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The method of characteristics is a bit strange here because the equation is underdetermined, so one can't expect to be able to specify a solution by fixing initial data for $u$ and $v$ along a surface …

You are correct that Cauchy-Kovalevskaya does not apply directly to this problem, but there are other theorems that give sufficient conditions, provided that you make certain basic regularity assumpti …

Here is another way to make the solutions of the equation locally unique: What you have is a bundle mapping $f$ from the bundle of $(n{-}1)$-forms on $\mathbb{R}^n$ to the bundle of $n$-forms on $\ma …

This isn't a solution, but it's too long for a comment. Before you try to apply Darboux' Method, you might want to clean up your system a bit. First, notice that this is an inhomogeneous linear syste …

Generally, you want there to be a non-characteristic transversal, i.e., a (let's say, smooth) curve $C$ in your domain $D$ such that each segment of each line $x=x_0$ in $D$ is connected and meets $C$ …

The standard method of constructing solutions is the following: First, observe that, if $(a,b)\in\mathbb{R}^n\times\mathbb{R}^n$ is any pair of vectors that satisfies $a\cdot b = 0$, and $h:\mathbb{R} …

Now that the question has been changed so extensively, the remarks that I made for the old version are no longer of any interest. Here is what I understand the problem to look like now: First, $\Omeg …

The answer is 'yes, a smooth, flat solution $f$ exists when $g$ is smooth and flat'. Here is one way to show this: I'll first do the case in which $g$ is even, i.e., $g(-p,-q)=g(p,q)$ and, for conve …

First of all, you have a sign wrong in your formula for the curvature. The curvature tensor you gave has positive constant sectional curvature +1 while you claim that you want negative sectional curv …

The answer is "No, it is not necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb{R}^2$". Here is a counterexample: Let $G(x,t) = x^2$ and let $h:\mathbb{R}\to\mathbb{R}$ be any smooth function …

There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example: First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual …

In general, the method of characteristics is not going to give you anything like the d'Alembertian solution of the wave equation unless $A$ and $B$ are simultaneously diagonalizable. For example, c …

First of all, your system seems to uncouple quite strongly. The first and third equations only involve the unknowns $v_1$ and $p_1$ and the second and fourth equations only involve $v_2$ and $p_2$, s …

Well, assuming that the matrix $\mathbf{B}$ is a real-analytic function of $t$, the local real-analytic theory gives you this result, which may or may not be useful to you: Start with a real-analyt …

This is not really an answer, just a sequence of comments that are all related, but are too long to put into a comment field. First, some good news: When $n=1$, there's always a (unique) solution f …