162

数理解析研究所講究録
第2082巻 2018年 162-167

Exact blow‐up profile for a heat equation with a nonlinear
boundary condition
Junichi Harada

The Education and Human Studies,
Akita University
We study the asymptotic behavior of positive blow‐up solutions for the heat equation.

\left{\begin{ar y}{l u_{\mathr {}=$\Delta$u,&x\in mathb{R}_+^{n},t\i (0,T)\ partil_{$\nu}=^{q},&x\inpartil\mathb{R}_+^{n},t\i(0,T)\ u=_{0},&x\in mathb{R}_+^{n},t=0 \end{ar y}\ight.

(P)\{x = (x', x_{n}) \in \mathbb{R}^{n-1} \times \mathbb{R};x_{n} > 0\}, \partial_{$\nu$} -\partial/\partial x_{n} and 1 < q < n/(n-2) . \mathrm{A}function u(x, t) is said to be x_{n} ‐mal symmetric, if u(x, t) depends only on |x'| and x_{n} for any
t \in [0, T) . We focus on x_{n} ‐mal symmetric functions throughout this report. Let u(x, t) be ablow‐up solution of (P) and T be its blow‐up time. If a limit
where \mathb {R}_{+}^{n}==U(x)=t\rightarrow T\mathrm{h}\mathrm{m}u(x, t)\in [0, \infty]
exists for any x\in \mathbb{R}_{+}^{n}\cup\partial \mathbb{R}_{+}^{n} , we call U(x) a blow‐up profile of u(x, t) .

U(x)\partial \mathb {R}_{+}^{n}\prime x_{n}A goal of this report is to determine the singularity of blow‐up profiles. The blow‐up profilefor (P) was first studied in [4, 5] for a one dimensional case. Indeed they constructed blow‐up
solutions satisfying

U(x)=Ax^{\frac{-1}{q-1}}

for x\in \mathbb{R}_{+} .

(1)

The author [6] extended their results to a multidimensional case and proved that if u(x, t) is apositive x_{n} ‐axial symmetric blow‐up solution, its blow‐up profile is given by

U(x)=A(1+o(1))(\cos $\theta$)^{\frac{-1}{q-1}|X|^{\frac{-1}{q-1}}}along x_{n}=|x|\cdot\cos $\theta$(2)

for any fixed $\theta$ \in [0, $\pi$/2 ). This profile function coincides with (1) if $\theta$= 0 . Unfortunately (2)does not hold on the boundary, since $\theta$= $\pi$/2 on the boundary. On the other hand, the boundary

singularity was studied in [7]. He obtained the following inequalities under some monotonicity
conditions.

\mathrm{k}_{1}(|\log|x| /|x|^{2})^{\frac{1}{2(q-1)} \leq U(x)\leq \mathrm{k}_{2}(|\log|x| /|x|^{2})^{\frac{1}{2(\mathrm{q}-1)} , x\in\partial \mathbb{R}_{+}^{n}, |x| <1.

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IJn this report, we improve this bound and consider more general situations.
Let u(x, t) be a positive x_{n} ‐axial symmetric solution which blows up at the origin. We

introduce self‐similar variables.

w(y, s)=(T-t)^{\frac{1}{2(q-1)}}u(\sqrt{T-t}\cdot y, t) , T-t=e^{-s}.
The rescaled function w(y, s) satisfies\left\{ begin{ar ay}{l} w_{s}=\triangle\mathrm{w}-\simeq\cdot\nablaw2-\frac{w}{2(q-1)},&y\in\mathb {R}_{+}^{n},s>-\logT,\ \partial_{\mathrm{v}w=w^{q},&y\in\partial\mathb {R}_{+}^{n},s>-\logT. \end{ar ay}\right.
It is known that \mathrm{w}(y, s) converges to the $\kappa$(y) as s \rightarrow \infty , where $\kappa$(y)positive bounded solution of=$\kappa$(y_{n}) is the unique\displaystyle \frac{d^{2} $\kap a$}{dy_{n}^{2} -\#_{2}\mathrm{p}_{\frac{d $\kap a$}{dy_{n} -\frac{ $\kap a$}{2(q-1)} =0(3)(see [3], [2]). To investigate the asymptotic behavior of solutions, we consider the linearization

around $\kappa$(y) .

