I am looking for a function $\varphi \in C^{\infty}(K)$ where $$K=\{(x,y,z) \in \mathbb{R}^3 \ | \ x^2+y^2 \leq 1/16 \ \ and \ \ 0 \leq z\leq 2\} \setminus B$$ with $B=\{(x,y,z) \in \mathbb{R}^3 \ | \ x^2+y^2+z^2 \leq h^2 \} \cup \{(x,y,z) \in \mathbb{R}^3 \ | \ x^2+y^2+(z-2)^2 \leq h^2 \}$ the union of two spheres of radius $1>h>\frac{1}{4}$. You will find a drawing of the set $K$ below.
The function $\varphi$ must verifiy
$$\partial_{11} \partial_{33} \varphi = 0 \ \ on \ \ K$$
with the following boundary conditions on the side $\Sigma_1$ and $\Sigma_2$, where we note $a=\sqrt{h^2-x^2-y^2}$
$$\varphi(x,y,a)= x a + f(y) \ \ and \ \ \partial_3 \varphi(x,y,a)=x$$ $$\varphi(x,y,2-a)= -x a + g(y) \ \ and \ \ \partial_3 \varphi(x,y,2-a)=x$$ for all $x,y$ such as $0 \leq x^2 + y^2 \leq \frac{1}{16}$, where $f,g$ are two given functions of $y$.
The difficulty here is to deal with the cross derivatives condition $\partial_{11} \partial_{33} \varphi=0$. I had to deal previously with the same kind of problem but with the condition $\partial_{3333} \varphi =0 \ \ on \ \ K$ and the construction was much more easier.
Does anyone have any ideas or advices that could help me on that problem ? Please feel free to ask me questions if you need more details.
I think such a function can be constructed.
For $(x,y,z)\in K$ with $0\leq z \leq h$ we set $$ \varphi(x,y,z)=xz+f(y) $$ and for $(x,y,z\in K$ with $2-h\leq z \leq 2$ we set $$ \varphi(x,y,z)=x(z-2)+g(y). $$ Then one can check that the boundary conditions on $\Sigma_1,\Sigma_2$ hold.
It remains to glue the functions together on the remaining subset of points $(x,y,z)\in K$ with $h\leq z \leq 2-h$. For that, we use a function $\psi\in C^\infty([0,2])$ such that $$ \psi(s)=0 \quad\text{if}\quad 0\leq s \leq h, \psi(s)=1 \quad\text{if}\quad 2-h\leq s \leq 2. $$ Such a function exists because $h<2-h$ (and can also be explicitly constructed, if necessary). Functions of this type are often called bump functions. Using the function $\psi$, we can define the final function $\varphi$ via $$ \varphi(x,y,z)=(xz + f(y))(1-\psi(z)) + (x(z-2)+g(y))\psi(z). $$ This function satisfies the boundary conditions due to the above observations. It remains to show that $\partial_{11}\partial_{33}\varphi=0$ on $K$. The function $\varphi$ has the structure $$ \varphi(x,y,z)= x\xi_1(z)+f(y)+\xi_2(y)\psi(z) $$ for suitable functions $\xi_1,\xi_2$. Then we have $\partial_{33}\varphi (x,y,z)=x\xi_1''(z)+\xi_2(y)\psi''(z)$. Because this is linear in $x$, it follows that $\partial_{11}\partial_{33}\varphi(x,y,z)=0$ is true.