數學分析 隐函數定理及其應用(第18章)

一.隐函數

1.概念:

注:①這裡隻表示存在着定義在 I I I上,值域包含于 J J J的函數 f f f，而不意味着 y y y能用 x x x的某一顯式來表示

2.存在性

3.隐函數定理

(1)隐函數存在唯一性定理:

定理18.1:若函數 F ( x , y ) F(x,y) F(x,y)滿足:

① F F F在以 P 0 ( x 0 , y 0 ) P_0(x_0,y_0) P0​(x0​,y0​)為内點的某一區域 D ⊂ R 2 D\sub R^2 D⊂R2上連續

② F ( x 0 , y 0 ) = 0 F(x_0,y_0)=0 F(x0​,y0​)=0(通常稱為初始條件)

③ F F F在 D D D上存在連續的偏導數 F y ( x , y ) F_y(x,y) Fy​(x,y)

④ F y ( x 0 , y 0 ) ≠ 0 F_y(x_0,y_0)≠0 Fy​(x0​,y0​)​=0

則

①存在點 P 0 P_0 P0​的某鄰域 U ( P 0 ) ⊂ D U(P_0)\sub D U(P0​)⊂D,在 U ( P 0 ) U(P_0) U(P0​)上方程 F ( x , y ) = 0 F(x,y)=0 F(x,y)=0唯一地決定了1個定義在某區間 ( x 0 − α , x 0 + α ) (x_0-α,x_0+α) (x0​−α,x0​+α)上的(隐)函數 y = f ( x ) y=f(x) y=f(x),使得當 x ∈ ( x 0 − α , x 0 + α ) x∈(x_0-α,x_0+α) x∈(x0​−α,x0​+α)時, ( x , f ( x ) ) ∈ U ( P 0 ) (x,f(x))∈U(P_0) (x,f(x))∈U(P0​)且 F ( x , f ( x ) ) ≡ 0 , f ( x 0 ) = y 0 F(x,f(x))\equiv0,f(x_0)=y_0 F(x,f(x))≡0,f(x0​)=y0​

(2)隐函數可微性定理:

定理18.2:設 F ( x , y ) F(x,y) F(x,y)滿足隐函數存在唯一性定理中的4個條件,又設在 D D D上還存在連續的偏導數 F x ( x , y ) F_x(x,y) Fx​(x,y),則由方程(1)所确定的隐函數 y = f ( x ) y=f(x) y=f(x)在其定義域 ( x 0 − α , x 0 + α ) (x_0-α,x_0+α) (x0​−α,x0​+α)上有連續偏導數,且 f ′ ( x ) = − F x ( x , y ) F y ( x , y ) ( 5 ) f'(x)=-\frac{F_x(x,y)}{F_y(x,y)}\qquad(5) f′(x)=−Fy​(x,y)Fx​(x,y)​(5)

(3)隐函數極值問題:

(4)多元隐函數的唯一存在與連續可微定理:

定理18.3:若

①函數 F ( x 1 , x 2 . . . x n , y ) F(x_1,x_2...x_n,y) F(x1​,x2​...xn​,y)在以點 P 0 ( x 1 0 , x 2 0 . . . x n 0 , y 0 ) P_0(x_1^0,x_2^0...x_n^0,y^0) P0​(x10​,x20​...xn0​,y0)為内點的區域 D ⊂ R n + 1 D\sub R^{n+1} D⊂Rn+1上連續

② F ( x 1 0 , x 2 0 . . . x n 0 , y 0 ) = 0 F(x_1^0,x_2^0...x_n^0,y^0)=0 F(x10​,x20​...xn0​,y0)=0

③偏導數 F x 1 , F x 2 . . . F x n , F y F_{x_1},F_{x_2}...F_{x_n},F_y Fx1​​,Fx2​​...Fxn​​,Fy​在 D D D上存在且連續

④ F y ( x 1 0 , x 2 0 . . . x n 0 , y 0 ) ≠ 0 F_y(x_1^0,x_2^0...x_n^0,y^0)≠0 Fy​(x10​,x20​...xn0​,y0)​=0

