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



Solution of a PDE and its uniqueness
Here is what you should try: Consider the function
$$
p(x,\lambda) = K\_0(x,0,\ldots,0)+\lambda\ K\_1(x,0,\ldots,0) + \cdots + \lambda^n\ K\_n(x,0,\ldots,0)
$$
For any analytic solution $\lambda=L(x)$ to this polynomial equation that is a simple root of $p(x,\lambda)=0$, then, by the Cauchy-Kovalevskya Theorem, there is a unique solution to your problem on an open neighborhood of $U\times \lbrace0\rbrace$ in $U\times I$ that satisfies $f\_{tt}(x,0)=L(x)$. Thus, if there is more than one simple root, you will have more than one solution. 


*Added comment:* Perhaps I should explain what is meant by 'simple root'. I mean that, first, $p\bigl(x,L(x)\bigr)\equiv0$ and, second, that, when one performs the polynomial division so as to get $p(x,\lambda) = \bigl(\lambda - L(x)\bigr)\ q(x,\lambda)$ where $q$ is analytic in $x$ and polynomial in $\lambda$, then $q\bigl(x,L(x)\bigr)$ is *nonvanishing* on $U$. Alternatively (i.e., a different way of saying the same thing), one could require that 
$p\bigl(x,L(x)\bigr)\equiv0$ while $p\_\lambda\bigl(x,L(x)\bigr)$ is nonvanishing on $U$, where $p\_\lambda$ means the partial derivative of $p$ with respect to $\lambda$. 


Meanwhile, if you have an analytic solution $\lambda=L(x)$ to this polynomial equation that is not a simple root, there may still be a solution (or there may not), but there are more conditions, and you will need the theory of singular analytic PDE to check them. For example, see R. Gérard and H. Tahara, *Singular nonlinear partial differential equations*, Aspects of Mathematics, Friedr. Vieweg & Sohn, Braunschweig, 1996.


