链式法则偏导数与波动方程：证明波动方程的通解结构

Hey there, let's walk through this proof step by step—we'll split it into two parts: first verifying that functions of the form $f(x,t)=g(x+ct)+h(x-ct)$ satisfy the wave equation, then proving every $C^2$ solution of the wave equation must have this form.

We just need to compute the second partial derivatives and substitute into the equation. Let's use the chain rule here:

• First, compute the spatial partial derivatives:

  • $f_x = g'(x+ct) \cdot \frac{\partial}{\partial x}(x+ct) + h'(x-ct) \cdot \frac{\partial}{\partial x}(x-ct) = g'(x+ct) + h'(x-ct)$
  • $f_{xx} = g''(x+ct) \cdot \frac{\partial}{\partial x}(x+ct) + h''(x-ct) \cdot \frac{\partial}{\partial x}(x-ct) = g''(x+ct) + h''(x-ct)$

• Next, compute the temporal partial derivatives:

  • $f_t = g'(x+ct) \cdot \frac{\partial}{\partial t}(x+ct) + h'(x-ct) \cdot \frac{\partial}{\partial t}(x-ct) = c \cdot g'(x+ct) - c \cdot h'(x-ct)$
  • $f_{tt} = c \cdot g''(x+ct) \cdot \frac{\partial}{\partial t}(x+ct) - c \cdot h''(x-ct) \cdot \frac{\partial}{\partial t}(x-ct) = c^2 g''(x+ct) + c^2 h''(x-ct)$

Now substitute $f_{xx}$ and $f_{tt}$ into the wave equation:
$c^2 f_{xx} = c^2\left(g''(x+ct) + h''(x-ct)\right) = f_{tt}$
Perfect, that's exactly the wave equation. So any function of this form is a solution.

We'll use a change of variables to simplify the wave equation. Let:
$\xi = x + ct$, $\eta = x - ct$

First, we need to express the second partial derivatives $f_{xx}$ and $f_{tt}$ in terms of $\xi$ and $\eta$. Using the chain rule for partial derivatives:

• Compute $f_x$:
  $f_x = f_\xi \cdot \frac{\partial \xi}{\partial x} + f_\eta \cdot \frac{\partial \eta}{\partial x} = f_\xi + f_\eta$

• Compute $f_{xx}$:
  $f_{xx} = \frac{\partial}{\partial x}(f_\xi + f_\eta) = f_{\xi\xi} \cdot \frac{\partial \xi}{\partial x} + f_{\xi\eta} \cdot \frac{\partial \eta}{\partial x} + f_{\eta\xi} \cdot \frac{\partial \xi}{\partial x} + f_{\eta\eta} \cdot \frac{\partial \eta}{\partial x}$
  Since $f$ is $C^2$, mixed partial derivatives are equal ($f_{\xi\eta}=f_{\eta\xi}$), so this simplifies to:
  $f_{xx} = f_{\xi\xi} + 2f_{\xi\eta} + f_{\eta\eta}$

• Compute $f_t$:
  $f_t = f_\xi \cdot \frac{\partial \xi}{\partial t} + f_\eta \cdot \frac{\partial \eta}{\partial t} = c f_\xi - c f_\eta$

• Compute $f_{tt}$:
  $f_{tt} = \frac{\partial}{\partial t}(c f_\xi - c f_\eta) = c\left(f_{\xi\xi} \cdot \frac{\partial \xi}{\partial t} + f_{\xi\eta} \cdot \frac{\partial \eta}{\partial t}\right) - c\left(f_{\eta\xi} \cdot \frac{\partial \xi}{\partial t} + f_{\eta\eta} \cdot \frac{\partial \eta}{\partial t}\right)$
  Substitute $\frac{\partial \xi}{\partial t}=c$ and $\frac{\partial \eta}{\partial t}=-c$, and again use $f_{\xi\eta}=f_{\eta\xi}$:
  $f_{tt} = c^2 f_{\xi\xi} - 2c^2 f_{\xi\eta} + c^2 f_{\eta\eta} = c^2\left(f_{\xi\xi} - 2f_{\xi\eta} + f_{\eta\eta}\right)$

Now plug $f_{xx}$ and $f_{tt}$ into the wave equation $c^2f_{xx}=f_{tt}$:
$c^2\left(f_{\xi\xi} + 2f_{\xi\eta} + f_{\eta\eta}\right) = c^2\left(f_{\xi\xi} - 2f_{\xi\eta} + f_{\eta\eta}\right)$
Since $c>0$, we can divide both sides by $c^2$:
$f_{\xi\xi} + 2f_{\xi\eta} + f_{\eta\eta} = f_{\xi\xi} - 2f_{\xi\eta} + f_{\eta\eta}$
Cancel out the identical terms on both sides, and we get:
$4f_{\xi\eta}=0 \implies f_{\xi\eta}=0$

Now we solve the equation $f_{\xi\eta}=0$:

1. First, integrate with respect to $\eta$: this gives $f_\xi = \phi(\xi)$, where $\phi(\xi)$ is an arbitrary $C^1$ function (since $f$ is $C^2$, the partial derivative $f_\xi$ is $C^1$).
2. Then integrate with respect to $\xi$: $f(\xi,\eta) = \int \phi(\xi) d\xi + \psi(\eta)$. Let $g(\xi) = \int \phi(\xi) d\xi$ (which is $C^2$ because $\phi$ is $C^1$) and $h(\eta)=\psi(\eta)$ (also a $C^2$ function). So we have:
   $f(\xi,\eta)=g(\xi)+h(\eta)$

Finally, substitute back $\xi=x+ct$ and $\eta=x-ct$, and we get:
$f(x,t)=g(x+ct)+h(x-ct)$
Which is exactly the form we needed to prove.

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