則

①存在點 P 0 P_0 P0​的某鄰域 U ( P 0 ) ⊂ D U(P_0)\sub D U(P0​)⊂D,在 U ( P 0 ) U(P_0) U(P0​)上方程 F ( x 1 , x 2 . . . x n , y ) = 0 F(x_1,x_2...x_n,y)=0 F(x1​,x2​...xn​,y)=0唯一地确定了1個定義在 Q 0 ( x 1 0 , x 2 0 . . . x n 0 ) Q_0(x_1^0,x_2^0...x_n^0) Q0​(x10​,x20​...xn0​)的某鄰域 U ( Q 0 ) ⊂ R n U(Q_0)\sub R^n U(Q0​)⊂Rn上的 n n n元連續(隐)函數 y = f ( x 1 , x 2 . . . x n ) y=f(x_1,x_2...x_n) y=f(x1​,x2​...xn​),使得當 ( x 1 , x 2 . . . x n ) ∈ U ( Q 0 ) (x_1,x_2...x_n)\in U(Q_0) (x1​,x2​...xn​)∈U(Q0​)時,有 ( x 1 , x 2 . . . x n , f ( x 1 , x 2 . . . x n ) ) ∈ U ( P 0 ) (x_1,x_2...x_n,f(x_1,x_2...x_n))∈U(P_0) (x1​,x2​...xn​,f(x1​,x2​...xn​))∈U(P0​)且 F ( x 1 , x 2 . . . x n , f ( x 1 , x 2 . . . x n ) ) ≡ 0 , y 0 = f ( ( x 1 0 , x 2 0 . . . x n 0 ) F(x_1,x_2...x_n,f(x_1,x_2...x_n))\equiv0,y^0=f((x_1^0,x_2^0...x_n^0) F(x1​,x2​...xn​,f(x1​,x2​...xn​))≡0,y0=f((x10​,x20​...xn0​)

② y = f ( x 1 , x 2 . . . x n ) y=f(x_1,x_2...x_n) y=f(x1​,x2​...xn​)在 U ( Q 0 ) U(Q_0) U(Q0​)上有連續偏導數 f x 1 , f x 2 . . . f x n f_{x_1},f_{x_2}...f_{x_n} fx1​​,fx2​​...fxn​​,且 f x 1 = − F x 1 F y , f x 2 = − F x 2 F y . . . f x n = − F x n F y f_{x_1}=-\frac{F_{x_1}}{F_y},f_{x_2}=-\frac{F_{x_2}}{F_y}...f_{x_n}=-\frac{F_{x_n}}{F_y} fx1​​=−Fy​Fx1​​​,fx2​​=−Fy​Fx2​​​...fxn​​=−Fy​Fxn​​​

4.隐函數的反函數:

二.隐函數組

1.概念:

2.隐函數組定理

(1)函數行列式:

(2)隐函數組定理:

定理18.4:若

① F ( x , y , u , v ) , G ( x , y , u , v ) F(x,y,u,v),G(x,y,u,v) F(x,y,u,v),G(x,y,u,v)在以點 P 0 P_0 P0​為内點的區域 V ⊂ R 4 V\sub R^4 V⊂R4上連續

② F ( x 0 , y 0 , u 0 , v 0 ) = G ( x 0 , y 0 , u 0 , v 0 ) F(x_0,y_0,u_0,v_0)=G(x_0,y_0,u_0,v_0) F(x0​,y0​,u0​,v0​)=G(x0​,y0​,u0​,v0​)(初始條件)

③在 V V V上 F , G F,G F,G具有1階連續偏導數

④ J = ∂ ( F , G ) ∂ ( u , v ) J=\frac{\partial(F,G)}{\partial(u,v)} J=∂(u,v)∂(F,G)​在點 P 0 P_0 P0​處不等于0

則

①存在點 P 0 P_0 P0​的某(4維空間)鄰域 U ( P 0 ) ⊂ V U(P_0)\sub V U(P0​)⊂V,在 U ( P 0 ) U(P_0) U(P0​)上方程組(1)唯一地确定了定義在點 Q 0 ( x 0 , y 0 ) Q_0(x_0,y_0) Q0​(x0​,y0​)的某(2維空間)鄰域 U ( Q 0 ) U(Q_0) U(Q0​)上的2個二進制隐函數 u = f ( x , y ) , v = g ( x , y ) u=f(x,y),v=g(x,y) u=f(x,y),v=g(x,y)使得 u 0 = f ( x 0 , y 0 ) , v 0 = g ( x 0 , y 0 ) u_0=f(x_0,y_0),v_0=g(x_0,y_0) u0​=f(x0​,y0​),v0​=g(x0​,y0​),且當 ( x , y ) ∈ U ( Q 0 ) (x,y)∈U(Q_0) (x,y)∈U(Q0​)時,有 ( x , y , f ( x , y ) , g ( x , y ) ) ∈ U ( P 0 ) F ( x , y , f ( x , y ) , g ( x , y ) ) ≡ 0 G ( x , y , f ( x , y ) , g ( x , y ) ) ≡ 0 (x,y,f(x,y),g(x,y))∈U(P_0)\\F(x,y,f(x,y),g(x,y))\equiv0\\G(x,y,f(x,y),g(x,y))\equiv0 (x,y,f(x,y),g(x,y))∈U(P0​)F(x,y,f(x,y),g(x,y))≡0G(x,y,f(x,y),g(x,y))≡0② f ( x , y ) , g ( x , y ) f(x,y),g(x,y) f(x,y),g(x,y)在 U ( Q 0 ) U(Q_0) U(Q0​)上連續

