数理方程PPT
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2. > .^‡
• (1)1˜aµu|>. = f1;
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∂u • (2)1 aµ ∂n |>. = f2; ∂u • (3)1naµ u + k ∂n | >. = f3.
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~µ Š’(3.1)1Ô K utt = a2 uxx , 0 < x < L, t > 0 u|x=0 = 0, ux|x=L = 0, (7) F0
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T2cos(α2) − T1cos(α1) T2sin(α2) − T1sin(α1) − ρg ds + F ds α1 α1 = 1 − + ··· 2! sin(α1) sin(α2) ≈ tan(α2), tan(α) ds = (dx)2 + (dy )2 =
1
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Ù©lCþ{
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^©lCþ{¦)½)¯K 1.¦ e >НK kŠÚ ’3.113K) (1). (9) )
∂2u ∂t2 u = a2 ∂ + f. ∂x2
2
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~µÅÄ•§
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a11 = 1, a12 = 0, a22 = −a2, = a122 − a11a22 = a2 > 0
~µ9D
u •§ ∂u = a2 ∂ + f. ∂t ∂x2
2
⇒ ÅÄ•§´V-. .
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¥§
R
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(4)
v tt = a2v xx, −∞ < x < +∞, t > 0, a2 = v (x, 0) = ϕ(x), v t(x, 0) = ψ (x).
R
1 . LC
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I(x, t) = e− L tw(x, t)§“\•§>ØI(x, t)•§¥§ wtt = a2wxx, −∞ < x < +∞, t > 0, a2 = w(x, 0) = ϕ(x), wt(x, 0) = ψ (x).
1 . LC
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(5)
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½)^‡ 1. Ð ©^‡ u ĵ
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u(x, 0) = ϕ(x),
9D Щ§ÝÜ©µ
∂u |t=0 = ψ (x). ∂t
u(x, 0) = ϕ(x).
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~µ Š’(3.1)1˜ K ½)¯K•µ utt = a2 uxx , 0 < x < L, t > 0 u| = 0, u|x=L = 0, (6) x=0
ut|t=0 = 0, u(x, 0) = ϕ(x).
h (−x L−C h x, 0 C
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Ù¥ϕ(x) =
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≤x≤C + L), C < x ≤ L.
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ên•§Ï"ESKÀù The last class reviews for the last examination of the mathematical equation for physics
October 27, 2014
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∆ = a122 − a11a22 =
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225 + 34 > 0, 4
•§´V- ." dKŒ• •§ A -‚•µ
(8)
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dy dx
2
+ 15
dy − 34 = 0. dx
)ƒ µ
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dy dx dy dx
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²Lz{$Ž §
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>ØÚ>6 •§•µ
uxx − LCutt − (LG + RC )ut − RGu = 0, CR = GL, I xx − LCI tt − (LG + RC )I t − RGI = 0. u(x, t) = e− L tv(x, t)§ “ \ • § > Øu(x, t)• §
u|t=0 =
SY
x, ut|t=0 = 0.
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‡©•§a. éu‡©•§
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ä
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∂ 2u ∂ 2u ∂u ∂u ∂ 2u + a22 2 + b1 + b2 + cu = f . a11 2 + 2a12 ∂x ∂x∂y ∂y ∂x ∂y
e > 0, V-.; 2 = a12 − a11a22 = 0, Ô.; < 0, ý ..
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9D •§ Ñ 3. Ñ!Ÿ…3z˜Ó%¥þ •§" )
§
á¥N
9D
dtž m S Ï LS16 \
¥Š S
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Uþ•
Q1 = −kur (r, t)4πr2dt.
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dtžmSÏLS26\¥Š S Uþ• Q2 = −kur (r + dr, t)4π (r + dr)2dt. dtžmS¥Š Sá UþQ3•
= −17, ⇒ = 2.
17x + y = 0, 2x − y = 0.
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ξ = 17x + y, K η = 2x − y, |Q| = ξx ξy 17 1 = = 0. ηx ηy 2 −1
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¤±§k
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a11 a12 a21 a22
= =
17 1 2 −1
1 − 15 2 15 − 2 −34 .
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u(x, t) = f1(17x + y ) + f2(2x − y ).
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= 0§éη È©µ ∂u = f (ξ ) " ∂ξ
2éξ È©µ
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u(ξ, η ) =
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f (ξ )dξ + f2(η )
= f 1 ( ξ ) + f 2 (η ).
éª(3)†à^¥Š½n§…-∆x → 0§
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µ
T (x)ux |x = ρutt.
qT = ρg (L − x)§‘\þª¥= u î Ä•§µ
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utt = g [(L − x)ux] |x = g
∂ ∂u ( L − x) ∂x ∂x
.
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dT = −ρg. dx é þ ª ' uxÈ © § ¿ | ^ 3x = 0? § Ü åT T Ð uu g-ρgL§ µ
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T = ρg (L − x).
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3Y²••Ü啵 (2)
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T (x + ∆x)sin(α2) − T (x)sin(α1) = ρ∆xutt.
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sin(α2) ≈ tan(α2) ≈ ux(x + ∆x), α2 → 0, sin(α1) ≈ tan(α1) ≈ ux(x), α1 → 0.
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¤±ª(2)Œ±z•µ
T (x + ∆x)ux(x + ∆x) − T (x)ux(x) = ρ∆xutt. (3) T (x)ux |x+∆x − T (x)ux |x = ρ∆xutt.
17 2 1 −1
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·17 0 53 + 152 ·17 53 + 152 0
? • § Œ ± z { • µ361uξη = 0 ⇒ uξη = 0§ ) ƒ :
u(ξ, η ) = f1(ξ ) + f2(η ).
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òξ, η
Lˆª“\þª¥§
•§ )•µ
⇒ T2 = T1 = T. = ρuttdx. = ρuttdx. = utt. = utt, = utt.
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