数理解析研究所講究録 1261 巻 2002 年 140-149

140
Gevrey Regularity of Solutions of Semilinear Hypoelliptic
Equations on the Plane

Nguyen Minh TRI\S 1. Introduction.
In this note we discuss the Gevrey regularity (in particular, the analyticity)of solutions of semilinear ellptic degenerate equations of Grushin’s type on.
Most of the results will appear in [1]. Some results are new and they are presented
here for the first time. We confine ourself with consideration of amodel equation.
Precisely, we will consider the folowing equation$\mathrm{R}^{2}$(1)$G_{k,\lambda}f+\Psi(x,$y, f,$\frac{\partial f}{\partial x},x^{k}\frac{\partial f}{\partial y})=0$

in adomain$\Omega\subset \mathrm{R}^{2}$

,

where
$G_{k,\lambda}= \frac{\partial^{2}}{\partial x^{2}}+x^{2k}\frac{\partial^{2}}{\partial y^{2}}+:\lambda x^{k-1}\frac{\partial}{\partial y}$

with
,
and
Let us define the following quantities$(x, y)\in\Omega\subset \mathrm{R}^{2}$$\lambda\in \mathbb{C},i=\sqrt{-1}$$k\in \mathrm{Z}_{+},\Omega$

is abounded domain in$R=(x^{k+1}+u^{k+1})^{2}+(k+1)^{2}(y-v)^{2},p= \frac{4x^{k+1}u^{k+1}}{R}$$\mathrm{R}^{2}$.

,$A_{+}=x^{k+1}+u^{k+1}+:(k+1)(y-v),A_{-}=x^{k+1}+u^{k+1}-:(k+1)(y-v)$ ,$M=A_{+}^{-\frac{k+\lambda}{2k+2}}A_{-}^{-\frac{k-\lambda}{2k+2}}$

here we take$z_{1}^{z_{2}}=e^{z_{2}\ln z_{1}}$$+i\varphi$$\mathrm{i}\mathrm{s}\mathrm{h}z_{1}=\mathrm{h}r$. First, we,forand if$-\pi<\varphi\leq\pi$ thenfind the uniform fundamental solution of, that$z_{1},$ $z_{2}\in \mathbb{C}$$z_{1}=re^{:}\varphi,$$\mathrm{w}\mathrm{i}\mathbb{I}$$G_{k,\lambda}$$G_{k,\lambda}F_{k,\lambda}(x, y, u,v)=\delta(x-u, y-v)$

in the following form
$F_{k,\lambda}(x,y,u, v)=F(p)M$,.

After some computations we arrive at$G_{k,\lambda}F_{k,\lambda}=16(k+1)^{2}u^{2k+2_{X}2k}[(u^{k+1}-x^{k+1})^{2}+(k+1)^{2}(y-v)^{2}]\mathrm{x}$$\mathrm{x}$$MR^{-3}F’(p)++4(k+1)x^{k-1}u^{k+1}[k(x^{2k+2}+u^{2k\dagger 2}+(k+1)^{2}(y-v)^{2})-$$-(6k+4)x^{k+1}u^{k+1}]MR^{-2}F’(p)++(\lambda^{2}-k^{2})x^{k-1}u^{k+1}MR^{-1}F(p)$[1] N. M. Tri, To appear in J. Math. Sci. Univ. Tokyo.

141
Therefore, if $F(p)$ satisfies the following hypergeometric equation$p(1-p)F’(p)+[c-(1+a+b)p]F’(p)-abF(p)=0$ ,

(2)

with

$a= \frac{k+\lambda}{2k+2}$,$b= \frac{k-\lambda}{2k+2}$,$c= \frac{k}{k+1}$, then formally we will have
$G_{k,\lambda}F_{k,\lambda}=0$.

The general solutions of.(2) are$F(p)=C_{1}F( \frac{k+\lambda}{2k+2},$ $\frac{k-\lambda}{2k+2}$,$\frac{k}{k+1},p)+C_{2}p^{\Gamma^{1}}+\mathrm{T}F(\frac{k+2+\lambda}{2k+2}$

where $F(a, b, c,p)$ is the Gauss hypergeometric function andplex constants [2].