③ f ( x , y ) , g ( x , y ) f(x,y),g(x,y) f(x,y),g(x,y)在 U ( Q 0 ) U(Q_0) U(Q0​)上有1階連續偏導數,且 ∂ u ∂ x = − 1 J ∂ ( F , G ) ∂ ( x , v ) , ∂ v ∂ x = − 1 J ∂ ( F , G ) ∂ ( u , x ) ∂ u ∂ y = − 1 J ∂ ( F , G ) ∂ ( y , v ) , ∂ v ∂ y = − 1 J ∂ ( F , G ) ∂ ( u , y ) ( 5 ) \begin{matrix}\frac{\partial u}{\partial x}=-\frac{1}{J}\frac{\partial(F,G)}{\partial(x,v)},\frac{\partial v}{\partial x}=-\frac{1}{J}\frac{\partial(F,G)}{\partial(u,x)}\\\frac{\partial u}{\partial y}=-\frac{1}{J}\frac{\partial(F,G)}{\partial(y,v)},\frac{\partial v}{\partial y}=-\frac{1}{J}\frac{\partial(F,G)}{\partial(u,y)}\end{matrix}\qquad(5) ∂x∂u​=−J1​∂(x,v)∂(F,G)​,∂x∂v​=−J1​∂(u,x)∂(F,G)​∂y∂u​=−J1​∂(y,v)∂(F,G)​,∂y∂v​=−J1​∂(u,y)∂(F,G)​​(5)

3.反函數組與坐标變換

(1)反函數組:

(2)反函數組定理:

定理18.5:設函數組(9)及其1階偏導數在某區域 D ⊂ R 2 D\sub R^2 D⊂R2上連續,點 P 0 ( x 0 , y 0 ) P_0(x_0,y_0) P0​(x0​,y0​)是 D D D的内點,且 u 0 = u ( x 0 , y 0 ) , v 0 = v ( x 0 , y 0 ) , ∂ ( u , v ) ∂ ( x , y ) ∣ P 0 ≠ 0 u_0=u(x_0,y_0),v_0=v(x_0,y_0),\frac{\partial(u,v)}{\partial(x,y)}|_{P_0}≠0 u0​=u(x0​,y0​),v0​=v(x0​,y0​),∂(x,y)∂(u,v)​∣P0​​​=0則在點 P 0 ′ ( u 0 , v 0 ) P_0'(u_0,v_0) P0′​(u0​,v0​)的某鄰域 U ( P 0 ′ ) U(P_0') U(P0′​)上存在唯一的1組反函數(10),使得 x 0 = x ( u 0 , v 0 ) , y 0 = y ( u 0 , v 0 ) x_0=x(u_0,v_0),y_0=y(u_0,v_0) x0​=x(u0​,v0​),y0​=y(u0​,v0​),且當 ( u , v ) ∈ U ( P 0 ′ ) (u,v)∈U(P_0') (u,v)∈U(P0′​)時,有 ( x ( u , v ) , y ( u , v ) ) ∈ U ( P 0 ) (x(u,v),y(u,v))∈U(P_0) (x(u,v),y(u,v))∈U(P0​)以及恒等式(11);此外,反函數組(10)在 U ( P 0 ′ ) U(P_0') U(P0′​)上存在連續的1階偏導數,且 ∂ x ∂ u = ∂ v ∂ y / ∂ ( u , v ) ∂ ( x , y ) , ∂ x ∂ v = − ∂ u ∂ y / ∂ ( u , v ) ∂ ( x , y ) ∂ y ∂ u = − ∂ v ∂ x / ∂ ( u , v ) ∂ ( x , y ) , ∂ y ∂ v = ∂ u ∂ x / ∂ ( u , v ) ∂ ( x , y ) ( 13 ) \begin{matrix}\frac{\partial x}{\partial u}=\frac{\partial v}{\partial y}/\frac{\partial(u,v)}{\partial(x,y)},\frac{\partial x}{\partial v}=-\frac{\partial u}{\partial y}/\frac{\partial(u,v)}{\partial(x,y)}\\\frac{\partial y}{\partial u}=-\frac{\partial v}{\partial x}/\frac{\partial(u,v)}{\partial(x,y)},\frac{\partial y}{\partial v}=\frac{\partial u}{\partial x}/\frac{\partial(u,v)}{\partial(x,y)}\end{matrix}\qquad(13) ∂u∂x​=∂y∂v​/∂(x,y)∂(u,v)​,∂v∂x​=−∂y∂u​/∂(x,y)∂(u,v)​∂u∂y​=−∂x∂v​/∂(x,y)∂(u,v)​,∂v∂y​=∂x∂u​/∂(x,y)∂(u,v)​​(13)