$\Delta \phi$-u2^{\cdot}\displaystyle \nabla $\phi$-\frac{ $\phi$}{2(q-1)}=- $\lambda \phi$ in \mathb {R}_{+}^{n},\partial_{ $\nu$} $\phi$=q$\kappa$^{q-1} $\phi$ on \partial \mathb {R}_{+}^{n} .

(4)Under x_{n} ‐axisl symmetric case setting, the eigenfunction is written as$\phi$_{k,l}(y)=h_{k}(|y'|)I_{l}(y_{n}) , k=0, 1, 2, \cdots , l\in \mathrm{N}.
Let $\lambda$_{k}^{\mathrm{R}^{n-1} be the kth eigenvalue corresponding h_{k}(|y'|) and $\lambda$_{l}^{\mathb {R}_{+} be the lth eigenvalue corre‐

sponding I_{l}(y_{n}) . The eigenvalue of (4) is given by $\lambda$_{k,l}

=

that

$\lambda$_{0,1} <0,$\lambda$_{1,1}=0,$\lambda$_{k,l}>0$\lambda$_{k}^{\mathb {R}^{n-1} +$\lambda$_{l}^{\mathb {R}+} +\displaystyle \frac{1}{2(q-1)} . It is known( k\geq 2 or l\geq 2 ).

Let L_{ $\rho$}^{2}(\mathb {R}_{+}^{n}) be a weighted Lebesgue space defined by L_{ $\rho$}^{2}(\mathb {R}_{+}^{n})=(\displaystyle \int_{\mathbb{R}_{+}^{n} w(y)^{2}e^{-|y|^{2}/4}dy)^{1/2}<\infty\} . We recall our previous result.

\{w \in L_{1\mathrm{o}\mathrm{c} ^{1}(\mathb {R}_{+}^{n});\Vert w\Vert_{L_{ $\rho$}^{P}(\mathb {R}_{+}^{n})=

Theorem 1 (Theorem 1.1 [7]). Let u(x, t) be a positive x_{n} ‐axial symmetric solution of(P) whichblows up at the origin. Then one of the following two cases occurs.

(c1) \Vert w(s) -( $\kappa$+\mathrm{c}_{1}s^{-1}$\phi$_{1,1})\Vert_{L_{ $\rho$}^{2}(\mathrm{R}_{+}^{n})}n, q=o(s^{-1}) , where \mathrm{c}_{1} < 0 is a constant depending only onor(c2) w(s)- $\kappa$ decays exponentially in L_{ $\rho$}^{2}(\mathb {R}_{+}^{n}) .
We now state our main result in this report.

Theorem 2. Let u(x, t) be a positive x_{n} ‐axial symmetric solution of (P) which blows up at
the origin.

(i) If w(y, s) behaves as (c1) in Theorem 1, then the blow‐up profile U(x) \in C(\mathbb{R}_{+}^{n}\cup\partial \mathbb{R}_{+}^{n}\backslash\{0\}) exists and satisfies

U(x)=\mathrm{k}(|\log|x| /|x|^{2})^{\frac{1}{2(\mathrm{q}-1)}}(1+o(1)) , x\in\partial \mathbb{R}_{+}^{n}, |x|<1,
where \mathrm{k} is a positive constant depending only on n and q.

(ii) If w(y, s) behaves as (c2) in Theorem 1, thenU(x)=\mathrm{k}|x|^{-\frac{k}{(\mathrm{q}-1)}}(1+o(1)) , x\in\partial \mathbb{R}_{+}^{n}, |x| <1
for some k\geq 2.

Generally a blow‐up solution behaves likew(s)= $\kappa$+\mathrm{c}_{1}e^{-$\lambda$_{kl^{S} }$\phi$_{k,l}+o(e^{-$\lambda$_{kl^{S} })in

L_{ $\rho$}^{2}(\mathb {R}_{+}^{n})164

for some k=0 , 1, 2,

and l\in \mathrm{N} . Unfortunately we have no idea about the case l\geq 2.

