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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

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I may be wrong, but at first sight I would say this looks pretty hopeless to answer. –  Kofi Jan 2 '13 at 9:10
    
why do you think so? –  hapchiu Jan 2 '13 at 9:13
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The Cauchy-Kowalesky theorem give you the answer: inf the initial data are non characteristic an analytic solution exists for small $|t|$. Moreover that analytic solution is the unique smooth solution. If the initial condition is characteristic, all bets are off. Thus you only need to verify if the initial conditions are characteristic or not. –  Liviu Nicolaescu Jan 2 '13 at 11:44
    
what do you mean by characteristic? –  hapchiu Jan 2 '13 at 12:38
1  
The Cauchy-Kovalevskya Theorem may not apply because of singularity issues if you can't solve the equation to make $f_{tt}$ an analytic function of the other variables. You may need the more general theory for singular PDE. (For example, see R. Gérard and H. Tahara, Singular nonlinear partial differential equations, Aspects of Math- ematics, Friedr. Vieweg & Sohn, Braunschweig, 1996.) –  Robert Bryant Jan 2 '13 at 13:38

1 Answer 1

up vote 7 down vote accepted

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.

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