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fixed dependencies

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nuknal
2024-10-24 15:46:01 +08:00
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vendor/github.com/nuknal/goNum/InterpSpline11.go generated vendored Normal file
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// InterpSpline11
/*
------------------------------------------------------
作者 : Black Ghost
日期 : 2018-12-8
版本 : 0.0.0
------------------------------------------------------
用节点处的一阶导数表示的三次样条插值函数,
一阶导数边界条件
n+1个点, n个区间
理论：
区间[x(i-1), xi]上的三次样条函数表达为：
(x-xi)^2 * [hi+2(x-x(i-1))]
Si(x) = -----------------------------y(i-1) +
hi^3
(x-x(i-1))^2 * [hi+2(xi-x)]
-----------------------------yi +
hi^3
(x-xi)^2 * (x-x(i-1))
-----------------------m(i-1) +
hi^2
(x-x(i-1))^2 * (x-xi)
-----------------------mi
hi^2
令 lambdai = h(i+1)/(hi+h(i+1))
Mi = 1-lambdai = hi/(hi+h(i+1))
y(i+1)-yi yi-y(i-1)
fi = 3(Mi---------- + lambdai-----------)
h(i+1) hi
(i = 1,...,n-1)
则mi可由n-1阶线性方程组求得（利用LEs_Chasing）：
|2 M1 || m1 | | f1-l1*m0 |
| l2 2 M2 || m2 | = | f2 |
| ........ || ... | | ... |
| l(n-2) 2 M(n-2)||m(n-2)| | f(n-2) |
| l(n-1) 2 ||m(n-1)| |f(n-1)-M(n-1)mn|
参考 李信真, 车刚明, 欧阳洁, 等. 计算方法. 西北工业大学
出版社, 2000, pp 116-123.
------------------------------------------------------
输入 :
A 数据点矩阵，(n+1)x3，第一列xi；第二列yi；
第三列y'i，且y'i只需给出y'0和y'n
输出 :
B 插值方程系数结果矩阵，从前到后对应从0到3阶，4xn
err 解出标志：false-未解出或达到步数上限；
true-全部解出
------------------------------------------------------
*/
package goNum
import (
"math"
)
// InterpSpline11 用节点处的一阶导数表示的三次样条插值函数, 一阶导数边界条件
func InterpSpline11(A Matrix) (Matrix, bool) {
/*
用节点处的一阶导数表示的三次样条插值函数, 一阶导数边界条件
输入 :
A 数据点矩阵，(n+1)x3，第一列xi；第二列yi；
第三列y'i，且y'i只需给出y'0和y'n
输出 :
B 插值方程系数结果矩阵，从前到后对应从0到3阶，4xn
err 解出标志：false-未解出或达到步数上限；
true-全部解出
*/
var err bool = false
n := A.Rows - 1
sol := ZeroMatrix(4, n)
BA := ZeroMatrix(n-1, n-1) //对角占优的三对角矩阵
BB := ZeroMatrix(n-1, 1) //解向量
BC := ZeroMatrix(n-1, 1) //值向量
//1解插值函数的一阶导数mi
//1.0.1第一行
if true { //限制变量使用范围
h1 := A.GetFromMatrix(1, 0) - A.GetFromMatrix(0, 0)
h2 := A.GetFromMatrix(2, 0) - A.GetFromMatrix(1, 0)
y0 := A.GetFromMatrix(0, 1)
y1 := A.GetFromMatrix(1, 1)
y2 := A.GetFromMatrix(2, 1)
l1 := h2 / (h1 + h2)
M1 := 1.0 - l1
f1 := 3.0 * (M1*(y2-y1)/h2 + l1*(y1-y0)/h1)
BA.SetMatrix(0, 0, 2.0)
BA.SetMatrix(0, 1, M1)
BC.Data[0] = f1 - l1*A.GetFromMatrix(0, 2)
}
//1.0.2其它行
for i := 2; i < n-1; i++ {
yi_1 := A.GetFromMatrix(i-1, 0)
yi := A.GetFromMatrix(i, 0)
yi1 := A.GetFromMatrix(i+1, 0)
hi := A.GetFromMatrix(i, 0) - A.GetFromMatrix(i-1, 0)
hi1 := A.GetFromMatrix(i+1, 0) - A.GetFromMatrix(i, 0)
lambdai := hi1 / (hi + hi1)
Mi := 1.0 - lambdai
fi := 3.0 * (Mi*(yi1-yi)/hi1 + lambdai*(yi-yi_1)/hi)
//赋予BA
BA.SetMatrix(i-1, i-2, lambdai)
BA.SetMatrix(i-1, i-1, 2.0)
