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24_SP_TA: fix, format
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@tiankaima
tiankaima committed Apr 16, 2024
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114 changes: 59 additions & 55 deletions 2bc0c8-2024_spring_TA/main.typ
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Original file line number Diff line number Diff line change
Expand Up @@ -10,7 +10,7 @@
#let dcases(..args) = {
math.cases(
gap: 0.6em,
math.display(..args)
math.display(..args),
)
}

Expand All @@ -20,20 +20,20 @@
]

#align(center)[
= 习题课 7 讲义
2024 Spring 数学分析 B2
= 习题课 7 讲义
2024 Spring 数学分析 B2

PB21000030 马天开
PB21000030 马天开
]

== 作业答案

#text(weight: "bold")[
- 4.1 P125 7(3)(4)(5)
- 4.3 P125 9 10(3)(4) 11(1)(3) 13 18 19 20 21
- 4.8 P156 2(1)(2)(3)(6)(7)(8)
- 4.10 P156 1(2)(3)(5)(6) 3 5 6 7
- 4.12 P166 1(1)(4)(5) 2(2)(5)(8) 3(2)(3) 4 6 7
- 4.1 P125 7(3)(4)(5)
- 4.3 P125 9 10(3)(4) 11(1)(3) 13 18 19 20 21
- 4.8 P156 2(1)(2)(3)(6)(7)(8)
- 4.10 P156 1(2)(3)(5)(6) 3 5 6 7
- 4.12 P166 1(1)(4)(5) 2(2)(5)(8) 3(2)(3) 4 6 7
]

=== 7(3)
Expand All @@ -46,7 +46,6 @@ $
(diff^2 f)/(diff y^2) &= e^(2x) dot 2 \
(diff^2 f)/(diff x diff y) &= e^(2x)(4+4y) \
)\
&dcases(
(diff f)/(diff x) &= 0 \
(diff f)/(diff y) &= 0 \
Expand Down Expand Up @@ -79,11 +78,11 @@ $
$
$
&=> [4x(x^2+y^2)-2a^2x]dif x +[4y(x^2+y^2)+2a^2y]dif y = 0 \
&=> (dif y)/(dif x) = -(4x(x^2+y^2)-2a^2x)/(4y(x^2+y^2)+2a^2y) \
&=> (dif y) / (dif x) = -(4x(x^2+y^2)-2a^2x) / (4y(x^2+y^2)+2a^2y) \
$

$
(dif y)/(dif x) = 0 space.quad => space.quad x = 0 "or" x^2+y^2=1/2a^2 \
(dif y) / (dif x) = 0 space.quad => space.quad x = 0 "or" x^2+y^2=1 / 2a^2 \
$

考虑 $x=0 => (x,y) = (0,0)$ 此处不可微, 舍去
Expand All @@ -93,7 +92,7 @@ dcases(
x^2+y^2=1/2a^2\
x^2-y^2=1/4a^2\
) space.quad => space.quad
y^2 = 1/8
y^2 = 1 / 8
$

接下来可以用更一般的做法判断是否是极大值/极小值, 我们这里推荐一种更加初等但是高效的做法:
Expand All @@ -118,12 +117,14 @@ $
$
&x^2+y^2+z^2-2x+2y-4z-10=0 \
=> & (2x-2)dif x +(2y+2)dif y+(2z-4)dif z=0 \
=> & (diff z)/(diff x) = -(x-1)/(z-2) space.quad (diff z)/(diff y) = -(y+1)/(z-2) \
=> & (diff^2 z)/(diff x^2) = -((x-1)^2+(z-2)^2)/(z-2)^3 space.quad (diff^2 z)/(diff y^2) = -((z-2)^2+(y+1)^2)/(z-2)^3 space.quad (diff^2 z)/(diff x diff y) = 0 \
=> & (diff z) / (diff x) = -(x-1) / (z-2) space.quad (diff z) / (diff y) = -(y+1) / (z-2) \
=> & (diff^2 z) / (diff x^2) = -((x-1)^2+(z-2)^2) / (z-2)^3 space.quad (diff^2 z) / (diff y^2) = -((z-2)^2+(y+1)^2) / (
z-2
)^3 space.quad (diff^2 z) / (diff x diff y) = 0 \
$

$
((diff z)/(diff x), (diff z)/(diff y)) = (0, 0) => (x, y) = (1, -1) \
((diff z) / (diff x), (diff z) / (diff y)) = (0, 0) => (x, y) = (1, -1) \
$

此时对应 $z_1 = 6, z_2=-2$, 我们分别在两个点的局部判断这是极大值/极小值, 即 $(1,-1,6)$ 和 $(1,-1,-2)$
Expand All @@ -141,11 +142,9 @@ $
#pagebreak()
=== 10(3)


$
u(x,y,z)&=sin(x)sin(y)sin(z)\
U(x,y,z,phi)&=sin(x)sin(y)sin(z)-phi dot.c(x+y+z-pi/2)\
&dcases(
(diff U)/(diff x) &= cos(x)sin(y)sin(z)-phi=0\
(diff U)/(diff y) &= sin(x)cos(y)sin(z)-phi=0\
Expand All @@ -157,53 +156,53 @@ $
可以解出