$C_{1}$,$\frac{k+2-\lambda}{2k+2}$,$C_{2}$,$\frac{k+2}{k+1},p)($are some com-

52. Case k is odd.
is odd, we note that $0\leq p\leq 1$ . Moreover, $p=1$ if and only if[3]$x=\pm u\neq 0$ , $y=v$ . If $u=0$ , $v=0$ then $p=0$ ;therefore, from the result of

Since

$k$

$G_{k,\lambda}F$($\frac{k+\lambda}{2k+2}$,$\frac{k-\lambda}{2k+2}$,)$\frac{k}{k+1},p$$M=- \frac{2^{2+\frac{1}{k+1}}\pi\Gamma(\frac{k}{k+1})}{\Gamma(\frac{k+\lambda}{2k+2})\Gamma(\frac{k-\lambda}{2k+2})}\delta(x, y)$we should choose.

$C_{1}=- \frac{\Gamma(\frac{k+\lambda}{2k+2})\Gamma(\frac{k-\lambda}{2k+2})}{2^{2+\frac{1}{k+1}}\pi\Gamma(\frac{k}{k+1})}$will be located at the one of $F(p)$ .
If $u\neq 0$ then the singularities ofOn the other hand, $F(p)$ , with $0\leq p\leq 1$ , has singularity only when $p=1$ . As
$parrow 1$ we have the following asymptotic expansions (see [2])$F_{k,\lambda}(x, y, u, v)$

$F$

$F$

(

(

$\frac{k+2+\lambda}{2k+2}$,,$\frac{k-\lambda}{2k+2}$,$\frac{k}{k+1},p)=-\frac{\Gamma(\frac{k}{k+1})}{\Gamma(\frac{k+\lambda}{2k+2})\Gamma(\frac{k-\lambda}{2k+2})}\log(1-p)+O(1)$$\frac{k+2-\lambda}{2k+2}$,$\frac{k+2}{k+1},p)=-\frac{\Gamma(\frac{k+2}{k+1})}{\Gamma(\frac{k+2+\lambda}{2k+2})\Gamma(\frac{k+2-\lambda}{2k+2})}\log(1-p)+O(1)$$\frac{k+\lambda}{2k+2}$,, p. 74.
[2] H. Bateman, and A. Erdelyi, 1953,. 6, 1999,[3] N. M. Tri, J. Math. Sci. Univ. Tokyo,$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{I}$$\mathrm{v}\mathrm{o}\mathrm{l}$$\mathrm{p}\mathrm{p}$. 437-452.

142
We expect that

$F_{k,\lambda}(x,$y,u, v) has singularity only when x $=u$ , y $=v$ . Since
when (x, $y)arrow(-u,$v), we should choose$p^{\frac{1}{k+1}}=(4R^{-1})^{\frac{1}{k+1}}xuarrow-1$$C_{2}=- \frac{\Gamma(^{k}2\mathrm{A}^{2}\mathrm{B}^{\lambda})k+2\Gamma(\frac{k+2-\lambda}{2k+2})}{2^{2+\frac{1}{k+1}}\pi\Gamma(\frac{k+2}{k+1})}$such that $F(p)$ has no singularity at $x=-u$ , $y=v$ . Note that the following
conditions(3)

A $\neq\pm[2N(k+1)+k]$ , A $\neq\pm[2N(k+1)+k+2]$ ,where N is anon-negative integer, guarantee that $C_{1},C_{2}<\infty$ and hence $F(p)$has logarithm growth (if u) at (x, $y)=(u,$ v).
$\neq 0$Definition. The parameter Ais $ca//ed$ admissible if A satisfies the condition (3).
Therefore, if Ais admissible then we expect that the function $F(p)M$ , or

