다변수 연쇄법칙

(a) We are told that

$$x(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u$$

Applying $\frac{\partial }{\partial u}$ to both equations gives

$$\begin{aligned} 2x(u,v)\frac{\partial x}{\partial u}(u,v) -\frac{\partial y}{\partial u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0 \\ 2x(u,v)\frac{\partial x}{\partial u}(u,v) +2y(u,v)\frac{\partial y}{\partial u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{aligned}$$

Setting $u=\frac{\pi}{2}$, $v=0$, $x\big(\frac{\pi}{2},0\big)=1$, $y\big(\frac{\pi}{2},0\big)=1$ gives

$$\begin{aligned} 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) -\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right)&=0 \\[0.1in] 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) +2\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{aligned}$$

Substituting $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=2\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)$, from the first equation, into the second equation gives $6\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi}$ so that $\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi}$ and $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}$.

(b) We are told that $z(u,v)=x(u,v)^4+y(u,v)^4$. So

$$\begin{aligned} \frac{\partial z}{\partial u}(u,v) &=4x(u,v)^3\ \frac{\partial x}{\partial u}(u,v) +4y(u,v)^3\ \frac{\partial y}{\partial u}(u,v) \end{aligned}$$

Substituting in $u=\frac{\pi}{2}$, $v=0$, $x\big(\frac{\pi}{2},0\big)=1$, $y\big(\frac{\pi}{2},0\big)=1$ and using the results of part (a),

$$\begin{aligned} \frac{\partial z}{\partial u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial x}{\partial u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial y}{\partial u}\!\left(\frac{\pi}{2},0\right) \\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right) \\ &=\frac{8}{\pi} \end{aligned}$$

This question uses bad (but standard) notation, in that the one symbol $f$ is used for two different functions, namely $f(u,v)$ and $f(x,y)=f(u,v)\big|_{u=x+y,v=x-y}$. A better wording is

• Let $f(u,v)$ and $F(x,y)$ be differentiable functions such that $F(x,y)=f(x+y,x-y)$. Find a constant, $\alpha$, such that

$$F_x(x,y)^2+F_y(x,y)^2=\alpha \big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\}$$

By the chain rule

$$\begin{aligned} \frac{\partial F}{\partial x}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial x}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial x}(x-y) \\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \frac{\partial F}{\partial y}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial y}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial y}(x-y) \\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{aligned}$$

Hence

$$\begin{aligned} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2 \\ & +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{aligned}$$

So $\alpha =2$ does the job.

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 $c$.