Advanced Engineering Mathematics 長榮大學科工系 105 級

Transcription

1 工程數學 Advanced Engineering Mathematics 長榮大學科工系 5 級 姓名 : 學號 : 工程數學 I 目錄 Part I: Ordinary Differential Equations (ODE / 常微分方程式 ) Chapter First-Order Differential Equations ( 一階 ODE) 3 Chapter Second-Order Differential Equations ( 二階 ODE) 9 Chapter 3 The Laplace Transform ( 拉氏轉換 ) 58 Chapter 4 Series Solutions ( 級數解 ) 86 Part II: Vector Analysis ( 向量分析 ) Chapter Vector Differential Calculus ( 向量微分 ) 9 工程數學 II Chapter Vector Integral Calculus ( 向量積分 ) Part III: Fourier Analysis ( 傅立葉分析 ) Chapter 3 Fourier Series ( 傅立葉級數 ) Chapter 4 Chapter 5 The Fourier Integral and Fourier Transforms ( 傅立葉積分與轉換 ) Special Functions, Orthogonal Expansions and Wavelets ( 特殊函數 正交展開與小波 ) - -

2 Chaptteerr IInttrroduccttiion (( 介紹 )) Advanced Engineering Mathematics may include the following topics ( 高等工程數學 之範疇 ): Ordinary Differential Equations ( 常微分方程式 ) (ˇ) Vector Analysis ( 向量分析 ) (ˇ) 3 Fourier Analysis ( 傅立葉分析 ) (ˇ) 4 Linear Algebra ( 線性代數 ) (3 學分 ; 選修 ) 5 Numerical Analysis ( 數值分析 ) (3 學分 ; 選修 ) 6 Complex Analysis ( 複變分析 ) (3 學分 ) 7 Partial Differential Equations ( 偏微分方程式 ) (3 學分 ; 研究所課程 ) Parrtt Orrdiinarry Diiffffeerreenttiiall Equattiionss(O DE) ( 常微分方程式 : 變數只有一個 ) A differential equation is an equation that contains( 包含 ) one or more derivatives( 導數 ) For example ( 例如 ), y ( x) y ( x) 4sin(3 x) ; 4 d ( ( t)) 4 dt e 4 An ordinary differential equation( 常微分方程式 ) is an equation that involves( 內含 ) only( 僅 ) total derivatives( 全導數 ), rather than( 而無 ) partial derivatives( 偏導數 ) For example, y ( x) y ( x) 4sin(3 x) ; (an ordinary differential equation; ODE) (a partial differential equation; PDE; 偏微分方程式 ) x y The order ( 階數 )of a differential equation is the order of its highest derivative ( 最高階導數之階數 ) For example, y ( x) y ( x) 4sin(3 x) ; (a second-order differential equation; 二階 ) 4 d ( ( t)) 4 dt e 4 (a fourth-order differential equation; 四階 ) - -

3 A solution ( 解 ) of a differential equation is any function that satisfies it ( 任何函數只要能夠滿足微分方程式之條件, 就稱之為 解 ) For example, y sin( x) is a solution of y 4 y ; y y x ln x x is a solution of y x We now begin a systematic development of ordinary differential equations by starting with the first-order case ( 先從一階微分方程式學起 ) Chaptteerr Fiirrsstt--Orrdeerr Diiffffeerreenttiiall Equattiionss (( 一階微分方程式 )) Chapter Second-Order Differential Equations ( 二階微分方程式 ) Chapter 3 The Laplace Transform ( 拉氏轉換 ) Chapter 4 Series Solutions ( 級數解 ) Prreelliimiinarry Concceepttss (( 初步概念 )) Some terminology ( 術語 ) and geometric insight ( 幾何含意 ) A first-order ( 一階 ) differential equation is an equation involving a first derivative, but no higher derivative ( 方程式中只含一階導數, 不含更高階導數 ) The most general form ( 通式 ) is F ( x, y, y) Note that ymust be present ( 出現 ), but x and/or y need not occur explicitly ( 無需外顯 ) that satis- A solution ( 解 ) of F( x, y, y) on an interval I ( 區間 I) is a function ( x) fies the equation for all x in I That is, ( 解可以只定義在某區間 I 內 ) F( x, ( x), x ( )) for all x in I A solution contained an arbitrary constant is called the general solution ( 通解 ) of the differential equation ( 包含任意常數的解, 稱之為 通解 ) - 3 -

4 For example, y ( x) x ln x cx is the general solution of y ( ( x) 是通解 ) x Each choice of the constant in the general solution yields a particular solution ( 特解 ) y For example, ( x) x ln x x is a particular solution of y ( ( x) 是一個特解 ) x A solution is explicit ( 外顯 ) if it is given as a function of independent variables ( 若一解能表示成獨立變數之函數, 則稱之為 外顯 ) x For example, y ke is an explicit solution of y y Note that this general solution is explicit, with y isolated on one side of an equation (y 單獨放在方程式的一邊 ) and a function of x on the other ( 而 x 放在另一邊 ) 3 By contrast ( 做一對照 ), consider y xy The general solution is y 3x y 8 e 4 x y 3 4 x e y k, which is implicit ( 內隱 ) In this example we are unable to solve the differential equation explicitly for y as a function of x while isolating y on one side (y 無法單獨放在方程式的一邊 ; 或 y 無法表示成單純的 x 函數形式 ) The graph of a particular solution of a first-order differential equation is called an integral curve of the equation ( 解的圖形 又稱為 微分方程式的積分曲線 ) Example - 4 -

5 Example If we specify that a particular solution is a solution passing through a particular point ( x, y ), then we have to find that particular integral curve passing through this point This is called an initial value problem ( 初始值問題 = 微分方程式 + 初始條件 ) ( 求初始值問題的解 = 找通過某一指定點的積分曲線 ) ( 一個特解 = 一條積分曲線 ) Thus, a first-order initial value problem ( 一階初始值問題 ) has the form F ( x, y, y) ; y( x ) y, in which x and y are given numbers ( x 與 y 為給定的數字 ) The condition y( x ) y is called an initial condition ( 初始條件 ) Consider the general first-order differential equation of the form ( 一階微分方程式之通式 ) F ( x, y, y) Suppose we can solve for yas y f ( x, y) Then a drawing of the plane, with short line segments of slope f ( x, y) drawn at selected point (x,y), is called a direction field of the differential equation ( 將代表切線斜率的小 線段一一畫在平面上, 稱為該微分方程式之 方向場 ) - 5 -

6 Example 5 Consider the equation y y ( 微分方程式 ) The general solution is y x k ( 通解 ) Homework for sec #, 7, 3, 7-6 -

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8 Seeparrabllee Diiffffeerreenttiiall Equattiionss (( 可分離型微分方程式 )) 3 Linear Differential Equations ( 線性微分方程式 ) 4 Exact Differential Equations ( 正合型微分方程式 ) Definition A first-order differential equation is called separable ( 可分離的 ) if it has the form ( 形式 ) y A( x) B( y) dy A( x) B( y) dy A ( x ) dx dx B( y) dy A ( B y x ) ( ) dx, which yields an integral equation for x and y ( 得到 x, y 的積分方程式 ) Example 7 Solve x y e y - 8 -

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10 T (F) t (min) Homework for sec #, 5,, 7, 9 - -