$F_{k,\lambda}(x, y, u, v)=- \frac{\Gamma(\frac{k+\lambda}{2k+2})\Gamma(\frac{k-\lambda}{2k+2})F(\frac{k+\lambda}{2k+2},\frac{k-\lambda}{2k+2},\frac{k}{k+1},p)}{2^{2+\frac{1}{k+1}}\pi\Gamma(\frac{k}{k+1})A^{\frac{\mathrm{t}+\lambda}{+2k+2}}A^{\frac{k-\lambda}{-2k+2}}}-$$- \frac{xu\Gamma(\frac{k+2+\lambda}{2k+2})\Gamma(\frac{k+2-\lambda}{2k+2})F(_{2k+2}^{k2\lambda}\mapsto,\frac{k+2-\lambda}{2k+2},\frac{k+2}{k+1},p)}{2^{2-\star_{1}}+\pi\Gamma(\frac{k+2}{k+1})A_{+}^{*_{+}\Rightarrow}A^{\frac{k+2-\lambda}{-2k+2}}k2\lrcorner\lambda}$,

will be our desired uniform fundamental solution. Indeed, we have

Theorem 1. Assume that Ais admissible. Then
$G_{k,\lambda}F_{k,\lambda}(x, y, u, v)=\delta(x-u,y-v)$.

Remark 1. Asimilar expression foris also given in [4].
Let us denote,, and. Noting that $F_{k,\lambda}(x,y,u,v)=F_{k,-\lambda}(u,v,x,y)$ , from Theorem 1we caneasily deduce$F_{k,0}$$X_{1}’= \frac{\partial}{\partial u}-iu^{k}\frac{\partial}{\partial v}$$X_{2}’= \frac{\partial}{\partial u}+iu^{k}\frac{\partial}{\partial v}$$G_{k,\lambda}’=X_{2}’X_{1}’+:(\lambda+$$k)u^{k-1_{\frac{\partial}{\partial v}}}$[4] R. Beals, Journ&\’Equations aux d&iv6es partielles, Saint-Jean-de-Monts, 1998,

Tl TO$\mathrm{p}\mathrm{p}$.

143
is a bounded
Proposition 1(Representation formula). Assume thatthen weadmissible
and Aisdomain with piece-wise smooth boundary, fhave$\Omega\subset \mathbb{R}^{2}$$\in C^{2}(\overline{\Omega})$(4)$f(x, y)= \int_{\Omega}F_{k,\lambda}(x, y, u, v)G_{k,\lambda}’f(u, v)dudv-$$- \int_{\partial\Omega}F_{k,\lambda}(x, y, u, v)B_{1}’(f(u, v),$ $k,$ $- \lambda)ds+\int_{\partial\Omega}f(u, v)B_{2}’(F_{k,\lambda}(x, y, u, v), k)ds$

,

where
$B_{1}’(f(u, v)$,$k,$ $-\lambda)=(\nu_{1}-iu^{k}\nu_{2})X_{2}’f(u,v)-i(-\lambda+k)u^{k-1}\nu_{2}f(u, v)$$B_{2}’(F_{k,\lambda}(x, y, u, v), k)=(\nu_{1}+iu^{k}\nu_{2})X_{1}’F_{k,\lambda}(x, y, u, v)$

and$\nu=(\nu_{1}, \nu_{2})$Now weis the unit outward normal vector on-statea$\mathrm{r}\mathrm{e}$$\mathrm{w}\mathrm{e}\mathrm{U}$

,

,

an.

-known theorem on hypoellipticity of$G_{k,\lambda}$

as follows

is hypoelliptic if and only if the hypergeometric equation (2)Theorem 2.
has no bounded solution on the interval $[0, 1]$ .
$G_{k,\lambda}$(5)$\frac{\partial^{2}f}{\partial x^{2}}=h-x^{2k}\frac{\partial^{2}f}{\partial y^{2}}-i\lambda x^{k-1}\frac{\partial f}{\partial y}$.

