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Let $\Omega \subset \mathbb{R}^{2}$ be an open set such that $\mathbf{0} \in \Omega$. Let $A := \Omega \setminus (\{0\}\times \mathbb{R})$, that is, $A$ is $\Omega$ with the $y$-axis removed.

Let $F: (A\cup \{\mathbf{0}\})\times\mathbb{R}^5 \to \mathbb{R}$, such that $F$ is real analytic on the open set $A \times \mathbb{R}^5$. Let us consider the partial differential equation on the set $A$

$$ \frac{\partial^2 \varphi}{\partial y^2} = F(x,y,\varphi, \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial^2\varphi}{\partial x\partial y}, \frac{\partial^2\varphi}{\partial x^2}) $$

with the initial conditions

$$ \varphi(x,0) = \frac{\partial \varphi}{\partial y}(x,0) = 0 $$

By Cauchy-Kovalevskaya this PDE has a solution in an open neighborhood of the $x$-axis inside the set $A$. Assume that the solution exists in all of $A$ and $F:D\rightarrow \mathbb{R}$, where $D = A \times \mathbb{R}^{5} \subset \mathbb{R}^{7}$. Furthermore assume that $F$ exists when $x=0$ and $y=0$.

Under what circumstances is it possible to extend this solution to a full neighbourhood of $\mathbf{0}$ ?

share|improve this question
    
to the first question: yes I am assuming that I remove the $y-$Axis. second question: Well $F$ is real analytic, but if one restricts all the data to the point $0$, then $F$ does exists there. –  Robert M. Oct 11 '13 at 8:19
    
so my question is: is it possible to extend it somehow to the y-axis? –  Robert M. Oct 11 '13 at 8:20
    
but i do not know if the equation makes sense everywhere. does it make sense ? –  Robert M. Oct 11 '13 at 8:53
    
well I meant that $F$ is defined on the whole real axis, i.e. the $x$-axis. Can one then extend it to the $y$-axis? –  Robert M. Oct 11 '13 at 10:12
    
ok i see. So $D$ has the form: $D=(A \cup \{ 0\}) \times \mathbb{R}^{5} \subset \mathbb{R}^{7}$. Thus $F:D\rightarrow \mathbb{R}$ is also real-analytic. –  Robert M. Oct 11 '13 at 10:30

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