数学物理方法(王元明)第二章2
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X (0) = A + B = 0 X (l ) = Ae β l + Be − β l = 0 A=B=0 X ( x) = 0 X ( x) = Ax + B X ′′ = 0 A=B=0 X ( x) = 0 X ( x) = A cos βx + B sin βx X ′′ + β 2 X = 0
思考 方程是非齐次的，是否可以用分离变量法？ 非齐次方程的求解思路
•用分解原理得出对应的齐次问题 •解出齐次问题 •求出任意非齐次特解 •叠加成非齐次解
数学物理方程与特殊函数
第2章分离变量法
∂ 2u ∂ 2u = a 2 2 + f ( x, t ), 0 < x < l, t > 0 ∂t 2 ∂x t >0 u (0, t ) = u (l , t ) = 0, ∂u ( x, 0) u ( x, 0) = ϕ ( x), = ψ ( x), 0 ≤ x ≤ l ∂t
数学物理方程与特殊函数
第2章分离变量法
四 非齐次方程的解法
求下列定解问题 ∂ 2u ∂ 2u = a 2 2 + f ( x, t ), 0 < x < l, t > 0 ∂t 2 ∂x t >0 u (0, t ) = u (l , t ) = 0, ∂u ( x, 0) u ( x, 0) = ϕ ( x), = ψ ( x), 0 ≤ x ≤ l ∂t
v n (0) = 0
n=0 n≠0
v0 (t ) = −
1
n 2π 2 v ′ (t ) + a 2 2 v n (t ) = 0 n l
ω
cos ωt + C
v 0 (t ) =
2 2 2n π −a t l2
1
ω
(1 − cos ωt )
vn (t ) = Ce
v n (t ) = 0
u=
1
ω
0< x<l X ′(l ) = 0
λ = −β 2 < 0
X ′′ − β 2 X = 0
X ′(0) = Aβ − Bβ = 0 A= B=0
X = Ae βx + Be − βx
X ′(l ) = Aβ e βl − Bβ e − βl
λ =0
λ =β2 >0
X =0 X = B0 X = Ax + B X ′′ = 0 X = A sin βx + B cos βx X ′′ + β 2 X = 0 X ′(0) = Aβ = 0 X ′(l ) = − Bβ sin βl = 0 2 nπ nπ βn = λn = β n2 = , n = 1,2,3,⋯ l l nπ X n = Bn cos x l
(1 − cos ωt )
数学物理方程与特殊函数
第2章分离变量法
例16 求下列定解问题 ∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t t>0 u (0, t ) = u (l , t ) = 0, ∂u ( x,0) 0≤ x≤l u ( x,0) = 0, ∂t = 0, 解：令 u ( x, t ) = X ( x )T (t )
∂u 2 ∂ u 0 < x < l, t > 0 ∂t = a ∂x 2 ∂u (0, t ) ∂u (l , t ) = = 0, t > 0 ∂x ∂x 0≤ x≤l u ( x, 0) = 0,
2
u ( x, t ) = X ( x)T (t ) T ′X = a 2TX ′′ T′ X ′′ = = −λ 2 aT X X ′′ + λX = 0
数学物理方程与特殊函数
2 ∂ 2V 2 ∂ V ∂t 2 = a ∂x 2 + f ( x, t ), 0 < x < l , t > 0, t > 0, V (0, t ) = V (l , t ) = 0, ∂V ( x, 0) V ( x, 0) = = 0, 0 ≤ x ≤ l , ∂t
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t λn = (nπ / l )2 , n = 1,2,3,⋯ t>0 u (0, t ) = u (l , t ) = 0, nπ ∂u ( x,0) X n ( x) = Bn sin x 0≤ x≤l u ( x,0) = 0, ∂t = 0, l ∞ 2 2 ∞ nπ nπ 2π 2aπ 2 n π
令： u ( x, t ) = V ( x, t ) + W ( x, t )
