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I found this method in this book for solving the Poisson equation with error converging with $\mathcal{O}(h^4)$. However, when I try to implement it for the 1D equation $-u''(x)=f(x)$, it only converges with around $\mathcal{O}(h^2)$. The math is as follows:

Using Taylor Series, we get $$\frac{1}{h^2}[-u_{i-1} + 2u_i - u_{i+1}] = -u''_i-\frac{h^2}{12}u^{(4)}_i + \mathcal{O}(h^4).$$ Substituting in $$-u^{(4)} = \frac{1}{h^2}[-u''_{i-1} + 2u''_i - u''_{i+1}] + \mathcal{O}(h^2)$$ we get $$\frac{1}{h^2}[-u_{i-1} + 2u_i - u_{i+1}] = f_i+\frac{1}{12}(f_{i-1} -2f_i +f_{1+1})_i + \mathcal{O}(h^4)$$ where $u_i =u(x_i)$ and $f_i = f(x_i)$.

This means that solving the system $$\frac{1}{h^2}Au = f-\frac{1}{12}Af$$ where $A$ is of the form $$ A = \left[ \begin{array}{c} 2&-1&0&0 \\-1&2&-1&0 \\ 0&-1&2&-1 \\ 0&0&-1&2\end{array}\right] $$ should result in an error that converges with $\mathcal{O}(h^4)$. But when I implement this (in Matlab), halving the step-size $h$ results in approximately quartering the error.

Is there something wrong with my math?

(If necessary, I can include the Matlab code here. But as of now I'm thinking the problem lies with the math.)

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I haven't checked the exact coefficients in your formulas, but there are typically two difficulties with this scheme:

  • Is your solution smooth enough to do the Taylor expansion to this order? For example, if your function $f$ is only once continuously differentiable, you can't expect the solution to be four times continuously differentiable.

  • What do you do with boundary conditions? That is, what do you use for $f_{i-1}$ at the left boundary, and similarly $f_{i+1}$ at the right?

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  • $\begingroup$ So, in the last few minutes that I've been playing with this, I've switched the function to $u(x) = \sin(2 \pi x)$, and now it works as intended (4th order error convergence), but before I was also using a $C^{\infty}$ function, $u(x) = xe^{x-1}-x$, and I can't figure out why that won't work. $\endgroup$ – jake Apr 15 '13 at 3:01
  • $\begingroup$ Oh, and the BCs are $u(0)=u(1)=0$. I should have mentioned that. $\endgroup$ – jake Apr 15 '13 at 3:03
  • $\begingroup$ @jake: The only issue I see is that your original $u$ seems incompatible with the right hand side of your equation system. $A$ is clearly designed with $u(0)=u(1)=0$ in mind. But you reuse this matrix for your right hand side. This is only OK if $f(0)=f(1)=0$ is true. But that doesn't seem right for $xe^{x-1}-x$ $\endgroup$ – sellibitze Apr 20 '19 at 11:12

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