Question: Let $A(z)$ be smooth with nonvanishing derivative $A'(z)$, and consider the system with first-order PDE's,
$$\frac{\partial u}{\partial x}=\frac{\partial K}{\partial x}(x,y)\frac{1}{A'(u)}$$
$$\frac{\partial u}{\partial y}=\frac{\partial K}{\partial y}(x,y)\frac{1}{A'(u)}$$
satisfying the initial data $u(x_0,y_0)=u_0$
Useing Frobenius theorem, show that a solution $u(x,y)$ exists, and find the algebraic equation that $u(x,y)$ satisfies
Answer:(Given as a hint) $A(u)-A(u_0)=K(x,y)-K(x_0,y_0)$
My attempt: I have found the $\mathbb{X}$ and $\mathbb{Y}$ to be:
$$\mathbb{X}=(1,0,\frac{\partial K}{\partial x}(x,y)\frac{1}{A'(z)})$$ and
$$\mathbb{Y}=(0,1,\frac{\partial K}{\partial y}(x,y)\frac{1}{A'(z)})$$
Computing $[\mathbb{X},\mathbb{Y}]$ I got the result $0$, so have the necessary condition for solution for differential equation and initial conditions to exist.
Point of contention: When I compute the flow $\Psi_s$ for $\mathbb{Y}$, I run into a problem.
We have: $x'=0,y'=1,z'=\frac{\partial K}{\partial y}(x,y)\frac{1}{A'(z)},x(0)=x_0,y(0)=y_0,z(0)=u_0$ giving:
$$x(s)=x_0,y(s)=s+y_0,z(s):\frac{\partial z}{\partial t}=\frac{\partial K}{\partial y}(x,y)\frac{1}{A'(z)}\Leftrightarrow\int_{u_0}^uA'(z)dz=\int_0^s\frac{\partial K}{\partial y}(x,y)ds\Leftrightarrow A(z)-A(u_0)=\frac{\partial K}{\partial y}(x,y)s$$
Where do I go from here, how do I obtain a function for $z(s)$?
Did I do this integral correctly? I am not sure what $A'(z)$ is differentiated with respect to $z$ or not.
I appear to be close to the answer and I feel that after I overcome this small obstacle I will be fine.
Progress: Multiplying by $A'(u)$ to simplify I get
$$\frac{\partial A(u)}{\partial x}=\frac{\partial k(x,y)}{\partial x}$$ $$\frac{\partial A(u)}{\partial y}=\frac{\partial k(x,y)}{\partial y}$$ with initial data $u(x_0,y_0)=u_0$
I compute the vector fields as follows: $\mathbb{X}=(1,0,\frac{\partial k(x,y)}{\partial x})$ and $\mathbb{Y}=(0,1,\frac{\partial k(x,y)}{\partial y})$ Calculating $[\mathbb{X},\mathbb{Y}]$ I get the result $0$ so the necessary condition is satisfied and the solution for $u(x,y)$ exists.
I now need to compute $(\Phi_x\circ\Psi_y)(x_0,y_0,z_0)$
For $\Psi_s$ the flow of vector field $\mathbb{Y}$:
We have: $x'=0,y'=1,z'=\frac{\partial K}{\partial y}(x,y),x(0)=x_0,y(0)=y_0,z(0)=u_0$
giving:
$$x(s)=x_0,y(s)=s+y_0,z(s):\frac{\partial z}{\partial s}=\frac{\partial K(x,y)}{\partial y}$$
How do I continue from here, seem further away from the answer than I previously was, don't know how to deal with this simplification.