2

How can one show that the forward Euler always under-estimates the exact solution provided that $αΔt<1$ and $αΔt≠0$. When

$$\frac{dy}{dt} =-αy$$

My solution

I found the exact solution to be as $$y(t)=e^{-αt}$$ and then showed $(1-αΔt)$ $\le$ $e^{-αt}$. Am I right?

Any help with be appreciated

etet112
  • 443
  • 2
    Are you supposed to show it underestimates it at every time, i.e. even after additional steps? Or just that the first step underestimates the solution? For the latter, you're basically good, except that the RHS of your equation should be $e^{-\alpha \Delta t}$, and except that you assumed that $y(0)=1$ when actually it wasn't given. – Ian Sep 18 '17 at 13:43

1 Answers1

2

Your solution is not correct.

Hint: The exact solution is $$y(t) = e^{-\alpha t} y(0).$$ The Forward-Euler scheme is given by the recurrence $y_{n+1} = ( 1 - \alpha \Delta t) y_n$ and initial condition $y_0 = y(0)$. This recurrence can be solved to obtain $$y_n = (1 - \alpha \Delta t)^n y(0).$$ Letting $x = \alpha \Delta t$, note that \begin{align*} y_n - y(n \Delta t) & = (1 - x)^n y(0) - e^{- n x} y(0) \\ & = \left( (1 - x)^n - e^{- n x} \right) y(0). \end{align*} Now, you have to show that $$(1 - x)^n - e^{- n x} \leq 0 \qquad \text{for } x \in (0,1) \text{ and } n \in \mathbb{N}. \tag{1}$$ Assuming $\text{(1)}$, you can conclude the following:

  • If $y(0) < 0$, the Forward-Euler scheme is an overestimator.
  • If $y(0) = 0$, the Forward-Euler scheme gives the exact (trivial) solution.
  • If $y(0) > 0$, the Forward-Euler scheme is an underestimator.
parsiad
  • 25,154