# Does this PDE have a solution?

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?

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

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.