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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?

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  • $\begingroup$ over which domain do you integrate $1/v(x)$. what is the integration variable? $\endgroup$
    – user35593
    Commented Mar 23, 2019 at 6:09

1 Answer 1

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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.

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  • $\begingroup$ So you say that assuming that $v(x)$ is Lipschitz then the solution of this PDE with given initial condition is unique? $\endgroup$
    – user135936
    Commented Mar 24, 2019 at 4:00
  • $\begingroup$ Yes, it is a general fact for a first-order scalar PDE, which is in fact equivalent to a system of ODE. Lipschitz-continuity is a sufficient condition to get a proper definition for the flow $\phi$ above, but it is not necessary: looking at a system of ODE $\dot x=v(x)$, it is enough to know that $v\in W^{1,1}$ with a bounded divergence to get a flow. The latter situation was studied by P.L. Lions \& R. DiPerna. $\endgroup$
    – Bazin
    Commented Mar 24, 2019 at 11:33