drawing a 2D-plane

Question

i need to draw this plane, but i don't know how

can anyone help me? its important i need it in my master report. i try to use this from an old example to start with but it didn't work

\begin{tikzpicture}
\begin{scope}
\clip[draw](48,1) to[closed,curve through={(50,3) .. (52,4) .. (51,6) ..(47,6) .. (46.6,5) .. (45,3.8) }] (48,1);
\node[draw,circle,fill=magenta,minimum width=2cm] (n1) at (50,3) {};
\end{scope}
\end{tikzpicture}

Answer 1

Something like this?

I used random steps decorator for the green path, and "manually" built the black path, using some key points and start/end angles. Also note the use of polar coordinates, which are more appropiate for this case than cartesian ones.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing,shapes.geometric}
\colorlet{myred}{red!60!black}
\colorlet{mygreen}{green!40!black}

\begin{document}
\begin{tikzpicture}[scale=0.7]
\node[draw=myred, ellipse, minimum width=3cm, minimum height=4.5cm] (e1) {};
\draw[myred, ->] (e1.-30) -- +(-20:2) node[right] {$\Omega_0$};

\node[draw=mygreen, ellipse, minimum width=3.8cm, minimum height=5.2cm,
     decoration={random steps, segment length=1.5mm}, decorate]
     (e2) {};
\draw[mygreen, ->] (e2.0) --  +(5:1.5) node[right] {$\widetilde\Omega_0$};
\draw[mygreen, <-] (e2.90) -- +(80:.5) node[right] {$v_\varepsilon=0$};


\draw (0:3) to[out=90,in=-65]      (45:4)
            to[out=180-65,in=0]    (90:5)
            to[out=180, in=180-70] (150:4.5)
            to[out=-70, in=90]     (190:3.2)
            to[out=-90, in=180-50] (234:5.1)
            to[out=-50, in=180]    (280:5.3)
            to[out=0, in=-45]      (-30:4)
            to[out=180-45, in=-90] (0:3) -- cycle;
\draw[->] (50:4.2) -- +(30:1) node[right] {$\Omega$};
\draw[<-] (85:5) -- +(85:1) node[right] {$u_\varepsilon=0$};

\end{tikzpicture}
\end{document}