第十六章斯特姆刘维尔问题 \color{blue}{第十六章斯特姆刘维尔问题}

问题引入: 问题引入:
(1−x 2 )y ′′ −2xy ′ +l(l+1)y=0→ddx [(1−x 2 )dydx ]+l(l+1)y=0 (1-x^2)y^{\prime \prime} - 2xy^{\prime} + l(l+1) y = 0 \to \dfrac{d}{dx}[(1-x^2) \dfrac{dy}{dx}] + l(l+1) y = 0
(1−x 2 )y ′′ −2xy ′ +[l(l+1)−m 2 1−x 2 ]y=0 (1-x^2)y^{\prime \prime} - 2xy^{\prime} + [l(l+1) - \dfrac{m^2}{1-x^2}]y = 0
→ddx [(1−x 2 )dydx ]−m 2 1−x 2 y+l(l+1)y=0 \to \dfrac{d}{dx}[(1-x^2)\dfrac{dy}{dx}] - \dfrac{m^2}{1-x^2} y + l(l+1) y = 0
x 2 y ′′ +xy ′ +[k 2 x 2 −n 2 ]y=0→ddx [(xdydx ]−n 2 x y+k 2 xy=0 x^2y^{\prime \prime} + xy^{\prime} + [k^2x^2 - n^2]y = 0 \to \dfrac{d}{dx}[(x\dfrac{dy}{dx}] - \dfrac{n^2}{x}y + k^2 xy = 0
ddx [k(x)dydx ]−q(x)y+λρ(x)y=0,a≤x≤b(1) \dfrac{d}{dx}[k(x) \dfrac{dy}{dx}] - q(x)y + \lambda \rho(x)y = 0, a \leq x \leq b \quad (1)

16.1.1S−L方程 \color{blue}{16.1.1 S-L方程}

1.定义: 1.定义:
ddx [k(x)dydx ]−q(x)y+λρ(x)y=0,a≤x≤b(1)−S−L方程 \dfrac{d}{dx} [k(x) \dfrac{dy}{dx}] - q(x) y + \lambda \rho(x) y = 0, a \leq x \leq b \quad (1) - S-L方程
k(x)≥0,q(x)≥0,ρ(x)≥0,λ−常数 k(x) \geq 0, q(x) \geq 0, \rho(x) \geq 0, \lambda - 常数

2.任意的二阶方程可化为S−L方程 2.任意的二阶方程可化为S-L方程
y ′′ (x)+p(x)y ′ (x)+h(x)y(x)=0(2) y^{\prime \prime}(x) + p(x)y^{\prime}(x) + h(x)y(x) = 0 \quad (2)
(2)⋅[k(x)=e ∫p(x)dx ]:ddx [e ∫p(x)dx dydx ]+e ∫p(x)dx h(x)y=0(3) (2)\cdot[k(x) = e^{\int p(x) dx}]:\dfrac{d}{dx}[e^{\int p(x) dx} \dfrac{dy}{dx}] + e^{\int p(x) dx} h(x) y = 0 \quad (3)
例:Hermit方程:y ′′ −2xy ′ +λy=0(4) 例:Hermit方程:y^{\prime \prime} - 2xy^{\prime} + \lambda y = 0 \quad (4)
p(x)=−2x,h(x)=λ→k(x)=e ∫−2xdx =e −x 2 p(x) = -2x, h(x) = \lambda \to k(x) = e^{\int -2x dx} = e^{-x^2}
(4)→ddx [e −x 2 y ′ ]+λe −x 2 y=0 (4) \to \dfrac{d}{dx}[e^{-x^2} y^{\prime}] + \lambda e^{-x^2}y = 0

