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 $\mathbb{R}^{|x|+|u^{(n)}|}$


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:

enter image description here

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 $f$.

enter image description here

  • $\begingroup$ Couldn't you find all solutions by solving the first order system \begin{align*} u_t &= v\\ v_t &= w\\ w_t &= u_x \end{align*} with initial data $(u,v,w)$ along $t = 0$? $\endgroup$ – Deane Yang Jul 28 '20 at 14:37
  • $\begingroup$ Your system is "equivalent" to the old one in terms of solutions. However the point here is not finding the solutions, but solving the apparent contradiction I have pointed out. Cheers $\endgroup$ – Michele Jul 28 '20 at 15:35
  • $\begingroup$ Why is this system not locally solvable? Doesn’t $u_{0,tx} = v_{0,x}$ follow from the assumption that the equations hold at $(x_0,t_0)$ up to second order. $\endgroup$ – Deane Yang Jul 28 '20 at 22:33
  • $\begingroup$ This is in the definitiion of $S(\Delta)$, but not in that of $V(\Delta)$. That is, there are 2nd order differential consequences (e.g. $u_{tx}=v_x$) that are not algebraic consequences. $\endgroup$ – Michele Jul 29 '20 at 7:21
  • 1
    $\begingroup$ Here, it appears that local solvability means that any second order jet that solves the system can be extended to a local smooth solution (which is not the standard definition of local solvability). To be a second order solution, the three equations and the first partial derivatives with respect to both $x$ and $t$ of the first two equations must hold. In particular, if a 2-jet is a solution at $(x_0,t_0)$, $u_{tx} = v_x$ then holds at $(x_0,t_0)$. So in fact this equation is a consequence of the jet being in $S(\Delta)$. $\endgroup$ – Deane Yang Jul 29 '20 at 16:28

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 $(x_0,t_0)$.


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