Let $U_0 \subset \mathbb{R}^3$ be an open neighborhood of $0$ and $X:U_0 \to\mathbb{R}^3$ a smooth vector field, such that $$X(x,y,z) = (X_1(x,y,z),1,0). $$

where $X_1:U_0 \to\mathbb{R}$, satisfies the following hypotheses:

  • $X_1(x,0,z) =0,$ $\forall$ $(x,0,z)$ $\in$ $U_0$.
  • $\frac{\partial}{\partial x}X_1(x,0,z)=\frac{\partial}{\partial y}X_1(x,0,z) =0$, $\forall$ $(x,0,z) \in U_0$
  • $\frac{\partial}{\partial y}X_1(x,0,z)\neq 0$ $\forall (x,0,z) \in U_0$
  • $X_1(0,0,0)=(0,0,0)$.

I need to find a change of coordinates $\varphi: W_0\subset \mathbb{R}^3 \to V_0\subset \mathbb{R}^3$ ($\varphi(0)=0$), such that $$Z(x,y,z) = \text{d}\varphi_{\varphi^{-1}(x,y,z)}X(\varphi^{-1}(x,y,z)) = (y,1,0) $$ or $$Z(x,y,z) = \text{d}\varphi_{\varphi^{-1}(x,y,z)}X(\varphi^{-1}(x,y,z)) = (-y,1,0). $$

The problem that I am facing does not allow me to mess up the z and y-axes in my coordinate system.

A natural way to solve this problem is trying to find this changing of coordinates in the following form $$\varphi(x,y,z) = (f(x,y,z),y,z).$$

(with $\frac {\partial f}{\partial x} (0) \neq 0$ and $f(0,0,0)=0$). If this coordinate change works, we would get

$$Z(\varphi(x,y,z)) = \text{d}\varphi(x,y,z) \cdot X(x,y,z)$$ $$(\pm y,1,0) = \left( \frac{\partial f}{\partial x}(x,y,z) X_1(x,y,z) + \frac{\partial f}{\partial y} (x,y,z) ,1,0 \right) .$$

And then my question appears, does anyone know if this PDE has a solution?

$$ \frac{\partial f}{\partial x}(x,y,z) X_1(x,y,z) + \frac{\partial f}{\partial y} (x,y,z)= \pm y,$$ $$f(0,0,0)=0,$$ $$\frac{\partial f}{\partial x}(0,0,0) \neq 0. $$

up vote 5 down vote accepted

The main idea is to use the characteristic method. It is motivated by the following equation: $$\langle \left(\nabla f\circ \gamma\right)(t), \gamma'(t) \rangle = g(t).$$ The curve $\gamma$ is called the characteristic cuve.

The first equality follows from the chain rule: remember that:

$$\frac{d}{dt}f\circ \gamma(t) = \langle \nabla f\circ \gamma(t),\gamma'(t)\rangle. $$

Note that $$\langle \left(\nabla f\circ \gamma\right)(t), \gamma'(t) \rangle = g(t) = \sum_{j=1}^3\frac{\partial f}{\partial x_i}\frac{dx^i}{dt},$$ therefore, one convert the PDE on a system of ODE's. After all, we isolate $t$ to recover the dependence of the coordinates.

Therefore, one must search for solutions for the problem:

$$\frac{dx}{dt} = X_1(x,y,z),$$ $$\frac{dy}{dt} = 1,$$ $$\frac{df}{dt} = \pm y,$$ and $\frac{dz}{dt}$ can be free.

Then, one has:

$$y(t) = t + c,$$ so $$\frac{df}{dt} = \pm (t+c),$$ and therefore, $$f(t) = \pm(\frac{t^2}{2} + ct) + d.$$

Once $t = y - c,$ we have:

$$f(x,y,z) = \pm((y-c)^2 + c(y-c)) + d.$$

Now, we want to relate our equation with $X_1.$ Once $t = y -c$, one has that $\frac{dx}{dt} = \frac{dx}{dy}$. Therefore, we have:

$$\frac{dx}{dy} = X_1(x,y,z).$$ This is a non autonomous ODE. To finish, convert this equation on an autonomous equation and impose your initial data, this must determine $x$ in terms of $y$. Use inverse theorem to obtain your function $f$ in terms of $X_1$. I can add more details if needed.

Note that the solution we have obtained if a function of $y$. Since you want to obtain a solution in terms of $x$ you could follow two ways:

$1)$ Instead of isolating $t = y-c$ you could just to plug $y = t+c$ on the equation for $x$, obtaining:

$$\frac{dx}{dt} = X_1(x,t+c,z).$$

This is an non autonomous ODE. You will obtain a solution $x(t)$. Therefore, you must isolate $t$ in terms of $x$, and then, plug this on the expression for $f(t)$ to obtain $f$ in terms of $x, X_1.$

$2)$ Follow the way I suggested at first. Solve the equation for $x(y)$. This is what I meant at first. But $1)$ seems more natural, although it leads to the same answer.

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