144
Therefore, differentiating$\frac{\partial^{n-2}}{\partial y^{n-g-1}\partial x^{g\neg-}}$both sides of (5) gives$\frac{\partial^{n}f}{\partial y^{n-j-1}\partial x^{j+1}}=\frac{\partial^{n-2}h}{\partial y^{n-j-1}\partial x^{j-1}}-$$- \sum_{=0}^{j}$$(\begin{array}{l}j\dot{l}\end{array})$$-i \lambda\sum_{\dot{|}=0}^{j}$$2k(2k-1) \cdots(2k-i+1)x^{2k-:}\frac{\partial^{n-\dot{l}}f}{\partial y^{n-j+1}\partial x^{\dot{g}-\dot{l}-1}}-$$(\begin{array}{l}ji\end{array})$$(k-1)(k-2) \cdots(k-i)x^{k-:-1}\frac{\partial^{n-\cdot-1}f}{\partial y^{n-j}\partial x^{j-\dot{|}-1}}.\in C(\Omega).\square$

Actualy, amore detailed examination of the proof of Theorem 2would show thatthe integral operators$K$${}^{t}K$:$h arrow K(h)(x,y)=\int_{\Omega}F_{k,\lambda}(x, y,u,v)h(u,v)dudv$:$h arrow {}^{t}K(h)(x,y)=\int_{\Omega}F_{k,\lambda}(u,v,x,y)h(u,v)dudv$

into. In other words, $K$ and are separately regular. Since
is afunction in the complement of the diagonal of, we conclude
that $K$ andvery
areregular.
Next, we introduce some notationsmap$F_{k,\lambda}$$C_{0}^{\infty}(\Omega)$$C^{\infty}(\Omega)$${}^{t}K$$C^{\infty}$$\Omega \mathrm{x}\Omega$${}^{t}K$$–t-=\{(\alpha,\beta,\gamma)\in \mathrm{Z}_{+}^{3} :\alpha+\beta\leq t, kt\geq\gamma\geq\alpha+(1+k)\beta-t\}$

Theorem 3. Assume that $m\geq 2k^{2}+6k+5$ . Let
equation (1) andarguments then so is $f$ .

$\Psi\in G^{\iota}$. Then$f\in \mathrm{G}8$

$f$

be a

. In particular,$\mathrm{H}_{lo\mathrm{c}}^{m}(\Omega)$$|.f\Psi$

is.

solution of theanalytic in its

Proof The proof of Theorem 3consists of Theorem 4and Theorem 5below. The
proof follows the scheme :. Cl
Theorem 4. Let be a-function of its arguments and $m\geq 2k^{2}+6k+5$ .
Assume thatis a solution of the equation (1) then.
Proof Theorem 4can be proved with the help of Proposition 2. El$f\in \mathbb{H}_{loc}^{m}\Rightarrow f\in C^{\infty}(\Omega)\Rightarrow f\in A(\Omega)$$\Psi$$f\in \mathbb{H}_{loc}^{m}(\Omega)$$C^{\infty}$$f\in C^{\infty}(\Omega)$

145. Then
Proposition 2. Let $m\geq 2k^{2}+6k+5$ . Assume that.
denote the set of pairs of multi-indiceslet
Next, put $r_{0}=2k+2$ . For
wheresuch that$f\in \mathbb{H}_{loc}^{m}(\Omega)$$\Psi(x, y, f,\partial\lrcorner, x\frac{\partial f}{\partial y}\partial xk)\in \mathbb{H}_{loc}^{m-1}(\Omega)$$\Gamma_{r}$$r\in \mathbb{Z}_{+}$$\Gamma_{r}=\Gamma_{r}^{1}\cup\Gamma_{f}^{2}$$(\alpha, \beta)$$\Gamma_{f}^{1}=\{(\alpha, \beta) :\alpha\leq r_{0},2\alpha+\beta\leq r\}$,{$\Gamma_{r}^{2}=$$(\alpha,$$\beta)$:a$\geq r_{0},$$\alpha+\beta\leq r-r_{0}$}.