\cdotsOur proof is a slight modification of arguments in a series of papers [8, 9, 10, 11, 12]. We
here focus on the asymptotic behavior of w(y, s) for large |y| . We introduce another rescale.

W(z, $\zeta$, s)=w(\mathrm{R}(s)z, $\zeta$, s)for z\in \mathbb{R}^{n-1},$\zeta$>0.

The function \mathrm{R}(s) (\displaystyle \lim_{s\rightarrow\infty}\mathrm{R}(s) = \infty) is determined by the long time behavior of \mathrm{w}(y, s) . \mathrm{A}goal is to determine the asymptotic behavior of W(z, $\zeta$, s) for |z|+| $\zeta$| <R as s\rightarrow\infty . We seethat

W(z, $\zeta$, s) satisfies\left\{ begin{ar ay}{l} W_{s}=\mathrm{R}^{-2}\triangle_{z}W-(\frac{1}{2}-\mathrm{R}\vec{\mathrm{R} )z\cdot\nabla_{z}W+W_{$\zeta\zeta$}-\frac{$\zeta$}{2}W_{$\zeta$}-mW,&z\in\mathb {R}^{n-1}, $\zeta$>0,s>-\logT,\ \partial_{$\nu$}W=W^{q},&z\in\mathb {R}^{n-1}, $\zeta$=0,s>-\logT. \end{ar ay}\right.
To investigate the long time behavior of W(z, $\zeta$, s) , we study a limiting problem of this equation.
For the case (i), since w(y, s)$\kappa$(y)+\mathrm{c}_{1}s^{-1}$\phi$_{1,1}(y)+o(s^{-1}) in L_{ $\rho$}^{2}(\mathb {R}_{+}^{n}) (see Theorem 1 (c1))
and $\phi$_{1,1}(y) =c(|y'|^{2}-2(n-1))I_{1} (yn), we take=\mathrm{R}(s)=\sqrt{s}.
A limiting equation is given by\left\{ begin{ar ay}{l -\frac{1}{2}z\cdot\nabla_{z}W+W_{$\zeta\zeta$}-4_{W_{$\zeta$}2-mW=0&\mathrm{f}\mathrm{o}\mathrm{r}z\in\mathb {R}^{n-1}, $\zeta$>0,\ \partial_{$\nu$}W=W^{2}&\mathrm{o}\mathrm{n}z\in\mathb {R}^{n-1}, $\zeta$=0. \end{ar ay}\right.

(5)

W(z, $\zeta$)=(1+\displaystyle \mathrm{b}|z|^{2})^{-m} $\kap a$(\frac{ $\zeta$}{\sqrt{1+\mathrm{b}|z|^{2} })(6)

We will see thatgives a solution of (5) (see below), where \mathrm{b}>0 is a free parameter. We choose \mathrm{b} to match (c1)in Theorem 1. Then W(z, $\zeta$, s) gives a better approximation of w(y, s) than $\kappa$+\mathrm{c}_{1}s^{-1}$\phi$_{1,1}.

From now, we derive (6) in a formal approach. Since z\cdot\nabla_{z}|z|^{2j}solutions in the form of

Substituting this into (5), we have=

2k|z|^{2_{J}} , we look forW(z, $\zeta$)=\displaystyle\sum_{$\gam a$=0}^{\infty}a_{J}($\zeta$)|z^{2j}.

a_{J}''-4_{a_{J}'}2-(m+j)a_{j}=0, $\zeta$>0.
Let A_{J}( $\eta$)=a_{J}( $\zeta$) with$\eta$=\mathrm{L}^{2}4 for $\eta$>0 . Then A_{J} satisfies$\eta$ A_{J}''+(\displaystyle \frac{1}{2}- $\eta$)A_{J}'-(m+j)A_{j}=0, $\eta$>0.

This is the confluent hypergeometric equation (see (8.1.1) p.

