solution du prolème de Dirichlet

[mona123]

solution du prolème de Dirichlet

I want to unswer the following problem please help me:


Let $G:=\left\{z\in \mathbb{C},\;-a<Re\;z<a,\;-b<Im\;z<b\right\}$, where $a,b>0$.


Suppose that $f:\partial G\to\mathbb{R}$ is a continuous function satisfifying $f(\bar z)=-f(z)$,$\;z\in\partial G$.


Show that the solution $u$ of the Dirichlet problem in $G$ with the boundary values $f$ satisfies $u(\bar z)=-u(z),\; z\in \bar G$. In particular $u(z)=0$ when $Im\;z=0,\;z\in \bar G$.


Thanks in advance.

 

[Deleted member 4993]

I want to unswer the following problem please help me:


Let $G:=\left\{z\in \mathbb{C},\;-a<Re\;z<a,\;-b<Im\;z<b\right\}$, where $a,b>0$.


Suppose that $f:\partial G\to\mathbb{R}$ is a continuous function satisfifying $f(\bar z)=-f(z)$,$\;z\in\partial G$.


Show that the solution $u$ of the Dirichlet problem in $G$ with the boundary values $f$ satisfies $u(\bar z)=-u(z),\; z\in \bar G$. In particular $u(z)=0$ when $Im\;z=0,\;z\in \bar G$.


Thanks in advance.

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[mona123]

This is what i wrote;

The solution $u$ of the Dirichlet problem in $G$ with the boundary values $f$ is given by: $u:\bar{G}\to \mathbb{R}$ with


$ u(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}z|^2} f ( e^{it}) \, dt, \qquad z \in {\mathbb{G}}. \tag{1}$


Then for all $z \in {\mathbb{G}}$


$ u(\bar z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}\bar z|^2} f ( e^{it}) \, dt=-\frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}\bar z|^2} f ( e^{-it}) \, dt, \qquad z \in \bar{\Bbb{G}}. $. car $f(z)=-f(\bar z)$ pour tout $z\in\partial G$
And i am blocked here.Help me please .

 

[JeffM]

This is what i wrote;

The solution \(\displaystyle u\) of the Dirichlet problem in \(\displaystyle \ G\) with the boundary values
f is given by: \(\displaystyle u:\bar{G}\to \mathbb{R}\) with


\(\displaystyle u(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}z|^2} f ( e^{it}) \, dt, \qquad z \in {\mathbb{G}}. \tag{1}\)


Then for all \(\displaystyle z \in {\mathbb{G}}\)


\(\displaystyle u(\bar z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}\bar z|^2} f ( e^{it}) \, dt=-\frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|1 - e^{-it}\bar z|^2} f ( e^{-it}) \, dt, \qquad z \in \bar{\Bbb{G}}\) because

\(\displaystyle f(z)=-f(\bar z)\) for every \(\displaystyle z\in\partial G\)
And i am blocked here.Help me please .

I have cleaned up the preceding post and translated the French.