[Math] Implicit Euler method and explicit Euler method

euler-mascheroni-constantnumerical methodsrunge-kutta-methods

I wanna know what is the difference between explicit Euler's method and implicit Euler's method. And is the local truncation error for both of them is $O(h)$ and the coefficient of the $O(h)$ term is $h/2$?

Best Answer

The error of both explicit and implicit Euler are $O(h)$. So

$$f(x-h) = f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots$$

and

$$f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$

So the backward Euler is

$$f(x) - f(x-h) = h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots$$

$$f'(x) = \frac{f(x) - f(x-h)}{h} + \frac{h}{2} f''(x) - \frac{h^2}{6} f'''(x) + \cdots$$

the backward Euler is first order accurate

$$f'(x) = \frac{f(x) - f(x-h)}{h} + O(h)$$

And the forward Euler is

$$f(x+h) - f(x) = h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$

the forward Euler is first order accurate

$$f'(x) = \frac{f(x+h) - f(x)}{h} + O(h)$$

We can do a central difference and find

$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) - (- h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots)$$

$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) + (h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots)$$

$$f(x+h) - f(x-h) = 2 h f'(x) + \frac{h^3}{6} f'''(x) + \cdots$$

$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} - \frac{h^2}{12} f'''(x) + \cdots$$

Therefore, the central difference is second order accurate.

$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} + O(h^2)$$

Related Solutions

[Math] Explicit formula for the implicit Euler method

With steplength $h$, the implicit Euler method for this equation would say $$y_{n+1}-y_n=hy_{n+1}^2.$$ Seeing this as a quadratic equation to solve for $y_{n+1}$, we get $$y_{n+1}=\frac{1\pm\sqrt{1-4hy_n}}{2h},$$ but you need to pick the correct sign. For concistency, you absolutely need to have $y_{n+1}$ converging to $y_n$ if you let $h\to0$, and that immediately eliminates the plus sign (which would have the right hand side going off to $\pm\infty$), leaving $$y_{n+1}=\frac{1-\sqrt{1-4hy_n}}{2h}$$ as the only sensible choice.

You may, and probably should, go further and verify that this is in fact consistent. This might involve taking a Taylor expansion of the square root, or perhaps my favourite trick for fractions of this kind: Multiplying numerator and denominator by $1+\sqrt{1-4hy_n}$. I'll stop here, as you do need to do some if this work on your own.

Explicit Euler method

This code implements the explicit Euler scheme.

In order to be implicit, there would be an instruction like

y = solve(z == y + h*f(x,z), z)

in the inner loop, where solve(equation, variable) is a suitable function to solve an equation (see here).

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[Math] Explicit formula for the implicit Euler method

Explicit Euler method

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