(1−x 2 )y ′′ −2xy ′ +l(l+1)y=0→ddx [(1−x 2 )dydx ]+l(l+1)y=0 (1-x^2)y^{\prime \prime} - 2xy^{\prime} + l(l+1)y = 0 \to \dfrac{d}{dx}[(1-x^2)\dfrac{dy}{dx}] + l(l+1) y = 0
(1−x 2 )y ′′ −2xy ′ +[l(l+1)−m 2 1−x 2 ]y=0 (1-x^2)y^{\prime \prime} - 2xy^{\prime} + [l(l+1) - \dfrac{m^2}{1-x^2}]y = 0
→ddx [(1−x 2 )dydx ]−m 2 1−x 2 y+l(l+1)y=0 \to \dfrac{d}{dx}[(1-x^2)\dfrac{dy}{dx}] - \dfrac{m^2}{1-x^2} y + l(l+1) y = 0
x 2 y ′′ +xy ′ +[k 2 x 2 −n 2 ]y=0→ddx [(xdydx )]−n 2 x y+k 2 xy=0 x^2y^{\prime \prime} + xy^{\prime} + [k^2x^2 - n^2]y = 0 \to \dfrac{d}{dx}[(x\dfrac{dy}{dx})] - \dfrac{n^2}{x}y +k^2xy = 0

16.1.2S−L方程的自然边界条件 \color{blue}{16.1.2 S-L方程的自然边界条件}

1.定义:为满足物理上的适定性，物理问题本身所应具有的边界条件。 1.定义:为满足物理上的适定性，物理问题本身所应具有的边界条件。

例:{Φ ′′ +n 2 Φ=0Φ(φ+2π)=Φ(φ)  例:\left \lbrace \begin{array}{l} \Phi^{\prime \prime} + n^2 \Phi = 0 \\ \Phi(\varphi + 2 \pi) = \Phi(\varphi) \end{array} \right.
{r 2 +2rR ′ −l(l+1)R=0r<RR(r)| r=0 →有限  \left \lbrace \begin{array}{l} r^2 + 2rR^{\prime} - l(l+1)R = 0 \quad r
{(1−x 2 )y ′′ −2xy ′ +l(l+1)y=0y| |x|=1 →有限  \left \lbrace \begin{array}{l}(1-x^2)y^{\prime \prime} - 2xy^{\prime} + l(l+1) y = 0 \\ y|_{|x| = 1} \to 有限 \end{array} \right.

2.S−L方程在以下情况下具有自然边界条件 2.S-L方程在以下情况下具有自然边界条件
S−L方程ddx [k(x)dydx ]−q(x)y+λρ(x)y=0,a≤x≤b(1) S-L方程\dfrac{d}{dx}[k(x) \dfrac{dy}{dx}] - q(x) y + \lambda \rho(x) y = 0, a \leq x \leq b \quad (1)
(1)当k(a)=0和或k(b)=0时,在边界x=a,x=b处,具有有限性自然边界条件. (1)当k(a) = 0 和或k(b) = 0时,在边界x = a, x = b处,具有有限性自然边界条件.
例:(1−x 2 )y ′′ −2xy ′ +l(l+1)y=0→ddx [(1−x 2 )dydx ]+l(l+1)y=0 例:(1-x^2)y^{\prime \prime} -2xy^{\prime} + l(l+1) y = 0 \to \dfrac{d}{dx}[(1-x^2)\dfrac{dy}{dx}] + l(l+1)y = 0
x 2 y ′′ +xy ′ +[k 2 x 2 −n 2 ]y=0→ddx [xdydx ]−n 2 x y+k 2 xy=0 x^2y^{\prime \prime} +xy^{\prime} + [k^2x^2 -n^2]y = 0 \to \dfrac{d}{dx}[x\dfrac{dy}{dx}] - \dfrac{n^2}{x}y + k^2xy = 0

(2)当k(a)=k(b)时,在边界x=a,x=b处,具有周期性边界条件. (2)当k(a) = k(b)时,在边界x = a, x = b处,具有周期性边界条件.
例:Φ ′′ +n 2 Φ=0→ddφ [1⋅dΦdφ ]+n 2 Φ=0 例:\Phi^{\prime \prime} + n^2 \Phi = 0 \to \dfrac{d}{d\varphi}[1 \cdot \dfrac{d\Phi}{d \varphi}] + n^2 \Phi = 0

