In recent years,many investigators paid much attention to boundary value problems for complex differential equations in different domains,such as unit disc,half plane,upper unit disc,ring,sector and even high dimension space [2-9]. In [6],the authors studied Schwarz problem in half disc and half ring,and Wang extended boundary conditions,discussing high order Schwarz problem for polyanalytic equation in half unit disc and a triangle [7, 8]. In [9],we also investigated a Schwarz problem for Cauchy-Riemann equation in a sector with angle $\theta=\frac{\pi}{\alpha},\ \alpha\geq 1$. In this article,we study a high order Schwarz problem for polyanalytic equation in a general sector with angle $\theta=\frac{\pi}{\alpha},\ \alpha\geq1/2$,giving explicit solvability expression.

Let $\Omega$ be a sector with angle $\theta=\frac{\pi}{\alpha}$ $\ (\alpha\geq 1/2)$,that is,$\Omega=\{ |z|<1,\ 0<\arg z<\frac{\pi}{\alpha},\ \alpha\geq 1/2\}$. Its boundary $\partial \Omega=[0, 1]\cup \Gamma\cup [\varpi,0]$ is oriented counter-clockwise,where $0,\ 1,\ \varpi=e^{i\theta}$ are three corner points and the oriented circular arc $\Gamma$ is given by $ \Gamma:\ \tau\longmapsto e^{i \tau},\ \ \tau\in\left[0,\ \frac{\pi}{\alpha}\right].$

Lemma 1.1 (see [9]) The Schwarz problem for Cauchy-Riemann equation in $\Omega$

$w_{\overline{z}}=f \ \text{in}\ \Omega,\ \ \ \mbox{ Re}w=\gamma \ \text{on}\ \partial \Omega,$

$\displaystyle\frac{\alpha}{\pi }\int_0^{\frac{\pi}{\alpha}}\mbox{ Im}w(e^{i\varphi})\mbox{ d}\varphi=c,\ \ c\in \mathbb{R}$

for $\ f\in L_p(\Omega;\mathbb{C}),\ p>2,\ \ \gamma\in C(\partial \Omega;\mathbb{R})$,is uniquely solvable by

$\begin{array}{ll}w(z)=&\displaystyle\frac{\alpha}{2\pi i}\int\limits_{\Gamma}\gamma(\zeta)\Bigg(\frac{\zeta^\alpha+z^\alpha}{\zeta^\alpha-z^\alpha} -\frac{\overline{\zeta^\alpha}+z^\alpha}{\overline{\zeta^\alpha}-z^\alpha}\Bigg)\frac{ d\zeta}{\zeta} +\displaystyle\frac{\alpha}{\pi i}\int\limits_{[\varpi,0]\cup[0, 1]}\gamma(\zeta)\Bigg(\frac{1}{\zeta^\alpha-z^\alpha} -\frac{z^\alpha}{1-z^\alpha\zeta^\alpha }\Bigg)\zeta^{\alpha-1}\mbox{ d}\zeta\\ &-\displaystyle\frac{\alpha}{\pi}\displaystyle\int\limits_{\Omega}\Bigg[f(\zeta)\Big(\frac{1}{\zeta^\alpha-z^\alpha} -\frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\Big)\zeta^{\alpha-1}-\overline{f(\zeta)}\Big(\frac{1}{\overline{\zeta^\alpha}-z^\alpha} -\frac{z^\alpha}{1-z^\alpha\overline{\zeta^\alpha}}\Big)\overline{\zeta^{\alpha-1}}\Bigg]\mbox{ d}\xi \mbox{ d}\eta+i c. \end{array}$

Let

$\begin{equation}\label{2.1} H(z,\zeta)=\left[\displaystyle\frac{1}{\zeta^\alpha-z^\alpha}-\frac{1}{\zeta^\alpha-\overline{z^\alpha}}-\frac{z^\alpha}{1-z^\alpha\zeta^\alpha }+ \frac{\overline{z^\alpha}}{1-\overline{z^\alpha}\zeta^\alpha}\right] \zeta^{\alpha-1},\ \ z,\zeta\in \Omega\end{equation}$

