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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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    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 $\xi(c)$?2011-04-11

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