Let $(M, \bar g)$, where $M=\mathbb{R}^n$, with coordinates $(x\_1, x\_2, .. x\_n)$, $f:\mathbb{R}^n \rightarrow \mathbb{R}$ and $\phi:\mathbb{R}^n \rightarrow \mathbb{R}$ with $n \geq 3$ and $f$ a positive function.


Let $g=\epsilon\_i \delta\_{ij}$ and $\bar g=\frac{1}{\phi^2}g$.


I denote $\phi\_{, x\_i x\_j}$, $\phi\_{, x\_i}$, $f\_{, x\_i x\_j}$ and $f\_{, x\_i}$ the second and the first order derivatives of $\phi$ and $f$ with respect to $x\_ix\_j$.


I consider smooth functions $\phi(\xi)$ and $f(\xi)$, where $\xi=\sum\_{i=1}^n\alpha\_ix\_i$, with $\alpha\_i \in \mathbb{R}$ and whenever $\sum\_{i=1}^n\alpha\_i^2\epsilon\_i \neq 0$, whitout loss of generality we may consider
$\sum\_{i=1}^n\alpha\_i^2\epsilon\_i=-1$.


So I mean:


$\phi\_{, x\_i x\_j}=\phi''\alpha\_i \alpha\_j$,


$\phi\_{, x\_i}= \phi' \alpha\_i$,


$f\_{, x\_i x\_j}=f'' \alpha\_i \alpha\_j$,


$f\_{, x\_i}=f'\alpha\_i$


I am trying to solve this system with these three PDE for *positive constant $\lambda$*:


**(I)** $(n-2)(f+1)\phi''-n\phi f''-2n\phi'f'=0$ for $i \neq j$.


**(II)** $-(f+1) \phi \phi'' + (n-1)(f+1) \phi'^2 - n\phi\phi' f'=\lambda (f+1)$ for $i=j$.


**(III)** $(f+1)\phi^2 f''-(n-2)(f+1)\phi \phi' f' + (n-1)\phi^2 f'^2=\lambda (f+1)^2$ for $i=j$.



PDE system solution on manifold with conformal metric
I don't know what the OP means by **canonical answer**. I assume that what's desired is to understand the solutions of the system of *three* second order ordinary differeential equations for *two* unknown positive functions $\phi(\xi)$ and $f(\xi)$.


At first glance, there's no reason to believe that any nontrivial solutions exist, since the system is overdetermined. One must first check compatibility conditions. Fortunately, this is easy to do, but to save writing (and typing), I'm going to simplify the notation: Instead of $\bigl(\xi,\phi(\xi),f(\xi)\bigr)$, I'll write $\bigl(t, u(t), v(t){-}1\bigr)$, and I'll also write $\lambda = nm^2/2>0$. The system of equations then becomes
$$\begin{aligned}
0 &= (n{-}2)\,v\,\ddot u-n\,u\,\ddot v - 2n\,\dot u\,\dot v\\
0&=-u\,v\,\ddot u - (n{-}1)\,v\,{\dot u}^2-n\,\dot u\,\dot v -\tfrac12 nm^2\,v\\
0&=v\,u^2\,\ddot v - (n{-}2)\,u\,v\,\dot u\,\dot v + (n{-}1)\,u^2{\dot v}^2-\tfrac12 nm^2\,v^2
\end{aligned}\tag1
$$
Solving the second and third equations for $\ddot u$ and $\ddot v$ and substituting them into the first equation, we find that the first equation can be replaced by a *first* order equation, namely
$$
0 = (n{-}2)\,v^2\,{\dot u}^2 - 2n\,u\,v\,\dot u\,\dot v + n\,u^2\,{\dot v}^2 - nm^2\,v^2 =: F(u,v,\dot u,\dot v).\tag2
$$
One then finds that, upon differentiating $F$ with respect to $t$ and then eliminating $\ddot u$ and $\ddot v$ using the second and third equations of (1), the resulting expression in $(u,v,\dot u,\dot v)$ is a multiple of $F(u,v,\dot u,\dot v)$. This shows that the combined system of equations (1) and (2) satisfies the compatibility conditions, so that the system has solutions, in fact, a 3-parameter family of them.


To describe these solutions more explicitly, note that the equations are $t$-autonomous and have a 2-parameter family of scaling symmetries. In particular, the equations are invariant under the $3$-parameter group of transformations of the form
$$
\Phi\_{a,b,c}(t,u,v) = (at{+}c,\,au,\,bv)
$$
where $a$ and $b$ are nonzero constants and $c$ is any constant.


In fact, the equation (2) implies that there is a function $p(t)$ such that
$$\begin{aligned}
\dot u &= \frac{2mn\,p\,(p{-}1)}
 {\bigl((n{-}2)\,p^2-2n\,p + n\bigr)}\\
\dot v &= \frac{m\,v\,\bigl((n{-}2)p^2-n\bigr)}
 {u\bigl((n{-}2)\,p^2-2n\,p + n\bigr)}
\end{aligned}\tag 3
$$
and then the second and third equations of (1) imply that $p$ must satisfy
$$
\dot p = \frac{m\bigl(n+2n\,p - (3n{-}2)\,p^2\bigr)}{u}\tag 4
$$
Conversely, the combined system of (3) and (4) gives the general solution of the original system.


This latter system is easily integrated by the usual separation of variables method: Eliminating $t$ yields a system of the form
$$
\frac{du}{u} = R(p)\,dp\qquad\text{and}\qquad
\frac{dv}{v} = S(p)\,dp
$$
where $R(p)$ and $S(p)$ are rational functions of $p$. Once $u$ and $v$ are written as elementary functions of $p$, then we can write
$$
dt = u\,T(p)\,dp,
$$
where $T$ is a rational function of $p$, so that $t$ can be written as a function of $p$ by quadrature. Thus, we have the integral curves in $(t,u,v,p)$-space in terms of explicit functions.