[1] ). One solution of this201

equation is give by the Kummer function M(m+j, \displaystyle \frac{1}{2}; $\eta$) , another is given by

U(m+j, \displaystyle \frac{1}{2}; $\eta$)=\frac{1}{ $\Gamma$(m+J)}\int_{0}^{\infty}e^{- $\eta$ t}t^{m+J-1}(1+t)^{\frac{1}{2}-(m+J)-1}dt(see (8.2. 1) p. 204 [1]). Since M(m+j, \displaystyle \frac{1}{2}; $\eta$) behaves like $\eta$^{m+J-\frac{1}{2} e^{ $\eta$} as $\eta$\rightarrow \infty (see (8. 1.8) \mathrm{p}.

203 [1]) , we take

A_{J}( $\eta$)=$\alpha$_{J}$\pi$^{-\frac{1}{2} $\Gamma$(m+j+\displaystyle \frac{1}{2})U(m+j, \frac{1}{2}, $\eta$),

where $\alpha$_{J} is a constant. Since U(a, c;0)=\displaystyle \frac{ $\Gamma$(1-c)}{ $\Gamma$(a+1-c)} if 1-c>0 (see (8.2.5) p. 205 [1]), it is clear
that

A_{J}(0)=$\alpha$_{J} .

(7)

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Form now, we determine $\alpha$_{J} . We substituteW(z, $\zeta$)=\displaystyle\sum_{=J0}^{\infty}A_{J}(_{4}^{\mathrm{L}^{2} )|z^{2_{J}
into a boundary condition in (5). Since U'(a, c;x)=-aU(a+1, c+1, x) (see (8.5.8) p. 212 [1])and U(a, c;x)=\displaystyle \frac{ $\Gamma$(\mathrm{c}-1)}{ $\Gamma$(a)}x^{1-c}(1+o(1) as x\rightarrow 0+\mathrm{i}\mathrm{f}1-c<0 (see p. 205 [1]), we get\displaystyle \frac{d}{d $\zeta$}U(m+j, \frac{1}{2}, \frac{$\zeta$^{2} {4})|_{ $\zeta$=0} = \frac{ $\zeta$}{2}U'(m+j, \frac{1}{2}, \frac{$\zeta$^{2} {4})|_{ $\zeta$=0}= -\displaystyle \frac{(m+J) $\zeta$}{2}U(m+j+1, \frac{3}{2}, $\zeta$_{\frac{2}{4} )|_{ $\zeta$=0}= -\displaystyle \frac{(m+J) $\zeta$}{2}\frac{ $\Gamma$(\frac{1}{2}) { $\Gamma$(m+J+1)}(\frac{$\zeta$^{2} {4})^{-\frac{1}{2} (1+o(1) |_{ $\zeta$=0}= -(m+j)\displaystyle \frac{ $\Gamma$(\frac{1}{2}) { $\Gamma$(m+j+1)}=-\frac{ $\Gamma$(\frac{1}{2}) { $\Gamma$(m+J)}.
Therefore it holds that\displaystyle \frac{d}{d $\zeta$}A_{J}(_{4}^{\mathrm{L}^{2} )|_{ $\zeta$=0=}-$\alpha$_{J}$\pi$^{-\frac{1}{2} $\Gam a$(m+j \frac{1}{2})_{\vec{ $\Gam a$(m+J}\overline{)} ^{ $\Gam a$(^{\underline{1} )}=-$\alpha$_{J\vec{ $\Gam a$(m+J)} ^{ $\Gam a$(m+J+^{\underline{1} )}.
From now we consider the case q=2 . Since

m=\displaystyle \frac{1}{2(q-1)}=\frac{1}{2} , it follows that\displayst le\partial_{$\nu$}W=-\partial_{$\zeta$}W|_{$\zeta$=0}=\sum_{J^=0}^{\infty}$\alpha$_{g}\frac{$\Gam a$(m+J \frac{1}2)}{$\Gam a$(m+J)}|z^{2j}=\sum_{J=0}^{\infty}$\alpha$_{J^\frac{$\Gam a$(}{$\Gam a$(}$\Gam a$\frac{j)}J |Z^{2_J} ^{1+}\overline{2}^{+}.