(2.1)

and a poly-Schwarz operator for $\Omega$ is

$\begin{equation}\label{2.2}\begin{array}{ll}\displaystyle &S_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z) \\ =&\displaystyle\sum\limits_{l=0}^{n-1}\frac{(-1)^l}{l!}\bigg\{\frac{\alpha}{2\pi i}\int\limits_{\Gamma}\gamma_l(\zeta)\left(\zeta-z+\overline{\zeta-z}\right)^l\left(\frac{\zeta^\alpha+z^\alpha}{\zeta^\alpha-z^\alpha}- \frac{\overline{\zeta^\alpha}+z^\alpha}{\overline{\zeta^\alpha}-z^\alpha}\right)\displaystyle\frac{\mbox{ d}\zeta}{\zeta}\\ & +\displaystyle\frac{\alpha}{\pi i}\int\limits_{[\varpi,0]\cup[0, 1]}\gamma_l(\zeta)\left(\zeta-z+\overline{\zeta-z}\right)^l\left(\frac{1}{\zeta^\alpha-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\right)\zeta^{\alpha-1}\mbox{ d}\zeta\bigg\},\ z\in\Omega. \end{array}\end{equation}$

(2.2)

Then we have the following result.

Lemma 2.1 For $\gamma_0,\gamma_1,\cdots,\gamma_{n-1}\in C(\partial\Omega,\mbox{ R})$,then

$\left\{\mbox{ Re}\displaystyle\frac{\partial^kS_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}]} {\partial\overline{z}^k}\right\}^+(t)=\gamma_k(t)\ \ \text{for}\ \ t\in \partial\Omega;\ \ k=0,1,2,\cdots,n-1$

and

$\displaystyle\frac{\partial^nS_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z)}{\partial\overline{z}^n} =0.$

Proof For $k=0,1,2,\cdots,n-1$,by (2.2),

$\begin{equation}\label{2.3}\begin{array}{ll} &\displaystyle\frac{\partial^kS_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z)}{\partial\overline{z}^k}\\ =&\displaystyle\sum\limits_{l=k}^{n-1}\frac{(-1)^{l-k}}{(l-k)!}\bigg\{\frac{\alpha}{2\pi i}\int\limits_{\Gamma}\gamma_l(\zeta)\left(\zeta-z+\overline{\zeta-z}\right)^{l-k}\left(\frac{\zeta^\alpha+z^\alpha}{\zeta^\alpha-z^\alpha}- \frac{\overline{\zeta^\alpha}+z^\alpha}{\overline{\zeta^\alpha}-z^\alpha}\right)\displaystyle\frac{\mbox{ d}\zeta}{\zeta}\\ &+\displaystyle\frac{\alpha}{\pi i}\int\limits_{[\varpi,0]\cup[0, 1]}\gamma_l(\zeta)\left(\zeta-z+\overline{\zeta-z}\right)^{l-k}\left(\frac{1}{\zeta^\alpha-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\right)\zeta^{\alpha-1}\mbox{ d}\zeta\bigg\}\\ =&\displaystyle\sum\limits_{l=k}^{n-1}\sum\limits_{j=0}^{l-k}\frac{(-1)^{l-k}}{j!(l-k-j)!}(-z-\overline{z})^{l-k-j} \bigg\{\displaystyle\frac{\alpha}{2\pi i}\int\limits_{\Gamma}\gamma_l(\zeta)\left(\zeta+\overline{\zeta}\right)^{j}\left(\frac{\zeta^\alpha+z^\alpha}{\zeta^\alpha-z^\alpha}- \frac{\overline{\zeta^\alpha}+z^\alpha}{\overline{\zeta^\alpha}-z^\alpha}\right)\displaystyle\frac{\mbox{ d}\zeta}{\zeta}\\ & +\displaystyle\frac{\alpha}{\pi i}\int\limits_{[\varpi,0]\cup[0, 1]}\gamma_l(\zeta)\left(\zeta+\overline{\zeta}\right)^{j}\left(\frac{1}{\zeta^\alpha-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\right)\zeta^{\alpha-1}\mbox{ d}\zeta\bigg\}.\end{array}\end{equation}$

(2.3)

