다변수 연쇄법칙 연습문제

1.OpenStax 3권 · 4.5절 · 34번대학 수준우리말 번역

다음 연습문제에서 다음 정보를 이용하세요. 함수 $f(x,y)$ 가 $f(tx,ty)={t}^{n}f(x,y)$ 를 만족하면 $n$ 차 동차함수라고 한다. $n$ 차 동차함수 에 대해서는 다음 등식이 항상 성립한다: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ . 주어진 함수가 동차함수임을 보이고 $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ 임을 확인하세요.

$f(x,y)=3{x}^{2}+{y}^{2}$

영어 원문 보기

For the following exercises, use this information: A function $f(x,y)$ is said to be homogeneous of degree $n$ if $f(tx,ty)={t}^{n}f(x,y).$ For all homogeneous functions of degree $n,$ the following equation is true: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$ Show that the given function is homogeneous and verify that $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$

$f(x,y)=3{x}^{2}+{y}^{2}$

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기APEX · 12.5절 · 4번

# 다변수 연쇄법칙증명·논술자동 분류

2.OpenStax 3권 · 4.5절 · 35번대학 수준우리말 번역

다음 연습문제에서는 이 정보를 이용하세요: 함수 $f(x,y)$ 가 $f(tx,ty)={t}^{n}f(x,y)$ 를 만족하면 $n$ 차 동차함수라고 합니다. 모든 $n$ 차 동차함수에 대해 다음 등식이 성립합니다: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ . 주어진 함수가 동차함수임을 보이고 $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ 임을 확인하세요.

$f(x,y)=\sqrt{{x}^{2}+{y}^{2}}$

영어 원문 보기

For the following exercises, use this information: A function $f(x,y)$ is said to be homogeneous of degree $n$ if $f(tx,ty)={t}^{n}f(x,y).$ For all homogeneous functions of degree $n,$ the following equation is true: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$ Show that the given function is homogeneous and verify that $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$

$f(x,y)=\sqrt{{x}^{2}+{y}^{2}}$

정답·풀이 보기

$f(tx,ty)=\sqrt{{t}^{2}{x}^{2}+{t}^{2}{y}^{2}}={t}^{1}f(x,y),$ $\frac{\partial f}{\partial y}=x\frac{1}{2}{({x}^{2}+{y}^{2})}^{-1/2}\,\times \,2x+y\frac{1}{2}{({x}^{2}+{y}^{2})}^{-1/2}\,\times \,2y=1f(x,y)$

원본 교재 풀이(영어)

$f(tx,ty)=\sqrt{{t}^{2}{x}^{2}+{t}^{2}{y}^{2}}={t}^{1}f(x,y),$ $\frac{\partial f}{\partial y}=x\frac{1}{2}{({x}^{2}+{y}^{2})}^{-1/2}\,\times \,2x+y\frac{1}{2}{({x}^{2}+{y}^{2})}^{-1/2}\,\times \,2y=1f(x,y)$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 33번

# 다변수 연쇄법칙증명·논술자동 분류

3.OpenStax 3권 · 4.5절 · 36번대학 수준우리말 번역

다음 연습문제에서는 이 정보를 이용하세요. 함수 $f(x,y)$ 가 $f(tx,ty)={t}^{n}f(x,y)$ 를 만족시키면 차수가 $n$ 인 동차함수라고 합니다. 차수가 $n$ 인 모든 동차함수에 대해 다음 방정식이 성립합니다: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ . 주어진 함수가 동차함수임을 보이고 $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$ 가 성립함을 확인하세요.

$f(x,y)={x}^{2}y-2{y}^{3}$

영어 원문 보기

For the following exercises, use this information: A function $f(x,y)$ is said to be homogeneous of degree $n$ if $f(tx,ty)={t}^{n}f(x,y).$ For all homogeneous functions of degree $n,$ the following equation is true: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$ Show that the given function is homogeneous and verify that $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$

$f(x,y)={x}^{2}y-2{y}^{3}$

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 1번

# 다변수 연쇄법칙증명·논술자동 분류

4.CLP-3 · 2.4절 · 4번대학 수준우리말 번역

열역학 교재에서는 다음 관계를 사용한다.

$\left(\frac{\partial y}{\partial x}\right) \left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial x}{\partial z}\right)=-1$

이 등식의 의미를 설명하고, 이것이 성립함을 증명하세요.