(8)

On the other hand, it holds from (7) that

W|_{ $\zeta$=0}^{2}($\alpha$_{0}+$\alpha$_{1}|z|^{2}+$\alpha$_{2}|z|^{4}+$\alpha$_{3}|z|^{6}+\cdots)(a_{0}+$\alpha$_{1}|z|^{2}+$\alpha$_{2}|z|^{4}+$\alpha$_{3}|z|^{6}+\cdots)= $\alpha$_{0}^{2}+2 $\alpha$ 0$\alpha$_{1}|z|^{2}+(2$\alpha$_{0}$\alpha$_{2}+$\alpha$_{1}^{2})|z|^{4}+(2$\alpha$_{0}$\alpha$_{3}+2$\alpha$_{1}$\alpha$_{2})|z|^{6}+(2$\alpha$_{0}$\alpha$_{4}+2$\alpha$_{1}$\alpha$_{3}+$\alpha$_{2}^{2})|z|^{8}+(2$\alpha$_{0}$\alpha$_{5}+2$\alpha$_{1}$\alpha$_{4}+2$\alpha$_{2}$\alpha$_{3})|z|^{10}+\cdots=

Therefore from (8) and (9), we obtain$\alpha$_{0_{$\Gam a$(_{\overline{2} \overline{)} ^{$\Gam a$(1)}\displaystyle\neg=$\alpha$_{0}^{2},$\alpha$_{1}\frac{$\Gam a$(2)}{$\Gam a$(\frac{3}{2}) =2$\alpha$_{0}$\alpha$_{1},$\alpha$_{2_{(}^{\frac{$\Gam a$}{$\Gam a$} \mathrm{T}_{\frac{3}{2} ^{\frac{)} } ^{(}=2$\alpha$_{0}$\alpha$_{2}+$\alpha$_{1}^{2},$\alpha$_{3_{ $\Gamma$(_{\overline{2} \ulcorner,)}^{ $\Gamma$ 4)}=2$\alpha$_{0}$\alpha$_{3}+2$\alpha$_{1}$\alpha$_{2}, $\alpha$ 4\displaystyle \frac{ $\Gamma$(5)}{ $\Gamma$(\frac{9}{2}) =2$\alpha$_{0}a_{4}+2$\alpha$_{1}$\alpha$_{3}+$\alpha$_{2}^{2},$\alpha$_{5}\displaystyle \frac{ $\Gam a$(5)}{ $\Gam a$(_{2}^{2}) =2$\alpha$_{0}$\alpha$_{5}+2$\alpha$_{1}$\alpha$_{4}+2$\alpha$_{2}$\alpha$_{3}.
It is clear that$\alpha$_{0}=\displaystyle\frac{$\Gam a$(1)}{$\Gam a$(\frac{1}{2})=$\pi$^{-\frac{1}{2} . For simplicity, we put $\alpha$_{J} =$\pi$^{-\frac{1}{2} $\beta$_{J} . Then we get(\displaystyle \frac{2.!}{\frac{3}{2}\frac{1}{2} -2)$\beta$_{2}=$\beta$_{1}^{2}, (_{$\tau$_{\frac{3 !}{2} $\tau$_{\overline{2} ,\overline{2} -2)$\beta$_{3}=2$\beta$_{1}$\beta$_{2}, (\frac{4'}{\frac{7}{2}.\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2} -2)$\beta$_{4}=2$\beta$_{1}$\beta$_{3}+$\beta$_{2}^{2},(_{\ovalbox{\t \smal REJECT}^{5!},\overline{2}\overline{2}\overline{2}\overline{2}\overline{2} -2)$\beta$_{5}=2$\beta$_{1}$\beta$_{4}+2$\beta$_{2}$\beta$_{3}.
Furthermore we put $\beta$_{J} =(2$\beta$_{1})^{j}$\gamma$_{J} . We easily see that\displaystyle \frac{8}{3}$\gamma$_{2}=1, \frac{6}{5}$\gamma$_{3}=$\gamma$_{2}, \frac{58}{35}$\gamma$_{4}=$\gamma$_{3}+$\gamma$_{2}^{2}, \frac{130}{63}$\gamma$_{5}=$\gamma$_{4}+2$\gamma$_{2}$\gamma$_{3}.
This implies$\gam a$_{2}=\displaystyle\frac{3}{8}=\frac{\frac{1}{2}.\frac{3}{2} {2!},$\gam a$_{3}=\displaystyle\frac{5}{16}=\frac{\frac{5}{2}.\frac{3}{2}\cdot\frac{1}{2} {3!},$\gam a$_{4}=\displaystyle\frac{35}{128}=\frac{\frac{7}{2}.\frac{5}{2}.\frac{3}{2}\cdot\frac{1}{2} {4!},$\gam a$_{5}=\displaystyle\frac{63}{256}=\ovalbox{\t\smal REJ CT}_{5!}2^{\underline{7}\underline{5}\underline{3}\underline{1}.

