Hallo,


I have the following PDE that I am trying to solve via the Cauchy-Kowalewski Theorem. But I have no idea how to do it or if its possible. Maybe one of you has an idea. Here is the problem: Let $U \subset \mathbb{C}^{n}$ be some open subset witch contains zero. Well you can shrink $U$ arbitralily if you wish. Let $z\_{j} = x\_{j} + i y\_{j}$ be the coordinates on $U$. I am looking for a real-valued analytic function $\beta$ defined on $U$ such that the equations are satisfied: $\frac{\partial \beta}{\partial x\_{j}} = F\_{j}(x,y,\beta)$ and $\frac{\partial \beta}{\partial y\_{j}} = G\_{j}(x,y,\beta)$,with $j = 1, ..., n$, where $F,G$ are also real analytic functions and with initial condition $\beta(x,0) = 0$, $\forall x \in U \cap \mathbb{R}^{n}$. 
Is it possible to solve such an system of equations via Cauchy-Kowalewski ? If yes, how? If no, why and is there any other method that can give me a solution? I hope that for a lot of answers and please excuse me if the question is too trivial. Thanks in advance.


Greeting
Andrei



Solving PDE with Cauchy - Kowalewski Theorem
You don't need the Cauchy-Kowalewski Theorem for your problem. In fact, real-analyticity is a red herring here. What you are asking for is a function $\beta(x,y)$ such that the graph $\bigl(x,y,\beta(x,y)\bigr)$ is an integral manifold of the $1$-form
$$
\theta = d\beta - F\_j(x,y,\beta)\ dx^j - G\_j(x,y,\beta)\ dy^j
$$
(sum on $j$ in both terms) defined on $\mathbb{R}^{2n+1} = \mathbb{C}^n\times \mathbb{R}$ (or some open neighborhood of $0$ in this space. This makes sense for smooth functions of course, and whether there is a solution or not doesn't depend on real-analyticity.


In fact, you have added the requirement that the $2n$-dimensional graph contain the $n$-dimension submanifold defined by $(x,y,\beta) = (x, 0, 0)$, and you can see from the above that $\theta$ vanishes on that graph if and only if the $F\_j$ satisfy $F\_j(x,0,0)\equiv0$, so this is certainly a necessary condition.


A sufficient condition, after that, would be, for example, that $d\theta\wedge\theta=0$, for then the Frobenius Theorem would apply. However, this is not necessary if all you are asking is that there be a solution to the specific 'initial value problem' you have posed. To get sufficient conditions, what you should do is use ODE to, for example, construct
$\beta\_1(x,y^1)$ satisfying the equation
$$
\frac{\partial\beta\_1}{\partial y^1} = G\_1(x,y^1,0,\ldots,0,\beta\_1)
$$
with the initial condition $\beta\_1(x,0) = 0$. Then you need to check that $\theta$ vanishes on the $(n{+}1)$-dimensional graph $\bigl(x,y^1,0,\ldots,0,\beta\_1(x,y^1)\bigr)$. Next, you construct $\beta\_2(x,y^1,y^2)$ by solving the equation
$$
\frac{\partial\beta\_2}{\partial y^2} = G\_2(x,y^1,y^2,0,\ldots,0,\beta\_2)
$$
with the initial condition $\beta\_2(x,y^1,0) = \beta\_1(x,y^1)$, and so on. At each stage, you'll get more conditions on the functions $G\_j$ and $F\_j$ in order for the constructed graph to be an integral of $\theta$. When you get to the end, these will be the necessary and sufficient conditions for this *particular* initial value problem.


