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.


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