(9)

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From this relation, we assume$\gamma$_{J}=\displaystyle \frac{-\perp^{-\underline{1} \equiv 2}{J!}=$\pi$^{-\frac{1}{2} \frac{ $\Gamma$(j+^{\underline{1} )}{J!}.
Since $\alpha$_{J} =$\pi$^{-1}(2$\beta$_{1})^{\mathrm{J} $\gamma$_{J}, W(z, $\zeta$) is written byW(z, $\zeta$)\displ\diaystslple\asystqrt{y$\lepi$\t}^i{m$\esGamU(a\$f(r1a)Uc{(\1fr}a{c2{1}}+2j,,\\ffrarca{1c{}21};{$2\}e;ta$\et$)}\unad$)erl|izn|e^{{$2\ja}lpha$}_{\mathrm{L}+\frac{$\alpha$_{1}{\sqrt{$\pi$} $\Gam a$(2)U(\frac{3}2,\frac{1}2;$\eta$)|z^{2}+\frac{1}\sqrt{$\pi$}\sum_{J=2}^{\infty}$\alpha$_{J}$\Gam a$(1+j)=\displaystyle\frac{1}{$\pi$}U(\frac{1}{2},\frac{1}{2};$\eta$)+\frac{$\beta$_{1}{$\pi$}U(\frac{3}{2},\frac{1}{2})$\eta$)|z^{21}+-$\Gam a$_{J}\sum_{=2}^{\infty}$\Gam a$(j+\frac{1}{2})U(\frac{1}{2}+j$\pi$,\frac{1}{2};$\eta$)(2$\beta$_{1})^{$\gam a$}|z^{2_{J}=\displaystyle\mathrm{B}_{$\pi$}^{1}\sum_{J^{=0}^{\infty}$\Gam a$(j+\frac{1}{2})U(\frac{1}{2}+j,\frac{1}{2};$\eta$)(2$\beta$_{1})^{\mathcal{J}|z^{2_{J}.
=(10)

We recall the following formula (see p. 219 [1]).

U(a,\displayst le\frac{1}2,\mathrm{L}^{2}4)=\infty$\Gam a$(a+_{\overline{2}\sqrt{$\pi$})$\Gam a$(a)\sum_{$\iota$=0}^{\infty}(-1)^{l}\frac{$\Gam a$(a+\frac{l}2)}{l!$\zeta$^{l}

.

(11)

We apply this formula in (10).

W(z, $\zeta$)=\displaystyle\frac{1}{$\pi$^{3}2 \sum_{J^{=0} ^{\infty}$\Gam a$(j+\frac{1}{2})U(\frac{1}{2}+j,\frac{1}{2};$\eta$)(2$\beta$_{1})^{\mathrm{J} |z^{2_{J}=\displayst le\frac{1} $\pi$}\sum_{=J0}^{\infty}\sum_{l=0}^{\infty}(-1)^{l}\frac{$\Gam a$(\frac{1}2+\frac{l}2+J)}{ !\cdotl!}(2$\beta$_{1})^{\mathrm{J}|z^{2_J}$\zeta$^{l}.
Since(1-x)^{-p}=1+\displaystyle \frac{ $\Gamma$(p+1)}{ $\Gamma$(p)}x+\frac{ $\Gamma$(p+2)}{2!\cdot $\Gamma$(p)}x^{2}+\cdots\frac{ $\Gamma$(p+n)}{n!\cdot $\Gamma$(p)}x^{n}+\cdots,