영어 원문 보기

Thermodynamics texts use the relationship

$\left(\frac{\partial y}{\partial x}\right) \left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial x}{\partial z}\right)=-1$

Explain the meaning of this equation and prove that it is true.

정답·풀이 보기

기본 가정은 세 양 $x$ , $y$ , $z$ 가 서로 독립이 아니라는 것이다. 이 중 어느 두 개가 주어지면 나머지 하나가 유일하게 정해진다. 이들이 관계식 $F(x,y,z)=0$ 을 만족한다고 가정하면, 이 식을 풀어서

$x$ 를 $y$ 와 $z$ 의 함수로 정할 수 있고(예를 들어 $x=f(y,z)$ ), 또한
$y$ 를 $x$ 와 $z$ 의 함수로 정할 수 있고(예를 들어 $y=g(x,z)$ ), 또한
$z$ 를 $x$ 와 $y$ 의 함수로 정할 수 있다 (예를 들어 $z=h(x,y)$ ).

예를 들어 $F(x,y,z) = xyz-1$ 이면

$F(x,y,z)=xyz-1=0$ 으로부터 $x=\frac{1}{yz}=f(y,z)$ 이고,
$F(x,y,z)=xyz-1=0$ 으로부터 $y=\frac{1}{xz}=g(x,z)$ 이고,
$F(x,y,z)=xyz-1=0$ 으로부터 $z=\frac{1}{xy}=h(x,y)$ 이다.

일반적으로 $F(x,y,z)=0$ 이 $x=f(y,z)$ 를 정한다는 것은

$F\big(f(y,z),y,z\big)=0 \tag{$*$}$

이 모든 $y$ 와 $z$ 에 대해 성립한다는 뜻이다. $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ 라 하자. $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ 에 ( $y$ 와 $z$ 를 독립변수로 보고) 연쇄 법칙을 적용하면

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z) &= \frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big) \end{aligned}$

이다.

등식 $(*)$ 는 모든 $y$ 와 $z$ 에 대해 $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ 임을 뜻한다. 따라서 등식 $(*)$ 를 $z$ 에 대해 미분하면

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z)&=\frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)=0 \\ &\implies \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)} {\frac{\partial F}{\partial x}\big(f(y,z),y,z\big)} \end{aligned}$

이 모든 $y$ 와 $z$ 에 대해 성립한다. 마찬가지로 $F\big(x,g(x,z),z\big)=0$ 을 $x$ 에 대해 미분하고 $F\big(x,y,h(x,y)\big)=0$ 을 $y$ 에 대해 미분하면

$\begin{aligned} \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,g(x,z),z\big)} {\frac{\partial F}{\partial y}\big(x,g(x,z),z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,h(x,y)\big)} {\frac{\partial F}{\partial z}\big(x,y,h(x,y)\big)} \end{aligned}$

를 얻는다.

$(x,y,z)$ 가 $F(x,y,z)=0$ 을 만족하는 임의의 점이면 (따라서 $x=f(y,z)$ , $y=g(x,z)$ , $z=h(x,y)$ 이다),

$\begin{aligned} \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\qquad \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \end{aligned}$

이고,

$\begin{aligned} \frac{\partial f}{\partial z}(y,z)\ \frac{\partial g}{\partial x}(x,z)\ \frac{\partial h}{\partial y}(x,y) &=-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \\ &=-1 \end{aligned}$

이다.

원본 교재 풀이(영어)

The basic assumption is that the three quantites $x$ , $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0$ , which can be solved to

determine $x$ as a function of $y$ and $z$ (say $x=f(y,z)$ ) and can alternatively be solved to
determine $y$ as a function of $x$ and $z$ (say $y=g(x,z)$ ) and can alternatively be solved to
determine $z$ as a function of $x$ and $y$ (say $z=h(x,y)$ ).

