热流问题 分离变量法 周期函数 正交性 Fourier 级数 第 12章 Fourier 分析 一维杆状物体的热流问题即在一个长度为1的长杆上，两端保持零度， 初始的温度分布为f(x),随着时间的演化，k是比热系数，求t时刻的温度分 布T(x,t). 0 X 该问题的数学模型为 gur -ror.lce,t) (1) T(0,t)=T(l,t)=0,(t>0)(边界条件) (2) T(x,0)=f(x),0<x<l (初始条件) (3) 11 返回 全屏 关闭 退出 1/15
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê 1 12 Ù Fourier ©Û \GÔN96¯K =3Ý l \þ, üà±"Ý, Щ§Ý©Ù f(x), Xmüz, k ´'9Xê, ¦ t §Ý© Ù T (x, t). O x x ` T¯KêÆ . ∂ 2T ∂x2 = k 2 ∂T ∂t , T = T (x, t) (1) T (0, t) = T (`, t) = 0, (t > 0) £>.^¤ (2) T (x, 0) = f(x), 0 < x < ` £Ð©^¤ (3) 1/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê ©lCþ{ mCþ Cþ´©l: T (x, t) = ϕ(x)ψ(t). òd\ (1) ª ϕ 00(x)ψ(t) = k 2ϕ(x)ψ 0 (t) = ϕ00(x) k2ϕ(x) = ψ0 (t) ψ(t) . (4) Ï x t ´ÕáCþ, ¤± (4) ªL² ϕ00(x) k2ϕ(x) = ψ0 (t) ψ(t) = −λ, Ù¥ λ ´~ê, â§ÝCzA:, k λ > 0. Ïd ϕ 00(x) + λk2ϕ(x) = 0, (5) ψ 0 (t) + λψ(t) = 0. (6) 2/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê k = 1 , (5) ªÏ) ϕ(x) = b sin(√ λx + c). d>.^ (2) c = 0, √ λ` = nπ, (n = 1, 2, · · ·) λ = λn = nπ ` 2 . (7) ò (7) \ (6) ª, ψn(t) = exp − nπ ` 2 t . Ïd, § (5), (6) k|): ϕn(x) = bn sin nπ ` x, ψn(t) = exp − nπ ` 2 t . ¤± Tn(x, t) = bn exp − nπ ` 2 t sin nπ ` x, n = 1, 2 · · · ´÷v (1), (2) ). 3/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê ϧ (1) ´5, ¤± (1) Ï) T (x, t) = X ∞ n=1 bn exp − nπ ` 2 t sin nπ ` x, (8) 2âЩ^ (3), Ak f(x) = X ∞ n=1 bn sin nπ ` x. (9) Ïd, 3~ê {bn} ¦ (9) ¤á, K (8) Ò´¤¦¯K). ¯K ¼ê f(x) ´Ä½±L« (9) ªmà?ê? XJ±k ùL«, @oN¦Xê bn ? 4/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê §12.1 ¼ê Fourier ?ê 12.1.1 ±Ï¼ê9¼ê±Ïòÿ ±Ï¼ê e f(x) 3 (−∞, +∞) þk½Â, 3 T > 0 ¦ f(x + T ) = f(x), ∀ x ∈ (−∞, +∞), K¡ f(x) ´±Ï¼ê, T ´ f(x) ±Ï. 1) e T ´ f(x) ±Ï, K nT (n ∈ Z) ´. 2) e f(x) 3k«mþÈ, T ´§±Ï, K Z a+T a f(x) dx = Z T 0 f(x) dx. 5/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê 3) e f(x) 3 (−`, `) þk½Â, K F(x) = f(x − 2n`), (2n − 1)` < x < (2n + 1)` f(`)+f(−`) 2 , x = (2n + 1)`, n = 0, ±1, ±2, · · · ´ (−∞, +∞) þ± 2` ±Ï¼ê. 4) e f(x) 3 [0, `) þk½Â, K F(x) = f(x), x ∈ [0, `] f(−x), x ∈ [−`, 0] ´ (−`, `) þó¼ê, U 3) ¥{±ò§òÿ (−∞, +∞) þ, ¤± 2` ±Ïó¼ê. 6/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê 5) e f(x) 3 (0, `) þk½Â, K F(x) = f(x), x ∈ (0, `) 0, x = 0 −f(−x), x ∈ (−`, 0) ´ (−`, `) þÛ¼ê, U 3) ¥{±ò§òÿ (−∞, +∞) þ, ¤± 2` ±ÏÛ¼ê. 7/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê 12.1.2 n¼ê5 R[−π, π] ´ [−π, π] þ Riemann ȼê N. éu f, g ∈ R[−π, π], ½Â hf, gi = 1 π Z π −π f(x)g(x) dx, (10) K hf, gi ´ R[−π, π] þSÈ: = 1 ◦ hf, fi > 0, hf, fi = 0 ⇐⇒ f a.e. ==== 0 5 2 ◦ hf, gi = hg, fi é¡5 3 ◦ hc1f1 + c2f2, gi = c1hf1, gi + c2hf2, gi 5 e hf, gi = 0, K¡ f g ´. 8/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê Uìþ¡½ÂSÈ, n¼êX 1, cos x,sin x, cos 2x,sin 2x, · · · , cos nx,sin nx, · · · ´¼êX, =Ù¥?¿üØÓ¼ê´. ¯¢þ, k 1 π Z π −π cos mx cos nx dx = δmn, (11) 1 π Z π −π sin mx sin nx dx = δmn, (12) 1 π Z π −π sin mx cos nx dx = 0, (13) Ù¥ δmn = 1, m = n 0, m 6= n. 9/15 kJ Ik J I £ ¶ '4 òÑ
96¯K ©lCþ{ ±Ï¼ê 5 Fourier ?ê 12.1.3 Fourier ?ê e¼ê f(x) 3 [−π, π] þ±Ðm¤n?ê, = f(x) = a0 2 + X ∞ n=1 an cos nx + bn sin nx , (14) Kâ?ênØ9n¼ê5, k hf(x), cos kxi = a0 2 h1, cos kxi + X ∞ n=1 anhcos nx, cos kxi + bnhsin nx, cos kxi = ak, k = 0, 1, · · · hf(x),sin kxi = a0 2 h1,sin kxi + X ∞ n=1 anhcos nx,sin kxi + bnhsin nx,sin kxi = bk, k = 1, 2, · · · 10/15 kJ Ik J I £ ¶ '4 òÑ