# Local solvability and Cauchy-Kovalevskaya theorem for PDEs

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 $$f$$.

• 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 $t = 0$? – Deane Yang Jul 28 at 14:37
• 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 – Michele Jul 28 at 15:35
• 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. – Deane Yang Jul 28 at 22:33
• 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. – Michele Jul 29 at 7:21
• 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 $S(\Delta)$. – Deane Yang Jul 29 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 $$(x_0,t_0)$$.