BA.SetMatrix(i-1, i, Mi)
BC.Data[i-1] = fi
}
//1.0.3最后一行
if true { //i=n-1
hn_1 := A.GetFromMatrix(n-1, 0) - A.GetFromMatrix(n-2, 0)
hn := A.GetFromMatrix(n, 0) - A.GetFromMatrix(n-1, 0)
yn_2 := A.GetFromMatrix(n-2, 1)
yn_1 := A.GetFromMatrix(n-1, 1)
yn := A.GetFromMatrix(n, 1)
lambdan_1 := hn / (hn_1 + hn)
Mn_1 := 1.0 - lambdan_1
fn_1 := 3.0 * (Mn_1*(yn-yn_1)/hn + lambdan_1*(yn_1-yn_2)/hn_1)
BA.SetMatrix(n-2, n-3, lambdan_1)
BA.SetMatrix(n-2, n-2, 2.0)
BC.Data[n-2] = fn_1 - Mn_1*A.GetFromMatrix(n, 2)
}
//1.1求解
soltemp, errtemp := LEs_Chasing(BA, BC)
if errtemp != true {
panic("Error in goNum.InterpSpline11: Solve Error with goNum.LEs_Chasing")
}
for i := 0; i < n-1; i++ {
BB.Data[i] = soltemp.Data[i]
}
//2求解Si(x)
S0 := ZeroMatrix(4, 1)
S1 := ZeroMatrix(4, 1)
S2 := ZeroMatrix(4, 1)
S3 := ZeroMatrix(4, 1)
for i := 1; i < n+1; i++ {
xi_1 := A.GetFromMatrix(i-1, 0)
xi := A.GetFromMatrix(i, 0)
yi_1 := A.GetFromMatrix(i-1, 1)
yi := A.GetFromMatrix(i, 1)
mi_1 := 0.0
mi := 0.0
if i == 1 {
mi_1 = A.GetFromMatrix(0, 2)
mi = BB.Data[i-1]
} else if i == n {
mi_1 = BB.Data[i-2]
mi = A.GetFromMatrix(n, 2)
} else {
mi_1 = BB.Data[i-2]
mi = BB.Data[i-1]
}
hi := xi - xi_1
temp0 := ZeroMatrix(4, 1)
temp1 := ZeroMatrix(4, 1)
//2.1 S0
temp0.Data[2] = 1.0
temp0.Data[1] = -2.0 * xi
temp0.Data[0] = xi * xi
for j := 3; j > 0; j-- {
temp0.Data[j] = 2.0 * temp0.Data[j-1]
temp1.Data[j-1] = (hi - 2.0*xi_1) * temp0.Data[j-1]
S0.Data[j] = (temp0.Data[j] + temp1.Data[j]) * yi_1 / math.Pow(hi, 3.0)
}
S0.Data[0] = temp1.Data[0] * yi_1 / math.Pow(hi, 3.0)
//2.1 S1
temp0 = ZeroMatrix(4, 1)
temp1 = ZeroMatrix(4, 1)
temp0.Data[2] = 1.0
temp0.Data[1] = -2.0 * xi_1
temp0.Data[0] = xi_1 * xi_1
for j := 3; j > 0; j-- {
temp0.Data[j] = -2.0 * temp0.Data[j-1]
temp1.Data[j-1] = (hi + 2.0*xi) * temp0.Data[j-1]
S1.Data[j] = (temp0.Data[j] + temp1.Data[j]) * yi / math.Pow(hi, 3.0)
}
S1.Data[0] = temp1.Data[0] * yi / math.Pow(hi, 3.0)
//2.2 S2
temp0 = ZeroMatrix(4, 1)
temp1 = ZeroMatrix(4, 1)
temp0.Data[2] = 1.0
temp0.Data[1] = -2.0 * xi
temp0.Data[0] = xi * xi
for j := 3; j > 0; j-- {
temp0.Data[j] = temp0.Data[j-1]
temp1.Data[j-1] = -1.0 * xi_1 * temp0.Data[j-1]
S2.Data[j] = (temp0.Data[j] + temp1.Data[j]) * mi_1 / math.Pow(hi, 2.0)
}
S2.Data[0] = temp1.Data[0] * mi_1 / math.Pow(hi, 2.0)
//2.3 S3
temp0 = ZeroMatrix(4, 1)
temp1 = ZeroMatrix(4, 1)
temp0.Data[2] = 1.0
temp0.Data[1] = -2.0 * xi_1
temp0.Data[0] = xi_1 * xi_1
for j := 3; j > 0; j-- {
temp0.Data[j] = temp0.Data[j-1]
temp1.Data[j-1] = -1.0 * xi * temp0.Data[j-1]
S3.Data[j] = (temp0.Data[j] + temp1.Data[j]) * mi / math.Pow(hi, 2.0)
}
S3.Data[0] = temp1.Data[0] * mi / math.Pow(hi, 2.0)
//2.4 Si(x)
for j := 0; j < 4; j++ {
sol.SetMatrix(j, i-1, S0.Data[j]+S1.Data[j]+S2.Data[j]+S3.Data[j])
}
}
err = true
return sol, err
}