First order partial differential equation I know there is a solution to this pde
$$\partial_{t} f(t,x)=  \partial_{x}(v(x)f(t,x))$$
$$ f(0,x)=g(x)$$
( Where $v$ and $g$ are known functions) 
which is given by 
$$ f(t,x)=\frac{1}{v(x)} h(t+\int \frac{1}{v(x)})$$ 
where $h(x)$ is determined by initial condition $g(x)$.
The question is if I use the method of characteristic would it give me the same solution with this initial condition?
 A: Your equation is linear, first-order and can be written as 
$$
\frac{\partial f}{\partial t}-v(x)\frac{\partial f}{\partial x}=v'(x) f.
\tag{$\ast$}$$
Using the characteristics of the vector field, you solve the (a priori non-linear) ODE, assuming say that $v$ is Lipschitz-continuous,
$$
\dot \phi(t,y)=-v(\phi(t,y)),\quad \phi(0,y)=y.
$$
Setting $w(t)=f(t,\phi(t,y))$, you see that if $f$ is a solution of your equation $(\ast)$, you have
$$
\dot w(t)=v'(\phi(t,y)) w(t),
\tag{$\ast\ast$}$$
 a linear scalar equation with solution
$
w(t)=w(0)\exp\bigl[\int_0^tv'(\phi(s,y))ds\bigr]
$. As a result,
$$
f(t,\phi(t,y))=g(\phi(t,y))\exp\bigl[\int_0^tv'(\phi(s,y))ds\bigr],
$$
and inverting the flow $\phi$ (at least for $t$ close to 0) with $x=\phi(t,y)$ equivalent to $y=\psi(t,x)$, we get
$$
f(t,x)=g(x)\exp\bigl[\int_0^tv'\bigr(\phi(s,\psi(t,x))\bigl)ds\bigr].
$$
Your solution is written as
$
F=\frac{1}{v(x)} h(t+\omega(x)), \quad \text{$\omega$ antiderivative of $1/v$}
$,
so that the initial condition forces
$g=h(\omega)/v$ and thus
$$
\tilde w(t)=\frac{h(t+\omega(\phi(t,y)))}{v(\phi(t,y))},
$$
is such that 
$\tilde w(0)=w(0)$ and satisfies the ODE $(\ast\ast)$. By uniqueness for this ODE, you get indeed the solution given by the method of characteristics.