附:证明
ddx [k(x)dy 1 dx ]−q(x)y 1 +λρ(x)y 1 =0,(2) \dfrac{d}{dx}[k(x)\dfrac{dy_1}{dx}] - q(x)y_1 +\lambda \rho(x)y_1 = 0 , \quad (2)
ddx [k(x)dy 2 dx ]−q(x)y 2 +λρ(x)y 2 =0,(3) \dfrac{d}{dx}[k(x)\dfrac{dy_2}{dx}] - q(x)y_2 + \lambda \rho(x)y_2 = 0 , \quad (3)
(3)⋅y 1 −(2)y 2 :y 1 ddx [k(x)dy 2 dx ]−y 2 ddx [k(x)dy 1 dx ]=0, (3)\cdot y_1 - (2)y_2: y_1\dfrac{d}{dx}[k(x)\dfrac{dy_2}{dx}] - y_2\dfrac{d}{dx}[k(x)\dfrac{dy_1}{dx}] = 0,
ddx [k(x)(y 1 y ′ 2 −y 2 y ′ 1 )]=0,→(y 1 y ′ 2 −y 2 y ′ 1 )=ck(x)  \dfrac{d}{dx}[k(x)(y_1y_2^{\prime} - y_2y_1^{\prime})] = 0, \to (y_1y_2^{\prime} - y_2y_1^{\prime}) = \dfrac{c}{k(x)}
(y 2 y 1  ) ′ =y 1 y ′ 2 −y 2 y ′ 1 y 2 1  =ck(x)y 2 1  ≠0(∴y 2 y 1  ≠c) (\dfrac{y_2}{y_1})^{\prime} = \dfrac{y_1y_2^{\prime} - y_2y_1^{\prime}}{y_1^2} = \dfrac{c}{k(x)y_1^2} \neq 0 (\therefore \dfrac{y_2}{y_1} \neq c)
y 2 =y 1 [∫ x x 0  ck(x)y 2 1  dx+c 1 ],当k(a)=0,k(b)=0时,y 2 →∞ y_2 = y_1[\int_{x_0}^{x} \dfrac{c}{k(x)y_1^2}dx + c_1],当k(a) = 0, k(b) = 0时,y_2 \to \infty
∴y 2 | x=a,b →有限 \therefore y_2|_{x=a,b} \to 有限

16.1.3S−L本征值问题 \color{blue}{16.1.3 S-L本征值问题}

1.定义:称 1.定义:称
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ddx [k(x)dydx ]−q(x)y+λρ(x)y=0,a≤x≤b(1)[αdydx +βy(x)+γk(x)] x=a,b =0 为S−L本征值问题 \left \lbrace \begin{array}{l}\dfrac{d}{dx}[k(x)\dfrac{dy}{dx}] - q(x) y + \lambda \rho(x) y = 0, a \leq x \leq b \quad (1) \\ [\alpha \dfrac{dy}{dx} + \beta y(x) + \gamma k(x)] _{x = a, b} = 0 \end{array} \right. 为S-L本征值问题

