線積分

在本章和下章中, 我們將討論多變函數的積分學. 多變數函數的積分有很多種. 本章先討論線積分與重積分, 而把面積分和三重積分留給下章.

令 $C: x=\gamma(t)$, $t \in [a, b]$ 為 $\mathbb{R} ^n$ 中的連續曲線, 若有 $a=t_0<t_1<\cdots<t_m=b$使 $\gamma(t)$ 局限在每個 [ t_i-1, t_i] 中時均有連續的導函數, (但當我們把 $\gamma(t)$ 看成整個 [a, b] 上的函數時, 不要求 $\gamma'(t_{i}), 0<i<m$, 存在), 則稱 C 為 $\mathbb{R} ^n$ 中的一條 piecewise continuously differentiable curve 或 piecewise smooth curve. 直觀地講, piecewise smooth curve 是說這曲線除了在有限個點處可能轉硬灣外, 別的地方都到處平滑.

令 D 為 $\mathbb{R} ^n$ 中的區域, $f(x)=(f_1(x), f_2(x), \cdots, f_n(x))$ 為定義於 $x\in D$ 的 $\mathbb{R} ^n$ 值函數. 再設 $C: x=\gamma(t), t \in [a, b]$為 D 中的 piecewise smooth curve. 則線積分 (line integral, curvilinear integral)

$$\int_c f(x)\,dx=\int_c f_1(x)\,dx_1+f_2(x)\,dx_2+\cdots+f_n(x)\,dx_n$$

之定義為

$$\int_a^b f(\gamma(t)), \gamma'(t)\,dt.$$

C 叫做該線積分的積分路 (path of integration).

線積分的物理意義如下: 若一質點在力場 f(x) 之影響下沿曲線 C 移動, 則其行經 C 後位能的變化便是

$$\int_c f(x)\cdot\,dx.$$

線積分的定義和曲線 C 的參數 t 的選擇並無關係. 事實上若 $h: [\alpha, \beta] \to [a, b]$為可微增函數, $\delta(s)=\gamma(h(s)), a=h(\alpha), b=h(\beta)$, 則

$$\int_{a}^{b} f(\gamma(t)) \cdot \gamma\prime(t)\,dt=\int_{\al... ...beta} f(\gamma(h(s))) \cdot \gamma\prime(h(s))h\prime(s))\,ds$$

$$= \int_{\alpha}^{\beta} f(\delta(s) \cdot \delta\prime(s))\,ds.$$

通常幾何上所說的曲線, 未必附帶就給出了參數. 依這裡的討論, 只要曲線的方向確定, 無論怎麼選參數, 得到的線積分都一樣. 所以我們也可以認定線積分的積分路只是業已選定方向的幾何曲線, 其參數可以任選.

例 1   在平面上考慮 C₁, C₂ 二線如下: C₁ 為連接 (0, 0) 和 (1, 1) 的直線線段, 即

$$x=t, \quad y=t, \quad t \in [0, 1]$$

C₂ 為連接 (0, 0) 至 (1, 0), 再連接 (1, 0) 至 (1, 1) 的折線:

$$\left\{ \begin{array}{ll} x=1+t, \\ y=0, \end{array} \right. \ t \in [-1, 0],$$

$$\left\{ \begin{array}{ll} x=1, \\ y=t, \end{array} \right. t \in [0, 1].$$

則

$$\int_{C_1} (2xy\,dx+y^{3}\,dy)=\int_{0}^1 (2t^2+t^{3})\,dt=\frac {11}{12}$$

$$\int_{C_2} (2xy\,dx+y^{3}\,dy)=\int_{0}^1 t^{3}\,dt=\frac {1}{4}$$

但另一方面我們有

$$\int_{C_1} 2xy\,dx+(x^2+y^2)\,dy=\int_{0}^1 4t^2\,dt=\frac {4}{3}$$

$$\int_{C_2} 2xy\,dx+(x^2+y^2)\,dy=\int_{0}^1 (1+t^2)\,dt=\frac {4}{3}$$

在前面的兩個積分裡, 線積分的值不但和積分路的端點有關, 並且和整條積分路 C 都有關, 而在後面的兩個例子裡兩個積分值是一樣的. 下定理指出線積分僅和 C 的端點有關的條件.