As an example, if $F(x,y,z) = xyz-1$ , then

$F(x,y,z)=xyz-1=0$ implies that $x=\frac{1}{yz}=f(y,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $y=\frac{1}{xz}=g(x,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $z=\frac{1}{xy}=h(x,y)$

In general, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

$F\big(f(y,z),y,z\big)=0 \tag{$*$}$

for all $y$ and $z$ . Set $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ . Applying the chain rule to $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ (with $y$ and $z$ independent variables) gives

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z) &= \frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big) \end{aligned}$

The equation $(*)$ says that $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ for all $y$ and $z$ . So differentiating the equation $(*)$ with respect to $z$ gives

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z)&=\frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)=0 \\ &\implies \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)} {\frac{\partial F}{\partial x}\big(f(y,z),y,z\big)} \end{aligned}$

for all $y$ and $z$ . Similarly, differentiating $F\big(x,g(x,z),z\big)=0$ with respect to $x$ and $F\big(x,y,h(x,y)\big)=0$ with respect to $y$ gives

$\begin{aligned} \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,g(x,z),z\big)} {\frac{\partial F}{\partial y}\big(x,g(x,z),z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,h(x,y)\big)} {\frac{\partial F}{\partial z}\big(x,y,h(x,y)\big)} \end{aligned}$

If $(x,y,z)$ is any point satisfying $F(x,y,z)=0$ (so that $x=f(y,z)$ and $y=g(x,z)$ and $z=h(x,y)$ ), then

$\begin{aligned} \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\qquad \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \end{aligned}$

and

$\begin{aligned} \frac{\partial f}{\partial z}(y,z)\ \frac{\partial g}{\partial x}(x,z)\ \frac{\partial h}{\partial y}(x,y) &=-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \\ &=-1 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 5번

# 다변수 연쇄법칙증명·논술기본자동 분류

5.CLP-3 · 2.4절 · 9번대학 수준우리말 번역

$f(x,y)$ 가 라플라스 방정식 $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$ 을 만족시킨다고 가정하세요. 합성함수 $g(s,t) = f (s - t, s + t)$ 에 대해서도 같은 식이 성립함을 보이세요. 즉, $\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0$ 임을 보이세요. $f(x,y)$ 는 매끄러운 함수여서 연쇄 법칙과 혼합 편도함수의 같음에 관한 클레로의 정리를 적용할 수 있다고 가정해도 됩니다.

영어 원문 보기

Assume that $f(x,y)$ satisfies Laplace's equation $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$ . Show that this is also the case for the composite function $g(s,t) = f (s - t, s + t)$ . That is, show that $\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0$ . You may assume that $f(x,y)$ is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

연쇄 법칙에 의해,

$\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}$

이고

$\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}$

이다. 인수를 생략하면

$\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}$

이므로 원하는 결과가 얻어진다.

원본 교재 풀이(영어)

By the chain rule,

$\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}$

and

$\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}$

Suppressing the arguments

$\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}$

as desired.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 37번

# 다변수 연쇄법칙증명·논술보통자동 분류

6.CLP-3 · 2.4절 · 12번대학 수준우리말 번역

$u(x,y)$ 가 다음과 같이 정의된다.

$u(x,y) = e^y\, F\big(xe^{-y^2}\big)$

여기서 $F(z)$ 는 임의의 함수이다.

(a) $F(z) = \ln(z)$ 일 때, $\frac{\partial u}{\partial x}$ 와 $\frac{\partial u}{\partial y}$ 를 구하세요. (b) 임의의 함수 $F(z)$ 에 대해 $u(x,y)$ 가 다음을 만족시킴을 보이세요.

$2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u$

영어 원문 보기

$u(x,y)$ is defined as

$u(x,y) = e^y\, F\big(xe^{-y^2}\big)$

for an arbitrary function $F(z)$ .

(a) If $F(z) = \ln(z)$ , find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ . (b) For an arbitrary $F(z)$ show that $u(x,y)$ satisfies

$2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u$

정답·풀이 보기

임의의 (미분가능한) 함수 $F$ 에 대해, 연쇄법칙과 곱의 미분법에 의해 다음이 성립한다.

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}$

(a) 특히 $F(z)=\ln(z)$ 일 때 $F'(z)=\frac{1}{z}$ 이므로

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) 일반적으로

$\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}$

원본 교재 풀이(영어)

For any (differentiable) function $F$ , we have, by the chain and product rules,

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}$

(a) In particular, when $F(z)=\ln(z)$ , $F'(z)=\frac{1}{z}$ and

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) In general

$\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 11번

# 다변수 연쇄법칙증명·논술보통자동 분류

7.CLP-3 · 2.4절 · 14번대학 수준우리말 번역

$w = f (xz, yz)$ 라고 하자. 여기서 $f$ 는 미분가능한 함수이다. 다음을 보여라.