∂ 2W ∂ 2W = a2 2 ∂t 2 ∂x W (0, t ) = W (l , t ) = 0, ∂W ( x, 0) W ( x, 0) = ϕ ( x), = ψ ( x) ∂t ∂ 2V ∂ 2V = a 2 2 + f ( x, t ), 0 < x < l , t > 0, ∂t 2 ∂x t > 0, V (0, t ) = V (l , t ) = 0, ∂V ( x, 0) V ( x, 0) = = 0, 0 ≤ x ≤ l , ∂t
v n (t ) ↔ Vn ( p)
2 2
nπ x l
f n (t ) ↔ Fn ( p)
2 ′ vn′ (t ) ↔ p 2Vn ( p ) − pvn (0) − v′ (0) = p Vn ( p) n
n 2π 2 p Vn ( p) + a Vn ( p) − Fn ( p) = 0 2 l
X (l ) = B sin β l = 0 λn = β n2 = (nπ / l )2
当 λ =0 时 当 λ = β 2 > 0时
X ( 0) = A = 0 β n = nπ / l , n = 1,2,3, ⋯ nπ X n ( x) = Bn sin x l
数学物理方程与特殊函数
第2章分离变量法
第2章分离变量法
nπ 令：V = ∑ vn (t ) sin 为什么？ x l n =1 ∞ ∞ 2 n 2π 2 nπ nπ ∑ v n′′ (t ) sin l x =∑ − a l 2 v n (t ) sin l x + f ( x, t ) n =1 n =1 ∞ 2 n 2π 2 nπ ∞ nπ = ∑ − a vn (t ) sin x + ∑ f n (t ) sin x 2 l l n =1 l n =1 ∞ 2 l nπ nπ f n (t ) = ∫ f ( x, t ) sin xdx f ( x, t ) = ∑ f n (t ) sin x 0 l l l n =1 n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l
∞
0 < x < l, t > 0 t>0 0≤ x≤l
nπ X n = Bn cos x, n = 0,1,2,3,⋯ l
∞
x
nπ u ( x,0) = ∑ vn (0) cos x=0 l n =0
′ v 0 (t ) = sin ωt
2 2 nπ 2 n π ∑ v n′ (t ) + a l 2 v n (t ) cos l x = sin ωt n=0
T ′ + a 2 λT = 0
数学物理方程与特殊函数
第2章分离变量法
X ′′ + λX = 0 T ′ + a 2 λT = 0
∂u (0, t ) = X ′(0)T (t ) = 0 ∂x ∂u (l , t ) = X ′(l )T (t ) = 0 ∂x
X ′′ + λX = 0 X ′(0) = 0,
数学物理方程与特殊函数
第2章分离变量法
例15 求下列定解问题
∂u ∂ 2u = a 2 2 + sin ωt 0 < x < l , t > 0 ∂x ∂t ∂u (0, t ) ∂u (l , t ) = = 0, t > 0 ∂x ∂x 0≤ x≤l u ( x,0) = 0, 解：先解对应的齐次问题
n≠2 n=2
n 2π 2 nπ nπ ′′ (t ) + a 2 2 vn (t ) = 0 vn (t ) = A cos a vn t + B sin a t l l l v n (t ) = 0 4π 2 2aπ ′′(t ) + a 2 2 v 2 (t ) = sin v2 t l l
数学物理方程与特殊函数
u = ∑ v n (t ) sin
n=0
l
x
∑ v ′′(t ) + a
n =1
n
l
2
v n (t ) sin x = sin x sin t l l l
nπ u ( x,0) = ∑ vn (0) sin x=0 l n =0
∞
vn (0) = 0
∂u ( x,0) ∞ nπ ′ = ∑ vn (0) sin x=0 ∂t l n =0 ′ vn (0) = 0
Vn ( p) =
1 p2 + a2 n π l2
2 2
Fn ( p)
k ↔ sin kt 2 2 p +k
nπa ↔ sin t 2 2 nπa l 2 2 n π p +a l2 l t nπa l nπa vn (t ) = sin t ⊗ f n (t ) = ∫0 f n (τ ) sin l (t − τ )dτ nπa nπa l 1 l
