max|y''| } $ #### Global Error Note $L$ denotes the [[Lipschitz Constant]]. $ \large \begin{align} z_{i+1} & = z_{i} + hf(t_{i},z_{i}) \\ |z_{i+1}-w_{i+1}| & =|z_{i}+hf(t_{i},z_{i})-(w_{i}+hf(t_{i},w_{i}))| \\ & =\left| (z_{i}-w_{i}) + h(f(t_{i},z_{i})-f(t_{i},w_{i})) \right| \\ & \leq |z_{i}-w_{i} + hL(z_{i}-w_{i})| \\ & = |(z_{i}-w_{i})(1+hL)| \\ |z_{i+1}-w_{i+1}| & = \epsilon _{i}(1+hL) \end{align} $ Assume that if $L<0$, $h\leq-\frac{1}{L}$, $1+hL\ge {0}$. $ \large \begin{align} \epsilon _{i+1} & = |z_{i+1}-y_{i+1}| \\ & = |z_{i+1}-w_{i+1}+w_{i+1}-y_{i+1}| \\ \\ & \leq |z_{i+1}-w_{i+1}| + |w_{i+1}-y_{i+1}| \\ \epsilon _{i+1} &\leq (1+hL)\epsilon _{i}+\frac{h^{2}}{2}|y''_\text{ext}| \end{align} $ $ \large \begin{align} \epsilon_{0} &= 0 \\ \\ \epsilon_{1} &\leq \frac{h^{2}}{2}|y''_\text{ext}| \\ \epsilon_{2} & \leq(1+hL) \frac{h^{2}}{2}|y''_\text{ext}| + \frac{h^{2}}{2}|y_\text{ext}''| \\ \epsilon_{3} & \leq (1+hL)^{2} \frac{h^{2}}{2}|y''_\text{ext}|+ (1+hL) \frac{h^{2}}{2}|y_\text{ext}''|+ \frac{h^{2}}{2}|y''_\text{ext}| \\ \vdots & \\ \epsilon _{i} &\leq \left( 1+(1+hL) + \cdots + (1+hL)^{i-1} \right) \frac{h^{2}}{2} |y''_\text{ext}| \end{align} $ $ \large \begin{align} \epsilon _{i} &\leq \frac{(1+hL)^{2}-1}{(1+hL)-1} \frac{h^{2}}{2} |y''_\text{ext}| \\ \epsilon(t_{i}) & = \frac{(1+hL)^{\frac{t_{i}-t_{0}}{h}}-1}{hL} \frac{h^{2}}{2} |y''_\text{ext}| \\ t_{i} & = t_{0} + ih \\ i &= \frac{t_{i}-t_{0}}{h} \\ \epsilon(t) &\leq \pa{ \\ e^{L(t-t_{0})}-1} \frac{h}{2|L|}|y''_\text{ext}| \end{align} $ We get $e$ from: $ \large \lim_{ h \to 0 } (1+hL)^{\frac{1}{h}} = \lim_{ h \to 0 } \left( \left( 1+hL \right) ^{\frac{1}{hL}} \right) ^{L} = e^{L} $ $\large \left( 1+hL \right) ^{\frac{t_{i}-t_{0}}{h}} \leq e^{L(t_{i}-t_{0})} $