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편미분방정식 맛보기

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

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 문제CLP-3 · 2.8절 · 11번우리말 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.

답 쓰고 채점
다 풀었으면 정답 확인

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

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