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12 3 Liineearr Diiffffeerreenttiiall Equattiionss (( 線性微分方程式 )) Linear ( 線性 ) and Non-Linear ( 非線性 ) dy y x dx (linear) 所有的項皆不出現 y 和 y 的導數的乘積或自乘 dy y x (non-linear) 任何一項有出現 y 和 y 的導數的乘積或自乘 dx 比較 Partial Differential Equations( 偏微分方程式 : 變數有兩個 ( 含 ) 以上 ) 例如 :Let x (, y) x y ( 二階線性 ) x y ( 一階線性 ) x y x y ( 一階非線性 ) x y Definition A first-order differential equation is called linear if it has the form y p( x) y q ( x) Multiply the differential equation by e p ( x ) dx (integrating factor; 積分因子 ) to get p ( e x) dx p ( x) dx p ( x) dx y p( x) e y q( x) e, which enables us to write Now integrate both sides to get d ) p ( x) dx p( x dx y( x) e q( x) e dx p( x) dx p( x) dx y( x) e q( x) e dx, where c is an arbitrary constant Finally, solve for y (x) : y( x) e q( x) e dx p( x) dx p( x) dx The function e p ( x ) dx is called an integrating factor( 積分因子 ) for the differential equation, because multiplication of the differential equation by this factor results in an equation that can be integrated to obtain the general solution - -

13 Example 4 Solve y y sin( x) 為線性 原式 ( e x ) dx 存在積分因子 e e x x e y e sin x e x y e x sin xdx x x y e e xdx x sin ( 該積分需操作兩次部分積分法 ) - 3 -

14 Q (pound) t (min) - 4 -

15 Homework for sec3 #, 3, 5, 3, 7-5 -

16 4 Exacctt Diiffffeerreenttiiall Equattiionss (( 正合型微分方程式 )) Definition 3 A function ( x, y) is a potential function ( 勢能函數 ) for the differential equation M ( x, y) N ( x, y) y, (not necessarily linear) if ( x, y) x M and ( x, y) y N - 6 -

17 Definition 4 The differential equation M ( x, y) N ( x, y) y ( 正合 ) when a potential function exists for it is said to be exact Theorem Test for Exactness ( 正合之測試 ) The differential equation M ( x, y) N ( x, y) y is exact M if and only if y M A simple verification: y N x N y x y x x y x y x Example 7 Solve y 3 xy 4 3x y 8e y 3 4 y 重寫上式為 xy (3 x y 8 e ) y 令 (, ) xy 3 M x y, N( x, y) 3 x y 8 e 比較 M y M 6xy y 與 N 6xy x N 該微分方程式為正合 x 4 y The General Solution of an Exact Differential Equation If M ( x, y) N ( x, y) y is an exact differential equation, then it means that there exists a potential function ( x, y) such that ( x, y) x M dy Then M ( x, y) N ( x, y) y x y dx d d c, and ( x, y) y N dx dy x y c: arbitrary const which is the general solution of the exact differential equation Solving Exact Differential Equations: M ( x, y) N ( x, y) y is exact there exists a potential function ( x, y) such that ( x, y) x M and ( x, y) y N - 7 -

18 Then we can solve for ( x, y) : x y dx M x y dx x (, ) (, ) and ( x, y) dy N x y dx y (, ) Compare the above two results to have a particular potential solution ( x, y) Then ( x, y) c is the general solution Note of Caution: If ( x, y) is a potential function for M ( x, y) N ( x, y) y, ( x, y) itself is not the solution The general solution is defined implicitly by the equation ( x, y) c Example 8 Solve x xy xy x y 3 (4 ) x x Example 9 Solve e sin( y) x ( e cos( y) ) y - 8 -

19 Homework for sec4 #, 3, 9, 5-9 -

20 5 IIntteegrrattiing Faccttorrss (( 積分因子 )) An integrating factor is such a function that a multiplication of this factor with the differential equation will result in an equation that can be integrated to obtain the general solution ( 與微分方程式相乘而能幫助其求解之函數, 稱之為 積分因子 ) Integrating Factor for a Separable Differential Equation:( 可分離型微分方程式之積分因子 ) For the separable differential equation A ( x) B( y) y, the general integrating factor is B( y) Integrating Factor for a Linear Differential Equation:( 線性微分方程式之積分因子 ) For the linear differential equation y p( x) y q ( x), the general integrating factor is e p ( x ) dx Integrating Factor for an Inexact ( 非正合 ) Differential Equation: ( 正合型微分方程式之積分因子 ) For the inexact differential equation M ( x, y) N ( x, y) y, the general integrating factor is ( x, y) provided ( M ) ( N ) y is exact or ( M ) ( N ) y x - -

21 Note of Caution: ( 重點提示 ) If we cannot find an integrating factor that is a function of just x or y, then we must try something else There is no template to follow, and often we must start with equation y ( M ) ( N ) x and be observant ( 找積分因子並無固定解法, 必須靠努力 觀察力 與帶點運氣 ) - -

22 Homework for sec5 #5, 7, 3, 5 - -

23 6 Diiffffeerreenttiiall Equattiionss off Speecciiall Forrmss (( 特殊類型之微分方程式 )) 6 Homogeneous Differential Equations ( 齊性微分方程式 ) Definition 6 A first-order differential equation of the form y y ' f x ( 成對的 x 與 y) is called a homogeneous differential equation x y For example, y sin( ) is homogeneous; y x y x y is not; y y is not; x y y / x y / x y is homogeneous ( x y ) / x y / x A homogeneous differential equation is always transformed ( 轉換 ) into a separable one by the transformation y ux (or y u ) x Differentiate y ux to get y ux u y Substitute it into y ' f to get x ux u f (u) u ( f ( u) u ) du dx, which is separable x f ( u) u x Example 5 Solve y xy ' y (solution: x x y ln x C ) - 3 -

24 6 The Bernoulli Differential Equation (Bernoulli 型微分方程式 ) Definition 7 A first-order differential equation of the form y ' p x y R x y with a real number, is called a Bernoulli equation The Bernoulli equation is separable if and is linear if About 696, Leibniz showed that the Bernoulli equation with ( 轉換 ) to a linear one under the change of variables: u y Compute y and y u u u Then ' y p x y R x y becomes u u pu R u u p u R u if u u ( ) pu ( ) R, which is linear can be transformed Example 7 Solve y ' y 3x y x 3 (solution: y 3 Cx 6x ) - 4 -

25 63 The Riccati Differential Equations (Riccati 型微分方程式 ) Definition 8 A first-order differential equation of the form y ' P x y Q x y R x, is called a Riccati differential equation If we can somehow obtain one solution, S (x), of a Riccati differential equation, then the change of variables y( x) S ( x) u( x) equation ( S x 為任意一個已知特解, u( x ) 為新的變數 ) transforms the Riccati equation to a linear Example 8 Solve y ' y y (solution: x x x By inspection, y S ( x) is a solution 3 K x y ) 3 K x - 5 -

26 Homework for sec6 #3, 5, 7, 9,, 3-6 -

27 7 Applliiccattiionss (( 應用 )) Kirchhof s Voltage Law( 高橋伏特定律 ) - The algebraic sum of the potential rises and drops around any closed loop in a circuit is zero ( 繞任何封閉電路一圈所得之電壓升降 之總和為零 ) RL Circuit ( 電阻 - 電感電路 ) Consider the circuit of figure 8 By Kirchhof s Voltage law,we have E ir L i Since L, the above equation leads to a linear one Solve this to obtain R E i i L L R t L E i( t) c e R RC Circuit ( 電阻 - 電容電路 ) Consider the circuit of figure 9 By Kirchhof s Voltage law,we have E ir q C or substituting i q to get E Rq q C Since R, the above equation leads to a linear one Solve this to obtain E q q RC R t RC q( t) EC c e - 7 -

