빌드라인 스쿨

문제 은행 · 미적분 · 대학 미적분 · 다변수 미분

편미분방정식 맛보기

열·파동·라플라스 방정식 같은 편미분방정식의 해를 확인한다

유형전체개념 확인계산그래프·표 해석활용증명·논술
난이도전체기본보통도전
풀이전체우리말 풀이 있는 문제만

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

3문제 · 1 / 1쪽

1.CLP-3 · 2.8절 · 11번대학 수준우리말 번역

점 (x,y,z)(x,y,z) 에서 원점 (0,0,0)(0,0,0) 까지의 거리는

r(x,y,z)=x2+y2+z2r(x,y,z) = \sqrt{x^2+y^2+z^2}

이다. nn 이 실수 상수일 때, 모든 (x,y,z)≠(0,0,0)(x,y,z)\ne (0,0,0) 에 대해 라플라스 방정식

uxx+uyy+uzz=0u_{xx}+u_{yy}+u_{zz}=0

을 만족하는 함수 u(x,y,z)=r(x,y,z)nu(x,y,z) = r(x,y,z)^n 을 모두 구하세요.

영어 원문 보기

The distance from the point (x,y,z)(x,y,z) to the origin (0,0,0)(0,0,0) is

r(x,y,z)=x2+y2+z2r(x,y,z) = \sqrt{x^2+y^2+z^2}

Find all functions u(x,y,z)=r(x,y,z)nu(x,y,z) = r(x,y,z)^n, with nn being a real constant, that obey Laplace's equation

uxx+uyy+uzz=0u_{xx}+u_{yy}+u_{zz}=0

for all (x,y,z)≠(0,0,0)(x,y,z)\ne (0,0,0).

답 쓰고 채점
정답·풀이 보기

편미분방정식에 대입하기 위해 uxxu_{xx}, uyyu_{yy}, uzzu_{zz} 를 계산한다.

u(x,y,z)=r(x,y,z)n=(x2+y2+z2)n/2ux(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂x(x2+y2+z2)=nx (x2+y2+z2)n/2−1uxx(x,y,z)=n (x2+y2+z2)n/2−1+nx (n/2−1)(x2+y2+z2)n/2−2(2x)=n (x2+y2+z2)n/2−1+n(n−2)x2(x2+y2+z2)n/2−2uy(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂y(x2+y2+z2)=ny (x2+y2+z2)n/2−1uyy(x,y,z)=n (x2+y2+z2)n/2−1+ny (n/2−1)(x2+y2+z2)n/2−2(2y)=n (x2+y2+z2)n/2−1+n(n−2)y2(x2+y2+z2)n/2−2uz(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂z(x2+y2+z2)=nz (x2+y2+z2)n/2−1uzz(x,y,z)=n (x2+y2+z2)n/2−1+nz (n/2−1)(x2+y2+z2)n/2−2(2z)=n (x2+y2+z2)n/2−1+n(n−2)z2(x2+y2+z2)n/2−2\begin{aligned} u(x,y,z) &= r(x,y,z)^n = \big(x^2+y^2+z^2\big)^{n/2} \\ u_x(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial x}\big(x^2+y^2+z^2\big) \\ &= nx\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{xx}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nx\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2x) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)x^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_y(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial y}\big(x^2+y^2+z^2\big) \\ &= ny\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{yy}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + ny\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2y) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)y^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_z(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial z}\big(x^2+y^2+z^2\big) \\ &= nz\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{zz}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nz\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2z) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)z^2 \big(x^2+y^2+z^2\big)^{n/2-2} \end{aligned}

따라서

uxx+uyy+uzz=3n(x2+y2+z2)n/2−1+n(n−2) (x2+y2+z2)(x2+y2+z2)n/2−2=3n(x2+y2+z2)n/2−1+n(n−2) (x2+y2+z2)n/2−1=[3n+n2−2n] (x2+y2+z2)n/2−1\begin{aligned} u_{xx}+u_{yy}+u_{zz} &=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ (x^2+y^2+z^2) \big(x^2+y^2+z^2\big)^{n/2-2} \\ &=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ &=[3n+n^2-2n]\ \big(x^2+y^2+z^2\big)^{n/2-1} \end{aligned}

이 값이 0이 될 필요충분조건은 다음과 같다.

n+n2=n(1+n)=0  ⟺  n=0,−1n+n^2=n(1+n)=0 \iff n=0,-1

원본 교재 풀이(영어)

In preparation for substituting into the PDE, we compute uxxu_{xx}, uyyu_{yy} and uzzu_{zz}.