證明. First assmume that $f(x)=\nabla F(x).$Then for every curve

$$C: x=\gamma(t), \quad t \in [a, b],$$

joing p to q we have

$$\begin{eqnarray*}\lefteqn{\int_{C}f(x) \cdot\,dx }\\ &=& \int_{a}^{b} \nabla F(... ...dt\\ &=& \int_{a}^{b} \frac {d}{\,dt}F(\gamma(t))\,dt=F(q)-F(p) \end{eqnarray*}$$

Conversely assume that the integral depends only on the end-points. We fix a point $p_{0} \in D.$ For any point $p \in D$ define

$$F(p)=\int_{C}f(x)\cdot\,dx$$

where C is any path in D joining p₀ to p. This function F is well-defined because the integral is independent of path. Let

$$p=(x_1, x_2, \cdots, x_{k-1}, x_{k}, x_{k+1}, \cdots x_{n}),$$

$$q=(x_1, x_2, \cdots, x_{k-1}, x_{k}+h, x_{k+1}, \cdots x_{n}),$$

and let C₀ be the curve

$$\gamma(t)=((x_1, x_2, \cdots, x_{k-1}, x_{k+ht}, x_{k+1}, x_{m}), \quad t \in [0, 1].$$

Then by th mean-value theorem,

$$\begin{eqnarray*}\lefteqn{\frac{1}{h}[F(q)-F(p)]=\frac{1}{h}\int_{c_{0}} f(x) \c... ..._n)h\,dt\\ &=& f_{k}(x_1, x_2, \cdots, x_{k}+rh, \cdots, x_{n}) \end{eqnarray*}$$

where $\tau \in$ (0, 1). Letting $h \to 0$, we get

$$\frac{\partial F}{\partial x_{k}}(x_1, x_2, \cdots, x_{n})=f_{k}(x_1, x_2, \cdots, x_{n}).$$

Q. E. D

在本定理中所討論呈 $\nabla F$ 形的函數, 叫作梯度場 (gradient field). 梯度場和以前講過的全微分有密切的關係: 若 $f=\nabla F$ 為 $\mathbb{R} ^n$ 中的梯度場, 則

$$dF=\sum_{i=1}^{n} f_i(x) \,dx_i.$$

在上面例1中, $2xy\,dy+(x^2+y^2)\,dy$ 確是一個正合積分, 因為

$$d(x^2y+\frac{1}{3}y^3)=2xy\,dx+(x^2+y^2)\,dy.$$

由本定理的前半段證明可知, 若 C 為連接 p 至 q 的 piecewise continuously differentiable curve 則

$$\int_C \nabla F \cdot\,dx=\int_C dF=F(q)-F(p).$$

下定理是梯度場的一個必要條件:

這定理本身是一個 trivial 的結果. H. Poincaré 問這定理的逆定理是否成立. 結果他得知逆定理之成立與否, 即上式是否保證 f 為梯度場, 和 D 的拓樸性質有關--即充要條件是 D 為 simply connected. 這是拓樸學的濫觴. 以下討論線積分的變數變換. 令 D 和 $\Delta$ 為 $\mathbb{R} ^n$ 中的兩個區域,

$$\Phi (x)=(\Phi_1 (x), \Phi_2 (x), \cdots, \Phi_n (x))$$

為自 D 至 $\Delta$ 的映像. 我們假定每個 $\Phi_i(x)$ 都連續可微. 又令

$$x=g(t)=(g_1(t), g_2(t), \cdots, g_{n}(t)), \qquad t \in [a, b]$$

描寫 D 中的一條 piecewise continuously differentiable 曲線 C. 這曲線經 $\Phi$ 映成 $\Delta$ 中的一條曲線 $\Gamma$:

$$\xi=\gamma(t)=(\gamma_1(t), \gamma_2(t), \cdots, \gamma_{n}(t))=\Phi (g(t)).$$

再設 $f=(f_1, f_2, \cdots, f_{n})$ 為 $\Delta$ 上的 $\mathbb{R} ^n$值的連續函數. 考慮線積分

$$\int_{\Gamma} f(\xi) \cdot d{\xi}=\int_a^b f(\gamma (t)) \cdot \gamma'(t)\,dt.$$

因 $\gamma_i (t)=\Phi_i (g_1 (t), g_2 (t), \cdots, g_n (t))$, 故

$$\gamma_i' (t)=\sum_{j=1}^{n} \frac{\partial \Phi_i}{\partial x_j} g_j' (t).$$

從而

$$\begin{eqnarray*}\lefteqn{\displaystyle\int_{\Gamma} f(\xi) \cdot d{\xi}=\displa... ...g(t))) \frac{\partial \Phi_i(g(t))}{\partial x_j} g_j'(t)\,dt . \end{eqnarray*}$$

令

$$F_j (x)=\sum_{i=1}^n f(\Phi (x)) \frac{\partial \Phi_i (x)}{\partial x_j}, \qquad j=1, 2, \cdots, n.$$

則有

$$\int_{\Gamma} f(\xi) \cdot d{\xi}=\int_C F(x) \cdot\,dx.$$

這便是線積分變數變換的公式.