n =1
n=2
4π 2 2aπ v ′′(t ) + a t v 2 (t ) = sin 2 2 l l
数学物理方程与特殊函数
第2章分离变量法
∂u ∂ 2u = a 2 2 + sin ωt ∂x ∂t ∂u (0, t ) ∂u (l , t ) = = 0, ∂x ∂x u ( x,0) = 0, ∞ nπ u = ∑ v n (t ) cos l n=0 =0
∞
数学物理方程与特殊函数
第2章分离变量法
V = ∑ vn (t ) sin
n =1
∞
n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l ∞ nπ ∂V ( x, 0) ∞ nπ V ( x, 0) = ∑ vn (0) sin x=0 ′ (0) sin = ∑ vn x=0 l l ∂t n =1 n =1 ′ vn (0) = 0 vn (0) = 0
第2章分பைடு நூலகம்变量法
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t t>0 u (0, t ) = u (l , t ) = 0, ∂u ( x,0) 0≤ x≤l u ( x,0) = 0, ∂t = 0, ∞ nπ v n (t ) = 0 n≠2 u = ∑ vn (t ) sin x
XT ′′ = a 2 X ′′T X ′′ 1 T ′′ = 2 = −λ X a T
u (0, t ) = X (0)T (t ) = 0 u (l , t ) = X (l )T (t ) = 0 X (0) = 0, X (l ) = 0
X ′′ + λX = 0, 0 < x < l X (l ) = 0 X (0) = 0,
X ′′ + λX = 0
T ′′ + λa 2T = 0
数学物理方程与特殊函数
第2章分离变量法
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t X ′′ + λX = 0, 0 < x < l u (0, t ) = u (l , t ) = 0, t>0 X (l ) = 0 ∂u ( x,0) X (0) = 0, 0≤ x≤l u ( x,0) = 0, ∂t = 0, 当 λ = − β 2 < 0 时 X ′′ − β 2 X = 0 X ( x) = Ae βx + Be − βx
思考 方程是非齐次的，是否可以用分离变量法？ 非齐次方程的求解思路
•用分解原理得出对应的齐次问题 •解出齐次问题 •求出任意非齐次特解 •叠加成非齐次解
数学物理方程与特殊函数
第2章分离变量法
∂ 2u ∂ 2u = a 2 2 + f ( x, t ), 0 < x < l, t > 0 ∂t 2 ∂x t >0 u (0, t ) = u (l , t ) = 0, ∂u ( x, 0) u ( x, 0) = ϕ ( x), = ψ ( x), 0 ≤ x ≤ l ∂t
数学物理方程与特殊函数
第2章分离变量法
四 非齐次方程的解法
求下列定解问题 ∂ 2u ∂ 2u = a 2 2 + f ( x, t ), 0 < x < l, t > 0 ∂t 2 ∂x t >0 u (0, t ) = u (l , t ) = 0, ∂u ( x, 0) u ( x, 0) = ϕ ( x), = ψ ( x), 0 ≤ x ≤ l ∂t
v n (0) = 0
n=0 n≠0
v0 (t ) = −
1
n 2π 2 v ′ (t ) + a 2 2 v n (t ) = 0 n l
ω
cos ωt + C
v 0 (t ) =
2 2 2n π −a t l2
1
ω
(1 − cos ωt )
vn (t ) = Ce
v n (t ) = 0
u=
1
ω
0< x<l X ′(l ) = 0
λ = −β 2 < 0
X ′′ − β 2 X = 0
X ′(0) = Aβ − Bβ = 0 A= B=0
X = Ae βx + Be − βx
X ′(l ) = Aβ e βl − Bβ e − βl
λ =0
λ =β2 >0
X =0 X = B0 X = Ax + B X ′′ = 0 X = A sin βx + B cos βx X ′′ + β 2 X = 0 X ′(0) = Aβ = 0 X ′(l ) = − Bβ sin βl = 0 2 nπ nπ βn = λn = β n2 = , n = 1,2,3,⋯ l l nπ X n = Bn cos x l
