I'm stuck on proving the following result:
Let $y\in \mathcal{C}^6([a,b])$ be the solution of $Ly\equiv -y''(x) +p(x)y'(x)+q(x)y(x) = r(x)$, $x\in (a,b)$, with boundary condition $y(a) = \alpha, y(b) = \beta$. Suppose there exists $p_*$ and $q^*$ constants such that for every $x\in (a,b)$ $$ |p(x)|\leq p_*, \qquad 0<q^*<|q(x)|$$ Let $y_j^h$ denote the approximate solution to this problem via "standard" finite-difference method. Prove that for $h<2/p_*$ we have the error expansion $$y(x_j)-y_j^h = e(x_j)h^2 + \mathcal{O}(h^4) $$ Where $e(x)$ is the solution to the following BVP $$(Le)(x) = \phi(x):= -\frac{1}{12}y^{(4)}(x) + \frac{1}{6}p(x)y^{(3)}(x),\qquad x\in(a,b);\qquad e(a) = e(b) =0 $$
Using the Taylor expansion i can deduce that $L^h[y]-L[y] = h^2L[e] + \mathcal{O}(h^4)$ (where $L^h$ is the finite difference operator relative to $L$). From here on, I'm not sure how to continue
Anyway , my professor told me to perform something similar to the proof of the theorem 7.4.10 from Stoer, Bulirsch - Introduction to Numerical Analysis (third edition).
My problem is that, since in this case $q(x)$ could take negative values i am not allowed to use the theorem (7.4.7)
How can i continue?
Thanks in advance.
Postscript: The sentence "standard" finite difference method refers to this
Observe that for any smooth function $$ \frac{y(x+h)-y(x-h)}{2h}=y'(x)+\frac{h^2}{6}y'''(x)+O(h^4) $$ and $$ \frac{y(x+h)-2y(x)+y(x-h)}{h^2}=y''(x)+\frac{h^2}{12}y^{(4)}(x)+O(h^4) $$ so that $$ L^h[y](x_i)=L[y](x_i)-\frac{h^2}{12}y^{(4)}(x_i)+p(x_i)\frac{h^2}{6}y'''(x_i)+O(h^4) $$ Now also $L[y](x_i)=r(x_i)=L^h[y^h](x_i)$, so that for the leading term in $y(x)-y^h(x)=e(x)h^2+O(h^4)$ one gets indeed $$ L^h[e](x)=-\frac1{12}y^{(4)}(x) +p(x)\frac1{6}y'''(x) $$ By construction the boundary conditions for $y$ and $y^h$ are the same, resulting in homogeneous boundary conditions for $e$.