$
P_0=(pi/6,pi/6,pi/6)\
P_1=(pi/2,0,0)\
P_2=(0,pi/2,0)\
P_3=(0,0,pi/2)\
P_0=(pi / 6,pi / 6,pi / 6)\
P_1=(pi / 2,0,0)\
P_2=(0,pi / 2,0)\
P_3=(0,0,pi / 2)\
$

分别代入 $u$ 可以得到 $u(P_0)=1/8, u(P_1)=0, u(P_2)=0, u(P_3)=0$

极大值极小值的判断不能直接从拉格朗日乘子法中得到, 应该通过如下方法判断:

$
u(x,y)=u(x,y,z)=u(x,y,pi/2-x-y)=sin(x)sin(y)cos(x+y)\
u(x,y)=u(x,y,z)=u(x,y,pi / 2-x-y)=sin(x)sin(y)cos(x+y)\
$

接下来继续处理 $Delta=(diff^2 u)/(diff x^2) dot.c (diff^2 u)/(diff y^2)-((diff^2 u)/(diff x diff y))^2$, 按照一般的二元函数的处理 (#strike[也许可以从头开始就按照这样的做法]), 最终可以获得结果.

#image("./imgs/6.png", width: 50%)

#box[
更一般的, 我们可以做如下处理:
更一般的, 我们可以做如下处理:

$
u&=sin(x)sin(y)sin(z)\
dif u&=cos(x)sin(y)sin(z)dif x + sin(x)cos(y)sin(z)dif y + sin(x)sin(y)cos(z)dif z\
$
$
u&=sin(x)sin(y)sin(z)\
dif u&=cos(x)sin(y)sin(z)dif x + sin(x)cos(y)sin(z)dif y + sin(x)sin(y)cos(z)dif z\
$
]

#align(center)[
#rect[
$
&x+y+z=pi/2 space.quad=>space.quad dif x + dif y + dif z = 0\
&x+y+z=pi / 2 space.quad=>space.quad dif x + dif y + dif z = 0\
$
]
]

$
=> & dif u = (cos(x)sin(y)sin(z) - sin(x)sin(y)cos(z))dif x + (sin(x)cos(y)sin(z) - sin(x)sin(y)cos(z))dif y\
=> & (diff u)/(diff x)=(cos(x)sin(y)sin(z) - sin(x)sin(y)cos(z))=0\
=> & (diff u)/(diff y)=(sin(x)cos(y)sin(z) - sin(x)sin(y)cos(z))=0\
=> & (diff^2 u)/(diff x^2)= -2sin(x)sin(y)sin(z) - 2cos(x)sin(y)cos(z)=-1\
=> & (diff^2 u)/(diff y^2) = -2sin(x)sin(y)sin(z) - 2sin(x)cos(y)cos(z)=-1\
=> & (diff^2 u)/(diff x diff y) = cos(x)cos(y)sin(z) - sin(x)cos(y)cos(z) - cos(x)sin(y)cos(z) - sin(x)sin(y)sin(z)=-1/2\
=> & (diff u) / (diff x)=(cos(x)sin(y)sin(z) - sin(x)sin(y)cos(z))=0\
=> & (diff u) / (diff y)=(sin(x)cos(y)sin(z) - sin(x)sin(y)cos(z))=0\
=> & (diff^2 u) / (diff x^2)= -2sin(x)sin(y)sin(z) - 2cos(x)sin(y)cos(z)=-1\
=> & (diff^2 u) / (diff y^2) = -2sin(x)sin(y)sin(z) - 2sin(x)cos(y)cos(z)=-1\
=> & (diff^2 u) / (diff x diff y) = cos(x)cos(y)sin(z) - sin(x)cos(y)cos(z) - cos(x)sin(y)cos(z) - sin(x)sin(y)sin(z)=-1 / 2\
$

所以有
$
Delta = A C - B^2 = 3/4 >0 space.quad A=-1<0
Delta = A C - B^2 = 3 / 4 >0 space.quad A=-1<0
$
正定, 最大值 $ u(pi/6,pi/6,pi/6)=1/8 $

Expand All @@ -221,15 +220,21 @@ $

$
&dif u = y z dif x + x z dif y + x y dif z\
\
&dif x + dif y + dif z = 0\
&x dif x + y dif y + z dif z = 0\
$

#align(center)[
#rect[
$
&dif x + dif y + dif z = 0\
&x dif x + y dif y + z dif z = 0\
$
]
]

在后两个方程中, 解出 $dif y, dif z$ 关于 $dif x$的表达式为:

$
dif y = (z-x)/(y-z) dif x quad dif z = (x-y)/(y-z) dif x
dif y = (z-x) / (y-z) dif x quad dif z = (x-y) / (y-z) dif x
$