(3)坐标變換:

三.幾何應用

1.平面曲線的切線與法線:

2.空間曲線的切線與法平面

3.曲面的切平面與法線:

四.條件極值

1.條件極值問題:

2.拉格朗日乘數法:

定理18.6:設在條件(2)的限制下,求函數(3)的極值問題,其中 f , φ k   ( k = 1 , 2... m ) f,φ_k\,(k=1,2...m) f,φk​(k=1,2...m)在區域 D D D上有連續的1階偏導數;若 D D D的内點 P 0 ( x 1 ( 0 ) . . . x n ( 0 ) ) P_0(x_1^{(0)}...x_n^{(0)}) P0​(x1(0)​...xn(0)​)是上述問題的極值點,且雅可比矩陣 [ ∂ φ 1 ∂ x 1 . . . ∂ φ 1 ∂ x n . . . . . . ∂ φ m ∂ x 1 . . . ∂ φ m ∂ x n ] ( 13 ) \left[\begin{matrix}\frac{\partialφ_1}{\partial x_1}&...&\frac{\partialφ_1}{\partial x_n}\\...&&...\\\frac{\partialφ_m}{\partial x_1}&...&\frac{\partialφ_m}{\partial x_n}\end{matrix}\right]\qquad(13) ⎣⎡​∂x1​∂φ1​​...∂x1​∂φm​​​......​∂xn​∂φ1​​...∂xn​∂φm​​​⎦⎤​(13)的秩為 m m m,則 ∃ m ∃m ∃m個常數 λ 1 ( 0 ) . . . λ m ( 0 ) λ_1^{(0)}...λ_m^{(0)} λ1(0)​...λm(0)​,使得 ( x 1 ( 0 ) . . . x n ( 0 ) , λ 1 ( 0 ) . . . λ m ( 0 ) ) (x_1^{(0)}...x_n^{(0)},λ_1^{(0)}...λ_m^{(0)}) (x1(0)​...xn(0)​,λ1(0)​...λm(0)​)為拉格朗日函數(12)的穩定點,即 ( x 1 ( 0 ) . . . x n ( 0 ) , λ 1 ( 0 ) . . . λ m ( 0 ) ) (x_1^{(0)}...x_n^{(0)},λ_1^{(0)}...λ_m^{(0)}) (x1(0)​...xn(0)​,λ1(0)​...λm(0)​)為 n + m n+m n+m個方程 { L x 1 = ∂ f ∂ x 1 + ∑ k = 1 m λ k ∂ φ k ∂ x 1 = 0 . . . L x n = ∂ f ∂ x n + ∑ k = 1 m λ k ∂ φ k ∂ x n = 0 L λ 1 = φ 1 ( x 1 . . . x n ) . . . L λ m = φ m ( x 1 . . . x n ) \begin{cases}L_{x1}=\frac{\partial f}{\partial x_1}+\displaystyle\sum_{k=1}^mλ_k\frac{\partialφ_k}{\partial x_1}=0\\...\\L_{x_n}=\frac{\partial f}{\partial x_n}+\displaystyle\sum_{k=1}^mλ_k\frac{\partialφ_k}{\partial x_n}=0\\L_{λ_1}=φ_1(x_1...x_n)\\...\\L_{λ_m}=φ_m(x_1...x_n)\end{cases} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​Lx1​=∂x1​∂f​+k=1∑m​λk​∂x1​∂φk​​=0...Lxn​​=∂xn​∂f​+k=1∑m​λk​∂xn​∂φk​​=0Lλ1​​=φ1​(x1​...xn​)...Lλm​​=φm​(x1​...xn​)​的解