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I'm asked to prove that the finite difference approximation $$u_{xx}(x_i) = \frac{u_{i+1} - 2u_i + u_{i-1}}{\Delta x^2}$$ gives a discretization error of order $\mathcal{O}(\Delta x^2).$

My attempt:

I realize I want prove that $$u_{xx}(x_i) - \frac{u_{i+1} - 2u_i + u_{i-1}}{\Delta x^2} = \mathcal{O}(\Delta x^2).$$

I try to rewrite $u_{xx}(x_i):$ $$u_{i+1} =u_i+\Delta x u_{x(i)} + \frac{\Delta x^2}{2}u_{xx(i)} + \mathcal{O}(\Delta x^3).$$ $$u_{xx(i)} = \frac{2u_{i+1} - 2u_i - 2\Delta x u_{x(i)} - 2\mathcal{O}(\Delta x^3)}{\Delta x^2}$$ and when I calculcate the above difference using this I get
$$\frac{u_{i+1}}{\Delta x^2} - 2\frac{u_{x(i)}}{\Delta x} - \frac{u_{i-1}}{\Delta x^2} - 2\frac{\mathcal{O}(\Delta x^3)}{\Delta x^2}.$$

I can't get further and it seems wrong. Is my approach wrong and if so how should I do it instead?

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  • $\begingroup$ The difference formula is even symmetric in $Δx$, as a result the error, that is, the difference to the constant $u_{xx}(x_i)$, has the same symmetry, thus is a function of $Δx^2$, so the error has to be $O(Δx^2)$ for $u$ sufficiently smooth. $\endgroup$ Commented Aug 16, 2019 at 14:53

1 Answer 1

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You should consider one more term in the expansion: $$ u_{i\pm 1}=u_i \pm u'\Delta x+\frac{\Delta x^2}2u''\pm\frac{\Delta x^3}{6}u'''+O(\Delta x^4). $$

Then you will see that this term will cancel out and you will end up with $\ldots +O(\Delta x^4)/\Delta x^2=\ldots +O(\Delta x^2)$

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