In addition to these line integrals, integrals with respect to the arc length are also of importance. Thus, assume that F(x) is a function defined for x varying in a domain $D \subset \mathbb{R} ^{n}$, and that $x=\gamma(t), t \in [a, b]$, is a piecewise continuously differentiable curve C in D. We denote by s(t) the arc length of the curve C from $\gamma(a)$ to $\gamma(t)$. Then the line integral of F with respect to the arc length is defined as

$$\int_{C}Fds=\int_{a}^{b}F(\gamma(t))\Vert\gamma'(t)\Vert\,dt.$$

Note that in this discussion we only need F to be defined and continuous on C. The previous type of line integrals may be expressed in terms of the line integral introduced here. Thus, let f(x) be an $\mathbb{R} ^{n}-valued$ function in D, and let T(t) be the unit vector tangent to C at $\gamma(t)$. Then

$$\int_{C}f(x)\cdot \,dx=\int_{C}f\cdot Tds.$$

習$\quad$題

1.

Let C be the unit circle in the plane. Evaluate

$$\int_{C}x \,dx+y\,dy$$

2.

(a)

Compute $$\int (2x\cos{y}-y)\,dx+(x\sinh{y}-y)\,dy$$, where C is the square with vertices $(\pm 1, \pm 1)$ traversed counterclockwise.

(b)

Evaluate the line integral

$$\int_{C}\frac{y\,dx-x\,dy}{x^2+y^2}$$

where C is the ellipse 4x²+y²=4 traversed counterclockwise.

3.

Evaluate the line integral

$$\int_{C}\frac{\,dx}{y}+\frac{\,dy}{x},$$

where C is the ellipse 4x²+y²=4, traversed counterclockwise.

4.

Compute $$\int_{C} yz\,dx+xyz^2\,dy+3xy^2z^{3}\,dz,$$where C is the curve $x=t, y=t, z=t, t\in [0, 1].$

5.

(a)

Compute $\int_{C}y^2\,dx,$ where C is the ellipse 2x²+y²=1 traversed counterclockwise.

(b)

Let f(x, y)=(y², 0). Prove that f is not a gradient field.

Note. f is a gradient field if $\exists F$ such that $f=\bigtriangledown F$.

6.

用 C 表示擺線 $x=t-\sin{t}, y=1-\cos{t}$ 在 $0\leq t\leq 2\pi$中的一段. 試求關於弧長的線積分 $$\int_{C} y^2\,ds$$ 的值.

7.

設 f(x, y)=(x/(x²+y²), y/(x²+y²)), 其中 $(x, y)\neq (0, 0)$.

(a)

試證 f(x, y) 為一 gradient field.

(b)

設曲線 C 之起點為 (1, 0), 終點為 (-2, 3), 其方程式為

y=x³+2x²-2x-1.

求線積分 $$\int_{C}f(x, y)\cdot d(x, y).$$

Remark. The function F(x, y)such that $\bigtriangledown F=f$ in Exercise 7 is called the logarithmic potential.

8.

Prove that the $\mathbb{R} ^{3}$-valued function

$$f(x, y, z)=(x^2+y^2+z^2)^{-3/2}(x, y, z), (x, y, z) \neq (0, 0, 0)$$

is a gradient field. Find a function F(x, y, z) such that

$$f(x, y, z)=\bigtriangledown F(x, y, z).$$

Remark, The function F(x, y, z)such that $\bigtriangledown F=f$ in Exercise 8 is called the Newtonian potential.

9.

Show that $(y+xy^2)\,dx-x\,dy$ is not an exact differential. But if

$$f(x, y)=\frac{1}{y^2},$$

then

$$f(x, y)(y+xy^2)\,dx-f(x, y)x\,dy$$

is an exact differential.
Remark. The function f(x, y)in Exercise 9 is called an integrating factor. We shall study some integrating factors in the Chapter7 of Part II of this course.