Define the following norm
$|f$

where

$|f$

,

,

$\Omega|,$$= \max_{\alpha(,\beta)\in\Gamma_{\mathrm{r}}}|\partial_{1}^{\alpha}\partial_{2}^{\beta}f$,,$\Omega|+$$\max_{\beta(\alpha,),\alpha\geq 1,\beta\geq 1\in \mathrm{r}_{P()\in\overline{\Omega}}}\max_{x,y}|\partial_{1}^{\alpha\dagger 2}\partial_{2}^{\beta}f|$$\Omega|=\max(x,y)\in\overline{\Omega}(|f|+|\frac{\partial f}{\partial x}|+|x^{k}\frac{\partial f}{\partial y}|)$.

. Thensolution of the equation (1) andbe a
Theorem 5. Let
is analytic in its arguments then so is .
$f\in G^{s}$ . In particular, if
Proof Theorem 5can be proved with the help of Proposition 3, Corollary 1,
Lemmas 2-4.
. Then there exist constants $C$, $D$ such that
Proposition 3. Assume thatfor every $H_{0}\geq 1$ , $H_{1}\geq CH_{0}^{2k+3}$ if$\Psi\in G^{\epsilon}$$C^{\infty}$

$f$

$f$

$\Psi$$\square$$\Psi\in G^{s}$

$|f$

,

$\Omega|_{d}\leq H_{0}H_{1}^{(d-r_{\mathrm{O}}-2)}(d-r_{0}-2)!^{s}$,$0\leq d\leq N+1$ , $r_{0}+2\leq N$

then$(x,y) \in\overline{\Omega}\max|\partial_{1}^{\alpha}\partial_{2}^{\beta}\Psi(x$for every$(\alpha, \beta)\in\Gamma_{N+1}$

, , ,
$y$

$f$

$\frac{\partial f}{\partial x}$,$x^{k} \frac{\partial f}{\partial y})|\leq DH_{0}H_{1}^{N-r_{\mathrm{O}}-1}(N-r_{0}-1)!^{\epsilon}$.

Corollary 1. Under the same hypotheses of Proposition 3with $d\leq N+1$ replacedby $d\leq N$ , then$\max_{x\in\overline{\Omega}}|\partial_{1}^{\alpha}\partial_{2}^{\beta}\Psi$for every(, , ,
$x$

$y$

$f$

$\frac{\partial f}{\partial x},x^{k}\frac{\partial f}{\partial y}$$(\alpha, \beta)\in\Gamma_{N+1}$)$|\leq D(|f,$ $\Omega|_{N+1}+H_{0}H_{1}^{N-\mathrm{r}_{0}-1}(N-r_{0}-1)!^{s})$.

is elliptic if $x\neq 0$ , it suffices to consider the case $(0, 0)\in\Omega$ andSince
asmall neighborhood of $(0, 0)$ . Let us define the distance$G_{k,\lambda}$$\rho((u, v)$, $(x,y))=\{$$\max\{|x^{k+1}-u^{k+1}|, (k+1)|y-v|\}$ ,for $xu\geq 0$$\max\{x^{k+1}+u^{k+1}, (k+1)|y-v|\}$ ,for $xu\leq 0$ .

$\Omega$is146
For two sets

$S_{1}$,$S_{2}$, the distance between them is defined as$\rho(S_{1}, S_{2})=\inf_{(x,y)\in S_{1},(u,v)\in S_{2}}\rho((x,y),$$(u,v))$.

Let $V^{T}(T\leq 1)$ be the cube with edges of size (in the metric) $2T$ which areparallel to the coordinate axes and centered at $(0, 0)$ . Denote by
the sub-cube
which is homothetic withand such that the distance between its boundary andthe boundary ofis . We shall prove by induction that if is small enoughthen there exist constants,
with $H_{1}\geq CH_{0}^{2k+3}$ such that$\rho$$V_{\delta}^{T}$$V^{T}$$V^{T}$$\delta$$T$$H_{0}$(6)|f,$H_{1}$

for $0\leq n\leq 6k+4$ ,$V_{\delta}^{T}|_{n}\leq H_{0}$

and

|f,

(7)