28 SSUMMARY 一階常微分方程式之三種基本類型 : No Type/ 類型 Standard form/ 標準式 Algorithm/ 求解步驟 separable y A( x) B( y) 變數分離 B ( y) dy A( x) dx 積分 ( x ) dx B ( y) p ( x) dx 存在一個積分因子 e linear y p( x) y q ( x) 3 exact M ( x, y) N ( x, y) y with M y N x p x) dx 微分方程式兩邊乘以積分因子 e 得 d p x dx ( ) p ( x) dx dx e y e q x ( ) (, 3 兩邊對 x 積分得 p ( x) dx p ( x) dx e y e q( x) dx 若為 exact, 則存在一 potential function x (, y) 令 M x ( x, y), 求解 x (, y ) 3 令 N y ( x, y), 求解 x (, y ) 4 比較 與 3 得 ( x, y), 則微分方程式之通解為 ( x, y) 一階常微分方程式之三種特殊類型 : No Type/ 類型 Standard form/ 標準式 Algorithm/ 求解步驟 y homogeneous y f ( x ) Bernoulli 3 Riccati y p( x) y R ( x) y y P( x) y Q ( x) y R(x) y 利用變數變換法, 令 u, 代回原式 x 將原式轉換成 separable 型 利用變數變換法, 令 u y, 代回原式 將原式轉換成 linear 型 找到任一特解 S(x) 利用變數變換法, 令 y( x) S( x), 代回原式 u( x) 3 將原式轉換成 linear 型 - 8 -

29 Chaptteerr Seeccond--Orrdeerr Diiffffeerreenttiiall Equattiionss (( 二階微分方程式 )) Prreelliimiinarry Concceepttss (( 初步概念 )) A second-order ( 二階 ) differential equation is an function that contains a second derivative, but no higher derivative The most general form ( 通式 ) is F ( x, y, yy, ) Note that ymust be present, but x, y, yneed not occur explicitly A solution ( 解 ) of F( x, y, yy, ) on an interval I is a function that satisfies the equation for all x in I That is, F( x, ( x), ( x), ( x)) for all x in I For example, ( x) 6cos(4 x) 7 sin(4x) is a solution of y 6y ; 3 ( x) x cos(ln x) is a solution of x y 5xy y The linear ( 線性 ) second-order differential equation has the standard form y p( x) y q( x) y f ( x) ( 所有的項皆不出現 y 和 y 的導數的乘積或自乘 ) Theeorry off Solluttiionss off y p( x) y q( x) y f ( x) Consider the simple, linear second-order differential equation y x We can write this as and integrate to obtain y x - 9 -

30 Integrate again y y dx xdx 6 x C 3 (6 ), y y dx x c dx x Cx K where C and K are arbitrary constants ( 任意常數 ) - 3 -

31 Recall that the general solution of a first-order differential equation contained one arbitrary constant, it seems nature that the solution of a second-order differential equation, involving two integrations, should contain two arbitrary constants ( 一階微分方程式之通解有一個任意常數 ; 二階微分方程式之通解有兩個任意常數 ) Two specific conditions are required to determine the two arbitrary constants or to specify a unique solution of a second-order differential equation ( 二階微分方程式之通解的兩 個任意常數, 需由兩個條件來決定 ) They could be obtained from ) initial value conditions ( 初始值條件 ) : conditions are specified at a starting point ( 條件給定在同一個點 x 上 ) ie y( x) C, y( x) C ; ) boundary value condition ( 邊界值條件 ) : conditions are specified at two isolated points ( 條件給定在兩個不同的點 x, x 上 ), ie y( x ) C, 3 y( x ) C4 The Homogeneous Equation y p( x) y q( x) y When f ( x), the resulting equation y p( x) y q( x) y is called homogeneous ( 齊性 ) Note that this term is unrelated to ( 與 無關 ) that of the first-order differential equation Theorem Superposition ( 疊加性 ) Any linear combination of solutions of y p( x) y q( x) y is also a solution ( 齊性微分方程式之解的任意線性組合, 仍舊是解 ) Proof: Let y and y are solutions of y p( x) y q( x) y, c and c are real numbers Substituting y x) c y ( x) c y ( ) into the differential equation, we obtain ( x ( c y c y ) p( x)( c y c y ) q( x)( c y c y ) c y p( x) y q( x) y ] c [ y p( x) y q( x) ] c [ y c c y( x) c y( x is also a solution Hence ) - 3 -

32 Definition Linear Dependence and Linear Independence Two function f(x) and g(x) are said to be linearly dependent ( 線性相依 ) on an interval I if there exists a constant c so that f ( x) cg ( x) for all x in I ( 函數間存在一常數倍 ) If f(x) and g(x) are not linearly dependent, then they are said to be linearly independent ( 線性獨立 ) 定義 The Wronskian of the two solutions y and y is defined( 定義 ) as y( x) y( x) W ( x) y ( x) y ( x) y ( x) y ( x) y( x) y( x) Theorem 3 Wronskian Test ( 兩函數是否線性相依之測試條件 ) The two solutions y and y are linearly dependent on I if and only if W ( x) for all x in I Example Example - 3 -

33 Example 3 Theorem 4 Let y and y be any two linearly independent solutions of y p( x) y q( x) y Then, any solution of this differential equation is a linear combination of y and y ( 對二階微分方程式, 只要找到其任意兩個線性獨立的解, 則該兩解的線性組合即為 通解 ) Proof: Let (x ) be any solution of y p( x) y q( x) y on the interval I The proof is to show that there must be numbers c and c such that ( x) c y c y Choose any x in I and let ( x ) A and ( x ) B By theorem, (x ) the unique solution of the initial value problem y p( x) y q( x) y ; y( x ) A and y ( x ) B is The two conditions y( x ) A y( x ) B ( x ) A ( x ) B c y( x ) c y ( x ) A c y( x ) c y( x ) B c c Ay ( x ) By ( x ) W ( x ) By ( ) x Ay( x ) W ( x ) where W x ) y ( x ) y ( x ) y ( x ) y ( ) ( x

34 Since y and y are linearly independent, it results that W ( x ) c and c can be determined by the preceding formulas Proved! in general Thus, Definition A Fundamental Set of Solutions and the General Solution Let y and y be two solutions of y p( x) y q( x) y Then ) y and y form a fundamental set of solutions ( 一組基礎解 ) if y and y are linearly independent ) If y and y form a fundamental set of solutions, we call the linear combination ( 線性組合 ) of y and y the general solution ( 通解 ) of the differential equation The homogeneous solution ( 齊性解 ) yh is used to denote the general solution of the homogeneous differential equation y p( x) y q( x) y That is, y h c y c y The Non-homogeneous ( 非齊性 ) Equation y p( x) y q( x) y f ( x) Theorem 5 Let yh the homogeneous solution of y p( x) y q( x) y and y p be a particular solution of y p( x) y q( x) y f ( x) Then, the general solution of y p( x) y q( x) y f ( x) can be expressed as y y h y p ( 非齊性通解 = 齊性通解 + 任一特解 )

35 Homework for sec #3, 5, 9,

36 3 Reeduccttiion off Orrdeerr (( 降階法 )) For a given y p( x) y q( x) y, we need two linearly independent solutions to form a fundamental set of solutions Reduction of order is a technique for finding a second solution, if we can somehow produce a first solution ( 若已知第一個解, 則可用 降階 法 嘗試找第二個解 ) Technique of Reduction of Order Suppose we know a first solution y ( 假定已知一個解 ) Let the second solution be y x) u( x) y ( ) ( x Compute y u y u y, y u y u y u y and substitute them into y p( x) y q( x) y to have ( u y u y u y ) p( x)( u y u y) q( x) u y u y u y py ) u ( y py qy ) ( On any interval in which y, we can write y py u u y Let v( x) u ( x) and y ( x) p( x) y g( x) y ( x) ( x), then v g( x) v The general solution of this linear first-order differential equation is g ( v x C e x ) ( ) dx We may take C since we need only one second solution y Thus ( ) y x) u ( x) y ( x) [ v( x) dx] y ( ) e g x dx dx] y ( x) ( x [ is found The general solution of y p x y q x y '' ' is c y c y We do not recommend memorizing formulas for g (x), v (x), u(x) and y ( ) x ( 不建 議記住 y ( ) 的公式 ) Instead, the following procedures are recommended: ( 建議直接操 x 作如下程序 ) ) Given a first solution y ) Substitute y u y into y p( x) y q( x) y 3) After some cancellations, solve the resulting linear first-order differential equation ( 經過一些消去後, 再求解剩下的一階線性微分方程式 )

