Hallo,

consider $f: U \times I \rightarrow \mathbb{R}$, where $U \subset \mathbb{R}^{n}$ and $0 \in I \subset \mathbb{R}$ be two open sets. I am looking for the solution $f$ of the following PDE $\sum_{i=0}^{n} (\frac{\partial^{2}f}{\partial t^{2}})^{i} K_{i}(x,t,f,\frac{\partial f}{\partial t}, \frac{\partial^{2} f}{\partial t \partial x_{j}}, \frac{\partial^{2} f}{\partial x_{k} \partial x_{j}}) = 0$, with initial condition $f(x,0) = \frac{\partial f}{\partial t}(x,0) = 0$. Here the $K_{i}$ are analytic functions depending of several variables and $f$ should also be analytic. The solution should not be on the whole $U \times I$, maybe on some smaller open set. Does there exists a solution? If yes, is this solution unique? When does a solution exists? I have read what bryant wrote and it seems now clearly to me. But I have a different suggestion (I am not 100 % convinced but maybe one can point out where I did a mistake). Here is what I thaught: Differentiate the whole PDE expression with respect to $t$, then one obtains: $\frac{\partial }{\partial t}(\sum_{i=0}^{n} (\frac{\partial^{2}f}{\partial t^{2}})^{i} K_{i}(x,t,f,\frac{\partial f}{\partial t}, \frac{\partial^{2} f}{\partial t \partial x_{j}}, \frac{\partial^{2} f}{\partial x_{k} \partial x_{j}})) = 0$. Thus one obtains $\sum_{i=1}^{n}i(\frac{\partial^{2}f}{\partial t^{2}})^{i-1} \cdot (\frac{\partial^{3}f}{\partial t^{3}})K_{i} + (\frac{\partial^{2}f}{\partial t^{2}}) \cdot \frac{\partial K_{i}}{\partial t} + (\frac{\partial^{2}f}{\partial t^{2}}) \cdot \frac{\partial K_{0}}{\partial t} = 0$. Now we write the PDE as: $(\frac{\partial^{3}f}{\partial t^{3}}) \cdot F + G = 0$, where $F = \sum_{i=1}^{n}i(\frac{\partial^{2}f}{\partial t^{2}})^{i-1}K_{i}$ and $G = \sum_{i=1}^{n}(\frac{\partial^{2}f}{\partial t^{2}}) \cdot \frac{\partial K_{i}}{\partial t} + (\frac{\partial^{2}f}{\partial t^{2}}) \cdot \frac{\partial K_{0}}{\partial t}$. Now if $F$ does not vanish on $\{t=0\}$ we consider the equation $(\frac{\partial^{3}f}{\partial t^{3}}) \cdot F + G = 0$ and can use Cauchy-Kovalewskaya with appropriate initial conditions. This was my guess. Is this right? Where is the mistake?

hapchiu