2.S−L本征值问题的性质: 2.S-L本征值问题的性质:
(1)有无穷多个本征值:λ 1 ≤λ 2 ≤⋯≤λ n ≤⋯ (1)有无穷多个本征值:\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n \leq \cdots
无穷多个本征函数:y 1 (x)y 2 (x)⋯y n (x)⋯ 无穷多个本征函数:y_1(x) y_2(x) \cdots y_n(x) \cdots
例:{ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0R(a)=0  例:\left \lbrace \begin{array}{l}\rho^2 R^{\prime \prime}(\rho) +\rho R^{\prime}(\rho) + (k^2 \rho^2 - n^2) R(\rho) = 0 \\ R(a) = 0 \end{array} \right.
本征值:k n m =x n m a ,m=1,2,⋯, 本征值:k_m^n = \dfrac{x_m^n}{a}, m =1, 2, \cdots,
本征函数:R m (kρ)=J n (x n m a ρ),m=1,2,⋯ 本征函数:R_m(k\rho) = J_n(\dfrac{x_m^n}{a}\rho), m = 1, 2, \cdots
(2)λ m ≥0,m=1,2,⋯,如:(k n m ) 2 ≥0 (2)\lambda_m \geq 0, m = 1, 2, \cdots, 如:(k_m^n)^2 \geq 0
(3)∫ b a ρ(x)y m (x)y ¯  n (x)dx=N 2 n δ mn  (3)\int_a^b \rho(x) y_m(x) \bar y_n(x) dx = N_n^2 \delta_{mn}
如:∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ=a 2 2 J 2 n+1 (k n l a)δ ml  如:\int_0^a \rho J_n(k_m^n \rho) J_n(k_l^n \rho) d \rho = \dfrac{a^2}{2} J_{n+1}^2(k_l^n a) \delta_{ml}
(4)f(x)=∑ m=1 ∞ c m y m (x)c m =1N 2 m  ∫ b a ρ(x)f(x)y ¯  m (x)dx (4) f(x) = \sum \limits_{m=1}^{\infty} c_my_m(x) \quad c_m = \dfrac{1}{N_m^2}\int_a^b \rho(x) f(x) \bar y_m(x) dx
如:f(ρ)=∑ m=1 ∞ c m J n (k n m ρ)c m =1a 2 2 J 2 n+1 (k n m a) ∫ a 0 ρf(ρ)J n (k n m ρ)dρ 如:f(\rho) = \sum \limits_{m=1}^{\infty} c_mJ_n(k_m^n \rho) \quad c_m = \dfrac{1}{\dfrac{a^2}{2}J_{n+1}^2(k_m^n a)} \int_0^a \rho f(\rho) J_n(k_m^n \rho) d\rho

附:证明性质(3) 附:证明性质(3)
ddx [k(x)dy m dx ]−q(x)y m +λ m ρ(x)y m =0,(4) \dfrac{d}{dx}[k(x)\dfrac{dy_m}{dx}] - q(x)y_m + \lambda_m \rho(x) y_m = 0, \quad (4)
ddx [k(x)dy ¯  n dx ]−q(x)y ¯  n +λ n ρ(x)y ¯  n =0,(5) \dfrac{d}{dx}[k(x)\dfrac{d \bar y_n}{dx}] - q(x) \bar y_n + \lambda_n \rho(x) \bar y_n = 0, \quad (5)
(4)⋅y ¯  n −(5)y m :(λ m −λ n )∫ b a ρ(x)y m y ¯  n dx (4)\cdot \bar y_n - (5)y_m: (\lambda_m - \lambda_n)\int_a^b \rho(x) y_m \bar y_n dx
=∫ b a y m ddx [k(x)y ¯  n dx ]dx−∫ b a y ¯  n ddx [k(x)dy m dx ]dx =\int_a^b y_m \dfrac{d}{dx}[k(x)\dfrac{\bar y_n}{dx}] dx - \int_a^b \bar y_n \dfrac{d}{dx}[k(x) \dfrac{d y_m}{dx}] dx
=[k(x)(y m y ¯  ′ n −y ¯  n y ′ m )] a b  = [k(x)(y_m \bar y_n^{\prime} - \bar y_n y_m^{\prime})]_b^a
1)第一类:y ¯  n (a)=0,y m (a)=0→右边=0 1)第一类:\bar y_n(a) = 0, y_m(a) = 0 \to 右边 = 0
2)第二类:y ¯  ′ n (a)=0,y ′ m (a)=0→右边=0 2)第二类:\bar y_n^{\prime}(a) = 0, y_m^{\prime}(a) = 0 \to 右边 = 0
3)第三类:y m y ¯  ′ n −y ¯  n y ′ m =y m y ¯  ′ n +hy m y ¯  ′ n −hy m y ¯  n −y ¯  n y ′ m  3)第三类:y_m\bar y_n^{\prime} - \bar y_n y_m^{\prime} = y_m \bar y_n^{\prime} + hy_m \bar y_n^{\prime} - hy_m \bar y_n - \bar y_n y_m^{\prime}
=y m (hy ¯  n +y ¯  ′ n )−y ¯  n (y ′ m +hy m )→右边=0 = y_m(h\bar y_n + \bar y_n^{\prime}) - \bar y_n(y_m^{\prime} + hy_m) \to 右边 = 0
4)当k(a)=0,k(b)=0时,→右边=0 4)当k(a) = 0, k(b) = 0时,\to 右边 = 0
m≠0:∫ b a ρ(x)y m y ¯  n dx=0 m \neq 0: \int_a^b \rho(x) y_m \bar y_n dx = 0
m=n:令m→n第一类:y ¯  n (a)=0,y m (a)≠0 m = n: 令m \to n 第一类: \bar y_n(a) = 0, y_m(a) \neq 0
∫ b a ρ(x)y m y ¯  n dx=∫ b a ρ(x)|y n | 2 dx=N 2 n  \int_a^b \rho(x) y_m \bar y_n dx = \int_a^b \rho(x) |y_n|^2 dx = N_n^2