Then we have

$\mbox{ Re}\left\{\displaystyle\frac{\partial^kS_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z)}{\partial\overline{z}^k}\right\} =\displaystyle\sum\limits_{l=k}^{n-1}\sum\limits_{j=0}^{l-k}\frac{(-1)^{l-k}(-z-\overline{z})^{l-k-j}}{j!(l-k-j)!}\frac{\alpha}{2\pi i}\int\limits_{\partial\Omega}\gamma_l(\zeta)\left(\zeta+\overline{\zeta}\right)^{j}H(z,\zeta){\mbox{ d}}\zeta,$

where $H(z,\zeta)$ is given by (2.1). From the proof in [9],

$\lim\limits_{z\rightarrow t,t\in \partial\Omega}\frac{\alpha}{2\pi i}\int\limits_{\partial\Omega}\gamma_l(\zeta)\left(\zeta+\overline{\zeta}\right)^{j}H(z,\zeta){\mbox{ d}}\zeta =\gamma_l(t)\left(t+\overline{t}\right)^{j}.$

Hence,

$\left\{\mbox{ Re}\displaystyle\frac{\partial^kS_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}]}{\partial\overline{z}^k}\right\}^+(t) =\displaystyle\sum\limits_{l=k}^{n-1}\sum\limits_{j=0}^{l-k}\frac{(-1)^{j}(t+\overline{t})^{l-k}\gamma_l(t)}{j!(l-k-j)!} =\gamma_k(t).$

Obviously,by (2.3),the second equation in Lemma 2.1 is also true.

On the other hand,we define a Pompeiu operator as follows

$\begin{equation}\label{2.4}\begin{array}{ll} T_n[f](z)=&\displaystyle\frac{(-1)^n\alpha}{\pi (n-1)!}\displaystyle\int\limits_{\Omega}\left(\zeta-z+\overline{\zeta-z}\right)^{n-1}\bigg[f(\zeta)\left(\frac{1}{\zeta^\alpha-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\right)\zeta^{\alpha-1}\\[2mm] &-\displaystyle\overline{f(\zeta)}\left(\frac{1}{\overline{\zeta^\alpha}-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\overline{\zeta^\alpha}}\right)\overline{\zeta^{\alpha-1}}\bigg]\mbox{ d}\xi\mbox{ d}\eta,\ \ z\in\Omega \end{array}\end{equation}$

(2.4)

with $f\in L_p(\Omega;C),\ p>2$,and

$\begin{equation}\begin{array}{ll} T_1[f](z)=&-\displaystyle\frac{\alpha}{\pi }\displaystyle\int\limits_{\Omega}\bigg[f(\zeta)\left(\frac{1}{\zeta^\alpha-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\zeta^\alpha}\right)\zeta^{\alpha-1}\\[2mm] &-\displaystyle\overline{f(\zeta)}\left(\frac{1}{\overline{\zeta^\alpha}-z^\alpha}- \frac{z^\alpha}{1-z^\alpha\overline{\zeta^\alpha}}\right)\overline{\zeta^{\alpha-1}}\bigg]\mbox{ d}\xi\mbox{ d}\eta. \end{array}\end{equation}$

(2.5)

Lemma 2.2 For $f\in L_p(\Omega;C),\ p>2$,

$\displaystyle\frac{\partial^nT_n[f](z)}{\partial\overline{z}^n}=f(z),\ \ z\in \Omega;\ \ \ \{\mbox{ Re}T_n[f]\}^+(t)=0,\ \ \ t\in\partial\Omega.$

Proof Since for $l=1,2,\cdots,n,$

$\begin{equation}\label{2.6}T_l[f](z)=\displaystyle\frac{(-1)^{l-1}}{(l-1)!}\sum\limits_{k=0}^{l-1}\left(\begin{array}{cc}l-1\\k\end{array}\right) (-z-\overline{z})^{l-k-1}T_1[(\zeta+\overline{\zeta})^kf](z),\ \ z\in\Omega,\end{equation}$

(2.6)

then by $\displaystyle\frac{\partial T_1[(\zeta+\overline{\zeta})^kf](z)}{\partial\overline{z}}=(z+\overline{z})^kf(z)$,we obtain