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

영어 원문 보기

Suppose that $w = f (xz, yz)$ , where $f$ is a differentiable function. Show that

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

정답·풀이 보기

먼저 $w(x,y,z)$ 의 1계 편도함수를 구한다. $u(x,y,z)=xz$ , $v(x,y,z)=yz$ 라 하면 $w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big)$ 이다. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}$

따라서

$\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}$

이므로 주어진 등식이 성립한다.

원본 교재 풀이(영어)

We'll first compute the first order partial derivatives of $w(x,y,z)$ . Write $u(x,y,z)=xz$ and $v(x,y,z)=yz$ so that $w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big)$ . By the chain rule,

$\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}$

So

$\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}$

as desired.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 22번

# 다변수 연쇄법칙증명·논술보통자동 분류

8.CLP-3 · 2.4절 · 19번대학 수준우리말 번역

$f(u,v)$ 를 두 변수의 미분가능한 함수라 하고, $z$ 를 $f(xz,yz) = 0$ 으로 음함수로 정의된 $x$ 와 $y$ 의 미분가능한 함수라 하자. 다음을 보이세요.

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

영어 원문 보기

Let $f(u,v)$ be a differentiable function of two variables, and let $z$ be a differentiable function of $x$ and $y$ defined implicitly by $f(xz,yz) = 0$ . Show that

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

정답·풀이 보기

함수 $z(x,y)$ 는 모든 $x$ 와 $y$ 에 대해

$f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}$

을 만족한다. 연쇄 법칙에 의해

$\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}$

이므로 $(*)$ 를 $x$ 에 대해, 그리고 $y$ 에 대해 미분하면

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

이고, 인수를 생략하여 나타내면

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

이다. 첫 번째 식을 $z_x$ 에 대해, 두 번째 식을 $z_y$ 에 대해 풀면

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

이므로

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

이다. 따라서 주어진 등식이 성립한다.

참고: 이는 물론 $x\,f_u+y\,f_v$ 가 0이 아니라는 가정 아래에서이다. 연쇄 법칙에 의해 이 가정은 $\frac{\partial }{\partial z}\big[f(xz,yz)\big]$ 가 0이 아니라는 가정과 같다. 이는 다시 $f(xz,yz)=0$ 을 $x$ 와 $y$ 의 함수로서 $z$ 에 대해 풀 수 있다는 진술과 거의 같지만, 완전히 같지는 않다.

원본 교재 풀이(영어)

We are told that the function $z(x,y)$ obeys

$f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}$

for all $x$ and $y$ . By the chain rule,

$\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}$

so differentiating $(*)$ with respect to $x$ and with respect to $y$ gives

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

or, leaving out the arguments,

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

Solving the first equation for $z_x$ and the second for $z_y$ gives

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

so that

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

as desired.

Remark: This is of course under the assumption that $x\,f_u+y\,f_v$ is nonzero. That is equivalent, by the chain rule, to the assumption that $\frac{\partial }{\partial z}\big[f(xz,yz)\big]$ is non zero. That, in turn, is almost, but not quite, equivalent to the statement that $f(xz,yz)=0$ is can be solved for $z$ as a function of $x$ and $y$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 18번

# 다변수 연쇄법칙증명·논술보통자동 분류

9.CLP-3 · 2.4절 · 23번대학 수준우리말 번역

(a) $f$ 를 실수 전체에서 정의된 임의의 미분가능한 함수라 하자. 평면 전체에서 다음과 같이 정의된 함수 $w$ 가

$w(x,y)=e^{-y}f(x-y)$

편미분방정식

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0$

을 만족시킴을 보이세요. (b) 방정식 $x=u^3-3uv^2$ , $y=3u^2v-v^3$ , $z=u^2-v^2$ 은 $z$ 를 $x$ 와 $y$ 의 함수로 정의한다. 점 $(x,y)=(2,11)$ 에 대응하는 점 $(u,v)=(2,1)$ 에서 $\frac{\partial z}{\partial x}$ 의 값을 구하세요.