37 Examples

38 Homework for sec3 #, 3,

39 4 Thee Conssttantt--Coeeffffiicciieentt Homogeeneeouss Liineearr Equattiion ( 常係數 齊性 線性方程式 ) The constant-coefficient, homogeneous, linear, second-order differential equation is of the standard form ( 標準式 ) as y Ay By, in which A and B are constants x Let y e with a constant Compute y them into ( 代入 ) y Ay By to get e x and y x e and substitute x x x e A e Be This can be true only if A B, which is called the characteristic equation ( 特徵方程式 ) of the differential equation Roots ( 根 ) of the characteristic equation 種情況 ): A A 4 B leads to three cases ( 有三 Case : A 4 B We have two distinct real roots ( 兩個不同的實數根 ), say A A 4 B A A 4 B and They leads to a fundamental set of solutions ( 形成一組基礎解 ), say x y e and x y e The general solution in this case is x x yh c e c e Example 6 Case : A 4 B We have a double root ( 一個重根 ), say A

40 So A x y e is one solution ( 得到第一個解 ) The second solution can be found with use of the technique of reduction of order ( 第二個解可利用降階法 求得 ) Try A x y u ( x) e and substitute into ( 代入 ) the differential equation to get A A A A A A A x x x x x x 4 A ue Au e u e A ( ue u e ) Bue Divide by ( 除以 ) A x e and rearrange terms to get Since A 4 B, it reduces to We can choose A u B ) u ( 4 u u u dx dx c u u dx c dx c x c u( x) x ( 選擇最簡單的一個有效解 ), and obtain a second solution A x y xy xe Since y and y are linearly independent and form a fundamental set of solutions ( y, y 為線性獨立且形成一組基礎解 ), the general solution is y h c A A A x x x e c xe e ( c x c ) Example 7 Case 3: A 4 B We have two distinct complex roots ( 兩個不同的複數根 ), say p iq with p A and q 4 A B The general solution can be written as h p iq x p iq x y c e c e (in complex form) px or yh e [ c3 cos( qx) c 4 sin( qx)] (in sinusoidal form) where c, c are complex constants ( 複數常數 ) while c 3, c4 are real constants ( 實數常數 ) - 4 -

41 Example 8 Relations between c, c and c 3, c 4 ( 係數間之關係式 ) By Euler s formula ( 尤拉公式 ) e ix cos x i sin x, we have y h c e c e ( p iq) x ( piq ) x px iqx iqx e ( ce c e ) e px [ c (cos( qx) i sin( qx)) c e px (cos( qx) i sin( qx))] [( c c ) cos( qx) i ( c c )sin( qx)] px Comparing it with yh e [ c3 cos( qx) c 4 sin( qx)], we will have the relations between c, c and c 3, c4 as c3 c c c 4 i ( c c ) or c c ( c 3 ( c 3 ic 4 ) ic ) 4 Example 9-4 -

42 Homework for sec4 #, 3, 5, 3, 5, 7-4 -

43 5 Eulleerr ss Equattiion ( 尤拉方程式 ) The homogeneous, linear, second-order differential equation y" Ay ' By x x, ( 非常係數 ) in which A and B are constants, is called Euler s equation It is define only on the half-lines x or x We will assume for this section that x Euler s equation can be transformed to a constant-coefficient one through a change of variables: ( 經由變數變換, 可將非常係數的尤拉方程式轉換成常係數者 ) t t x e and y( x) y( e ) Y ( t) Let Compute d dy ( t) y ( x) dt dx Y ( t) dt dx Y ( t) x, d d y ( x) dx [ Y ( t) x ] Y ( t) ( ) [ x dx Y ( t )] x Y t) ( ) [ Y ( t) ] x x [ Y ( t) Y ( t x x ( )] Substitute into Euler s equation to get ( ) A ( )] B Y t Y t Y ( t) Y ( t) [ x x x x [ Y ( t) ( A ) Y ( t) BY ( t)] x For x, it is required that Y ( t) ( A ) Y ( t) BY ( t) ( 常係數 ; 有固定解法 ) This is a constant-coefficient homogeneous linear second-order equation for Y (t) Solve this equation, then let t ln x in the solution Y (t) to obtain y (x) Example

44 Example Example

45 Homework for sec5 #, 5, 9, 3, 7,

46 6 Thee Non--Homogeeneeouss Equattiion y '' p x y ' q x y f x (( 非齊性方程式 )) Consider the non-homogeneous, linear, second-order differential equation ' y '' p x y q x y f x The general solution is y yh y p, where yh is the homogeneous solution ( 齊性解 ) and y p is any particular solution ( 任意一特解 ) of the differential equation 解法雙步驟 Step :Find h y ( 求齊性解 ) : it is to solve y p x y q x y '' ' Step :Find y p ( 求特解解 ) : two methods are introduced 6 The Method of Variation of Parameters ( 參數變換法 ) 6 The Method of Undetermined Coefficients ( 未定係數法 ) 6 The Method of Variation of Parameters ( 參數變換法 ) Suppose we can find a fundamental set of solutions y and y equation ( 假定已知齊性微分方程式的一組基礎解 ) for the homogeneous Try y p uy vy ( 利用試誤法, 嘗試找看看無限多個特解中的任何一個 ) y p u+ y uy + v+ y vy Let ( 令 ) u+ y v= y () y p uy + vy y p u y uy v y vy " ' ' " ' ' " Substitute into the non-homogeneous equation ( 代入非齊性方程式 ) u ' y ' uy " v ' y ' vy " p uy ' vy ' q uy vy f x u y " py ' qy vy " py ' qy u ' y ' v ' y ' f x u ' y ' v ' y ' f x ()

47 y ' y u ' y y ' y y ' y f x y ( 消去 v) ' u y f ( x) y y yy u udx y v ' y y ' y y ' y f x ( 消去 u) v y f ( x) y y yy v vdx For any non-vanishing pair of u and v, we can use them to find y p ( 若能找到任何一組 非全為零的 u, v, 則表示已找到所要的 y p ) Example

48 Example

49 6 The Method of Undetermined Coefficients ( 未定係數法 ; 僅適用於常係數者 ) Consider the constant-coefficient ( 常係數 ) non-homogeneous equation y '' Ay ' By f x Guess y p according to the form of f (x) ( 根據 f (x) 的形式來猜測 y p ) Here is a list ( 清單 ) of f (x) to try to y p f (x) initial guess for y p P n (x) try Q n (x) ax ce try ax de cos(bx) or sin(bx ) try c cos( bx) d sin( bx) P ) ax n ( x e try Q ( x) e n ax P n ( x) cos( bx) or P n ( x)sin( bx) try Q ( x) cos( bx) R ( x)sin( bx) ax ax ax ax Pn ( x) e cos( bx) or Pn ( x) e sin( bx) try Qn ( x) e cos( bx) R n ( x) e sin( bx) n n

50 Example 7 Example 8-5 -

51 Example Example A non-constant-coefficient equation 63 The Principle of Supposition ( 疊加原理 ) The Principle of Superposition Consider the equation Suppose pj Then we claim that y p x) y q( x) y f ( x) f ( x) f ( ) ( N x y is a particular solution of y p( x) y q( x) y f ( x), j=,,, N y p y y y is a particular solution of the original differential equation p p pn j Example 5-5 -

