I'm doing exercises in the first chapter of Serge Alinhac's "Hyperbolic Partial Differential Equations", and while most of the other exercises for this chapter are straightforward, I'm stuck on Q3:
Let $X$ be a non-vanishing field in $\mathbb R^n$ and $a\in C^1(\mathbb R^n)$ a complex function. Explain how the study of the equation $Xu+au = f$ can be (locally) reduced to the study of the equation $Xv=g$.
Here, Serge is using the notation $Xu \triangleq X\cdot\nabla u$; $u$ is a scalar function (presumably, due to $a$, taking values in $\mathbb C$). I would assume $f,g$ can be taken to be smooth.
Things I've tried:
The "(locally)" that appears in the question suggests to me a change of variables. But if I say let tildes mean composition with some nice $\phi$, e.g. $\tilde u=u\circ \phi$ and so on, all I end up with is $(\tilde X \cdot ((\nabla \phi)^{-1}\circ \phi)\nabla \tilde u) + \tilde a \tilde u = \tilde f$ which doesn't seem helpful.
I tried to consider the real and imaginary parts. (Earlier in the book, it is remarked that $\partial_t + i\partial_x$ is the Cauchy-Riemann operator; the real and imaginary parts of $\partial_tu + i\partial_xu =0$ form precisely the standard system called the Cauchy-Riemann equations) But I don't see any way to continue from the resulting system. If say $u=v+iw$, $f=g+ih$, and $a=b+ic$, then the two equations you get for $(v,w)$ are $$ Xv+bv - cw = g, \\ Xw + cv + bw = h.$$
I tried to use what I know from basic ODE theory, namely integrating factors. If I could solve $Xz = az$, with $z\neq 0$ everywhere, then by Leibniz rule, $$ X (uz) = zXu + uX z = z Xu + au z $$ Then $$Xu + au = f\iff zXu + auz = fz \iff X(uz) = fz$$ Of course, one takes $uz=v$ and $fz=g$, to match the question's notation, so this feels like the most promising approach. But I don't get what the complex part of $a$ is supposed to do for me, and the book did not consider solving things like $Xz = az$ yet, nor am I 100% sure its always possible to solve this equation.
Are any of these a step in the right direction, and does anyone have some pointers?
(edit after sleeping) So in the case where $X=\partial_1$ and $a$ takes purely imaginary values, it works out like this. You start the solution of $Xz=az$ at whatever $z_0$ say $z_0=1$. The solution is $z(x) = z_0 \exp(\int_0^{x_1} a(s,x_2,\dots,x_n) ds)$. My stronger assumption on $a$ means that $$\partial_1\left( \frac{|z|^2}2 \right) = \Re(\partial_1 z\overline z)= \Re ( a |z|^2) = 0$$ so that $|z|^2$ is a constant, and in particular $z\neq 0 $ everywhere.
My lack of formal training in differential geometry is showing, but I suppose from try #1 above, the nonvanishing of $X$ allows me to reduce to this case? I guess I need to solve $\partial_1 = \tilde X \cdot ((\nabla \phi)^{-1}\circ \phi)$?
And also, my hunch now is that the result is false if $a$ can take more general complex values...
I suppose that Alignac provides details for showing that $Xv=g$ admits smooth local solutions in a neighborhood of a point where $X$ is non-zero? (e.g. by integrating along the flow of X).
After that you pose and solve locally, as suggested in comments, $XA=a$ and then $XB=e^{A}f$ (there is no need to separated into real and imaginary parts). After that $u=Be^{-A}$ provides a local solution to the problem since $$ X u + a u = (X + a) (B e^{-A}) = (XB) e^{-A}-B a e^{-A} +a B e^{-A} = f $$
The non-vanishing of $X$ ensures that the problems are equivalent. Otherwise you run into counter examples like: $$ x u'(x) + u(x) = 1 $$ which does have a smooth solution ($u\equiv 1$) but for which $x A'(x)=1$ does not have a solution in a neighborhood of zero.