analytical solution of inhomogeneous transport PDE with arbitrary time-dependent velocity

Question

I am trying to find an analytic expression for the solution of the transport PDE: \begin{align} \newcommand{\pdrev}[2]{\dfrac{\partial #1}{\partial #2}} \newcommand{\drev}[2]{\dfrac{d #1}{d #2}} &\pdrev{u}{t} + v(t)\pdrev{u}{x} = \lambda(\theta(t)-u), \quad x\in[0,L],\quad v(t) \in (0,\infty) \\ &u(x=0,t)=u_{in}(t) \end{align} until now, I have tried the method of characteristics, however there is a rigor gap in my solution where I have to assume a constant $v$ to relate the constants of the solutions of the ODEs produced from the characteristics, my final solution is \begin{align} &u(x^*,t)=\theta(t)-\lambda e^{-\lambda t}\int\limits_{t-\tau}^{t}e^{\lambda s} \theta'(s)ds + \Big[u_{in}(t-\tau)-\theta(t-\tau)\Big]e^{-\lambda\tau(x,t)} \\ \end{align} where $\tau(x^*,t)$ is the velocity dependent time needed to cross a distance $x^*$ of the domain and defined by the implicit integral \begin{align} &x^*=\int\limits_{t-\tau}^{t}v(s)ds \end{align} can anyone help me with the derivation

Answer 1

Let's apply the method of characteristics

• $\frac{dt}{ds} = 1$, letting $t(0) = t_0$, we know $t = s+t_0$. The chain rule gives $\frac{d}{ds} = \frac{d}{dt}$.
• $\frac{dx}{ds} = v(t)$, letting $x(0) = 0$, we know $x = \int_{0}^{s} v(\sigma+t_0)\, d\sigma$
• $\frac{du}{ds} = \lambda (\theta - u)$, letting $u(0) = u_{in}(t_0)$, we know $u = u_{in}(t_0)\, e^{-\lambda s} + \lambda\int_0^s e^{\lambda (\sigma - s)} \theta(\sigma + t_0)\, d\sigma$.

It remains to substitute $s$, $t_0$ in the expression of $u$ by using the expressions of $t$, $x$. However, to be able to do so, we need to invert the mapping $\Phi: (s,t_0) \mapsto (t,x)$ defined by $$ \Phi(s,t_0) = \big(s+t_0 , V(s + t_0) - V(t_0)\big) = (t,x) \, , $$ where $V: s\mapsto \int^s v(\tau) \, d\tau$ is an antiderivative of $v \geq 0$. From the sign of $v$, we deduce that $V$ is continuous and monotonously increasing. Thus, we can define an inverse $V^{-1}$. By substitution of $s = t-t_0$ in the expression of $x$, we find $V(t_0) = V(t)- x$, so that finally, $$ \Phi^{-1}(t,x) = \left(t-V^{-1}(V(t)-x) , V^{-1}(V(t)-x) \right) = (s,t_0) \, . $$ In the case where $v$ is constant, we find $(s,t_0) = (x/v, t-x/v)$.