关于偏导数作用于含常导数PDE表达式的合理性与化简咨询

Great question—this hinges on resolving ambiguity in variable definitions and how partial derivatives interact with dependent variables like $x = x(t)$. Let’s break this down step by step:

1. The Ambiguity in the Original Expression #

The core issue is that $x = x(t)$ blurs the line between independent and dependent variables:

• In standard Eulerian notation (used most often in PDEs for fluid dynamics, heat transfer, etc.), $x$ is an independent spatial coordinate, and $t$ is an independent time coordinate. When you write $x = x(t)$, you’re describing the trajectory of a single particle moving through this fixed spatial grid.
• The partial derivative $\frac{\partial}{\partial x}$ is defined to hold $t$ constant. But if $x = x(t)$, $t$ and $x$ are no longer independent—this creates a conflict in how we interpret the derivative.

2. Calculating the Derivative Properly #

Let’s apply the product rule carefully, accounting for the dependency $t = t(x)$ (since $x = x(t)$ is invertible for most physical trajectories):
$$
\frac{\partial}{\partial x} \left( \frac{dx}{dt} \cdot u(x,t) \right) = \frac{\partial}{\partial x}\left( \frac{dx}{dt} \right) \cdot u(x,t) + \frac{dx}{dt} \cdot \frac{\partial u}{\partial x}
$$
Now, $\frac{dx}{dt}$ is a function of $t$, which in turn is a function of $x$ (via $x = x(t)$). Using the chain rule for the first term:
$$
\frac{\partial}{\partial x}\left( \frac{dx}{dt} \right) = \frac{d}{dt}\left( \frac{dx}{dt} \right) \cdot \frac{dt}{dx} = \frac{d^2x}{dt2} \cdot \frac{1}{\frac{dx}{dt}}
$$
Strictly speaking, the partial derivative $\frac{\partial u}{\partial x}$ holds $t$ fixed, but if we’re tying $t$ to $x$ via $x(t)$, this partial derivative isn’t the right tool to describe changes along the trajectory—we’d need to account for $u$’s time dependence too.

3. When Is It Equal to $\frac{dx}{dt} \cdot \frac{\partial u}{\partial x}$? #

The two expressions are only equivalent in very specific cases:

• If the trajectory has constant velocity: $\frac{d^2x}{dt2} = 0$, which eliminates the first term from the product rule.
• And if $u$ has no explicit time dependence ($\frac{\partial u}{\partial t} = 0$), so there’s no cross-term from the $x$-$t$ dependency.

In most general scenarios (e.g., accelerating trajectories, time-dependent $u$), the two expressions are not equal.

4. Fixing the Original Expression #

If your goal is to describe the rate of change of $u$ along the trajectory $x(t)$, you should use the total (material) derivative instead—it properly accounts for both $u$’s explicit time dependence and the particle’s movement through space:
$$
\frac{du}{dt} = \frac{\partial u}{\partial t} + \frac{dx}{dt} \cdot \frac{\partial u}{\partial x}
$$

内容的提问来源于stack exchange，提问作者FokkerPlankInquisitor