52 64 Higher-Order Differential Equation ( 高階微分方程式 ) A higher-order differential equation can be converted to a system of first-order differential equations For example, consider the six-order differential equation (6) (4) y 4 y y 5y Method : The corresponding characteristic equation is Method : Define new variables 69 i, i, 73 4i z y, z y, z y 3, z y 4, (4) z5 y, (5) z6 y Then the six-order differential equation can be converted to a system of six first-order differential equations: z z, z z 3, z 3 z 4, z 4 z 5, z 5 z 6, z 4 z 6 z5 z 5 To find solutions, powerful matrix techniques can be invoked ( 可以使用強而有力的矩陣計算技術求解之 ) - 5 -

53 Homework for sec6 #, 7,, 7,

54 7 Applliiccattiionss (( 應用 )) mass/spring system ( 重物 / 彈簧系統 ) RCL circuit ( 電阻 - 電容 - 電感之電路 ) c k f ( t) R y y y is analog to( 類比於 ) q( t) q( t) q( t) E( t) m m m L LC L displacement( 位移 ) y to charge( 電荷 ) q velocity( 速度 ) y to current( 電流 ) i driving force( 外力 ) f ( t) to electromotive force( 電動勢 ) E( t) mass( 質量 ) m to inductance( 電感 ) L damping constant( 阻滯係數 ) c to resistance( 電阻 ) R spring modulus( 彈簧模數 ) k to reciprocal of capacitance( 電容倒數 ) C Step : Solve c k y y y for y h m m The corresponding characteristic equation is ( 對應之特徵方程式為 ) It is obtained c m m c 4km c k m m Case : c 4 km (overdamping; 過阻滯 ) We have two negative real roots and t t yh ce ce, (see Fig5) which is a transient solution ( 過渡解 ) because lim y t h

55 Case : c 4 km (critical damping; 臨界阻滯 ) We have a negative double root t t yh ce cte, (see Fig6) which is a transient solution because lim y t h Case 3: c 4 km (underdamping; 下阻滯 ) We have two complex roots, say, p iq where c p, m 4km c q m pt h y e [ c cos( qt) c sin( qt)], (see Fig7) which is a transient solution too because lim y t h c k A Step : Solve y y y cos( t ) for y p m m m Note that the driving force f ( t) A cos( t ) is assigned ( 給定 ) for convenience, and is called the input frequency ( 輸入頻率 ) To find y p, we can use the method of undetermined coefficient ( 未定係數法 ), mentioned in Sec6, and let y a cos( t ) bsin( t ) a and b can be solved to get p A( k m ) Ac y p cos( t ) sin( t ) ( k m ) c ( k m ) c k let, which is called the nature frequency( 自然頻率 ) m ma( ) A c cos( ) sin( t ), m c m c t ( ) ( ) A steady-state solution ( 定常狀態解 ; 穩定解 ) is obtained for y p

56 Resonance ( 共振 ) In the absence of damping ( c ; 當阻滯係數為零時 ), an interesting phenomenon called resonance can occur ( 可能發生 共振 之有趣現象 ) A c y p cos( t ) m( ) The closer are the natural and input frequencies, the larger is the amplitude of the cos( t ) term in the solution ( 當自然頻率 與輸入頻率 越接近, 則振動的幅度 A 就會越大 ), that is, m( ) lim y p (see Figure )

57 - 57 -

58 Chaptteerr 3 Thee Lapllaccee Trranssfforrm (( 拉氏轉換 )) 3 Deeffiiniittiion and Bassiicc Prropeerrttiieess (( 定義與基本特性 )) A transform( 轉換 ) is a device that converts one type of problem into another type, presumably easier to solve ( 把一種問題轉換成令一種問題的機制, 稱之為 轉換, 通常 是把較難的轉換成較簡單的 ) In case of the Laplace transform, a process we can diagram as follows ( 圖示如下 ): initial value problem( 初始值問題 ) algebra problem( 代數問題 ) ( 困難 ) ( 容易 ) solution of the initial value problem solution of the algebra problem ( 初始值問題的解 ) ( 代數問題的解 ) Definition 3 Laplace Transform( 拉普拉斯轉換 ; 拉氏轉換 ) The Laplace transform of f is a function defined by ( 定義函數之拉氏轉換 ) st L [ f ]( s) e f ( t) dt, for all real s such that this integral converges ( 以積分能夠收斂來定義 s 的存在範圍 ) It is convenient to use the notation ( 為了方便, 使用以下記號 ), F( s) L [ f ]( s), G( s) L [ g]( s), H ( s) L [ h]( s), Examples

59 Examples 3 A Laplace transform is rarely computed by referring directly to the definition and integrating ( 函數之拉氏轉換很少直接從積分的定義來求得 ) Instead we use tables of Laplace transforms of commonly used functions ( 一般是利用查表的方式得到 ) Table 3 Table of Laplace Transforms of Some Functions ( 常用函數之拉氏轉換表 ) f (t) F(s) s t s n! s 3 t n (n=,, 3, ) n 4 e at sa a s a 5 sin(at) s s a 6 cos(at) a s a 7 sinh(at) s s a 8 cosh(at) Theorem 3 Linearity ( 線性特性 ) of the Laplace Transform L[ f ( t) g( t)] L [ f ( t)] L [ g( t)], where and are real numbers ( 實數 ) Proof: L[ f g ]( s) st [ ( ) ( )] st e f t g t dt e f ( t) dt e st g( t) dt F ( s) G ( s), which is proved

60 Definition 3 Piecewise Continuity ( 函數之成段連續性 ) f (t) is piecewise continuous on [a, b] if there are points a t t t such that all of the following one-side limits are finite: ( 不連續點的個數為有限大 ) n b lim f ( t), lim f ( t), lim f ( t) ta where j=,, 3,, n tt j tt j and lim f ( t), tb This means that f (t) each of which f (t) is continuous on [a, b] except perhaps at finitely many points, at has finite one-side limits from within the interval Theorem 3 Existence ( 存在 ) of the Laplace Transform If ) f (t) then is piecewise continuous on [, ) and ) there are numbers M and b such that e st f ( t) dt converges for all s>b f bt ( t) M e for t>, Many functions satisfy these conditions, including polynomials, sin(at), cos(at), e at and so on The conditions of the theorem are sufficient ( 充足 ), but not necessary ( 但非必要 ), for a function to have a Laplace transform Definition 33 Inverse Laplace Transform ( 反拉普拉斯轉換 ) A function g is called an inverse Laplace transform of G if L [ g] G In this event, we write g L [ G] For example, L at [ ] e, sa L [ ] sin t s Note that for a given G, there will be many functions whose Laplace transform is G ( 對一 函數 G 而言, 有很多函數的拉氏轉換會等於 G) 例如, L, [ e t ] s and L [ h( t)] s for t e, h( t), t 3 t 3-6 -

61 Which one do we call the inverse Laplace transform of s? ( 哪一個才是 s 的反拉氏轉換呢?) One answer is provided bylerch s theorem, which states that two continuous functions having the same Laplace transform must be equal ( 有相同拉氏轉換的兩個連續性函數 必定相等 ) Theorem 33 Lerch s Theorem If f and g are continuous on [, ) and L[ f ] L [ g], then f = g Theorem 34 Linearity of the Inverse Laplace Transform ( 反拉氏轉換之線性特性 ) L [ F( s) G ( s)] f ( t) g ( t), where and are real numbers - 6 -

