关于偏微分方程$u_{t}+cu_x=f(x+ct)$求解推导的技术咨询

Hey Danxe, let's walk through each of your questions clearly—they all center on the method of characteristics, which is the standard approach for solving first-order linear PDEs like the one you're working with.

(1) Deriving $\frac{dx}{dt}=c$ (and the general case $\frac{dy}{dx}=\frac{b}{a}$) #

The core idea here is to convert the PDE into an ordinary differential equation (ODE) by looking at how $u$ changes along specific curves called characteristic curves. Let's start with your specific equation:
$$u_t + c u_x = f(x+ct)$$

Think of $u$ as a function of $t$ and $x(t)$ (i.e., $u$ depends on $t$, and $x$ itself is a function of $t$). Using the chain rule for total derivatives:
$$\frac{du}{dt} = u_t + u_x \frac{dx}{dt}$$

Compare this to your PDE: if we set $\frac{dx}{dt}=c$, the total derivative simplifies exactly to the left-hand side of your equation:
$$\frac{du}{dt} = u_t + c u_x = f(x+ct)$$
Now we've turned the PDE into an ODE for $u$ along the curve defined by $\frac{dx}{dt}=c$—that's why we define this characteristic curve.

For the general case $a u_x + b u_y = g(x,y)$:
We use the same chain rule logic, but now treat $u$ as a function of $x(y)$ (so $x$ depends on $y$). The total derivative of $u$ with respect to $y$ is:
$$\frac{du}{dy} = u_x \frac{dx}{dy} + u_y$$
Rearrange the original PDE to match this form: divide both sides by $a$ to get $u_x + \frac{b}{a} u_y = \frac{g}{a}$. If we set $\frac{dx}{dy}=\frac{b}{a}$, the total derivative becomes:
$$\frac{du}{dy} = u_x \frac{b}{a} + u_y = \frac{g}{a}$$
Which again converts the PDE into an ODE along the characteristic curve defined by $\frac{dx}{dy}=\frac{b}{a}$. Alternatively, you'll often see this written as the characteristic direction: $\frac{dx}{a} = \frac{dy}{b}$, which is equivalent to $\frac{dy}{dx}=\frac{b}{a}$.

(2) The purpose of $x(h,0)=h$ #

The equation $\frac{dx}{dt}=c$ gives us the family of characteristic curves, but it's a first-order ODE with infinitely many solutions (one for each constant of integration). The condition $x(h,0)=h$ is an initial condition that pins down exactly which characteristic curve corresponds to each starting point.

Let's solve $\frac{dx}{dt}=c$: the general solution is $x(t) = c t + C$, where $C$ is a constant. When $t=0$, we set $x(0)=h$, so $C=h$. That means each characteristic curve is defined by $x(t) = c t + h$, or rearranged, $h = x - c t$.

This is crucial because it lets us link the value of $u$ at time $t$ and position $x$ back to the initial value of $u$ at time $t=0$ and position $h=x-ct$. Without this initial condition, we couldn't uniquely determine the solution to the PDE—we'd just have a general form without tying it to the problem's starting state.

(3) Unfinished next steps? #

You mentioned "后续……" didn't complete your question. If you share the remaining steps you're confused about (like solving the ODE for $u$, or applying initial conditions to get the final solution), I can walk through those with you too.

内容的提问来源于stack exchange，提问作者Danxe