# Second Order forward finite difference scheme

Show that

$d^2u/dx^2(x_i)=[(-u_{i+3})+(4u_{i+2})-(5u_{i+1})+2u_i]/h^2 +O(h^2)$

provided all terms in the expression are well defined is a second order finite difference scheme for second order derivative.

I know how to approach this question. I know I use the taylor expression and everything but I don't know which formula to use. For example, I know that when finding a first order derivative I use

$u'(x)=au(x+2h)+bu(x+h)+cu(x)+du(x-h)+eu(x-2h)$.

I've already solved a question using this but I don't know what the formula is for $u''(x)$.

Would it be something like

$u''(x)=au(x+3h)+bu(x+2h)+cu(x+h)+du(x)+eu(x-3h)+fu(x-2h)+g(x-h) ?$

I'm pretty sure this is wrong because I don't think I should have this many unknowns. What equation should I be using?

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I assume you meant to use subscripts in the above such as $u_{i+2}$ instead of $ui+2$ –  par Oct 3 '13 at 15:41
Yes I did. I just didn't know how to do that here. –  Ruth Gutierrez Oct 3 '13 at 15:42
Click edit and take a look at the syntax for future reference. I am currently working on an answer to your question. –  par Oct 3 '13 at 15:44

Substitute a smooth solution $u$ into the finite difference scheme to get \begin{align*} & \phantom{=}\frac{-u\left(x+3h\right)+4u\left(x+2h\right)-5u\left(x+h\right)+2u\left(x\right)}{h^{2}}\\ & =-\frac{1}{h^{2}}\left[u\left(x\right)+3hu^{\prime}\left(x\right)+\frac{9}{2}h^{2}u^{\prime\prime}\left(x\right)+\frac{9}{2}h^{3}u^{\prime\prime\prime}\left(x\right)+O\left(h^{4}\right)\right]\\ & \qquad+\frac{4}{h^{2}}\left[u\left(x\right)+2hu^{\prime}\left(x\right)+2h^{2}u^{\prime\prime}\left(x\right)+\frac{4}{3}h^{3}u^{\prime\prime\prime}\left(x\right)+O\left(h^{4}\right)\right]\\ & \qquad-\frac{5}{h^{2}}\left[u\left(x\right)+hu^{\prime}\left(x\right)+\frac{1}{2}h^{2}u^{\prime\prime}\left(x\right)+\frac{1}{6}h^{3}u^{\prime\prime\prime}\left(x\right)+O\left(h^{4}\right)\right]\\ & \qquad+\frac{2}{h^{2}}u\left(x\right).\\ & =\frac{u\left(x\right)}{h^{2}}\underbrace{\left[-1+4-5+2\right]}_{0}+\frac{u^{\prime}\left(x\right)}{h}\underbrace{\left[-3+8-5\right]}_{0}\\ & \qquad+u^{\prime\prime}\left(x\right)\underbrace{\left[-\frac{9}{2}+8-\frac{5}{2}\right]}_{1}+u^{\prime\prime\prime}\left(x\right)h\underbrace{\left[-\frac{9}{2}+\frac{16}{3}-\frac{5}{6}\right]}_{0}+O\left(h^{2}\right)\\ &=u^{\prime\prime}\left(x\right)+O\left(h^2\right) \end{align*} as desired.

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Oh so here I don't have to use a formula to find the coefficients like you can do for a first order derivative? –  Ruth Gutierrez Oct 3 '13 at 16:42
The case for a first-order derivative and that of a second-order derivative are the same. The coefficients in the discretization (i.e. -1, 4, -5 and 2) are already given to you. You are required to show that if you substitute a smooth solution (as I did above), the error in the approximation is $O\left(h^2\right)$. –  par Oct 3 '13 at 17:16
Oh ok. Yes I see. It makes sense. Thank you so much. –  Ruth Gutierrez Oct 3 '13 at 17:36
If it is satisfactory, could you accept my answer? Click the checkmark. –  par Oct 3 '13 at 22:09