# Particular order for a BVP problem with finite differences

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.

• $|x|^{5/2}$ is twice continuously differentiable on $[-1,1]$. The first and second derivatives at $x=0$ are both zero. More generally, the solution is $C^{2,1/2}$, so $u''$ is $C^{0,1/2}$ (also known as Holder(1/2) continuous). I think this is the detail you need. – Ian Jan 10 '18 at 18:36
• Okay, but the fourth derivative (which is the one I need) of $|x|^{5/2}$ doesn't exists globally in $[-1,1]$. I'd like to find something about the local error $\tau_{i}^{2}$. – VoB Jan 10 '18 at 19:53
• The fourth derivative is what you would use to use the standard Taylor estimate, but really what's going on there is an estimate of $\frac{f(x+h)+f(x-h)-2f(x)}{h^2}-f''(x)$. That estimate is related primarily to the regularity of $f''(x)$. If $f''$ is $C^2$ then it is $O(h^2)$, but otherwise it is gated by the regularity of $f''$ itself, which in this case is $C^{0,1/2}$. – Ian Jan 10 '18 at 19:55
• Also the main reason I give this nitpick is because your actual title makes it sound like the solution here is weak, when in fact it is not: all derivatives actually used in the equation exist in the solution. Just the higher derivatives that you want to use for estimates of numerical error fail to exist. – Ian Jan 10 '18 at 19:57
• Anyway, to proceed you generally have $f(x+h)=f(x)+hf'(x)+h^2f''(x)/2+O(h^{5/2})$ for your true solution. This would imply that $f''(x)$ is approximated with an $O(h^{1/2})$ error by your difference scheme. The subtle question as usual is whether the $O(h^{5/2})$ terms in $f(x+h)$ and $f(x-h)$ cancel out. If they do, then you can get better error than order 1/2, which appears to be happening based on your plots. – Ian Jan 10 '18 at 20:02