# Proving the solution of a first order partial differential equation.

I am trying to solve the next partial differential equation, for some region in $$\Re^3$$:

$$-\frac{\partial \rho}{\partial t}=\frac{\partial \rho}{\partial x} v_{x}+\frac{\partial \rho}{\partial y} v_y$$ where $$\rho(t,x,y)$$ is a scalar-function of $$(t,x,y)$$, $$v_x(x,y)$$ is a scalar function of $$(x,y)$$ and $$v_y(x,y)$$ is also a scalar-function of $$(x,y)$$.

If we define the vector field $$\vec{V}=(1,v_x,v_y)$$ then the equation has the form: $$L_{\vec{V}}\rho=0$$ where $$L_{\vec{V}}$$ is the derivative in the direction of the vector field. This means that $$\rho$$ is a first integral of $$\vec{V}$$. So suppose we can found a function $$g= (g_x(t,x,y),g_y(t,x,y))$$ such that: $$\frac{\partial}{\partial t}(g_x(t,x,y))=v_x(g_x(t,x,y),g_y(t,x,y)) \quad \frac{\partial}{\partial t}(g_y(t,x,y))=v_y(g_x(t,x,y),g_y(t,x,y))$$

i.e. $$g$$ is the solution of the differential equation $$(\dot{\alpha},\dot{\beta})=(v_x(\alpha,\beta),v_y(\alpha,\beta))$$. Then by the argument of a first integral one possible solution for the differential equation is: $$\rho(t,x,y)=\rho_0(g_x(-t,x,y),g_y(-t,x,y))$$ where $$\rho_0(a,b)$$ is a scalar function of $$(a,b)$$ (similar to some initial conditions).

Until this point everything is good, but when I try to plug the solution back to the partial differential equation to check the solution I have some problems. For example:

$$\frac{\partial}{\partial t} (\rho)= \frac{\partial \rho_0}{\partial a} (g_x(-t,x,y),g_y(-t,x,y)) \frac{\partial}{\partial t} (g_x(-t,x,y))+ \frac{\partial \rho_0}{\partial b} (g_x(-t,x,y),g_y(-t,x,y)) \frac{\partial}{\partial t} (g_y(-t,x,y))$$ $$=(-\frac{\partial \rho_0}{\partial a} v_x - \frac{\partial \rho_0}{\partial b} v_y)|_{(g_x(-t,x,y),g_y(-t,x,y))}$$

where everything is evaluated in $$(g_x(-t,x,y),g_y(-t,x,y))$$. But when I try the other derivatives: $$\frac{\partial \rho}{\partial x}= \frac{\partial}{\partial x}[\rho_0(g_x(x,y),g_y(x,y))]$$ $$=\frac{\partial \rho_0}{\partial a}|_{(g_x(-t,x,y),g_y(-t,x,y))} \frac{\partial g_x}{\partial x}|_{(-t,x,y)} +\frac{\partial \rho_0}{\partial b}|_{(g_x(-t,x,y),g_y(-t,x,y))} \frac{\partial g_y}{\partial x}|_{(-t,x,y)}$$ and $$\frac{\partial \rho}{\partial y}= \frac{\partial}{\partial y}[\rho_0(g_x(x,y),g_y(x,y))]$$ $$=\frac{\partial \rho_0}{\partial a}|_{(g_x(-t,x,y),g_y(-t,x,y))} \frac{\partial g_x}{\partial y}|_{(-t,x,y)} +\frac{\partial \rho_0}{\partial b}|_{(g_x(-t,x,y),g_y(-t,x,y))} \frac{\partial g_y}{\partial y}|_{(-t,x,y)}$$

Introducing this values for the partial differential equation I would expect that, if the solution is to hold for arbitrary initial funtion, then: $$v_x|_{(g_x(-t,x,y),g_y(-t,x,y))}= \frac{\partial g_x}{\partial x}|_{(-t,x,y)} v_x (x,y)+ \frac{ \partial g_x}{\partial y}|_{(-t,x,y)} v_y(x,y)$$ $$v_y|_{(g_x(-t,x,y),g_y(-t,x,y))}= \frac{\partial g_y}{\partial x}|_{(-t,x,y)} v_x (x,y)+ \frac{ \partial g_y}{\partial y}|_{(-t,x,y)} v_y(x,y)$$

Or equivalently: $$\frac{\partial}{\partial t} (g_x)|_{(t,x,y)}= \frac{\partial g_x}{\partial x}|_{(t,x,y)} v_x (x,y)+ \frac{ \partial g_x}{\partial y}|_{(-t,x,y)} v_y(x,y)$$ $$\frac{\partial}{\partial t} (g_y)|_{(t,x,y)}= \frac{\partial g_y}{\partial x}|_{(-t,x,y)} v_x (x,y)+ \frac{ \partial g_y}{\partial y}|_{(-t,x,y)} v_y(x,y)$$

This is the part where I get stuck. In summary I just want to plug back a solution and check if it is actually a solution, but I do not know if some how to prove some relationship. My question is: Is there a formal way to prove the last part for an arbitrary $$g$$? Or perhaps I committed a mistake in my deduction or reasoning, if so, where?

Thank you so much.