62 Homework for sec3 #3, 5, 7,, 3, 5-6 -

63 3 Solution of Initial Value Problems Using the Laplace Transform ( 使用拉氏轉換求解初始值問題 ) The Laplace transform is a powerful tool ( 強效的工具 ) for solving some kinds of initial value problems ( 求解某些初始值問題 ) This technique depends on the following formula about the Laplace transform of a derivative Theorem 35 Laplace Transform of a Derivative ( 一階導數之拉氏轉換 ) Suppose f is continuous on [, ) and f is piecewise continuous on [, k] for every positive k Suppose also that lim[ e sk f ( k)] k for s> Then L [ f ] s F ( s) f () Proof: k st L[ f ] lim e f ( t) dt k st st (let u e, dv f ( t) dt, then du se dt, v f (t) ) k k = st k st lim [ e f ( t)] se f ( t) dt= sk k st lim e f ( k) f () s e f ( t) dt k = f () s L [ f ] = s F( s) f (), which is proved Theorem 36 Laplace Transform of a Higher Derivative ( 高階導數之拉氏轉換 ) Suppose f, f,, f n are continuous on [, ) and f (n) is piecewise continuous sk ( j) on [, k] for every positive k Suppose also that lim[ e f ( k)] k for s> and j=,, 3,, n- Then s ( ) [ n n n n ( n) ( n) L f ] = F( s) s f () s f () sf () f () In case n=, L [ f ] = s F ( s) f () In case n=, L [ f ] = s F( s) s f () f ( )

64 Use the Laplace transform to solve initial value problems Examples 33 Solve y 4 y ; y() t Examples 34 Solve y 4 y 3y e ; y(), y() Examples 35 Solve y y t ; y(), y()

65 Homework for sec3 #, 3, 5, 7,

66 33 Shifting Theorems and the Heaviside Function ( 平移定理與 Heaviside 函數 ) Theorem 37 First Shifting Theorem ( 第一平移定理 ;s 變數之平移 ) at L [ e f ( t)] F ( s a ) Proof: L at at st ( sa ) t [ e f ( t)] e e f ( t) dt e f ( t) dt F ( s a ), which is proved at s Examples 36 Find L[ e cos( bt)]( s) with known L [cos( bt)] s b Examples 37 Find L [ t ] s 3 7t 3 6 L[ t e ]( s) with known 4-4 Examples 38 Find L [ ] s 4 s Definition 34 Heaviside Function The Heaviside function H is defined by ( t) if if t H t

67 The Heaviside function may be thought of as a flat switching function, of when t, and on when t Functions with jump discontinuities ( 含跳躍式不連續點的函數 ) can be treated very efficiently using the unit step function or the Heaviside function For examples, if t a H ( t a ), (see Figure 3) if t a if t a H ( t a ) g ( t) (see Figure 3) g ( t) if t a Definition 35 Pulse ( 脈波 ) A pulse is a function of the form in which a b H ( t a) H ( t b ), Multiplying a function g by this pulse has the effect of leaving g(t) switching off ( 關閉或 歸零 ) for t a and t b, and turned on ( 開啟或維持不變 ) for a t b

68 For examples, t t [ H ( t ) H ( t )] e e if if if t t t Theorem 38 Second Shifting Theorem ( 第二平移定理 ;t 變數之平移 ) as L [ H ( t a ) f ( t a )] e F ( s) Proof: st st L[ H ( t a ) f ( t a )] e H ( t a ) f ( t a ) dt e f ( t a ) dt (let t a, then t a, dt d) a s ( a ) e f ( ) de as s e f ( ) de as F ( s), which is proved \Examples 3 Compute L [ H ( t a )] as as L[ H ( t a )] L [ H ( t a ) ] e L[] e s for t Examples 3 Compute L [ g( t)], where g( t) t for t L L L [ g] [( t ) H ( t )] [(( t ) 4( t ) 5) H ( t )]

69 Examples 33 Compute L 3 s se [ ] s 4 Examples 34 Solve the initial value problem y 4 y f ( t) ; y() y(), where for t 3 f ( t) t for t 3 Analysis of RC Circuits (RC 電路之分析 ) Examples 36 & 37 The governing equation ( 控制方程式 ) for an RC circuit is q( t) q( t) E( t), () RC R which is linear

70 Method : By the formula derived in Sec 3, we can get ( ) RC t ( ) RC t RC t RC t q t e E t e dt e ( H ( t ) H ( t 3)) e dt R R Method : Use the Laplace transform: L[()] L[ q( t) q( t)] L [ ( H ( t ) H ( t 3))] RC R sq s q Q s e e, q() s 3 s ( ) () RC ( ) R ( s s ) s 3s ( s RC ) Q( s) R s ( e e ) Q ( s ) ( e e ) [ ]( e e ) s 3 s RC RC s 3 s R s( s ) R s s RC RC s 3 s C[ s ]( e e ) s RC C [ e e e e ] s s 3 s 3 s s s s s RC RC q ( t ) L [ Q ( s )] C L [ e e e e ] s s 3 s 3 s s s s s RC RC C { L [ e ] L [ e ] L [ e ] L [ e ]} s s 3 s 3 s s s s s RC RC ( t) ( t3) RC RC C [ H ( t ) H ( t ) e H ( t 3) H ( t 3) e ] ( t) ( t3) RC RC C { H ( t )[ e ] H ( t 3)[ e ]} ( t) ( t3) RC RC out ( ) C ( ) { ( )[ ] ( 3)[ ]} E t q t H t e H t e out out ( t) ( t3) RC RC E ( t) E ( t) E ( t) [ H ( t ) e H ( t 3) e ] - 7 -

71 Homework for sec33 #, 3, 5, 7,, 5, 7-7 -

72 34 Convolution ( 摺積 ) Definition 36 Convolution The convolution of two functions f (x) and g(x) is defined by ( 兩函數間摺積之定義 ) f g ( f g)( t) f ( t ) g( ) d t for t Theorem 39 Convolution Theorem ( 摺積定理 ) L[ f g ] L [ f ] L [ g] It means that the Laplace transform of the convolution of f and g is equal to the product of the individual transforms of f and g ( 函數間取摺積後再拉氏轉換, 等於 個別函數拉氏轉換後之相乘 ) - 7 -

73 Theorem 3 Let L [ F] f and L [ G] g Then L [ F G ] f g Example

74 Theorem 3 The convolution operation is commutative ( 摺積運算具備交換性 ) It means that t f g = g f Proof: f g = f ( t ) g( ) d (let z t, then t z and d dz ) = t t f ( z) g( t z )( dz ) = f ( z) g( t z ) dz = g f Examples

75 Homework for sec34 #, 5, 7, 9,

76 35 Unit Impulses( 單位脈衝 ) and the Dirac Delta Function An impulse ( 脈衝 ) is intuitively understood as a force of large magnitude applied over an instant of time A pulse ( 脈波 ) can be defined by (see fig 33) ( t ) [ H ( t) H ( t )] This is a pulse of magnitude ( 大小 ) and duration ( 延續時間 ) The (Dirac s) delta function, as a kind of impulse, is defined by ( t ) lim [ H ( t) H ( t )] Then the shifted ( 平移的 ) delta function ( t a ) is zero except for t a And ( t a) lim [ H ( t a) H ( t a )] The Laplace transform of the shifted delta function L [ ( t a )] = st e { lim [ H ( t a ) H ( t a )] }dt e = lim st [ H ( t a ) H ( t a )] dt = lim a st e a dt * st = lim e, = as e In case a, we have L [ t ( )] = * a t a

77 Theorem 3 Filtering Property of the Delta Function (Delta 函數之過濾特性 ) Let a and let f ( t) be integrable for t [, ) and be continuous at t a Then f ( t) ( t a ) dt f ( a) Proof: f ( t) ( t a ) dt f ( t) lim [ H ( t a ) H ( t a )] dt lim f ( t)[ H ( t a ) H ( t a )] dt a lim f ( t) dt a By the mean value theorem for integrals, there is some t between a and a such that Then a f ( t) dt f ( t) a a lim f ( t) dt lim f ( t) lim f ( t) f ( a) a f ( t) ( t a ) dt f ( a) is proved If we apply the filtering property in theorem 3, we get st e ( t a ) dt e as, which is consistent with ( 與 吻合 ) the definition of the Laplace transform of the delta function Further, if we change notation in the filtering property and write it as f ( t) ( t a ) dt f ( a), which is equivalent to f f It means that the convolution of any function with the delta function is invariant ( 不變的 )

