다변수 연쇄법칙

Recall that $u(x,t)=v\big(\xi(x,t),\eta(x,t)\big)$. By the chain rule

$$\begin{aligned} \frac{\partial u}{\partial x}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x} \\ &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \\ \frac{\partial u}{\partial t}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t} \\ &= -c\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +c\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$$

Again by the chain rule

$$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) &= \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] + \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big)\Big]\\ &= \frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\& + \frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\ &=\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) +2\frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial^2 u}{\partial t^2}(x,t) &=-c \frac{\partial }{\partial t} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] +c \frac{\partial }{\partial t} \Big[ \frac{\partial v}{\partial\eta} \big(\xi(x,t),\eta(x,t)\big)\Big] \\ &=-c \Big[\frac{\partial^2 v}{\partial \xi^2} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big] \\& +c \Big[\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big]\\ &=c^2\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) -2c^2\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) +c^2\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$$

so that

$$\frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t) =4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)$$

Hence

$$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 \text{ for all }(x,t) &\iff 4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)=0\text{ for all }(x,t)\\ &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta) \end{aligned}$$

(b) Now $\frac{\partial^2 v\,}{\partial\xi\partial\eta}\big(\xi,\eta\big) =\frac{\partial}{\partial\xi} \Big[\frac{\partial v}{\partial\eta}\Big]=0$. Temporarily rename $\frac{\partial v}{\partial \eta}=w$. The equation $\frac{\partial w}{\partial \xi}(\xi,\eta)=0$ says that, for each fixed $\eta$, $w(\xi,\eta)$ is a constant. The value of the constant may depend on $\eta$. That is, $\frac{\partial v}{\partial \eta}(\xi,\eta)=w(\xi,\eta) =H(\eta)$, for some function $H$. (As a check, observe that $\frac{\partial }{\partial \xi}H(\eta)=0$.) So the derivative of $v$ with respect to $\eta$, (viewing $\xi$ as a constant) is $H(\eta)$.

Let $G(\eta)$ be any function whose derivative is $H(\eta)$ (i.e. an indefinite integral of $H(\eta)$). Then $\frac{\partial }{\partial \eta}\big[v(\xi,\eta)-G(\eta)]=H(\eta)-H(\eta)=0$. This is the case if and only if, for each fixed $\xi$, $v(\xi,\eta)-G(\xi,\eta)$ is a constant, independent of $\eta$. That is, if and only if

$$v(\xi,\eta)-G(\eta)=F(\xi)$$

for some function $F$. Hence

$$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta)\\ &\iff v(\xi,\eta)=F(\xi)+G(\eta)\text{ for some functions $F$ and $G$}\\ &\iff u(x,t)=v\big(\xi(x,t),\eta(x,t)\big) =F\big(\xi(x,t)\big)+G\big(\eta(x,t)\big)\\ &\phantom{\iff u(x,t)\,}=F(x-ct)+G(x+ct) \end{aligned}$$

(c) We'll give the interpretation of $F(x-ct)$. The case $G(x+ct)$ is similar. Suppose that $u(x,t)=F(x-ct)$. Think of $u(x,t)$ as the height of water at position $x$ and time $t$. Pick any number $z$. All points $(x,t)$ in space time for which $x-ct=z$ have the same value of $u$, namely $F(z)$. So if you move so that your position is $x=z+ct$ (i.e. you move the right with speed $c$) you always see the same wave height. Thus $F(x-ct)$ represents a wave moving to the right with speed $c$.

Similarly, $G(x+ct)$ represents a wave moving to the left with speed $c$.