we obtain

W(z, $\zeta$)=\displayst le\frac{1} $\pi$}\sum_{J^=0}^{\infty}\sum_{l=0}^{\infty}\frac{$\Gam a$(\frac{1}2+\frac{l}2+J)}{J!l}(2$\beta$_{1})^{j}|z^{2_{J}$\zeta$^{l}=\displaystle\frac{1}$\pi}\sum_{l=0}(-1)_{l!}^$\iota$_{-\mapsto^{+)}$\zeta$^{l}\sum_{J}^$\Gam a$(+)}(2$\beta$_{1}|z^{2}) \mathrm{J} ^{$\Gam a$(^{\underlin{1}\underlin{$\iota$}\underlin{1}\underlin{} \infty\inftyJ^{=0}\infty!$\Gam a$(\frac{1}2+\frac{$\iota$^{J} 2)=\displaystyle\frac{1}{$\pi$}\sum_{l=0}^{\infty}(-1)^{l}\frac{$\Gam a$(\frac{1}{2}+\frac{l}2}){l!}$\zeta$^{l}(1-2$\beta$_{1}|z^{2})^{-(\frac{1}{2}+\frac{l}2})=\displayst le\frac{1} $\pi$}(1-2$\beta$_{1}|z^{2})^{-\frac{1}2}\sum_{l=0}^{\infty}(-1)^{l \underline{$\Gam a$} \mapsto(^{\underline{1}l!+^{\underline{l})(\frac{$\zeta$}{\sqrt{1-2$\beta$_{1}|z^{2} )^{l}.

We set$\eta$_{1}=\displaystyle \frac{$\zeta$^{2} {4(1-2$\beta$_{1}|z^{2}) . We again use (11) to getW(z, $\zeta$)=\displaystyle\frac{1}{$\pi$}(1-2$\beta$_{1}|z^{2})^{-\frac{1}{2} \sum_{l=0}^{\infty}(-1)^{l}\frac{$\Gam a$(\frac{1}{2}+\frac{l} 2}) {l!}(\frac{$\zeta$}{\sqrt{1-2$\beta$_{1}|z^{2} )^{l}= \displaystyle \frac{1}{ $\pi$}(1-2$\beta$_{1}|z|^{2})^{-\frac{1}{2} U(\frac{1}{2}, \frac{1}{2})$\eta$_{1}).

We put -2$\beta$_{1} =\mathrm{b} . We finally obtainW(z, $\zeta$)=\displaystyle \frac{1}{ $\pi$}(1+\mathrm{b}|z|^{2})^{-\frac{1}{2} U(\frac{1}{2}, \frac{1}{2})\tilde{ $\eta$}_{1}) , \overline{ $\eta$}_{1}=\frac{$\zeta$^{2} {4(1+\mathrm{b}|z|^{2})}.

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Let $\kappa$( $\zeta$) be given in (3). Since $\kap a$( $\zeta$)=\displaystyle \frac{1}{ $\pi$}U(\frac{1}{2}, \frac{1}{2}, $\varsigma$_{\frac{2}{4} ) (see (3.6) in [5]), we obtainW(z, $\zeta$)=(1+\displaystyle \mathrm{b}|z^{2})^{-\frac{1}{2} $\kap a$(\frac{ $\zeta$}{\sqrt{1+\mathrm{b}|z^{2} ).

By a direct computation, we can check this actually satisfies (5) with q=2 . By the same way,

we find thatW(z, $\zeta$)=(1+\mathrm{b}|z|^{2})^{-m} $\kappa$()\overline{\sqrt{1+\mathrm{b}|z|^{2} }.
gives a solution of (5) for any q> 1.
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