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Let $\alpha \geq 1$. Suppose that for each $c\geq c_0>0$ there exists a point $\xi (c) \in ]0,1[$ s.t. the BVP:

$$\begin{cases} [x^\alpha u^\prime (x)]^\prime +c\ u(x)=0 &\text{, in } ]\xi(c),1[ \\ u(\xi(c))=1,\ u(1)=0 \end{cases}$$

has a positive, strictly decreasing, convex solution in $[\xi (c),1]$ with $u^\prime (\xi (c))=-c\ \xi^{1-\alpha}(c)$; moreover, assume that $\xi (c_0)=0$ and $\xi(\cdot)$ is strictly increasing in $[c_0,+\infty[$ and $\displaystyle \lim_{c\to c_0} \xi(c)=0$.

The question is:

Is it possible to prove an estimate of the type:

$$\tag{1} \xi(c) \leq K\ (c-c_0)^\beta$$

with $K,\beta >0$ suitable constants?

N.B.: (1) says that the point $\xi(c)$ cannot approach $\xi (c_0)=0$ too slowly when $c$ approaches $c_0$.

AFAIK, an estimate of type (1) holds in the case $\alpha=1$ with $\beta=1/2$, and it can be recovered using the power series expansion of Bessel functions. So I was wondering if there's an analogous result for $\alpha \neq 1$ and, in the positive case, if there are references where I can read it.

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Are you sure you have $\xi(c)$ strictly decreasing? It should be increasing, no? – Willie Wong Apr 11 '11 at 0:28
Is the existence of such a $\xi(c)$ an assumption of the problem or a result that can be proven? – GWu Apr 11 '11 at 0:34
@Willie Wong: Yes, $\xi (c)$ is increasing; I'm going to edit that. @GWu: The existence of $\xi (c)$ can be proved; but the proof I know is not useful to prove the estimate. – Pacciu Apr 11 '11 at 1:25
Another comment: by assumption that $u$ is strictly decreasing and convex, you see that for $c = c_0$ ($\xi(c_0) = 0$), you must have that $u'(0) \leq -1$. This would contradict the requirement $u'(\xi(c)) = -c \xi^{1-\alpha}(c)$ if $\alpha < 1$. So it is necessary that $\alpha \geq 1$ for your boundary conditions to be satisfied. – Willie Wong Apr 11 '11 at 7:50
You also cannot have solutions for $\alpha \geq 2$. Notice that the boundary value condition requires $(x^\alpha u')|_{x = \xi(c_0)} = -c \xi(c_0) = 0$. Which means that we can write $u'(x) = -x^{-\alpha} \int_0^x cu(z) dz$. Using that $u$ is continuous and bounded away from zero, we have that $u'(x) \leq - C x^{1-\alpha}$. Integrating again gives a contradiction: if $\alpha \geq 2$, $u'$ is not integrable near 0, and $u$ cannot be bounded. So can you please clarify or include your existence proof for $\xi(c)$? – Willie Wong Apr 11 '11 at 10:56

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