I'm solving this BVP problem.
$u''(x) + u^2(x) = \frac{15}{4} |x|^{1/2} + |x|^5$, $u(-1)=u(1)=1$,
$x \in A=(-1,1)$.
The solution is $u(x)=|x|^{5/2}$. What can we say about the convergence order to the analytical solution?
I have to solve this problem with finite centred differences, of order two. My discretization for $$u''(x_i)=\frac{u(x_{i+1}) - 2u(x_i) + u(x_{i-1})}{h^2} + \tau^{2}_{i},\text{ with }\tau^{2}_{i} = \frac{h^2 u^{(4)}(\bar{x_i})}{12},$$ for an $\bar{x_i} \in (x_{i-1},x_{i+1})$.
Since $x \in A$, the analytical solution $u(x)$ is not globally differentiable in $A$. There should be the problem. Since the local error $\tau^{2}_{i}$ depends on the value of $u^{(4)}(\bar{xi})$, I computed it and get $u^{(4)}(x)=\frac{-15}{(16 x^{3/2})}$, assuming $x$ is real.
Particularly, for $x$ really close to $0$, it gets really high values, so the local error is no more a power of $h^2$, but it's scaled for a "non trascurable" constant and therefor it's no more of the second order.
Here is what I get numerically. The first image is the output of the BVP problem, while the second one is the error-graphic ($m$ is the number of nodes, so $h=\frac{2}{(m-1)}$)
Now I would like to say that the order is $\frac{3}{2}$, but I don't know how to go on. Maybe I should expand with Taylor the denominator(?).
What would you say?
P.S I'm Italian, hope to have written in a comprehensible way.