目录
- Scherk曲面
Scherk曲面
证明形如 z = f ( x ) + g ( y ) z=f(x)+g(y) z=f(x)+g(y)的极小曲面,若非平面,则除相差一个常数外,它可以写成
z = 1 a log cos a y cos a x . z=\dfrac{1}{a}\log\dfrac{\cos ay}{\cos ax}. z=a1logcosaxcosay.
此曲面称为 Scherk 曲面.
证明:设方程 z = f ( x ) + g ( y ) z=f(x)+g(y) z=f(x)+g(y)定义了一个极小曲面 S . S. S.显然,它有参数表达式
r ( u , v ) = ( u , v , f ( u ) + g ( v ) ) . \mathbf{r}(u,v)=(u,v,f(u)+g(v)). r(u,v)=(u,v,f(u)+g(v)).
直接计算,有
r u = ( 1 , 0 , f ′ ( u ) ) , r v = ( 0 , 1 , g ′ ( v ) ) . \mathbf{r}_u=(1,0,f'(u)),\quad\mathbf{r}_v=(0,1,g'(v)). ru=(1,0,f′(u)),rv=(0,1,g′(v)).
故
E = 1 + f ′ ( u ) 2 , F = f ′ ( u ) g ′ ( v ) , G = 1 + g ′ ( v ) . E=1+f'(u)^2,\:F=f'(u)g'(v),\:G=1+g'(v). E=1+f′(u)2,F=f′(u)g′(v),G=1+g′(v).
由
r u ∧ r v = ( − f ′ ( u ) , − g ′ ( v ) , 1 ) , \mathbf{r}_u\wedge\mathbf{r}_v=(-f'(u),-g'(v),1), ru∧rv=(−f′(u),−g′(v),1),
知
n = 1 1 + f ′ ( u ) 2 + g ′ ( v ) 2 ( − f ′ ( u ) , − g ′ ( v ) , 1 ) . \mathbf n=\dfrac{1}{\sqrt{1+f'(u)^2+g'(v)^2}}(-f'(u),-g'(v),1). n=1+f′(u)2+g′(v)21(−f′(u),−g′(v),1).
而
r u u = ( 0 , 0 , f ′ ′ ( u ) ) , r u v = 0 , r v v = ( 0 , 0 , g ′ ′ ( v ) ) , \mathbf{r}_{uu}=(0,0,f''(u)),\:\mathbf{r}_{uv}=\mathbf{0},\:\mathbf{r}_{vv}=(0,0,g''(v)), ruu=(0,0,f′′(u)),ruv=0,rvv=(0,0,g′′(v)),
故
L = f ′ ′ ( u ) 1 + f ′ ( u ) 2 + g ′ ( v ) 2 , M = 0 , N = g ′ ′ ( v ) 1 + f ′ ( u ) 2 + g ′ ( v ) 2 . L=\frac{f''(u)}{\sqrt{1+f'(u)^2+g'(v)^2}},\quad M=0,\quad N=\frac{g''(v)}{\sqrt{1+f'(u)^2+g'(v)^2}}. L=1+f′(u)2+g′(v)2f′′(u),M=0,N=1+f′(u)2+g′(v)2g′′(v).
由于曲面是极小的,
H = L G − 2 M F + N E 2 ( E G − F 2 ) = f ′ ′ ( u ) ( 1 + g ′ ( v ) 2 ) + g ′ ′ ( v ) ( 1 + f ′ ( u ) 2 ) 2 ( 1 + f ′ ( u ) 2 + g ′ ( v ) 2 ) 3 2 ) = 0. H=\frac{LG-2MF+NE}{2(EG-F^2)}=\frac{f''(u)(1+g'(v)^2)+g''(v)(1+f'(u)^2)}{2(1+f'(u)^2+g'(v)^2)^{\frac{3}{2}})}=0. H=2(EG−F2)LG−2MF+NE=2(1+f′(u)2+g′(v)2)23)f′′(u)(1+g′(v)2)+g′′(v)(1+f′(u)2)=0.
这等价于
f ′ ′ ( u ) ( 1 + g ′ ( v ) 2 ) + g ′ ′ ( v ) ( 1 + f ′ ( u ) 2 ) = 0. f''(u)(1+g'(v)^2)+g''(v)(1+f'(u)^2)=0. f′′(u)(1+g′(v)2)+g′′(v)(1+f′(u)2)=0.
继而,等价于
f ′ ′ ( u ) 1 + f ′ ( u ) 2 = − g ′ ′ ( v ) 1 + g ′ ( v ) 2 . \frac{f''(u)}{1+f'(u)^2}=-\frac{g''(v)}{1+g'(v)^2}. 1+f′(u)2f′′(u)=−1+g′(v)2g′′(v).
由于上面等式两边是关于不同变量的函数,故它们只能为同一常数,记为 a . a. a.
若 S S S不是平面,则必有 a ≠ 0. a\neq0. a=0.
否则,由 f ′ ′ ( u ) = 0 = g ′ ′ ( v ) f^{\prime\prime}(u)=0=g^{\prime\prime}(v) f′′(u)=0=g′′(v),
有 f ( u ) = k u + m f(u)=ku+m f(u)=ku+m, g ( v ) = l v + n g(v)=lv+n g(v)=lv+n.
从而 , S ,S ,S 是平面.
现在
d ( arctan f ′ ( u ) ) = d ( a u ) , d ( arctan g ′ ( v ) ) = d ( − a v ) . d(\arctan f'(u))=d(au),\quad d(\arctan g'(v))=d(-av). d(arctanf′(u))=d(au),d(arctang′(v))=d(−av).
故
arctan f ′ ( u ) = a u + b , arctan g ′ ( v ) = − a v − c , \arctan f'(u)=au+b,\quad\arctan g'(v)=-av-c, arctanf′(u)=au+b,arctang′(v)=−av−c,
其中 b , c b,c b,c 为常数.从而,
f ( u ) = − 1 a log cos ( a u + b ) + d , g ( v ) = 1 a log cos ( a v + c ) + e , f(u)=-\frac{1}{a}\log\cos(au+b)+d,\quad g(v)=\frac{1}{a}\log\cos(av+c)+e, f(u)=−a1logcos(au+b)+d,g(v)=a1logcos(av+c)+e,
其中 d , e d,e d,e 为常数.所以,
z = f ( x ) + g ( y ) = 1 a log cos ( a y + c ) cos ( a x + b ) + d + e . z=f(x)+g(y)=\dfrac{1}{a}\log\dfrac{\cos(ay+c)}{\cos(ax+b)}+d+e. z=f(x)+g(y)=a1logcos(ax+b)cos(ay+c)+d+e.