u(x,y,z)=r(x,y,z)n=(x2+y2+z2)n/2ux(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂x(x2+y2+z2)=nx (x2+y2+z2)n/2−1uxx(x,y,z)=n (x2+y2+z2)n/2−1+nx (n/2−1)(x2+y2+z2)n/2−2(2x)=n (x2+y2+z2)n/2−1+n(n−2)x2(x2+y2+z2)n/2−2uy(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂y(x2+y2+z2)=ny (x2+y2+z2)n/2−1uyy(x,y,z)=n (x2+y2+z2)n/2−1+ny (n/2−1)(x2+y2+z2)n/2−2(2y)=n (x2+y2+z2)n/2−1+n(n−2)y2(x2+y2+z2)n/2−2uz(x,y,z)=n2 (x2+y2+z2)n/2−1 ∂∂z(x2+y2+z2)=nz (x2+y2+z2)n/2−1uzz(x,y,z)=n (x2+y2+z2)n/2−1+nz (n/2−1)(x2+y2+z2)n/2−2(2z)=n (x2+y2+z2)n/2−1+n(n−2)z2(x2+y2+z2)n/2−2\begin{aligned} u(x,y,z) &= r(x,y,z)^n = \big(x^2+y^2+z^2\big)^{n/2} \\ u_x(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial x}\big(x^2+y^2+z^2\big) \\ &= nx\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{xx}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nx\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2x) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)x^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_y(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial y}\big(x^2+y^2+z^2\big) \\ &= ny\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{yy}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + ny\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2y) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)y^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_z(x,y,z)&= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \frac{\partial }{\partial z}\big(x^2+y^2+z^2\big) \\ &= nz\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ u_{zz}(x,y,z)&= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nz\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2z) \\ &=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)z^2 \big(x^2+y^2+z^2\big)^{n/2-2} \end{aligned}

So

uxx+uyy+uzz=3n(x2+y2+z2)n/2−1+n(n−2) (x2+y2+z2)(x2+y2+z2)n/2−2=3n(x2+y2+z2)n/2−1+n(n−2) (x2+y2+z2)n/2−1=[3n+n2−2n] (x2+y2+z2)n/2−1\begin{aligned} u_{xx}+u_{yy}+u_{zz} &=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ (x^2+y^2+z^2) \big(x^2+y^2+z^2\big)^{n/2-2} \\ &=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ &=[3n+n^2-2n]\ \big(x^2+y^2+z^2\big)^{n/2-1} \end{aligned}

This is zero if and only if

n+n2=n(1+n)=0  ⟺  n=0,−1n+n^2=n(1+n)=0 \iff n=0,-1

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.8절 · 10번
# 편미분방정식 맛보기계산도전자동 분류
2.CLP-3 · 2.8절 · 13번대학 수준우리말 번역

u(x,y)u(x,y) 가 편미분방정식

α(x,y) ux(x,y)+β(x,y) uy(x,y)=0\alpha (x,y)\,u_x(x,y) +\beta (x,y)\,u_y(x,y)=0

을 만족한다고 하자. 여기서 α(x,y)\alpha (x,y) 와 β(x,y)\beta (x,y) 는 주어진 함수이다. (X(t),Y(t))\big(X(t),Y(t)\big) 를 xyxy 평면 위의 곡선(이러한 곡선을 이 편미분방정식의 특성곡선이라고 한다.)이라 하고, 이 곡선이

dXdt(t)=α(X(t),Y(t))dYdt(t)=β(X(t),Y(t))\begin{aligned} \frac{\mathrm{d}X}{\mathrm{d}t}(t)&=\alpha \big(X(t),Y(t)\big) \\ \frac{\mathrm{d}Y}{\mathrm{d}t}(t)&=\beta \big(X(t),Y(t)\big) \end{aligned}

을 만족한다고 하자. uu 가 그 곡선을 따라 상수임을 보이세요. 즉, u(X(t),Y(t))u\big(X(t),Y(t)\big) 가 tt 에 무관함을 보이세요.

영어 원문 보기

Suppose that u(x,y)u(x,y) obeys the PDE

α(x,y) ux(x,y)+β(x,y) uy(x,y)=0\alpha (x,y)\,u_x(x,y) +\beta (x,y)\,u_y(x,y)=0

where α(x,y)\alpha (x,y) and β(x,y)\beta (x,y) are given functions. Let (X(t),Y(t))\big(X(t),Y(t)\big) be a curve (Such curves are called characteristics of the PDE.) in the xyxy-plane that obeys

dXdt(t)=α(X(t),Y(t))dYdt(t)=β(X(t),Y(t))\begin{aligned} \frac{\mathrm{d}X}{\mathrm{d}t}(t)&=\alpha \big(X(t),Y(t)\big) \\ \frac{\mathrm{d}Y}{\mathrm{d}t}(t)&=\beta \big(X(t),Y(t)\big) \end{aligned}

Show that uu is constant along that curve. That is, show that u(X(t),Y(t))u\big(X(t),Y(t)\big) is independent of tt.

