Last modified: December 31, 2024

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Euler's Method

Euler's Method is a numerical technique applied in the realm of initial value problems for ordinary differential equations (ODEs). The simplicity of this method makes it a popular choice in cases where the differential equation lacks a closed-form solution. The method might not always provide the most accurate result, but it offers a good trade-off between simplicity and accuracy.

Mathematical Formulation

Consider an initial value problem (IVP) represented as:

$$u' = f(t, u),$$ $$u(t_0) = u_0.$$

This IVP can be solved by Euler's method, where the method employs the following approximation:

$$u_{n+1} = u_n + h*f(t_n, u_n),$$

where: - $u_{n+1}$ is the approximate value of $u$ at $t = t_n + h$, - $u_n$ is the approximate value of $u$ at $t = t_n$, - $h$ is the step size, - $f(t_n, u_n)$ is the derivative of $u$ at $t = t_n$.

Derivation

Let's start with the Taylor series:

$$ u(t+h)=u(t)+h u'(t) + O(h^2) $$

We may alternatively rewrite the above equation as follows:

$$ u(t+h)=u(t)+ h f(u(t),t)+ O(h^2).$$

Which is roughly equivalent to:

$$ u(t+h)=u(t)+ h f(u(t),t)$$

Algorithm Steps

  1. Start with initial conditions $t_0$ and $u_0$.
  2. Calculate $u_{n+1}$ using the formula: $u_{n+1} = u_n + h*f(t_n, u_n)$.
  3. Repeat the above step for a given number of steps or until the final value of $t$ is reached.

Example

$$ u'(t)=u(t),$$

$$ u(0)=1$$

$$u(0.1)=?$$

Let's choose the step value: $h = 0.05$

We start at $t=0$:

$$ u(0.05) \approx u(0)+0.05u'(0) $$

$$ u(0.05) \approx1+0.05u(0) $$

$$ u(0.05) \approx1+0.05 \cdot 1 $$

$$ u(0.05) \approx 1.05 $$

Now that we know $u(0.05)$, we can calculate the second step:

$$ u(0.1) \approx u(0.05)+0.05u'(0.05) $$

$$ u(0.1) \approx1.05+0.05u(0.05) $$

$$ u(0.1) \approx1.05+0.05 \cdot 1.05 $$

$$ u(0.1) \approx 1.1025 $$

Advantages

Limitations

Table of Contents

    Euler's Method
    1. Mathematical Formulation
    2. Derivation
    3. Algorithm Steps
    4. Example
    5. Advantages
    6. Limitations