Transport equation

Remarks

For $g \in C^1(\R^n)$ the initial value problem has the unique $C^1$ solution $u(x, t) = g(x - tb)$. The reason is that $u$ is constant along each line $s \mapsto (x + sb, t + s)$, since the derivative of $u$ along it is $\partial_t u + b \cdot \nabla u = 0$. The inhomogeneous equation $\partial_t u + b \cdot \nabla u = f$ has the solution $u(x, t) = g(x - tb) + \int_0^t f(x + (s - t) b, s)\, ds$. With a variable velocity field $b(x, t)$ one also calls $\partial_t u + b \cdot \nabla u = 0$ a transport equation, and $\partial_t u + \nabla \cdot (u\, b) = 0$ the continuity equation; the two agree when $\nabla \cdot b = 0$.

Example: $n = 1$, $b = 2$, $g(x) = e^{-x^2}$, and $u(x, t) = e^{-(x - 2t)^2}$. Check: $\partial_t u = 4(x - 2t) u$ and $\partial_x u = -2(x - 2t) u$, so $\partial_t u + 2 \partial_x u = 0$, and $u(x, 0) = g(x)$.

Non-example: $u(x, t) = e^{-t} \sin x$ with $n = 1$ is not a solution for any $b$. Indeed $\partial_t u + b\, \partial_x u = e^{-t}(-\sin x + b \cos x)$, which is not identically zero.