정답·풀이 보기

연쇄 법칙에 의해

ddtu(X(t),Y(t))=ux(X(t),Y(t)) dXdt(t)+uy(X(t),Y(t)) dYdt(t)=α(X(t),Y(t)) ux(X(t),Y(t))+β(X(t),Y(t)) uy(X(t),Y(t))\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big) &=u_x\big(X(t),Y(t)\big)\,\frac{\mathrm{d}X}{\mathrm{d}t}(t) +u_y\big(X(t),Y(t)\big)\,\frac{\mathrm{d}Y}{\mathrm{d}t}(t) \\ &=\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) \end{aligned}

그런데 α(x,y) ux(x,y)+β(x,y) uy(x,y)=0\alpha (x,y)\,u_x(x,y) +\beta (x,y)\,u_y(x,y)=0 에 x=X(t)x=X(t), y=Y(t)y=Y(t) 를 대입하면

α(X(t),Y(t)) ux(X(t),Y(t))+β(X(t),Y(t)) uy(X(t),Y(t))=0\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) =0

이므로

ddtu(X(t),Y(t))=0\frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big)=0

이다.

원본 교재 풀이(영어)

By the chain rule,

ddtu(X(t),Y(t))=ux(X(t),Y(t)) dXdt(t)+uy(X(t),Y(t)) dYdt(t)=α(X(t),Y(t)) ux(X(t),Y(t))+β(X(t),Y(t)) uy(X(t),Y(t))\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big) &=u_x\big(X(t),Y(t)\big)\,\frac{\mathrm{d}X}{\mathrm{d}t}(t) +u_y\big(X(t),Y(t)\big)\,\frac{\mathrm{d}Y}{\mathrm{d}t}(t) \\ &=\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) \end{aligned}

But evaluating α(x,y) ux(x,y)+β(x,y) uy(x,y)=0\alpha (x,y)\,u_x(x,y) +\beta (x,y)\,u_y(x,y)=0 at x=X(t)x=X(t), y=Y(t)y=Y(t) gives

α(X(t),Y(t)) ux(X(t),Y(t))+β(X(t),Y(t)) uy(X(t),Y(t))=0\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) =0

so

ddtu(X(t),Y(t))=0\frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big)=0

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.8절 · 11번
# 편미분방정식 맛보기증명·논술도전자동 분류
3.CLP-3 · 2.8절 · 14번대학 수준우리말 번역

(a) u(x,y)u(x,y) 가 편미분방정식 3ux(x,y)+6uy(x,y)=u(x,y)3u_x(x,y) + 6u_y(x,y)=u(x,y) 를 만족한다고 하자. v(X,Y)=u(X,Y+2X)v(X,Y) = u(X, Y+2X) 로 정의할 때, vv 가 만족하는 편미분방정식을 구하세요. (b) u(x,y)u(x,y) 가 편미분방정식 xux(x,y)+yuy(x,y)=u(x,y)xu_x(x,y) + yu_y(x,y)=u(x,y) 를 만족한다고 하자. v(X,Y)=u(X,XeY)v(X,Y) = u(X, Xe^Y) 로 정의할 때, vv 가 만족하는 편미분방정식을 구하세요.

영어 원문 보기

(a) Suppose that u(x,y)u(x,y) obeys the PDE 3ux(x,y)+6uy(x,y)=u(x,y)3u_x(x,y) + 6u_y(x,y)=u(x,y). Define v(X,Y)=u(X,Y+2X)v(X,Y) = u(X, Y+2X). Find a PDE that vv obeys. (b) Suppose that u(x,y)u(x,y) obeys the PDE xux(x,y)+yuy(x,y)=u(x,y)xu_x(x,y) + yu_y(x,y)=u(x,y). Define v(X,Y)=u(X,XeY)v(X,Y) = u(X, Xe^Y). Find a PDE that vv obeys.