$\begin{eqnarray*} \displaystyle\frac{\partial T_l[f](z)}{\partial\overline{z}}&=&\displaystyle\frac{(-1)^{l-1}}{(l-1)!}\sum\limits_{k=0}^{l-1} \left(\begin{array}{cc}l-1\\k\end{array}\right)\bigg\{\left(\displaystyle\frac{\partial}{\partial\overline{z}}(-z-\overline{z})^{l-k-1}\right) T_1[(\zeta+\overline{\zeta})^kf](z) \\&&+(-z-\overline{z})^{l-k-1}\displaystyle\frac{\partial T_1[(\zeta+\overline{\zeta})^kf](z)}{\partial\overline{z}}\bigg\}\\ &=&\displaystyle\frac{(-1)^{l-2}}{(l-2)!}\sum\limits_{k=0}^{l-2} \left(\begin{array}{cc}l-2\\k\end{array}\right)(-z-\overline{z})^{l-k-2}T_1[(\zeta+\overline{\zeta})^kf](z)=T_{l-1}[f](z). \end{eqnarray*}$

Therefore,$\displaystyle\frac{\partial^nT_n[f](z)}{\partial\overline{z}^n}=\displaystyle\frac{\partial T_1[f](z)}{\partial\overline {z}}=f(z)$. What's more,from (2.6),

$\mbox{ Re}\{T_n[f](z)\}=\displaystyle\frac{(-1)^{n-1}}{(n-1)!}\sum\limits_{k=0}^{n-1} \left(\begin{array}{cc}n-1\\k\end{array}\right)(-z-\overline{z})^{n-k-1}\mbox{ Re}\{T_1[(\zeta+\overline{\zeta})^kf](z)\}.$

By the result in [9],$\mbox{ Re}\{T_1[(\zeta+\overline{\zeta})^kf]\}^+(t)=0,\ \ t\in\partial\Omega$,thus for $t\in\partial\Omega,$ $\{\mbox{ Re}T_n[f]\}^+(t)=0.$

Theorem 2.1 The Schwarz problem for polyanalytic equation in $\Omega$,

$\begin{equation}\label{2.7} \left\{\begin{array}{ll} \partial^n_{\overline{z}}w(z)=f(z),& z\in\Omega,\ \ f\in L_p(\Omega,\mbox{ C}),\ p>2;\\[2mm] \left\{\mbox{ Re}(\partial^k_{\overline{z}}w)\right\}^+(t)=\gamma_k(t),&t\in\partial\Omega,\ \gamma_k\in C(\partial\Omega,\ \mbox{ R}),\ k=0,1,\cdots,n-1 \end{array}\right. \end{equation}$

(2.7)

is solvable by

$w(z)=S_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z)+T_n[f](z) +i\sum\limits_{k=0}^{n-1}(z+\overline{z})^kc_k,$

where $c_k\in \mbox{ R}$,$S_n,\ T_n$ are given by (2.2) and (2.4),respectively.

Proof Let

$w_0=S_n[\gamma_0,\gamma_1,\cdots,\gamma_{n-1}](z)+T_n[f](z).$

Obviously,from Lemmas 2.1 and 2.2,we know $w_0$ satisfies the boundary condition (2.7). We write the solution $w(z)$ as

$w(z)=w_0(z)+U(z),$

then

$\left\{\begin{array}{ll} \partial^n_{\overline{z}}U(z)=0,& z\in\Omega;\\[2mm] \left\{\mbox{ Re}(\partial^k_{\overline{z}}U)\right\}^+(t)=0,&t\in\partial\Omega,\ k=0,1,\cdots,n-1. \end{array}\right.$

From the first equation and [1],the polyanalytic function $U(z)$ in $\Omega$ can be expressed as

$\begin{equation}\label{2.8} U(z)=\sum\limits_{k=0}^{n-1}\displaystyle(z+\overline{z})^kf_k(z),\ \ \ z\in \Omega,\end{equation}$

(2.8)

where $f_k$ are analytic functions. Putting (2.8) into the second equation,we get $\{\mbox{ Re}f_k\}^+(t)=0,$ $ t\in \partial\Omega$. Thus,by Lemma 1.1,$f_k(z)=ic_k$ with $c_k$ being real numbers.Then we complete the proof.