[自動轉寄] [程式] 有關mathematica求解的一個問題

作者: af2486 (我喜歡旅行) 看板: Statistics
標題: [程式] 有關mathematica求解的一個問題
時間: Thu Jul 29 17:26:11 2010

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[軟體程式類別]:Mathematica
請填入軟體程式類別 例如SAS、SPSS、R、EVIEWS...等

[程式問題]:
在用mathematica牛頓法求解時，會先給個固定值讓程式在那值附近求解(一連串的多組解)。
不過後來發現往往後面求出的解會跑掉，因此現在要設定每次求解都以上一個算出來的值
當作新的求解範圍。不過我不知道該用什麼指令好，有人能給個方向嗎?

xiLimit = 15.;
Hankel1[m_, x_] :=
If[Abs[x] < xiLimit, BesselJ[m, x] + I BesselY[m, x],
Sqrt[2/(Pi*x)] Exp[I*(x - m/2*Pi - Pi/4)]];
BesselJp[m_, x_] :=
If[Abs[x] < xiLimit,
D[BesselJ[m, x1],
x1] /. {x1 -> x}, -Sqrt[2/(Pi*x)] Sin[(x - m/2*Pi - Pi/4)]];
BesselYp[m_, x_] :=
If[Abs[x] < xiLimit, D[BesselY[m, x1], x1] /. {x1 -> x},
Sqrt[2/(Pi*x)] Cos[(x - m/2*Pi - Pi/4)]];
Hankel1p[m_, x_] :=
If[Abs[x] < xiLimit, D[Hankel1[m, x1], x1] /. {x1 -> x},
I*Sqrt[2/(Pi*x)] Exp[I*(x - m/2*Pi - Pi/4)]];
kr[kz_, k_] := Sqrt[k^2 - kz^2];
\[Alpha]r[kz_, k_] := Sqrt[kz^2 - k^2];

f[kz_, opt_] := Module[{sub},
k0 = \[Omega] Sqrt[\[Epsilon]0 \[Mu]0] /. opt;
k1 = \[Omega] Sqrt[\[Epsilon]1 \[Mu]1] /. opt;
k2 = \[Omega] Sqrt[\[Epsilon]2 \[Mu]2] /. opt;
r0 = r0 /. opt;
r1 = r1 /. opt;
m = m /. opt;
kr0 = kr[kz, k0];
kr1 = kr[kz, k1];
\[Alpha]r2 = \[Alpha]r[kz, k2];
BJ00 = BesselJ[m, kr[kz, k0] r0];
BJ10 = BesselJ[m, kr[kz, k1] r0];
BJ11 = BesselJ[m, kr[kz, k1] r1];
BY10 = BesselY[m, kr[kz, k1] r0];
BY11 = BesselY[m, kr[kz, k1] r1];
BJp00 = BesselJp[m, kr[kz, k0] r0];
BJp10 = BesselJp[m, kr[kz, k1] r0];
BJp11 = BesselJp[m, kr[kz, k1] r1];
BYp10 = BesselYp[m, kr[kz, k1] r0];
BYp11 = BesselYp[m, kr[kz, k1] r1];
HA21 = Hankel1[m, I \[Alpha]r[kz, k2] r1];
HAp21 = Hankel1p[m, I \[Alpha]r[kz, k2] r1];
M = {{BJ00, 0, -BJ10, -BY10 , 0, 0, 0, 0},
{0, BJ00, 0, 0, -BJ10, -BY10, 0, 0},
{(\[Omega] \[Epsilon]0)/(kr0 r0) BJp00, (m kz)/(kr0^2 r0^2)
BJ00, -((\[Omega] \[Epsilon]1)/(kr1 r0))
BJp10, -((\[Omega] \[Epsilon]1)/(kr1 r0)) BYp10, -((m kz)/(
kr1^2 r0^2)) BJ10, -((m kz)/(kr1^2 r0^2)) BY10, 0, 0},
{(m kz)/(kr0^2 r0^2) BJ00, (\[Omega] \[Mu]0)/(kr0 r0)
BJp00, -((m kz)/(kr1^2 r0^2)) BJ10, -((m kz)/(kr1^2 r0^2))
BY10, -((\[Omega] \[Mu]1)/(kr1 r0))
BJp10, -((\[Omega] \[Mu]1)/(kr1 r0)) BYp10, 0, 0 },
{0, 0, BJ11, BY11 , 0, 0, -HA21, 0},
{0, 0, 0, 0, BJ11, BY11, 0, -HA21},
{0, 0, (\[Omega] \[Epsilon]1)/(kr1 r1)
BJp11, (\[Omega] \[Epsilon]1)/(kr1 r1) BYp11, (m kz)/(
kr1^2 r1^2) BJ11, (m kz)/(kr1^2 r1^2) BY11, (
I \[Omega] \[Epsilon]2)/(\[Alpha]r2 r1) HAp21, (
m kz)/(\[Alpha]r2^2 r1^2) HA21},
{0, 0, (m kz)/(kr1^2 r1^2) BJ11, (m kz)/(kr1^2 r1^2)
BY11, (\[Omega] \[Mu]1)/(kr1 r1) BJp11, (\[Omega] \[Mu]1)/(
kr1 r1) BYp11, (m kz)/(\[Alpha]r2^2 r1^2) HA21, (
I \[Omega] \[Mu]2)/(\[Alpha]r2 r1) HAp21}
} /. opt;
equ = Det[M];
Return[equ];
]

Do[
n0 = Sqrt[
1 + (0.68671749* lambda^2)/(lambda^2 -
0.072675189^2) + (0.43481505* lambda^2)/(lambda^2 -
0.11514351^2) + (0.89656582* lambda^2)/(lambda^2 -
10.002398^2)];
n1 = Sqrt[
1 + (0.6961663* lambda^2)/(lambda^2 - 0.0684043^2) + (0.4079426*
lambda^2)/(lambda^2 - 0.1162414^2) + (0.8974794*
lambda^2)/(lambda^2 - 9.896161^2)];
n2 = 1.0;
opt = {m -> 1, \[Omega] -> 2 Pi/lambda, r0 -> 4.1,
r1 -> 62.5, \[Epsilon]0 -> n0^2, \[Mu]0 -> 1., \[Epsilon]1 ->
n1^2, \[Mu]1 -> 1., \[Epsilon]2 -> n2^2, \[Mu]2 -> 1., kz -> y};
sol2 = FindRoot[f[y*(2 Pi/lambda), opt] == 0, {y, 1.4435}];
Print[FullForm[Re[sol2[[1, 2]]]]],
{lambda, 1.54, 1.56, 0.001}]

以上是部分程式碼，從1.54~1.56每次加0.001在1.4435附近求解，會求出21組解。現在想要
每算完一組解就以這個解當新的值繼續往下算；例如說1.54算出1.4480196760169795，
1.541就以這個值往下算，之後的以此類推。請問這樣程式碼要怎改比較好呢?謝謝!

[軟體熟悉度]:
低(1~3個月)

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◆ From: 140.115.41.12
推 chungyuandye:沒有f的定義  07/30 10:54
※ 編輯: af2486 來自: 111.252.204.176 (07/30 23:00)
→ af2486:已附上完整程式碼  07/30 23:02