数学物理方程课件例题与习题

t =0
=0
解:设 u(x, t) = v (x, t) + W (x) 代入方程, 可得 4 8 W ′′ − = 0 9 9 W = x2 W x = 0 = 0, W x = π = π 2 4 v tt = v xx (0 < x < π , t > 0) 9 利用叠加原理, 得 v x = 0 = 0, v x =π = 0
∴ u( x , t ) = cos(3at ) sin( 3 x )
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⎧1 ， x ∈ [0 , 1] 1 f ( x) = ∑ sin( 2k − 1)πx = ⎨ π k =1 2k − 1 ⎩0 , x ∉ [0 , 1]
4
∞
1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1
1
X = B sin λ x
0.5 0 -0.5 -1 0 0.2 0.4 0.6 0.8 1
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cos (π x)
0.2
0.4 x
0.6
0.8
1
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固有值问题V ⎧ X ′′ + λ X = 0, 0 < x < L ⎨ ⎩ X ( 0) = 0, [ X ′ + hX ] x = L = 0 X ( x ) = A cos λ x + B sin λ x 通解: X(0)=0 A=0 X = B sin λ x [ X ′ + hX ] x = L = 0 λ cos λ L + h sin λ L = 0 λ v tan λ L = − 令 v = λ L tan v = − h Lh 2 y
u = f (ξ ) + g(η )
u = f ( x − y ) + g(9 x − y )
其中, f 和 g 是任意函数
1/16
例1.分离变量法求解 波动方程定解问题 解:利用公式
∞
⎧ utt = a 2 u xx , (0 < x < π , t > 0 ) ⎪ ⎨ u x = 0 = 0, u x = π = 0 ⎪ ⎩ u t = 0 = sin( 3 x ), ut t = 0 = 0
2
u t = 0 = sin x + 2 sin 3 x u x = 0 = 0, u x =π = 0
方程的Fourier解
∞
u( x , t ) = ∑ Bn e
n =1 n
∞
− ( na ) 2 t
sin nx
初始条件
∑B
n =1
sin nx = sin x + 2 sin 3 x
Bn = 0
2 ∑ 3 nDn sin(nx ) = 0 n =1
Dn = 0
v ( x , t ) = cos 2t × sin 3 x
u( x , t ) = v + W = cos 2t × sin 3 x + x
2
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热传导方程
第一类边界
ut = a u xx (0 < x < π , t > 0)
1.5 1
0.5
0 0 0.2 0.4 0.6 0.8 1
1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1
3/16
例6 P.51
4 8 utt = u xx − ( 0 < x < π , t > 0) 9 9 u x = 0 = 0, u x =π = π 2
u t = 0 = sin 3 x + x 2 , ut
s in(π x) 1
0.5
0 0 0.2 0.4 x 0.6 0.8 1
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⎧ X ′′ + λ X = 0, 0 < x < L 固有值问题II ⎨ ⎩ X ( 0 ) = 0, X ′( L ) = 0
( 2n + 1) π λn = 4 L2
2
1 0.5 0 -0.5 -1 0 0.2 0.4 x 0.6 0.8 1
− ( 3a )2 t
B1 = 1, B3 = 2,
( n ≠ 1, 3 )
u( x , t ) = e
− a 2t
sin x + 2e
sin 3 x
6/16
习题3.1第1题(1)
utt = a 2 u xx (0 < x < L, t > 0)
u
x=0
= 0, u
x= L t =0
=0 = x( L − x )
cos ((1+1/2) π x) 1 0.5 0 -0.5 -1 0 0.2 0.4 x 0.6 0.8 1
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固有值问题IV
nπ λn = 2 L
2
1 0.5 0 -0.5 -1 0
2
⎧ X ′′ + λ X = 0, 0 < x < L ⎨ ⎩ X ′( 0 ) = 0, X ′( L ) = 0 nπ X n ( x ) = An cos x L
u t = 0 = 0, ut
解: 固有值问题
nπ 2 固有值 λ n = ( ) L