78 Examples 3 Examples

79 - 79 -

80 Homework for sec35 #, 3, 7, 9-8 -

81 36 Laplace Transform Solution of Systems ( 拉氏轉換求解聯立方程組 ) The Laplace transform can be use in solving systems of equations involving derivatives and integrals Examples 33 Examples

82 Apply Kirchhoff s laws to each circuit loop to get - 8 -

83 37 Differential Equations with Polynomial Coefficients ( 帶有多項式為係數之微分方程式 ) is differen- Theorem 33 Let L[ f ( t)] F ( s) for s b and suppose that F( s) tiable Then L [ t f ( t)] F( s) It means that the Laplace transform of t f ( t) is the negative of the derivative of F( s ) Further, for any positive integer n, we have ( ) L [ t n f ( t)] ( ) n F n ( s) Example

84 Theorem 34 Let L [ f ( t)] F ( s) If f ( t) all t, then lim F( s) s is piecewise continuous and finite for This result will enable us to solve the following initial value problem Example

85 - 85 -

86 Chaptteerr 4 Seerriieess Solluttiionss (( 級數解 )) Sometimes a series solution is a good strategy ( 好的策略 ) for solving an initial value problem It may reveal important information ( 透露重要訊息 ) about the behavior of the solution ( 關於解的表現 ) for example, ) whether it passes through the origin ( 是否通過原點 ), ) whether it is an even or odd function or ( 是否為奇函數或偶函數 ) 3) whether the function is increasing or decreasing on a given interval ( 是否漸增或漸減 ) 45 Appendix on Power Series ( 次方級數之附錄 ) 45 Convergence of Power Series ( 次方級數之收斂 ) Definition 44 Power Series ( 次方級數 ) A power series about x is a series of the form a ( ) n n x x n where x is the center of the series and an are the constant coefficients Theorem 45 If a ( ) n n x x converges at x x, n then the series converges for all x such that x x x x Definition: Radius of Convergence ( 收斂半徑 ) A positive number r is called the radius of convergence of the series if the power series converges for x x r and diverges for x x r Note that the series may or may not converge for x x r

87 Theorem 46 Ratio Test ( 比例測試 ) b Suppose bn and lim n n b n L Then the series b n n converges absolutely for L and diverges for L If L, then this test allows no conclusion We can determine convergence or divergence at the endpoints of this interval by direct substitution Examples 47 Find the open interval of convergence of ( ) n n ( x ) n n ( n )9 Compute bn n lim lim ( x ) ( x ) n b n 9( n ) 9 n 453 Taylor and Maclaurin Series Expansions (Taylor 及 Maclaurin 級數展開 ) The Taylor series expansion ( 泰勒級數展開 ) for f ( x) about x representation is the power series f ( x) a ( x x ) n n n where a f x n ( n) n! ( ) Proof: The constants an can be determined as following: let f ( x) a a ( x x ) a ( x x ) a ( x x ) n n then f ( x) a a ( x x ) 3 a ( x x ) na ( x x ) n 3 n ( ) 3 3( ) ( ) ( ) n n f x a a x x n n a x x f ( x) n! a ( x x ) ( x x ) ( n) ( n)! ( n)! n!! Substitute x x into the above equations to get f ( x) a a f ( x ) f ( x) a a f ( x ) f ( x) a a f ( x) ( n) f ( x) n! a n a f x n ( n) n! ( )

88 Example: 48 Expand g( x) ln( x ) in a power series about x The resulting series about x f x f x x x is called the Taylor series for f ( x) ( n) n ( ) ( )( ) n n! Not every function has a Taylor series expansion about a point If such an expansion exists, we call the function analytic ( 可解析的 ) there If x, the Taylor series expansion is often called a Maclaurin series expansion f ( x) n n a x n where a n ( n) n! f () And the resulting series Maclaurin series for f ( x) about x ( n) n f ( x) f () x n n! is called the For example, e x n sin x x n! n n n ( ) x (n )! n

89 4 Power Series Solutions of Initial Value Problems ( 初始值問題之次方級數解 ) Definition 4 Analytic Function ( 可解析函數 ) A function f (x) is analytic at x if f (x) has a power series representation in some open interval about x ( 若一函數在某一點可表示成次方級數, 則稱該函數在該點為 可解析 ), that is, in some interval ( x h, x h ) f ( x) n a ( x x n n ) Theorem 4 If p( x) and q( x) are analytic at x, then the initial value problem y p( x) y q( x) ; y( x) y has a solution that is analytic at x It means that y( x) can be expressed as n f ( x) an ( x x ) in some interval ( x h n x h ), Examples 4 Find the power series solution of the problem x, () 4 y e y x y Theorem 4 guarantees an analytic solution at x, that is, y( x) n n n a x n y n! ( n) () x n y () y () x y () x y () x 3! 3!

90 Theorem 4 If p( x ), q( x) and f ( x) are analytic at x, then the initial value problem y p( x) y q( x) y f ( x) ; y( x) has a unique solution that is analytic at x A, y( x) B It also means that y( x) can be expressed as f ( x) a n n ( x x n ) in some interval ( x h, x h ) Similarly, we have a f x n ( n) n! ( ) Examples 43 Find the power series solution of the problem x y xy e y 4 ; y(), y() 4 Theorem 4 guarantees an analytic solution at x, that is, y( x) n n n a x n y n! ( n) () x n y () y () x y () x y () x 3! 3! - 9 -

91 4 Power Series Solutions Using Recurrence Relations ( 利用遞迴關係式求微分方程式之次方級數解 ) Another way to generate coefficients is to develop a recurrence relation ( 遞迴關係式 ), which allow us to produce coefficients once certain preceding ones are known A recurrence relation is particularly suited to computer generation of coefficients Example 45 Solve y x y to get an analytic solution about x Let y( x) n n a x n Then we have n ( ) nanx, n y x Substitute them into the differential equation to get n n n y ( x) n( n ) a x n n n ( n ) anx x a nx n n n n n ( n ) anx anx n n n n n n n n ( n )( n ) a x a x n a 6 a x [( n )( n ) a a ] x 3 n n n a a3 ( n )( n ) an an, n,3, 4, The last relation implies an an, n=, 3, 4, ( n )( n ) or a4 a, a5 a, a6, a7, a8 a 4 a, Then y( x) a x n a a x x x a x a x x x n n a x a x a ( x x ) a ( x a x x )

92 Part II Vector Analysis ( 向量分析 ) Vector differential calculus extends out ability to analyze motion problems from real line to curves and surfaces in 3-space ( 向量微分之學習有助於將運動問題的分析從一直線擴展到三度空間的曲線及曲面 ) Tools such as the directional derivative, divergence and curl of a vector, gradient ( 可應用於方向導數 散度 旋度及梯度等物理量之計算 ) play significant roles in many applications Vector integral calculus generalizes integration to curves and surfaces in 3-space ( 向量積分則將積分問題延伸到對三度空間的曲線及曲面 ) This will pay many dividends, including the computation of quantities such as mass, center of mass, work, and flux of a vector field ( 可應用於質量 質心 做功 流量等物理量之計算 ), as well as physical interpretations of vector operations Chapter Vector Differential Calculus ( 向量微分 ) Vector Function of One Variable ( 單變數向量函數 ) Definition Vector Function of One Variable ( 單變數向量函數之定義 ) A vector function of one variable is a vector, each component of which is a function the same single variable ( 單變數向量函數是一種向量, 該向量的每一分量是同一個變數的函數 ) Typically, such a function has the appearance F ( t) x ( t) i y ( t) j z ( t) k, in which x( t ), y( t ), and z( t) are the component functions of F( t) A vector function is continuous or differentiable ( 連續或可微分 ) at t if each component function is continuous or differentiable at t - 9 -