영어 원문 보기

(a) Let $f$ be an arbitrary differentiable function defined on the entire real line. Show that the function $w$ defined on the entire plane as

$w(x,y)=e^{-y}f(x-y)$

satisfies the partial differential equation:

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0$ (b) The equations $x=u^3-3uv^2$ , $y=3u^2v-v^3$ and $z=u^2-v^2$ define $z$ as a function of $x$ and $y$ . Determine $\frac{\partial z}{\partial x}$ at the point $(u,v)=(2,1)$ which corresponds to the point $(x,y)=(2,11)$ .

정답·풀이 보기

(a) 곱의 미분법과 연쇄 법칙에 의해

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

이므로

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

이다. 따라서 성립한다.

(b) $x=u^3-3uv^2$ , $y=3u^2v-v^3$ 을 $x$ , $y$ 가 주어진 매개변수이고 $u$ , $v$ 가 미지수인 두 방정식으로 생각하자. 문제는 적어도 $(u,v)=(2,1)$ , $(x,y)=(2,11)$ 근처에서 이 두 방정식을 $u$ , $v$ 에 대해 $x$ , $y$ 로 풀 수 있음을 암묵적으로 알려 준다. 즉, 함수 $u(x,y)$ 와 $v(x,y)$ 가 다음 식으로 정해진다는 것을 알려 준다.

$x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3$

등식 $x=u(x,y)^3-3u(x,y)v(x,y)^2$ 의 양변에 $\frac{\partial }{\partial x}$ 를 취하면

$\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}$

이다. 다음으로 $y=3u(x,y)^2v(x,y)-v(x,y)^3$ 의 양변에 $\frac{\partial }{\partial x}$ 를 취하면

$\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}$

이다. $x=2$ , $y=11$ , $u=2$ , $v=1$ 을 대입하면

$\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}$

이다. 두 번째 식에서 $\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)$ 이다. 이를 첫 번째 식에 대입하면

$1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)$

이므로 $\frac{\partial u}{\partial x}(2,11)=\frac{1}{25}$ 이고 $\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75}$ 이다. 또한 문제에서 $z(x,y)=u(x,y)^2-v(x,y)^2$ 임을 알 수 있다. 따라서

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) By the product and chain rules

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

Hence

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

as desired.

(b) Think of $x=u^3-3uv^2$ , $y=3u^2v-v^3$ as two equations in the two unknowns $u$ , $v$ with $x$ , $y$ just being given parameters. The question implicitly tells us that those two equations can be solved for $u$ , $v$ in terms of $x$ , $y$ , at least near $(u,v)=(2,1)$ , $(x,y)=(2,11)$ . That is, the question implicitly tells us that the functions $u(x,y)$ and $v(x,y)$ are determined by

$x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3$

Applying $\frac{\partial }{\partial x}$ to both sides of the equation $x=u(x,y)^3-3u(x,y)v(x,y)^2$ gives

$\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}$

Then applying $\frac{\partial }{\partial x}$ to both sides of $y=3u(x,y)^2v(x,y)-v(x,y)^3$ gives

$\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}$

Substituting in $x=2$ , $y=11$ , $u=2$ , $v=1$ gives

$\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}$

From the second equation $\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)$ . Substituting into the first equation gives

$1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)$

so that $\frac{\partial u}{\partial x}(2,11)=\frac{1}{25}$ and $\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75}$ . The question also tells us that $z(x,y)=u(x,y)^2-v(x,y)^2$ . Hence

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 20번

# 다변수 연쇄법칙증명·논술보통자동 분류

10.CLP-3 · 2.4절 · 26번대학 수준우리말 번역

파동 방정식

$\frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\,\frac{\partial^2u}{\partial t^2} =0$

은 파동과 같은 현상을 다루는 많은 모형에서 나타난다. $u(x,t)$ 와 $v(\xi,\eta)$ 가 다음의 변수 변환으로 서로 연결되어 있다고 하자.

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

(a) $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ 일 필요충분조건이 $\ \frac{\partial^2 v}{\partial\xi\partial\eta}=0$ 임을 보이세요. (b) $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ 일 필요충분조건이 어떤 함수 $F$ 와 $G$ 에 대해 $\ u(x,t)=F(x-ct)+G(x+ct)\$ 임을 보이세요. (c) $\ F(x-ct)+G(x+ct)\$ 를 진행파의 관점에서 해석하세요. $u(x,t)$ 를 $x$ 축을 따라 진행하는 파동의, 위치 $x$ 와 시각 $t$ 에서의 높이라고 생각하세요.

