Local solvability and Cauchy-Kovalevskaya theorem for PDEs

Question

I am trying to understand the exact implications between local solvability and a general version of the Cauchy-Kovalevskaya (CK) theorem, in the context of PDEs.

Let $\Delta(x,u^{(n)})=0$ be a system of PDEs of order $n$. Here $x$ is the vector of independent variables, and $u^{(n)}$ is the vector of dependent variables and all their derivatives up to order $n$. According to Olver ("Applications of Lie groups to Differential Equations", 2/e, Ch. 2, Def. 2.70), $\Delta$ is locally solvable if the variety it induces on $\mathbb{R}^{|x|+|u^{(n)}|}$

$V(\Delta):=\{(x_0,u_0^{(n)}):\Delta(x_0,u_0^{(n)})=0\}$

coincides with its solution variety

$S(\Delta):={\small \{(x_0,u_0^{(n)})\in V(\Delta) : \text{ $\exists$ an analytic solution $U$ of $\Delta$ in a neighborhood of $x_0$ s.t. } U^{(n)}(x_0)=u_0^{(n)} \}}$

Olver shows that if $\Delta$ is in Kovalevskaya form then it is locally solvable (Corollary 2.74, p. 163; this is indeed an easy consequence of CK theorem). He then states that the same result still holds when $\Delta$ is in general Kovalevskaya form:

I am struggling to convince myself of the validity of this statement. For instance, consider the 2nd order system $\Delta$ in the independent variables $t,x$ and the dependent variables $u,v$: \begin{align*} u_t & = v\\ v_{tt} &= u_x\,. \end{align*} From what I gather, this system is not locally solvable. Indeed, there are differential consequencences that are not captured algebraically by the two equations above, such as $u_{tx}=v_x$. So there are points $(x_0,u_0^{(2)})\in V(\Delta)$ s.t. (with obvious notation) $u_{0,tx}\neq v_{0,x}$, hence not in $S(\Delta)$.

Yet, $\Delta$ is in general Kovalevskaya form, is it?

Edit: for reference, I paste Olver's original definition. Note that $pr^{(n)}f$ denotes the $n$-th prolongation of $f$.

Answer 1

The system you wrote down First, let's assume everything is smooth. \begin{align*} u_t &= v\\ v_{tt} &= u_x\\ \end{align*} is equivalent to the first order system \begin{align*} u_t &= v\\ v_t &= w\\ w_t &= u_x \end{align*} in the sense that $(u,v)$ is a solution to the first system if and only if $(u,v,w)$ is a solution to the second system, where we set $w = v_t$.

By Cauchy-Kovalevski, given any analytic functions $u_0(x), v_0(x), w_0(x)$, there exists a unique analytic solution $(u,v,w)$ to the second system such that $u(x,0) = u_0(x), v(x,0) = v_0(x), w(x,0) = w_0(x)$. This is equivalent to saying that given any analytic functions $u_0, v_0, w_0$, there exists a unique analytic solution $(u,v)$ to the first system such that $u(x,0) = u_0(x)$, $v(x,0) = v_0(x)$, and $v_t(x,0) = w(x)$, which is what Olver asserts. The fact that $u_{tx} = v_x$ is a consequence of the equations and need not be specified in the initial data.

As for Olver's definition of local solvability, an element of $V(\Delta)$ for the original system consists of a $2$-jet $(x_0,t_0,u(x_0,t_0), u_x(x_0,t_0), u_t(x_0,t_0), v(x_0,t_0), v_x(x_0,t_0), v_t(x_0,t_0))$ that satisfies the system up to second order. On other words, at $(x_0,t_0)$, \begin{align*} u_t &= v\\ u_{tx} &= v_x\\ u_{tt} &= v_t\\ v_{tt} &= u_x \end{align*} In particular, if a $2$-jet lies in $V(\Delta)$, then $u_{tx} = v_x$ does hold for that $2$-jet at $(x_0,t_0)$. Given such a $2$-jet, you can extend $u$, $v$, and $v_t$ arbitrarily to initial data along $t = 0$ and solve the system as described above. In particular, the initial data is assumed to satisfy $u_{tx} = v_x$ at $(x_0,t_0)$.