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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 …

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 …

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 …

Another way to understand this 'transformation' is to think in terms of differential forms. Let $(u,v,\eta,\zeta)$ be coordinates on $\mathbb{R}^4$ and consider the pair of $2$-forms $$ \Upsilon_1 = …

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} …

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 …

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 …

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 …

Thanks for the clarification; I wasn't familiar with this terminology. I assume that the coefficients $a_i$, $b_i$, $c_i$, $d_i$, $e_i$, and $f_i$ are specified functions of $x_1,x_2,h_1,h_2$. (Let …

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 …

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 …

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 …

In this generality, it is not at all clear what you mean by 'any general rule'. Of course, there is a general rule: Compute the rank of the appropriate matrix or linear map. However, computing that …

No. Note that $V$ has dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is never ellipti …

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 …