$V_{\delta}^{T}|_{n} \leq H_{0}(\frac{H_{1}}{\delta})^{n-r_{0}-2}(n-\prime 0 -2)!^{\delta}:=Q_{n-1}$

for $n\geq 6k+4$ , and sufficiently smal. Hence the desired conclusion follows. (6)follows easily ffom thesmoothness assumption on . Assume that (7) holds
for $n=N$ . We shall prove it for $n=N- l- 1$ . Put $\delta’=\delta(1-1/N)$ , $\delta’=\delta(1-4/N)$ .
Fix
and then define $\sigma=\rho((x, y),$
and
. Let
denote the cube with center at $(x, y)$ and edges of length
which are parallelto the coordinate axes, andthe boundary of. Note that,
, . Let $E_{1},E_{3}(E_{2}, E_{4})$ be edges ofandwhich are parallel to$Ox(Oy)$ respectively. We have to estimate
for all,, andfor all,. Let us abbreviate,,as ,, respectively.
$\delta$$C^{\infty}$$f$$(x, y)\in V_{\delta}^{T}$$\partial V^{T})$$\tilde{\sigma}=\sigma/N$$V_{\overline{\sigma}}(x,y)$$2\mathrm{a}$$S_{\overline{\sigma}}(x,y)$$V_{\overline{\sigma}}(x,y)$$V_{\overline{\sigma}}(x, y)\subset V_{\delta}^{T}$$\sigma\geq\delta$$S_{\overline{\sigma}}(x, y)$$\max_{(x,y)\in V_{\delta}^{T}}|_{\gamma}\partial_{\alpha,\beta}(\partial_{12}^{\alpha_{1\# 1}}f)|$$(\alpha,\beta,\gamma)\in---1$$\Gamma_{N+1}$$(\alpha_{1},\beta_{1})\in\Gamma_{N+1}$$\max_{(x,y)\in V_{\delta}^{T1}}(\partial_{1}^{2+\alpha_{1}}\#_{2}1f)|$$\alpha_{1}\geq 1,\beta_{1}\geq 1$$\frac{\partial^{\alpha}}{\partial u^{\alpha}}$$\frac{\partial^{\beta}}{\partial v^{\beta}}$$\frac{\partial^{\alpha+\beta}}{\partial u^{\alpha}\partial v^{\beta}}$$\partial_{1l}^{\alpha}$

Lemma 2. Assume that
there exists a constant $C$ such thatand$(\alpha,\beta,\gamma)\in---1$$(\alpha_{1},\beta_{1})\in\Gamma_{N+1}$$(\mathrm{a}\mathrm{i},\mathrm{A})\in$$\partial_{v}^{\beta},\partial_{u}^{\alpha}\partial_{v}^{\beta}$. Then if$\alpha_{1}\geq 1,\beta_{1}\geq$$1$$(x,y)\in V_{\delta}^{T}\mathrm{m}\mathrm{a}\mathrm{x}|\gamma\partial\alpha,\beta(\partial_{12}^{\alpha_{1\theta_{f(}^{1}x,y))|}}\leq C$

Lemma 3. Assume that($(\alpha,\beta, \gamma)\in---1$$T^{\frac{1}{k+1}}|f$,, $|_{N+1}+Q_{N}(T^{\frac{1}{k+1}}+ \frac{1}{H_{1}})$$V_{\delta}^{T}$. Then there exists a constant$(x,y)\in V_{\delta}^{T}\mathrm{m}\mathrm{a}\mathrm{x}|\gamma\partial\alpha,\beta(\partial_{2}^{N+1}f(x, y))|\leq C(T^{\Leftrightarrow \mathrm{T}|f}$

,

$C$

).

such that$V_{\delta’}^{T},|_{N+1}+Q_{N}(T^{\frac{1}{k+1}}+ \frac{1}{H_{1}}))$.

147
Lemma 4. Assume that$(\alpha, \beta, \gamma)\in--1-$. Then there exists a constant,$(x,y)\in V_{\delta}^{T}\mathrm{m}\mathrm{a}\mathrm{x}|\gamma\alpha\partial,\beta(\partial_{1}^{N-r\mathrm{o}+1}f(x, y))|\leq C(T^{\frac{1}{k+1}}|f$Lemma 5. Assume thata constant $C$ such that$(\alpha_{1}, \beta_{1})\in\Gamma_{N+1}\backslash \Gamma_{N}$,$C$such that, $|_{N+1}+Q_{N}(T^{\frac{1}{k+1}}+ \frac{1}{H_{1}}))$ .