16.1.4例题 \color{blue}{16.1.4 例题}

1.已知S−L问题:{X ′′ (x)+λX(x)=0(1)X(0)=0,X(l)=0(2)  1.已知S-L问题:\left \lbrace \begin{array}{l}X^{\prime \prime}(x) + \lambda X(x) = 0 \quad (1) \\ X(0) = 0, X(l) = 0 \quad (2) \end{array} \right.
求:1)k(x)=?,k(0)=?,k(l)=?,q(x)=?,ρ(x)=? 求:1)k(x) = ?, k(0) = ?, k(l) = ?, q(x) = ?, \rho(x) = ?
2)λ=?,本征函数=?,N l =?, 2) \lambda = ?,本征函数 = ?, N_l = ?,
3)将f(x)=x∈[0,l]按上述本征函数展开 3)将f(x) = x \in [0, l]按上述本征函数展开
解:1)k(x)=1,k(0)=k(l)=1,q(x)=0,ρ(x)=1; 解:1)k(x) = 1, k(0) = k(l) = 1, q(x) = 0, \rho(x) = 1;
2)λ=(nπl ) 2 ,n=1,2,⋯;X n (x)=sinnπxl  2) \lambda = (\dfrac{n \pi}{l})^2, n = 1, 2, \cdots; X_n(x) = \sin \dfrac{n \pi x}{l}
N 2 l =∫ l 0 sin 2 nπxl dx=12 ∫ l 0 [1−cos2nπxl ]dx=l2  N_l^2 = \int_0^l \sin^2 \dfrac{n \pi x}{l} dx = \dfrac{1}{2} \int_0^l[1 - \cos \dfrac{2n \pi x}{l}] dx = \dfrac{l}{2}
3)x=∑ n=1 ∞ c n sinnπxl ,c n =1N 2 l  ∫ l 0 xsinnπxl dx=2l(−1) n+1 nπ  3)x = \sum \limits_{n=1}^{\infty} c_n \sin \dfrac{n \pi x}{l}, c_n = \dfrac{1}{N_l^2}\int_0^l x \sin \dfrac{n \pi x}{l} dx = \dfrac{2l(-1)^{n+1}}{n \pi}

2.已知H n (x)=(−1) n e x 2  d n dx n  e −x 2  (1) 2.已知 H_n(x) = (-1)^n e^{x^2} \dfrac{d^n}{dx^n} e^{-x^2} \quad (1)
试证:(1)e 2tx−t 2  =∑ n=0 ∞ H n (x)n! t n (2) 试证:(1)e^{2tx - t^2} = \sum \limits_{n=0}^{\infty} \dfrac{H_n(x)}{n!} t^n \quad (2)
(2){H ′ n (x)=2nH n−1 (x)(3)H n+1 (x)−2xH n (x)+2nH n−1 (x)=0(4)  (2) \left \lbrace \begin{array}{l}H_n^{\prime}(x) = 2n H_{n-1}(x) \quad (3) \\ H_{n+1}(x) - 2xH_n(x) + 2nH_{n-1}(x) = 0 \quad (4) \end{array} \right.
证明:令e 2tx−t 2  =∑ n=0 ∞ a n (x)t n ,则证明:令e^{2tx -t^2} = \sum \limits_{n=0}^{\infty} a_n(x)t^n, 则
a n (x)=1n! d n dt n  e 2tx−t 2  | t=0 =1n! e x 2  d n dt n  e −(x 2 −2tx+t 2 ) | t=0  a_n(x) = \dfrac{1}{n!} \dfrac{d^n}{dt^n} e^{2tx-t^2} |_{t=0} = \dfrac{1}{n!} e^{x^2} \dfrac{d^n}{dt^n} e^{-(x^2-2tx+t^2)}|_{t=0}
= ξ=t−x 1n! e x 2  d n dξ n  e −ξ 2  | ξ=−x =1n! e x 2  d n d(−x) n  e −x 2   \stackrel{\xi=t-x}{=} \dfrac{1}{n!}e^{x^2} \dfrac{d^n}{d \xi^n} e^{-\xi^2} |_{\xi = -x} = \dfrac{1}{n!} e^{x^2} \dfrac{d^n}{d(-x)^n}e^{-x^2}
∴a n (x)=H n (x)n!  \therefore a_n(x) = \dfrac{H_n(x)}{n!}

