数学物理方程非齐次边界条件的处理

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
nπ X n = cos x, n = 0,1,2,3,L l
X =0 X = Ax + B X =1 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 2 βn = λn = β n = , n = 1,2,3,L l l
2 2 2
Vn ( p) =
1
2
Fn ( p)
例8 求下列定解问题
∂u ∂ 2u = a 2 2 + sin ωt 0 < x < l, t > 0 ∂t ∂u (0, t ) ∂x∂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
A(t ) =
u 2 (t ) − u1 (t ) W ( x, t ) = x + u1 (t ) l
u 2 (t ) − u1 (t ) l
∂ 2u ∂ 2u = a2 2 , 0 < x < l, t > 0 2 ∂x ∂t t >0 u = V +W u (0, t ) = u1 (t ), u (l , t ) = u2 (t ), ∂u ( x,0) u ( x,0) = ϕ ( x), = ψ ( x), 0 < x < l W ( x, t ) = u 2 (t ) − u1 (t ) x + u (t ) 1 ∂t l
n 2π 2 v′′ (t ) + a 2 2 vn (t ) = 0 n l
nπa nπa nπa ′ vn (t ) = A' (t ) sin t+ A(t ) cos t l l l nπa nπa nπa + B′(t ) cos t− B(t ) sin t l l l nπa nπa 令 A' (t ) sin t + B′(t ) cos t =0 l l
V = ∑ vn (t ) sin
n =1 ∞
∞
nπ x l
n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l
nπ V ( x, 0) = ∑ vn (0) sin x=0 l n =1
vn (0) = 0
∂V ( x, 0) ∞ nπ ′ = ∑ vn (0) sin x=0 ∂t l n =1
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
四 非齐次边界条件的处理
∂ 2u ∂ 2u = a2 2 , 0 < x < l, t > 0 2 ∂x ∂t t >0 u (0, t ) = u1 (t ), u (l , t ) = u2 (t ), ∂u ( x,0) u ( x,0) = ϕ ( x), ∂t = ψ ( x), 0 < x < l
2 ∂ 2W ∂ 2W a − 2 = 0, 2 ∂x ∂t W (0, t ) = u1 (t ), W (l , t ) = u2 (t )
a 2W ′′ = 0, W (0) = u1 , W (l ) = u 2
例7 求下列定解问题 ∂ 2u ∂ 2u = a2 2 , 0 < x < l, t > 0 2 ∂x ∂t t >0 u (0, t ) = 0, u (l , t ) = q, ∂u ( x,0) u ( x,0) = 0, ∂t = 0, 0 < x < l 解：令 u ( x, t ) = V ( x, t ) + W ( x) 设 W ( x) = Ax + B W (0) = 0 q W= x 得 l 2 ∂ 2V 2 ∂ V , 2 =a 2 ∂x ∂t V (0, t ) = 0,V (l , t ) = 0, q ∂u ( x,0) V ( x,0) = − l x, ∂t = 0,
=
nπa ∫0
l
t
nπa f n (τ ) sin (t − τ ) d τ l
nπ V = ∑ vn (t ) sin x l n =1
∞
vn =
nπa ∫0
l
tFra Baidu bibliotek
f n (τ ) sin
nπa (t − τ )dτ l
故原定解问题的解为
V ( x, t ) = ∑ [
n =1
∞
nπa ∫0
l
t
nπa nπ f n (τ ) sin (t − τ )dτ ] sin x l l
由（3）（4）式可得，
（3） （4）
nπa A(t ) = ∫ f n (t ) cos tdt nπa l nπa −l B (t ) = ∫ f n (t ) sin tdt nπa l l
nπa t l nπa τdτ +C) vn (t) = sin t(∫ fn (τ ) cos 0n a π l l nπa t −l nπa +cos τdτ + D) t(∫ fn (τ )sin 0n a l π l
T ′ + a 2 λT = 0
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,
0< x<l X ′(l ) = 0
∂V ( x, 0) ∞ nπ ′ = ∑ vn (0) sin x=0 ∂t l n =1