$V_{\delta}^{T}$$\alpha_{1}\geq 1,\beta_{1}\geq 1$. Then there exists$(x,y) \in V_{\delta}^{T}\mathrm{m}\mathrm{a}\mathrm{x}|(\partial_{1}^{\alpha_{1}+2}\partial_{2}^{\beta_{1}}f(x, y))|\leq C(T^{\tau_{+}^{1}\overline{1}|f,V_{\delta’}^{T}|_{N+1}+Q_{N}(T^{\frac{1}{k+1}}},+\frac{1}{H_{1}}))$.

\S 3. Case k is even.
A. First, we consider the case $\lambda=2N(k+1)$ , where $N$ is an integer. In
this case we will prove asimilar result as in 52 by establishing the explicit uniform fundamental solutions of $G_{k,2N(k+1)}$ . Let us maintain the notations used for$p,A_{+}$ ,,from the very beginning of the paper (now, of course, with,,change sign whenan even ). If $(u, v)\neq(0, v)$ is fixed then the real parts of
may have singulari$(x, y)$ passes through $(-u,v)$ . Therefore,
complexnumber A. But
ties alone the half-line $(x, v)$ with $x\leq-u$ for an arbitrary
is smooth
if A $=2N(k+1)$ then it is not difficult to see that.
alone the half-Hue $(x, v)$ with $x<-u$ , that is
is even and $u\neq 0$ we have $-\infty\leq p\leq 1$ . More precisely,
Moreover, when$parrow 1$ when $(x,y)arrow(u,v)$ , and $parrow-\infty$ when $(x, y)arrow(-u,v)$ . If $N<0$ and(see [2],$parrow-\infty$ then from the asymptotic expansions of hypergeometric functions
as in the beginning of
p. 63)we should choose the expressions for constants Ci,$F_{k,2N(k+1)}(\cdot,\cdot,u,v)\in$
the paper (with Areplaced by $2N(k+1)$ ). And we will have
, with$A_{-}$$M,F_{k,\lambda}$$\ldots$$A_{+}$$k$$A_{-}$$M=A_{+}^{-\frac{k+\lambda}{2k+2}}A_{-}^{-\frac{k-\lambda}{2k+2}}$$M=A_{+}^{-\frac{k+\lambda}{2k+2}}A_{-}^{-\frac{k-\lambda}{2k+2}}$$M(\cdot, \cdot, u, v)\in C^{\infty}(\mathbb{R}^{2}\backslash \{(u_{1}v), (-u,v)\})$$k$$C_{2}$$C^{\infty}(\mathrm{R}^{2}\backslash (u, v))$

$F_{k,2N(k+1)}(-u,v, u, v)=0$.

Similar conclusions hold for $F_{k,2N(k+1)}(x, y, u, v)$ when $N>0$ . If $N=0$ then, with
$F_{k,0}(\cdot, \cdot,u,v)\in C^{\infty}(\mathbb{R}^{2}\backslash (u, v))$

$F_{k,0}(-u,v, u, v)=- \frac{\cot\frac{k\pi}{2k+2}}{4u^{k}}$.

148. Assume that $m\geq 2k^{2}+6k+5$ , $\lambda=2N(k+1)$ , andsolution of the equation (1). Then
. In particular, if is

Theorem 6. Let$\Psi\in G^{s}$isanalytic in its arguments then so is$f$$a\mathbb{H}_{loc}^{m}(\Omega)$

Proof Almost$f$$f\in G^{s}$.

