Re: [理工] [工數]-成大98-工科PDE
※ 引述《hihaka2001 (hihaka)》之銘言:
: Utt=Uxx+F(x,t)
: U(0,t)=U(L,t)=0
: U(x,0)=p(x)
: Ut(x,0)=0
吃光光 表演要舞台
令U(x t) = ψ(x t) + φ(x t)
代回pde中
ψ (x t) + φ (x t) = ψ (x t) +φ (x t) +F(x t)
tt tt xx xx
令ψ (x t) = ψ (x t)
tt xx
φ (x t) = φ (x t) +F(x t)
tt xx
U(0 t) = ψ(0 t) + φ(0 t) = 0 φ(0 t) = ψ(0 t)=0
U(L,t) = ψ(L t) + φ(L t) = 0 φ(L t)= ψ(L t)=0
Ut(x,0)= ψt(x 0) + φt(x 0) = 0 ψt(x 0) = φt(x 0)=0
U(x,0)=ψ(x 0) + φ(x 0) = p(x) 令ψ(x 0)=p(x) φ(x 0)=0
(1)
nπx ∞
先解φ(x t) 特徵函數{sin── }n=1
L
∞ nπx ∞ nπx
φ(x t)= Σ an(t)sin ── F(x t)=Σ qn(t)sin──
n=1 L n=1 L
代回PDE中
∞ nπ 2 nπx
Σ [an(t)" +(──) an(t) -qn(t)]sin── =0
n=1 L L
由an(0) = an'(0)
t L nπ
an(t) = ∫ ── sin──(t-τ)qn(τ) dτ
0 nπ L
(2) nπx ∞
解ψ(x t) 特徵涵數 {sin── }n=1
L
∞ nπx
ψ(x t)=Σ bn(t)sin ──
n=1 L
代回pde中
∞ nπ 2 nπx
Σ [bn(t)" +(──) bn(t) ]sin── =0
n=1 L L
nπ nπ
bn(t)= Ancos──t + Bnsin──t
L L
∞ nπ nπ nπx
ψ(x t)=Σ [Ancos──t + Bnsin──t]sin ──
n=1 L L L
ψt(x 0) = 0 =Bn
∞ nπx
ψ(x 0) = p(x) = Σ Ansin ──
n=1 L
2 L nπx
An = ──∫ p(x)sin──dx
L 0 L
∞ 2 L nπx nπ nπx
ψ(x t) = Σ [ ──∫ p(x)sin──dx] cos──t sin ──
n=1 L 0 L L L
∞ t L nπ nπx
φ(x t)= Σ [∫ ── sin──(t-τ)qn(τ) dτ]sin ──
n=1 0 nπ L L
U(x t) = ψ(x t) + φ(x t)
獻醜了= =
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◆ From: 59.105.159.190
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (02/01 01:22)
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