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편미분방정식 맛보기

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

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 문제CLP-3 · 2.8절 · 10번우리말 번역

v(x,y)v(x,y) 를 조화함수라 하자. 즉, v(x,y)v(x,y) 는 vxx+vyy=0v_{xx}+v_{yy}=0 을 만족한다. aa, bb, cc, dd 는 상수라 하자. 벡터 <a,b>\left< a,b\right> 와 <c,d>\left< c,d\right> 의 길이가 같고 서로             ‾\underline{\ \ \ \ \ \ \ \ \ \ \ \ } (빈칸에 알맞은 말을 쓰세요) 이면, u(x,y)=v(ax+by , cx+dy)u(x,y)=v(ax+by\,,\,cx+dy) 도 조화함수임을 보이세요.

다 풀었으면 정답 확인

u(x,y)=v(ax+by , cx+dy)u(x,y)=v(ax+by\,,\,cx+dy) 에 대하여 uxx+uyyu_{xx}+u_{yy} 를 계산한다. 연쇄 법칙에 의해

u(x,y)=v(ax+by , cx+dy)ux(x,y)=a vx(ax+by , cx+dy)+c vy(ax+by , cx+dy)uy(x,y)=b vx(ax+by , cx+dy)+d vy(ax+by , cx+dy)uxx(x,y)=a2 vxx(ax+by , cx+dy)+ac vxy(ax+by , cx+dy)+ca vyx(ax+by , cx+dy)+c2 vyy(ax+by , cx+dy)uyy(x,y)=b2 vxx(ax+by , cx+dy)+bd vxy(ax+by , cx+dy)+db vyx(ax+by , cx+dy)+d2 vyy(ax+by , cx+dy)\begin{aligned} u(x,y) &= v(ax+by\,,\,cx+dy)\\ u_x(x,y) &= a\,v_x(ax+by\,,\,cx+dy)+c\,v_y(ax+by\,,\,cx+dy) \\ u_y(x,y) &= b\,v_x(ax+by\,,\,cx+dy)+d\,v_y(ax+by\,,\,cx+dy) \\ u_{xx}(x,y) &= a^2\,v_{xx}(ax+by\,,\,cx+dy)+ ac\,v_{xy}(ax+by\,,\,cx+dy) \\ & + ca\,v_{yx}(ax+by\,,\,cx+dy) +c^2\,v_{yy}(ax+by\,,\,cx+dy) \\ u_{yy}(x,y) &= b^2\,v_{xx}(ax+by\,,\,cx+dy)+ bd\,v_{xy}(ax+by\,,\,cx+dy) \\ & + db\,v_{yx}(ax+by\,,\,cx+dy) +d^2\,v_{yy}(ax+by\,,\,cx+dy) \end{aligned}

이므로 다음을 얻는다.

uxx+uyy=(a2+b2)vxx(ax+by , cx+dy)+(c2+d2)vyy(ax+by , cx+dy)+2(ac+bd)vxy(ax+by , cx+dy)\begin{aligned} u_{xx}+u_{yy} &=(a^2+b^2) v_{xx}(ax+by\,,\,cx+dy) +(c^2+d^2) v_{yy}(ax+by\,,\,cx+dy) \\& +2(ac+bd) v_{xy}(ax+by\,,\,cx+dy) \end{aligned}

  • a2+b2=c2+d2a^2+b^2=c^2+d^2 이면, 즉 <a,b>\left< a,b\right> 와 <c,d>\left< c,d\right> 의 길이가 같으면, vxx+vyy=0v_{xx}+v_{yy}=0 이므로 우변의 첫째 줄은 0이다.
  • <a,b>⋅<c,d>=ac+bd=0\left< a,b\right>\cdot\left< c,d\right>=ac+bd=0 이면, 즉 <a,b>\left< a,b\right> 와 <c,d>\left< c,d\right> 가 서로 수직이면, 우변의 둘째 줄은 0이다.

따라서 <a,b>\left< a,b\right> 와 <c,d>\left< c,d\right> 의 길이가 같고 서로 수직이면 uxx+uyy=0u_{xx}+u_{yy}=0 이다. 빈칸에 알맞은 말은 “수직”이다.

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