정답·풀이 보기

(a) u(x,y)u(x,y) 가 편미분방정식

3ux(x,y)+6uy(x,y)=u(x,y)3u_x(x,y) + 6u_y(x,y)=u(x,y)

을 만족한다고 하자.

v(X,Y)=u(X,Y+2X)v(X,Y) = u(X, Y+2X) 로 정의하자. 그러면 연쇄 법칙에 의해

vX(X,Y)=∂∂X[u(X,Y+2X)]=ux(X,Y+2X)+2uy(X,Y+2X)=13{3ux(X,Y+2X)+6uy(X,Y+2X)}=13u(X,Y+2X)=13v(X,Y)\begin{aligned} v_X(X,Y)&=\frac{\partial }{\partial X}\big[ u(X, Y+2X)\big] \\ &=u_x(X, Y+2X) +2u_y(X, Y+2X) \\ &=\frac{1}{3}\big\{3u_x(X, Y+2X) + 6u_y(X, Y+2X)\big\} \\ &=\frac{1}{3} u(X, Y+2X) \\ &=\frac{1}{3} v(X,Y) \end{aligned}

이다.

(b) v(X,Y)=u(X,XeY)v(X,Y) = u(X, Xe^Y) 로 정의하자. 그러면 연쇄 법칙에 의해

vX(X,Y)=∂∂X[u(X,XeY)]=ux(X,XeY)+eYuy(X,XeY)\begin{aligned} v_X(X,Y)&=\frac{\partial }{\partial X}\big[ u(X, Xe^Y)\big] \\ &=u_x(X, Xe^Y) +e^Yu_y(X, Xe^Y) \end{aligned}

이다.

이제 xux(x,y)+yuy(x,y)=u(x,y)xu_x(x,y) + yu_y(x,y)=u(x,y) 이면, x=Xx=X, y=XeYy=Xe^Y 에서의 값을 생각할 때

Xux(X,XeY)+XeYuy(X,XeY)=u(X,XeY)Xu_x(X, Xe^Y) + Xe^Yu_y(X, Xe^Y)=u(X, Xe^Y)

임에 주목하자.

따라서

vX(X,Y)=1X{Xux(X,XeY)+XeYuy(X,XeY)}=1Xu(X,XeY)=1Xv(X,Y)\begin{aligned} v_X(X,Y) &=\frac{1}{X}\big\{X u_x(X, Xe^Y) +X e^Y u_y(X, Xe^Y) \big\} \\ &=\frac{1}{X} u(X, Xe^Y) \\ &=\frac{1}{X} v(X,Y) \end{aligned}

이다.

원본 교재 풀이(영어)

(a) Suppose that u(x,y)u(x,y) obeys the PDE

3ux(x,y)+6uy(x,y)=u(x,y)3u_x(x,y) + 6u_y(x,y)=u(x,y)

Define v(X,Y)=u(X,Y+2X)v(X,Y) = u(X, Y+2X). Then, by the chain rule,

vX(X,Y)=∂∂X[u(X,Y+2X)]=ux(X,Y+2X)+2uy(X,Y+2X)=13{3ux(X,Y+2X)+6uy(X,Y+2X)}=13u(X,Y+2X)=13v(X,Y)\begin{aligned} v_X(X,Y)&=\frac{\partial }{\partial X}\big[ u(X, Y+2X)\big] \\ &=u_x(X, Y+2X) +2u_y(X, Y+2X) \\ &=\frac{1}{3}\big\{3u_x(X, Y+2X) + 6u_y(X, Y+2X)\big\} \\ &=\frac{1}{3} u(X, Y+2X) \\ &=\frac{1}{3} v(X,Y) \end{aligned}

(b) Define v(X,Y)=u(X,XeY)v(X,Y) = u(X, Xe^Y). Then, by the chain rule,

vX(X,Y)=∂∂X[u(X,XeY)]=ux(X,XeY)+eYuy(X,XeY)\begin{aligned} v_X(X,Y)&=\frac{\partial }{\partial X}\big[ u(X, Xe^Y)\big] \\ &=u_x(X, Xe^Y) +e^Yu_y(X, Xe^Y) \end{aligned}

Now notice that if xux(x,y)+yuy(x,y)=u(x,y)xu_x(x,y) + yu_y(x,y)=u(x,y), then, evaluating at x=Xx=X and y=XeYy=Xe^Y gives

Xux(X,XeY)+XeYuy(X,XeY)=u(X,XeY)Xu_x(X, Xe^Y) + Xe^Yu_y(X, Xe^Y)=u(X, Xe^Y)

So

vX(X,Y)=1X{Xux(X,XeY)+XeYuy(X,XeY)}=1Xu(X,XeY)=1Xv(X,Y)\begin{aligned} v_X(X,Y) &=\frac{1}{X}\big\{X u_x(X, Xe^Y) +X e^Y u_y(X, Xe^Y) \big\} \\ &=\frac{1}{X} u(X, Xe^Y) \\ &=\frac{1}{X} v(X,Y) \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.8절 · 3번
# 편미분방정식 맛보기계산도전자동 분류