93 The derivative ( 導數 ) of a vector function is the vector function formed by differentiating each component For example, F( t) [ x( t) i ] [ y( t) j] [ z( t) k ] x( t) i y ( t) j z ( t) k We may think of F( t) as an arrow extending from the origin to ( x( t ), y( t ), z( t ) ) In this case, F( t) is called a position vector ( 位置向量 ) for the curve (See Figure ) The derivative of a position vector ( 位置向量的導數 ) is the tangent vector ( 切線向量 ) to this curve (See Figure 3) F( t t) F ( t) F( t) lim t t x ( t t) x ( t) y( t t) y ( t) z( t t) z ( t ) lim i j k t t t t x ( t ) i y ( t ) j z ( t ) k Example Let the position vector F( t) t i sin( t) j t k Then the tangent vector is F ( t) ti cos( t) j tk At the origin ( t ), we have F () j At the point t or (x, y, z)=(, sin(), ), we have F() i cos() j k

94 From calculus, we know that the length of a curve ( 曲線之長度 ) given parametrically by x x( t), y y( t), z z ( t) for a t b is length = tb ta tb ds ( dx) ( dy) ( dz) ta b ( ) ( ) ( ) dt a dx dy dz dt dt dt b ( ( )) ( ( )) ( ( )) a x t y t z t dt where b a F( t) dt F( t) ( x( t)) ( y( t)) ( z( t)) is the length of the tangent vector The result shows that the length of a curve is the integral of the length of the tangent vector over the curve ( 曲線之長度 等於 切線向量長度沿著曲線之積分 ) Example Consider the curve given by the parametric equations ( 參數方程式 ) x cos( t), y sin( t), t z 3 for 4 t 4 The position vector for the curve is F ( t) cos( t) i sin( t) j tk The tangent vector at any point ( 任一點之切線向量 ) is F ( t) sin( t) i cos( t) j k The length of the tangent vector ( 切線向量之長度 ) is F( t) sin ( t) cos ( t) 9 3 Then the length of the curve ( 曲線之長度 ) is 3 3 Length = ( ) F t dt 3 dt Unit Tangent Vector ( 單位切線向量 ) Sometimes it is convenient to write the position vector of a curve in such a way that the tangent vector at each point has length Such a tangent vector is called a unit tangent vector

95 Let F ( t) x ( t) i y ( t) j z ( t) k for a t b And define the arc length function ( 弧長函數 ) t s( t) F( t) dt a s( t) is the length of the part of the curve from its initial point ( x( a ), y( a ), z( a ) ) to ( x( t ), y( t ), z( t ) ) Also note that s( a) and s( b) L, which is the total length of the curve By the fundamental theorem of calculus ( 根據微積分基本定理 ), we have ds F ( t) dt Then define the new position vector of the curve G( s) F ( t( s)) x ( t( s)) i y ( t( s)) j z ( t( s)) k G( s) is a position vector for the same curve as F( t) As t varies from a to b, F( t) sweeps out the same curve as G( s), as s varies from to L However, G( s) has the advantage that the tangent vector G( s) always has length d df( t) dt G ( s) F( t( s)) F ( t) F ( t) ds dt ds ds dt F( t) which is a unit vector Then the unit tangent vector, T ( t), T ( t) G ( s) F ( t) F( t) Example 3 Consider again the helix position vector ( 螺旋形位置向量 ) F ( t) cos( t) i sin( t) j tk for 4 t 4 The tangent vector at any point is F ( t) sin( t) i cos( t) j k The length of the tangent vector is F( t) sin ( t) cos ( t) 9 3 Therefore, the length function along this curve is Then we have s( t) F ( ) d d ( t 4 ) t t t( s) s

96 Homework for sec #9,

97 向量函數之應用 :Velocity( 速度 ), Acceleration( 加速度 ) Definition Velocity( 速度 ), Speed( 速率 ) The velocity v( t) of a particle ( 粒子或質點 ) at time t is defined to be v( t) F ( t) ( 速度等於位置向量的一階導數 ) The speed v( t) of the particle at time t is the magnitude of the velocity ds v( t) v( t) F ( t) ( 速率等於速度的大小 ) dt Definition 3 Acceleration ( 加速度 ) The acceleration a( t) of the particle is the rate of change of the velocity with respect to time: a( t) v( t) F ( t) ( 加速度等於速度的一階導數 ) Example 4 Consider the position vector of a curve F( t) sin( t) i ˆ e t ˆ j t ˆ k Find v( t), v( t ), a( t) and T ( t)

98 Definition 4 Curvature ( 曲度 ) The curvature of a curve is the magnitude of the rate of change of the unit tangent with respect to arc length along the curve ( 曲度等於單位切線向量對弧長之微分之大小 ) ( s) dt ds Note that dt dt r( s) ds ds ( s) ( 曲度等於單曲率半徑之倒數 ) Check: The curvature along a circle is equal to the inverse of its radius ( 一圓之曲度等於其半徑之倒數 ) Consider the position vector of a circle of radius r, ( 考慮一半徑為 r 之圓之位置向量 ) F( ) r cos( ) iˆ r sin( ) ˆj and F ( ) r sin( ) iˆ r cos( ) ˆj F ( ) r sin( ) iˆ r cos( ) ˆj It is found T ( ) sin( ) iˆ cos( ) ˆj and F ( ) r T ( ) cos( ) iˆ sin( ) ˆj Define the arc length function ( 弧長函數 ) s( ) rd r Then dt dt d dt ( ) ( cos( ) i sin( ) j) ds d ds dds d r r The result is consistent with the known fact ( 結果與已知事實相吻合 ) If the unit tangent vector T T ( s), then computing dt ( s ) ds is straightforward ( 直接 的 ) However, if T T ( u) ( u is an arbitrary parameter), then by the chain rule dt ( u) dt du dt dt ( u) T ( u) ds du ds ds du du df( u) du du F( u)

99 ( u) T ( u) F( u) This is a handy formula ( 方便手算的公式 ) because it does not require introduction of the arc length function s( t ) ( 因為不需要引進弧長函數 ) Example 5 Consider the position vector of a line ( 考慮一直線之位置向量 ) F( t) ( a bt) iˆ ( c dt ) ˆj ( e ht) kˆ Find T ( t) and ( t) Example 6 Consider the position vector of a circle in polar coordinates as ( 考慮一以極座標表示之圓之位置向量 ) F( ) 4cos( ) iˆ 3ˆj 4sin( ) kˆ Find T ( ) and ( )

100 Definition 5 Unit Normal Vector ( 單位法線向量 ) Use arc length s as parameter on the curve, the unit normal vector N( s) is defined by N ( s) T ( s), ( s) provided that ( s) dt dt ds T ( s) Note that N( s) dt dt ds ( s) ( 單位法線向量等於單位切線向量之微小變化量除以自己的大小 ) Notes: ) For unit vectors, N( s) T ( s) ) Since T ( s) T ( s) T ( s) is a constant, we have d T ( s) T ( s) dt ( s) T ( s) T ( s) dt ( s) T ( s) dt ( s) T ( s) dt ( s) it means that T ( s) dt ( s) Since N ( s) // dt ( s), so N ( s ) T ( s ) It means that the unit normal vector N( s) is orthogonal to ( 正交於 ) the unit tangent vector T ( s) - -

101 Theorem Tangential and Normal Components of Acceleration ( 加速度之切線 與法線方向之分量 ) dv v a att a N N T N dt r Proof: Since v vt, then d d ( ) a v vt T v dv T v ds dt dt dt dt dt dt dt ds dv dv v T v v N T N dt r dt r Unit Binormal Vector ( 單位雙垂線向量 ) The binormal vector B T N is also a unit vector and is orthogonal to T and N The triple T, N, B form a right-handed rectangular coordinate system We can in effect put an x, y, z coordinate system at any point on a curve, with the positive x, y, z axes along T, N, B, respectively - -