∞
⎧ X ′′ + λ X = 0, 0 < x < L ⎨ ⎩ X ( 0 ) = 0, X ( L ) = 0
固有函数
n = 1,2,……
nπ X n = sin x L
nπ nπ nπ u( x , t ) = ∑ [C n cos( at ) + Dn sin( at )]sin( x) L L L n =1 ∞ nπ 利用初值条件 ∑ C n sin( x) = 0 L n =1
u( x , t ) = ∑ [C n cos(ant ) + Dn sin(ant )] sin( nx )
⎧0, n ≠ 3 C n = ∫ sin( 3ξ ) sin( nξ )dξ = ⎨ π 0 ⎩1, n = 3 2 1 Dn = ∫0 0 sin(nξ )dξ = 0 nπa 2
π
n =1
2
( 2n + 1)π X n ( x ) = Bn sin x 2L s in((1+1/2) x)
π
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⎧ X ′′ + λ X = 0, 0 < x < L 固有值问题III ⎨ ⎩ X ′( 0 ) = 0, X ( L ) = 0 2 2 ( 2n + 1)π ( 2n + 1) π X n ( x ) = An cos x λn = 2L 4 L2
8/16
习题3.2第1题
ut = a 2 u xx (0 < x < L, t > 0)
u
x=0
= 0, u
x= L
=0
u t =0 = x( L − x )
解: 固有值问题
nπ 2 固有值 λ n = ( ) L
∞wenku.baidu.com
⎧ X ′′ + λ X = 0, 0 < x < L ⎨ ⎩ X ( 0 ) = 0, X ( L ) = 0
习题2.4,2题(求方程通解) (1) uxx +10 uxy +9 uyy = 0 解:特征方程
dy =1 dx dy =9 dx
λ2 − 10λ + 9 = 0
y = x + C1
(λ − 1)(λ − 9) = 0
ξ = x− y η = 9x − y
y = 9 x + C2
原方程化简为 uξη = 0
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Cn = 0
L
nπa nπ ∑ L Dn sin( L x ) = x( L − x ) n =1 nπa nπ 2 L
∞
L
Dn =
L∫
0
x ( L − x ) sin
L
xdx
nπ L L nπ ∫0 x( L − x ) sin L xdx = nπ ∫0 ( L − 2 x ) cos L xdx L 2 L nπ L 3 xdx = −2( = 2( ) ∫ sin ) [cos nπ − 1] 0 nπ L nπ 4 L3 Dn = [1 − ( −1) n ] 所以 4 a ( nπ ) n πa nπ 4 L3 ∞ [1 − ( −1) n ] u( x , t ) = t sin sin x 4 ∑ 4 aπ n=1 n L L
v
t =0
= sin 3 x , v t
t =0
=0
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2 2 v ( x , t ) = ∑ [C n cos( nt ) + Dn sin( nt )]sin( nx ) 3 3 n =1
∞
∑C
n =1 ∞
∞
n
sin( nx ) = sin 3 x
⎧0, n ≠ 3 Cn = ⎨ ⎩1, n = 3
v
⎛ vk ⎞ λk = ⎜ ⎟ (k = 1,2,·······) ⎝ L⎠ 1 ( k − )π < v k < kπ 2
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⎧ X ′′ + λ X = 0 ⎨ ⎩ X (0) = 0, X (1) + X ′(1) = 0
tan λ = − λ
λ3
·····
λ1
λ2
2.0288 4.9132 7.9787 11.0855
固有函数
n = 1,2,……
nπ X n = sin x L
nπ nπ u( x , t ) = ∑ An exp(− at ) sin( x) L L n =1 ∞ 利用初值条件 ∑ An sin( nπ x ) = x( L − x ) L n =1
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nπ 2 L 4 L2 n An = ∫ x ( L − x ) sin xdx = 3 3 [1 − ( −1) ] 0 L L nπ
[1 − ( −1) ] nπ nπ exp(− x) at ) sin( u( x , t ) = 3 ∑ 3 π n =1 L n L 4L
2 ∞ n
1
1
0.5
0
0 1 0.5 0 1 0 0.5
-1 1 0.5 0 1 0 0.5
10/16
⎧ X ′′ + λ X = 0, 0 < x < L 固有值问题I ⎨ ⎩ X ( 0 ) = 0, X ( L ) = 0 n 2π 2 nπ λn = 2 X n ( x ) = Bn sin x L L