ddx (2):2te 2tx−t 2  =∑ n=0 ∞ H ′ n (x)n! t n  \dfrac{d}{dx}(2): 2te^{2tx-t^2} = \sum \limits_{n=0}^{\infty} \dfrac{H_n^{\prime}(x)}{n!} t^n
2∑ n=0 ∞ H n (x)n! t n+1 =∑ n=0 ∞ H ′ n (x)n! t n  2 \sum \limits_{n=0}^{\infty} \dfrac{H_n(x)}{n!} t^{n+1} = \sum \limits_{n=0}^{\infty} \dfrac{H_n^{\prime}(x)}{n!} t^n
t n :2H n−1 (x)(n−1)! =H ′ n (x)n!  t^n: 2 \dfrac{H_{n-1}(x)}{(n-1)!} = \dfrac{H_n^{\prime}(x)}{n!}
∴H ′ n (x)=2nH n−1 (x) \therefore H_n^{\prime}(x) = 2nH_{n-1}(x)

ddx (2):2(x−t)e 2tx−t 2  =∑ n=0 ∞ H n (x)(n−1)! t n−1  \dfrac{d}{dx}(2): 2(x-t)e^{2tx -t^2} = \sum \limits_{n=0}^{\infty} \dfrac{H_n(x)}{(n-1)!}t^{n-1}
2(x−t)∑ n=0 ∞ H n (x)n! t n =∑ n=0 ∞ H n (x)(n−1)! t n−1  2(x-t)\sum \limits_{n=0}^{\infty} \dfrac{H_n(x)}{n!} t^n = \sum \limits_{n=0}^{\infty} \dfrac{H_n(x)}{(n-1)!} t^{n-1}
t n :2xH n (x)n! −2H n−1 (x)(n−1)! =H n+1 (x)n!  t^n: 2x \dfrac{H_n(x)}{n!} - 2\dfrac{H_{n-1}(x)}{(n-1)!} = \dfrac{H_{n+1}(x)}{n!}
2xH n (x)−2nH n−1 (x)−H n+1 (x)=0 2xH_n(x) - 2nH_{n-1}(x) - H_{n+1}(x) = 0

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数学物理方法 16 斯特姆刘维尔问题

第十六章斯特姆刘维尔问题 \color{blue}{第十六章斯特姆刘维尔问题}

16.1.1S−L方程 \color{blue}{16.1.1 S-L方程}

16.1.2S−L方程的自然边界条件 \color{blue}{16.1.2 S-L方程的自然边界条件}

16.1.3S−L本征值问题 \color{blue}{16.1.3 S-L本征值问题}

16.1.4例题 \color{blue}{16.1.4 例题}

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数学物理方法 16 斯特姆刘维尔问题

第十六章斯特姆刘维尔问题 \color{blue}{第十六章 斯特姆刘维尔问题}

16.1.1S−L方程 \color{blue}{16.1.1 S-L方程}

16.1.2S−L方程的自然边界条件 \color{blue}{16.1.2 S-L方程的自然边界条件}

16.1.3S−L本征值问题 \color{blue}{16.1.3 S-L本征值问题}

16.1.4例题 \color{blue}{16.1.4 例题}

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第十六章斯特姆刘维尔问题 \color{blue}{第十六章斯特姆刘维尔问题}