Verification of solution to a transport equation

Question

Given $T > 0$ and some smooth function $\psi: \mathbb{R} \times (0,T) \to \mathbb{R} $ with compact support, let $v$ be the solution of the transport equation

$$
v_t + bv_x = \psi \qquad \text{ in } \mathbb{R} \times (0,T),
$$

with $v = 0$ when $t = T$ where $b(x,t)$ is some known function. I wish to verify that

$$
v(x,t) = – \int_t ^ T \psi(f(s),s) \,ds,
$$

where f satisfies $f(t) = x$, $\dot{f}(s) = b(f(s),s)$ is the solution to the problem. I think I can calculate the partial derivatives of $v$ okay but can't seem to get enough cancellations to get what I want. Any help would be appreciated!

Answer 1

The notation is confusing. Indeed, one has implicitly assumed that we are following the solution along a characteristics curve $x=f(s)$, $t=s$. The method of characteristics gives

• $\frac{\text d t}{\text d s}=1$, letting $t(0)=0$, we have $t(s)= s$.
• $\frac{\text d x}{\text d s}=b(x(s),s)$, letting $x(s)=f(s)$, we have $\dot f(s) = b(f(s),s)$.
• $\frac{\text d v}{\text d s}=\psi(x(s),s)$, letting $v(T)=0$, we have $v(s) = \int_T^s \psi(f(\sigma),\sigma)\,\text d\sigma$.

The other way round, we have $\frac{\text d v}{\text d s} = \psi(f(s),s)$ along the characteristics. Moreover, $\frac{\text d v}{\text d s} = \frac{\partial v}{\partial t} \frac{\text d t}{\text d s} + \frac{\partial v}{\partial x} \frac{\text d x}{\text d s}$ by the chain rule, so that $$ \frac{\partial v}{\partial t} + b(x,t)\frac{\partial v}{\partial x} = \psi(x,t) \, . $$