# What can we say about transport equation?

If we have a transport equation, i.e.

$$u_t + \vec{b} \cdot D_x u=0,$$

is it true that at some point the particular directional derivative of $u$ becomes $0$, i didn't understand why? As Evans says that this property can be exploited to get the solution, but I didn't get how. Any help will be appreciated. Thank you.

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Please let me know why down vote to this question ? –  Theorem Jul 9 '12 at 13:54
Here I assume you are talking about the transport equation on $\Omega \subset \mathbb{R}^n \times \mathbb{R}$, so that $$u: \Omega \longrightarrow \mathbb{R}.$$ The transport equation can be rewritten as $$(\vec{b},1) \cdot D_{(x,t)} u = 0,$$ which is the directional derivative of $u$ in the direction of $(\vec{b},1)$. So the transport equation tells us that $u$ is constant along lines parallel to $(\vec{b}, 1)$ in $\mathbb{R}^{n+1}$.
As stated, however, the transport equation is not well-posed, as any constant function is a solution. One way to fix this is to introduce an initial condition: $$\begin{cases} u_t + \vec{b} \cdot D_x u = 0, \\ u(x,0) = g(x) \end{cases}$$ for some $g: \mathbb{R}^n \longrightarrow \mathbb{R}$. Now given $(x,t) \in \Omega)$, let $(x_0,0)$ be the point in $\mathbb{R}^n \times \{0\}$ on the line through $(x,t)$ parallel to $(\vec{b},t)$. Since $u$ is constant on such lines, we have that $$u(x,t) = u(x_0,0) = g(x_0) = g(x - t\vec{b}).$$