다변수 미분 전체 문제

(a) In polar coordinates $x=r\cos\theta$, $y=r\sin\theta$

$$\begin{aligned} \frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2} &=\frac{2r^2\cos^2\theta + r^3\cos^2\theta\sin\theta - r^3\cos\theta\sin^2\theta + 2r^2\sin^2\theta}{r^2} \\ &=2+ r\big[\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big] \end{aligned}$$

As

$$r\big|\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big| \le 2r \rightarrow 0\text{ as }r\rightarrow 0$$

we have

$$\lim_{(x,y)\rightarrow(0,0)} \frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}=2$$

(b) Since

$$\begin{aligned} \frac{x^2y^2 -2 x^2y + x^2} {(x^2 + y^2-2y+1)^2} =\frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} \end{aligned}$$

and, in polar coordinates centred on $(0,1)$, $x=r\cos\theta$, $y=1+r\sin\theta$,

$$\frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} =\frac{r^4\cos^2\theta\sin^2\theta}{r^4} =\cos^2\theta\sin^2\theta$$

we have that the limit does not exist. For example, if we send $(x,y)$ to $(0,1)$ along the line $y=1$, so that $\theta=0$, we get the limit $0$, while if we send $(x,y)$ to $(0,1)$ along the line $y=x+1$, so that $\theta=\frac{\pi}{4}$, we get the limit $\frac{1}{4}$.

As

$$\begin{aligned} u_t(x,y,z,t) &=-\frac{3}{2}\frac{1}{t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} +\frac{1}{4\alpha \,t^{7/2}}(x^2+y^2+z^2) e^{-(x^2+y^2+z^2)/(4\alpha t)} \\ u_x(x,y,z,t) &=-\frac{x}{2\alpha \,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} \\ u_{xx}(x,y,z,t) &=-\frac{1}{2\alpha \,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} +\frac{x^2}{4\alpha ^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} \\ u_{yy}(x,y,z,t) &=-\frac{1}{2\alpha \,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} +\frac{y^2}{4\alpha ^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} \\ u_{zz}(x,y,z,t) &=-\frac{1}{2\alpha \,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} +\frac{z^2}{4\alpha ^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} \end{aligned}$$

we have

$$\begin{aligned} \alpha \big(u_{xx} + u_{yy} + u_{zz} \big) &=-\frac{3}{2\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} +\frac{x^2+y^2+z^2}{4\alpha \,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\alpha t)} =u_t \end{aligned}$$

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 $c$.