📂Partial Differential Equations

Transport Equation

Definition¹

Below is referred to as a transport equation.

$$ \begin{equation} u_{t} + b \cdot Du=0\quad \text{in }\mathbb{R}^n \times (0,\ \infty) \end{equation} $$

• $b=(b_{1}, b_2, \cdot, b_{n}) \in \mathbb{R}^n$ is a fixed vector
• $u=u(x,t)$ is $u:\mathbb{R}^n \times [0,\infty) \rightarrow \mathbb{R}$
• $x=(x_{1}, \cdots , x_{n})\in \mathbb{R}^n$
• $t \ge 0$ is time
• $Du=D_{x}u=(u_{x_{1}}, \cdots ,u_{x_{n}})$ is the gradient of $u$ with respect to the spatial variable $x$

Explanation

Assume $u \in C^1$ is a solution to $(1)$. Then, along the line in direction $(b,1)$ passing through a fixed point $(x,t)$, on $(x+sb,\ t+s)=(x,\ t)+s(b,\ 1)$, $u$ is constant. That is, $u(x+sb,\ t+s)$ is independent of $s$. This can be verified as follows. Let’s define $z$ as follows.

$$ z(s):=u(x+sb,\ t+s)\quad (s \in \mathbb{R}) $$

It is sufficient to show $\dfrac{dz(s)}{ds}=0$.

$$ \begin{align*} \dot{z}(s) &= \dfrac{dz}{ds} \\ &= \dfrac{\partial u}{\partial x}\dfrac{d x}{d s} + \dfrac{\partial u}{\partial t}\dfrac{d t}{d s} \\ &= \dfrac{\partial u(x+sb,\ t+s)}{\partial x}\cdot\dfrac{d(x+sb)}{ds}+u_{t}(x+sb,\ t+s) \\ &= Du(x+sb,\ t+s) \cdot b +u_{t}(x+sb,\ t+s) \\ &= 0 \end{align*} $$

Since $u$ satisfies $(1)$, the last equality holds.

See Also

• Solving initial value and inhomogeneous problems