$\Psi$

the arguments used for the case when is odd can be applied
here. Therefore, we only give the sketch of the proof. Instead of the distance in\S 2 we use the following metric$\mathrm{a}\mathbb{I}$$k$$\rho$$\tilde{\rho}((u,v)$, $(x, y))= \max\{|x^{k+1}-u^{k+1}|, (k+1)|y-v|\}.\square$

B. In this sub-section we will present some computations for finding the fundamental solutions ofwith source at the origin $(0, 0)$ for Aother than the values$2N(k+1)$ considered in sub-section A. Make the following change of variables$G_{k,\lambda}$

x

Then

$G_{k,\lambda}$

sign$=\rho|\sin\theta|^{\frac{1}{k+1}}$

sign$(\sin\theta)$,y$= \frac{\rho^{k+1}}{k+1}\cos\theta$,$\theta\in(-\pi, \pi)$.

will be transformed into$(\sin\theta)|\sin\theta|^{\mathrm{p}^{-}\mp 1}k1(\mathrm{s}$ $\dot{\mathrm{m}}\theta\frac{\partial^{2}}{\partial\rho^{2}}+(k+1)^{2}\rho^{-2}\sin\theta\frac{\partial^{2}}{\partial\theta^{2}}+$(Acoe $\theta+(k+1)\sin\theta$)$\rho^{-1}\frac{\partial}{\partial\rho}+(k+1)\rho^{-2}(k\cos\theta-\cdot\lambda\sin\theta)\frac{\partial}{\partial\theta})$.

If we seek the fundamental solution in the form $F_{k,\lambda}(x, y)=\rho^{-k}F(\theta)$ thenmust satisfy the following equation$F(\theta)$$(k+1)^{2}\sin\theta F’(\theta)+(k+1)(k\cos\theta-i\lambda\sin\theta)F’(\theta)-$(8)$-ik\lambda$

coe$\theta F(\theta)=0$.

The general solutions of (8) are$F(\theta)=$$(C_{3}+C_{4} \int_{0}^{\theta}|\sin s|^{-\frac{k}{k+1}}e^{-\mathrm{i}^{\lambda}}k\dot{.}+1ds)$$e^{\frac{\dot{l}\lambda\theta}{k+1}}$

,

where
and
are some complex constants. Among aU these solutions, weare interested in finding anon-trivial periodic solution. When $\lambda=2N(k+1)-$
this case was considered in sub-section A-the periodic solution is,and the function $F_{k,\lambda}(x, y)=\rho^{-k}F(\theta)$ serves as afundamental solution. When
$C_{3}$$C_{4}$$F(\theta)=e^{\frac{\lambda\theta}{k+1}}.$

149. But inequationthethis case, we have $F_{k,\lambda}(x, y)=\rho^{-k}F(\theta)$ is anon-smooth solution of
fails in this case. If$G_{k,\lambda}f(x, y)=0$ (see [3]); hence, hypoellipticity for$\lambda\neq 2N(k+1)$ and $\lambda\neq(2N+1)(k+1)$ then we should choose$\lambda=(2N+1)(k+1)$ then the periodic solution again is$F(\theta)=e^{\frac{i\lambda\theta}{k+1}}$$G_{k,\lambda}$$$C_{3}=\ovalbox{\tt\small
REJECT} iC_{4}($$e^{\frac{\pi\lambda}{k+1}} \dot{.}\int_{0}^{\pi}|\sin s|^{-\frac{k}{k+1}}e^{-\frac{\lambda\epsilon}{k+1}}.ds$2$e^{-\frac{\pi\lambda}{k+1}} \dot{.}\int_{-\pi}^{0}|\sin s|^{-\frac{k}{k+1}}e^{-\frac{\lambda s}{k+1}}.ds)$$\sin\frac{\pi\lambda}{k+1}$

to obtain the only periodic solution. In this case, the function
will be our desired fundamental solution.

$F_{k,\lambda}(x,y)=\rho^{-k}F(\theta)$

Address:
Institute of Mathematics, P. O. Box 631, Boho 10000, Hanoi, Vietnam$\mathrm{e}$-mail:triminh@thevinh.ncst.ac.vn

andDivision of Applied Mathematics, Korea Advanced Institute of Scienceand Technology,373-1 Kusong-dong, Yusong-ku, Taejon 305-701, S. Korea$\mathrm{e}$-mail:triminh@amath.kaist.ac.kr