接下来可以类似处理得到 $(dif u)/(dif x) , (dif^2 u)/(dif x^2)$, 按照一元函数的极值点处理即可.
Expand All @@ -252,7 +257,7 @@ $
首先考虑 $(diff U)/(diff x),(diff U)/(diff y),(diff U)/(diff z)$ 中 $mu x,mu y,mu z$的对称性, 我们将这三项对应加起来:

$
&(diff U)/(diff x) + (diff U)/(diff y) + (diff U)/(diff z)\
&(diff U) / (diff x) + (diff U) / (diff y) + (diff U) / (diff z)\
&= x z + y z + x y - 3 lambda - 2 mu (x+y+z)\
&= x z + y z + x y - 3 lambda = 0
$
Expand All @@ -262,7 +267,7 @@ $
$
(x+y+z)^2&=x^2+y^2+z^2+2(x y+y z+z x)\
&=1+2(x y+y z+z x)=0 \
=> & x y+y z+z x = -1/2
=> & x y+y z+z x = -1 / 2
$

因此得到 $lambda = - 1/6$, 接下来只需要关心前两项 $(diff U)/(diff x), (diff U)/(diff y)$:
Expand All @@ -275,8 +280,8 @@ x z + 1/6 - 2 mu y = 0 \
y^2 z + 1/6 y - 2 mu x y = 0 \
x^2 z + 1/6 x - 2 mu y x = 0 \
)\
&=> [z(x+y) + 1/6](x-y) = 0 \
&=> [(x+y)^2 - 1/6](x-y) = 0 \
&=> [z(x+y) + 1 / 6](x-y) = 0 \
&=> [(x+y)^2 - 1 / 6](x-y) = 0 \
$

接下来我们分别讨论 $x=y$ 和 $(x+y)^2=1/6$. 其实他们反映的是一种情况的两个对称.
Expand All @@ -285,16 +290,16 @@ $

$
x=y => z=-2x\
x^2 + y^2 + z^2 = 6 x^2 = 1 => x = plus.minus sqrt(6)/6 \
P_1(sqrt(6)/6,sqrt(6)/6, -sqrt(6)/3) quad P_2(-sqrt(6)/6, -sqrt(6)/6, sqrt(6)/3)
x^2 + y^2 + z^2 = 6 x^2 = 1 => x = plus.minus sqrt(6) / 6 \
P_1(sqrt(6) / 6,sqrt(6) / 6, -sqrt(6) / 3) quad P_2(-sqrt(6) / 6, -sqrt(6) / 6, sqrt(6) / 3)
$

- $(x+y)^2=1/6$
- $(x+y)^2=1 / 6$

$
(x+y)^2=1/6 => z^2 = 1/6 \
=> x^2 + y^2 = 5/6 \
(x-y)^2 = 9/6 \
(x+y)^2=1 / 6 => z^2 = 1 / 6 \
=> x^2 + y^2 = 5 / 6 \
(x-y)^2 = 9 / 6 \
$

$
Expand All @@ -305,8 +310,8 @@ dcases(
$

$
&P_3(sqrt(6)/3, -sqrt(6)/6, -sqrt(6)/6) quad &P_4(-sqrt(6)/3, sqrt(6)/6, sqrt(6)/6) \
&P_5(sqrt(6)/6, -sqrt(6)/3, sqrt(6)/6) quad &P_6(-sqrt(6)/6, sqrt(6)/3, -sqrt(6)/6) \
&P_3(sqrt(6) / 3, -sqrt(6) / 6, -sqrt(6) / 6) quad &P_4(-sqrt(6) / 3, sqrt(6) / 6, sqrt(6) / 6) \
&P_5(sqrt(6) / 6, -sqrt(6) / 3, sqrt(6) / 6) quad &P_6(-sqrt(6) / 6, sqrt(6) / 3, -sqrt(6) / 6) \
$

因此 $ u_max = sqrt(6)/18 quad u_min = -sqrt(6)/18 $
Expand All @@ -323,9 +328,8 @@ $
即便没有这样的几何直观, 处理起来也是固定的模式, 先讨论内部的极值点、再讨论边界上的条件极值.

$
(diff z)/(diff x) = 2x quad (diff z)/(diff y) = -2y \
(diff^2 z)/(diff x^2) = 2 quad (diff^2 z)/(diff y^2) = -2 quad (diff^2 z)/(diff x diff y) = 0
(diff z) / (diff x) = 2x quad (diff z) / (diff y) = -2y \
(diff^2 z) / (diff x^2) = 2 quad (diff^2 z) / (diff y^2) = -2 quad (diff^2 z) / (diff x diff y) = 0
$

Hessian 矩阵总是负定的, 函数在${(x,y)mid("|") x^2+y^2 <4}$内部不存在极值点. (事实上也说明在任何区域内部都不存在极值点)
Expand All @@ -339,7 +343,7 @@ Hessian 矩阵总是负定的, 函数在${(x,y)mid("|") x^2+y^2 <4}$内部不存
max_(overline(Omega)) u = max_(partial Omega) u
$

- *极值原理*: 非调和函数的最值总在边界上取得.
- *极值原理*: 非常数调和函数的最值总在边界上取得.

$
max_(Omega) u = max_(partial Omega) u \
Expand Down

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