′ vn (0) = 0
′ ′ vn′ (t ) ↔ p 2Vn ( p) − pvn (0) − vn (0)= p 2Vn ( p)
f n (t ) ↔ Fn ( p)
2 2
n π Vn ( p) − Fn ( p) = 0 p Vn ( p) + a 2 l
解：令 u ( x, t ) = V ( x, t ) + W ( x, t )
W (0, t ) = u1 (t ) W (l , t ) = u 2 (t )
设： W ( x, t ) = A(t ) x + B(t )
W (0, t ) = B(t ) = u1 (t )
W (l , t ) = A(t )l + B (t ) = u 2 (t )
V = ∑ vn (t ) sin
n =1 ∞
∞
nπ x l
n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l
nπ V ( x, 0) = ∑ vn (0) sin x=0 l n =1
vn (0) = 0
v n (t ) ↔ Vn ( p)
2 2
nπa nπa nπa 2 nπa ′ vn′ (t ) = A' (t ) cos t −( ) A(t ) sin t l l l l nπa nπa nπa 2 nπa ′(t ) sin − B t −( ) B(t ) cos t （2） l l l l 2 2 2 n π 把（2）代入（1）式，得 ′ v n′ (t ) + a v n (t ) − f n (t ) = 0
求解
n π ′ v n′ (t ) + a v n (t ) − f n (t ) = 0 2 l
2 2 2
常 数
（1）
易 法
变
先求解上式的齐次方程的通解
nπa nπa vn (t ) = A sin t + B cos t l l 齐次方程的解 nπa nπa vn (t ) = A(t ) sin t + B(t ) cos t l l nπa nπa nπa ′ vn (t ) = A' (t ) sin t+ A(t ) cos t l l l nπa nπa nπa + B′(t ) cos t− B(t ) sin t l l l
nπ X n = cos x l
∂u ∂ 2u = a 2 2 + sin ωt 0 < x < l, t > 0 ∂t ∂u (0, t ) ∂x∂u (l , t ) = = 0, t > 0 ∂x ∂x 0< x<l u ( x,0) = 0, ∞ nπ 设 u = ∑ vn (t ) cos x （4） l n =0 把（4）式代入（1）与（3）式 4 1 3
nπa nπa l A′(t ) cos t − B′(t ) sin t = f n (t ) l l nπa
l2
vn (t ) = A(t ) sin
nπa nπa t + B(t ) cos t l l
nπa nπa A' (t ) sin t + B′(t ) cos t =0 l l nπa nπa l A′(t ) cos t − B′(t ) sin t = f n (t ) l l nπa
W (l ) = q
∂ 2V ∂ 2V = a 2 2 + a 2W ′′ ∂t 2 ∂x 0 < x < l, t > 0 t >0 0≤ x≤l
五 非齐次方程的解法
求下列定解问题
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
nπ p +a l2 1 l nπa k ↔ sin t 2 2 ↔ sin kt nπa l 2 2 n π p2 + k 2 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） （2） （3）
nπ （4） x l n =1 ∞ nπ （5） 其中 f (t ) = 2 l f ( x, t ) sin nπ xdx f ( x, t ) = ∑ f n (t ) sin x n l ∫0 l l n =1
设 V = ∑ vn (t ) sin
∞
把（4）（5）代入（1）中 ∞ ∞ 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 n 2π 2 ′ v n′ (t ) + a 2 2 v n (t ) − f n (t ) = 0 l
∂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
′ vn (0) = 0
nπa t l nπa vn (t) = sin t(∫ fn (τ ) cos τdτ +C) 0n a π l l nπa t −l nπa t(∫ fn (τ )sin +cos τdτ + D) 0n a l π l
C =0 D=0
nπa t l nπa nπa t −l nπa vn (t) = sin t∫ fn (τ ) cos t∫ fn (τ )sin τdτ + cos τdτ 0n a 0n a l π l l π l