Euler's Method

A numerical method for approximating the values of a solution $y(t)$ of a Differential Equation, by discretizing the differential equation into a difference equation that can be used to find the approximate solution $y_n$ through recursive definition.

Discrete Derivative

$$(Dy)(t_n)=\frac{y_{n+1}-y_n}{h}=\frac{y(t_{n+1})-y(t_n)}{h}$$

Example

$$\begin{array}{c}{y}^{\prime }=2y-1\\ y(0)=1\end{array}$$

This is the differential equation we are starting with. Let $h$ be the time step between approximations. It will be $\frac{1}{10}$ in this example. Use the initial condition to find $y_0$

$$y(0)=1⟹y(t_0)=y_0=1$$

Equate the discrete derivative and the true derivative

$$\begin{array}{c}\frac{y_{n+1}-y_n}{h}=y^{\prime }\\ \frac{y_{n+1}-y_n}{h}=2y_n-1\\ y_{n+1}=h(2y_n-1)+y_n\\ y_{n+1}=(2h+1)y_n-h\end{array}$$

We have a recursive definition of $y_n$ for arbitrary $n$ Now let’s find the error at $t_1$ Find the discrete $y_1$

$$\begin{array}{c}{y}_1=\left(2\frac{1}{10}+1\right)y_0=\frac{1}{10}\\ y_1=\left(2\frac{1}{10}+1\right)\cdot 1-\frac{1}{10}\\ y_1=\frac{11}{10}\end{array}$$

Find the exact solution $\varphi (t)$ to the original system

$$\begin{array}{c}{\varphi }^{\prime }=2\varphi -1\\ \frac{{\varphi }^{\prime }}{2\varphi -1}=1\\ \int \frac{{\varphi }^{\prime }}{2\varphi -1}dt=\int 1dt\\ \int \frac{d\varphi }{2\varphi -1}=t+C\\ \int \frac{d\varphi }{2\varphi -1}=t+C\\ \text{Let }u=2\varphi -1⟹du=2d\varphi \\ \frac{1}{2}\ln (2\varphi -1)=t+C\\ \ln (2\varphi -1)=2t+C\\ 2\varphi -1=C{e}^{2t}\\ \varphi =C{e}^{2t}+\frac{1}{2}\\ \varphi (0)=1⟹C+\frac{1}{2}=1⟹C=\frac{1}{2}\\ \therefore \varphi (t)=\frac{1}{2}{e}^{2t}+\frac{1}{2}\end{array}$$

$E_1={\varphi }_1-y_1=\frac{1}{2}{e}^2+\frac{1}{2}-\frac{11}{10}\approx 3.095$

Find $E_n={\varphi }_n-y_n$

$${\varphi }_n=\frac{1}{2}\exp (2t_n)+\frac{1}{2}$$ $$y_n=(2h+1)y_{n-1}-h$$ $$\frac{1}{2}\exp (2t_n)+\frac{1}{2}-(2h+1)y_{n-1}+h$$

todo some taylor series stuff happens here

$$E_n=\frac{1}{2}{h}^2{\varphi }_n^{\prime \prime }$$