(1 − cos ωt )
数学物理方程与特殊函数
第2章分离变量法
例16 求下列定解问题 ∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t t>0 u (0, t ) = u (l , t ) = 0, ∂u ( x,0) 0≤ x≤l u ( x,0) = 0, ∂t = 0, 解：令 u ( x, t ) = X ( x )T (t )
∂u 2 ∂ u 0 < x < l, t > 0 ∂t = a ∂x 2 ∂u (0, t ) ∂u (l , t ) = = 0, t > 0 ∂x ∂x 0≤ x≤l u ( x, 0) = 0,
2
u ( x, t ) = X ( x)T (t ) T ′X = a 2TX ′′ T′ X ′′ = = −λ 2 aT X X ′′ + λX = 0
数学物理方程与特殊函数
2 ∂ 2V 2 ∂ V ∂t 2 = a ∂x 2 + f ( x, t ), 0 < x < l , t > 0, t > 0, V (0, t ) = V (l , t ) = 0, ∂V ( x, 0) V ( x, 0) = = 0, 0 ≤ x ≤ l , ∂t
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t λn = (nπ / l )2 , n = 1,2,3,⋯ t>0 u (0, t ) = u (l , t ) = 0, nπ ∂u ( x,0) X n ( x) = Bn sin x 0≤ x≤l u ( x,0) = 0, ∂t = 0, l ∞ 2 2 ∞ nπ nπ 2π 2aπ 2 n π
令： u ( x, t ) = V ( x, t ) + W ( x, t )
∂ 2W ∂ 2W = a2 2 ∂t 2 ∂x W (0, t ) = W (l , t ) = 0, ∂W ( x, 0) W ( x, 0) = ϕ ( x), = ψ ( x) ∂t ∂ 2V ∂ 2V = a 2 2 + f ( x, t ), 0 < x < l , t > 0, ∂t 2 ∂x t > 0, V (0, t ) = V (l , t ) = 0, ∂V ( x, 0) V ( x, 0) = = 0, 0 ≤ x ≤ l , ∂t
v n (t ) ↔ Vn ( p)
2 2
nπ x l
f n (t ) ↔ Fn ( p)
2 ′ vn′ (t ) ↔ p 2Vn ( p ) − pvn (0) − v′ (0) = p Vn ( p) n
n 2π 2 p Vn ( p) + a Vn ( p) − Fn ( p) = 0 2 l
X (l ) = B sin β l = 0 λn = β n2 = (nπ / l )2
当 λ =0 时 当 λ = β 2 > 0时
X ( 0) = A = 0 β n = nπ / l , n = 1,2,3, ⋯ nπ X n ( x) = Bn sin x l
数学物理方程与特殊函数
第2章分离变量法
第2章分离变量法
nπ 令：V = ∑ vn (t ) sin 为什么？ x l n =1 ∞ ∞ 2 n 2π 2 nπ nπ ∑ v n′′ (t ) sin l x =∑ − a l 2 v n (t ) sin l x + f ( x, t ) n =1 n =1 ∞ 2 n 2π 2 nπ ∞ nπ = ∑ − a vn (t ) sin x + ∑ f n (t ) sin x 2 l l n =1 l n =1 ∞ 2 l nπ nπ f n (t ) = ∫ f ( x, t ) sin xdx f ( x, t ) = ∑ f n (t ) sin x 0 l l l n =1 n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l
∞
0 < x < l, t > 0 t>0 0≤ x≤l
nπ X n = Bn cos x, n = 0,1,2,3,⋯ l
∞
x
nπ u ( x,0) = ∑ vn (0) cos x=0 l n =0
′ v 0 (t ) = sin ωt
2 2 nπ 2 n π ∑ v n′ (t ) + a l 2 v n (t ) cos l x = sin ωt n=0