참고: 낯선 기호 $\xi$ 와 $\eta$ 에 당황하지 마세요. 이들은 그리스 문자 중 해롭지 않은 두 글자일 뿐이며, 각각 “크사이”와 “에타”라고 읽습니다.

영어 원문 보기

The wave equation

$\frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\,\frac{\partial^2u}{\partial t^2} =0$

arises in many models involving wave-like phenomena. Let $u(x,t)$ and $v(\xi,\eta)$ be related by the change of variables

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

(a) Show that $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ if and only if $\ \frac{\partial^2 v}{\partial\xi\partial\eta}=0$ . (b) Show that $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ if and only if $\ u(x,t)=F(x-ct)+G(x+ct)\$ for some functions $F$ and $G$ . (c) Interpret $\ F(x-ct)+G(x+ct)\$ in terms of travelling waves. Think of $u(x,t)$ as the height, at position $x$ and time $t$ , of a wave that is travelling along the $x$ -axis.

Remark: Don't be thrown by the strange symbols $\xi$ and $\eta$ . They are just two harmless letters from the Greek alphabet, called “xi” and “eta” respectively.

정답·풀이 보기

$u(x,t)=v\big(\xi(x,t),\eta(x,t)\big)$ 임을 기억하자. 연쇄 법칙에 의해

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

연쇄 법칙을 다시 적용하면

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

이고

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

이므로

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

이다. 따라서

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

(b) 이제 $\frac{\partial^2 v\,}{\partial\xi\partial\eta}\big(\xi,\eta\big) =\frac{\partial}{\partial\xi} \Big[\frac{\partial v}{\partial\eta}\Big]=0$ 이다. 잠시 $\frac{\partial v}{\partial \eta}=w$ 라고 놓자. 등식 $\frac{\partial w}{\partial \xi}(\xi,\eta)=0$ 은, 각각의 고정된 $\eta$ 에 대해 $w(\xi,\eta)$ 가 상수임을 뜻한다. 이 상수의 값은 $\eta$ 에 따라 달라질 수 있다. 즉 어떤 함수 $H$ 에 대해 $\frac{\partial v}{\partial \eta}(\xi,\eta)=w(\xi,\eta) =H(\eta)$ 이다. (확인해 보면 $\frac{\partial }{\partial \xi}H(\eta)=0$ 이다.) 따라서 $v$ 의 $\eta$ 에 대한 도함수는 ( $\xi$ 를 상수로 보면) $H(\eta)$ 이다.

$G(\eta)$ 를 도함수가 $H(\eta)$ 인 임의의 함수 (즉 $H(\eta)$ 의 부정적분)라고 하자. 그러면 $\frac{\partial }{\partial \eta}\big[v(\xi,\eta)-G(\eta)]=H(\eta)-H(\eta)=0$ 이다. 이것은 각각의 고정된 $\xi$ 에 대해 $v(\xi,\eta)-G(\xi,\eta)$ 가 $\eta$ 와 무관한 상수일 때, 그리고 그때에 한해 성립한다. 즉 다음과 필요충분조건이다.

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

(어떤 함수 $F$ 에 대해). 따라서

$\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{ 모든 }(\xi,\eta)\\ &\iff v(\xi,\eta)=F(\xi)+G(\eta)\text{ 어떤 함수 $F$ 와 $G$ 에 대해}\\ &\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) $F(x-ct)$ 의 해석을 제시하겠다. $G(x+ct)$ 의 경우도 비슷하다. $u(x,t)=F(x-ct)$ 라고 하자. $u(x,t)$ 를 위치 $x$ , 시각 $t$ 에서의 물의 높이라고 생각하자. 임의의 수 $z$ 를 잡자. 시공간에서 $x-ct=z$ 를 만족하는 모든 점 $(x,t)$ 는 $u$ 의 값이 $F(z)$ 로 모두 같다. 그러므로 위치가 $x=z+ct$ 가 되도록 움직이면 (즉 속력 $c$ 로 오른쪽으로 움직이면) 항상 같은 파동의 높이를 보게 된다. 따라서 $F(x-ct)$ 는 속력 $c$ 로 오른쪽으로 진행하는 파동을 나타낸다.

마찬가지로 $G(x+ct)$ 는 속력 $c$ 로 왼쪽으로 진행하는 파동을 나타낸다.

원본 교재 풀이(영어)

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 $c$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 17번

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