T ′ + a 2 λT = 0
数学物理方程与特殊函数
第2章分离变量法
X ′′ + λX = 0 T ′ + a 2 λT = 0
∂u (0, t ) = X ′(0)T (t ) = 0 ∂x ∂u (l , t ) = X ′(l )T (t ) = 0 ∂x
X ′′ + λX = 0 X ′(0) = 0,
数学物理方程与特殊函数
第2章分离变量法
例15 求下列定解问题
∂u ∂ 2u = a 2 2 + sin ωt 0 < x < l , t > 0 ∂x ∂t ∂u (0, t ) ∂u (l , t ) = = 0, t > 0 ∂x ∂x 0≤ x≤l u ( x,0) = 0, 解：先解对应的齐次问题
n≠2 n=2
n 2π 2 nπ nπ ′′ (t ) + a 2 2 vn (t ) = 0 vn (t ) = A cos a vn t + B sin a t l l l v n (t ) = 0 4π 2 2aπ ′′(t ) + a 2 2 v 2 (t ) = sin v2 t l l
数学物理方程与特殊函数
u = ∑ v n (t ) sin
n=0
l
x
∑ v ′′(t ) + a
n =1
n
l
2
v n (t ) sin x = sin x sin t l l l
nπ u ( x,0) = ∑ vn (0) sin x=0 l n =0
∞
vn (0) = 0
∂u ( x,0) ∞ nπ ′ = ∑ vn (0) sin x=0 ∂t l n =0 ′ vn (0) = 0
Vn ( p) =
1 p2 + a2 n π l2
2 2
Fn ( p)
k ↔ sin kt 2 2 p +k
nπa ↔ sin t 2 2 nπa l 2 2 n π p +a l2 l t nπa l nπa vn (t ) = sin t ⊗ f n (t ) = ∫0 f n (τ ) sin l (t − τ )dτ nπa nπa l 1 l
n =1
n=2
4π 2 2aπ v ′′(t ) + a t v 2 (t ) = sin 2 2 l l
数学物理方程与特殊函数
第2章分离变量法
∂u ∂ 2u = a 2 2 + sin ωt ∂x ∂t ∂u (0, t ) ∂u (l , t ) = = 0, ∂x ∂x u ( x,0) = 0, ∞ nπ u = ∑ v n (t ) cos l n=0 =0
∞
数学物理方程与特殊函数
第2章分离变量法
V = ∑ vn (t ) sin
n =1
∞
n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l ∞ nπ ∂V ( x, 0) ∞ nπ V ( x, 0) = ∑ vn (0) sin x=0 ′ (0) sin = ∑ vn x=0 l l ∂t n =1 n =1 ′ vn (0) = 0 vn (0) = 0
第2章分பைடு நூலகம்变量法
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t t>0 u (0, t ) = u (l , t ) = 0, ∂u ( x,0) 0≤ x≤l u ( x,0) = 0, ∂t = 0, ∞ nπ v n (t ) = 0 n≠2 u = ∑ vn (t ) sin x
XT ′′ = a 2 X ′′T X ′′ 1 T ′′ = 2 = −λ X a T
u (0, t ) = X (0)T (t ) = 0 u (l , t ) = X (l )T (t ) = 0 X (0) = 0, X (l ) = 0
X ′′ + λX = 0, 0 < x < l X (l ) = 0 X (0) = 0,
X ′′ + λX = 0
T ′′ + λa 2T = 0
数学物理方程与特殊函数
第2章分离变量法
∂ 2u ∂ 2u 2π 2aπ = a 2 2 + sin x sin t 0 < x < l, t > 0 2 l l ∂x ∂t X ′′ + λX = 0, 0 < x < l u (0, t ) = u (l , t ) = 0, t>0 X (l ) = 0 ∂u ( x,0) X (0) = 0, 0≤ x≤l u ( x,0) = 0, ∂t = 0, 当 λ = − β 2 < 0 时 X ′′ − β 2 X = 0 X ( x) = Ae βx + Be − βx