200 lines
6.2 KiB
Go
200 lines
6.2 KiB
Go
// InterpSpline12
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-12-8
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版本 : 0.0.0
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------------------------------------------------------
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用节点处的一阶导数表示的三次样条插值函数,
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二阶导数边界条件
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n+1个点, n个区间
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理论：
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区间[x(i-1), xi]上的三次样条函数表达为：
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(x-xi)^2 * [hi+2(x-x(i-1))]
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Si(x) = -----------------------------y(i-1) +
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hi^3
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(x-x(i-1))^2 * [hi+2(xi-x)]
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-----------------------------yi +
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hi^3
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(x-xi)^2 * (x-x(i-1))
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-----------------------m(i-1) +
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hi^2
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(x-x(i-1))^2 * (x-xi)
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-----------------------mi
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hi^2
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令 lambdai = h(i+1)/(hi+h(i+1))
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Mi = 1-lambdai = hi/(hi+h(i+1))
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y(i+1)-yi yi-y(i-1)
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fi = 3(Mi---------- + lambdai-----------)
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h(i+1) hi
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(i = 1,...,n-1)
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则mi可由n+1阶线性方程组求得（利用LEs_Chasing）：
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|2 1 || m0 | | f0 |
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|l1 2 M1 || m1 | | f1 |
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| l2 2 M2 || m2 | = | f2 |
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| ........ || ... | | ... |
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| l(n-1) 2 M(n-1)||m(n-1)| |f(n-1)|
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| 1 2 || mn | | fn |
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参考 李信真, 车刚明, 欧阳洁, 等. 计算方法. 西北工业大学
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出版社, 2000, pp 116-123.
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------------------------------------------------------
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输入 :
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A 数据点矩阵，(n+1)x3，第一列xi；第二列yi；
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第三列y''i，且y''i只需给出y''0和y''n
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输出 :
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B 插值方程系数结果矩阵，从前到后对应从0到3阶，4xn
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err 解出标志：false-未解出或达到步数上限；
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true-全部解出
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------------------------------------------------------
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*/
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package goNum
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import "math"
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// InterpSpline12 用节点处的一阶导数表示的三次样条插值函数，二阶导数边界条件
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func InterpSpline12(A Matrix) (Matrix, bool) {
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/*
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用节点处的一阶导数表示的三次样条插值函数，二阶导数边界条件
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输入 :
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A 数据点矩阵，(n+1)x3，第一列xi；第二列yi；
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第三列y'i，且y'i只需给出y'0和y'n
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输出 :
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B 插值方程系数结果矩阵，从前到后对应从0到3阶，4xn
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err 解出标志：false-未解出或达到步数上限；
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true-全部解出
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*/
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var err bool = false
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n := A.Rows - 1
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sol := ZeroMatrix(4, n)
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BA := ZeroMatrix(n+1, n+1) //对角占优的三对角矩阵
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BB := ZeroMatrix(n+1, 1) //解向量
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BC := ZeroMatrix(n+1, 1) //值向量
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//1解插值函数的一阶导数mi
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//1.0.1第一行
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if true { //限制变量使用范围
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h1 := A.GetFromMatrix(1, 0) - A.GetFromMatrix(0, 0)
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y0 := A.GetFromMatrix(0, 1)
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y1 := A.GetFromMatrix(1, 1)
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f0 := 3.0*(y1-y0)/h1 - h1*A.GetFromMatrix(0, 2)/2.0
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BA.SetMatrix(0, 0, 2.0)
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BA.SetMatrix(0, 1, 1.0)
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BC.Data[0] = f0
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}
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//1.0.2其它行
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for i := 1; i < n; i++ {
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yi_1 := A.GetFromMatrix(i-1, 0)
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yi := A.GetFromMatrix(i, 0)
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yi1 := A.GetFromMatrix(i+1, 0)
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hi := A.GetFromMatrix(i, 0) - A.GetFromMatrix(i-1, 0)
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hi1 := A.GetFromMatrix(i+1, 0) - A.GetFromMatrix(i, 0)
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lambdai := hi1 / (hi + hi1)
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Mi := 1.0 - lambdai
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fi := 3.0 * (Mi*(yi1-yi)/hi1 + lambdai*(yi-yi_1)/hi)
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//赋予BA
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BA.SetMatrix(i, i-1, lambdai)
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BA.SetMatrix(i, i, 2.0)
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BA.SetMatrix(i, i+1, Mi)
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BC.Data[i] = fi
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}
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//1.0.3最后一行
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if true { //i=n
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hn := A.GetFromMatrix(n, 0) - A.GetFromMatrix(n-1, 0)
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yn_1 := A.GetFromMatrix(n-1, 1)
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yn := A.GetFromMatrix(n, 1)
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fn := 3.0*(yn-yn_1)/hn + hn*A.GetFromMatrix(n, 2)/2.0
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BA.SetMatrix(n, n-1, 1.0)
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BA.SetMatrix(n, n, 2.0)
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BC.Data[n] = fn
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}
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//1.1求解
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soltemp, errtemp := LEs_Chasing(BA, BC)
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if errtemp != true {
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panic("Error in goNum.InterpSpline12: Solve Error with goNum.LEs_Chasing")
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}
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for i := 0; i < n+1; i++ {
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BB.Data[i] = soltemp.Data[i]
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}
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//2求解Si(x)
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S0 := ZeroMatrix(4, 1)
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S1 := ZeroMatrix(4, 1)
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S2 := ZeroMatrix(4, 1)
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S3 := ZeroMatrix(4, 1)
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for i := 1; i < n+1; i++ {
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xi_1 := A.GetFromMatrix(i-1, 0)
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xi := A.GetFromMatrix(i, 0)
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yi_1 := A.GetFromMatrix(i-1, 1)
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yi := A.GetFromMatrix(i, 1)
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mi_1 := BB.Data[i-1]
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mi := BB.Data[i]
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hi := xi - xi_1
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temp0 := ZeroMatrix(4, 1)
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temp1 := ZeroMatrix(4, 1)
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//2.1 S0
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temp0.Data[2] = 1.0
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temp0.Data[1] = -2.0 * xi
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temp0.Data[0] = xi * xi
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for j := 3; j > 0; j-- {
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temp0.Data[j] = 2.0 * temp0.Data[j-1]
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temp1.Data[j-1] = (hi - 2.0*xi_1) * temp0.Data[j-1]
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S0.Data[j] = (temp0.Data[j] + temp1.Data[j]) * yi_1 / math.Pow(hi, 3.0)
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}
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S0.Data[0] = temp1.Data[0] * yi_1 / math.Pow(hi, 3.0)
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//2.1 S1
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temp0 = ZeroMatrix(4, 1)
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temp1 = ZeroMatrix(4, 1)
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temp0.Data[2] = 1.0
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temp0.Data[1] = -2.0 * xi_1
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temp0.Data[0] = xi_1 * xi_1
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for j := 3; j > 0; j-- {
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temp0.Data[j] = -2.0 * temp0.Data[j-1]
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temp1.Data[j-1] = (hi + 2.0*xi) * temp0.Data[j-1]
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S1.Data[j] = (temp0.Data[j] + temp1.Data[j]) * yi / math.Pow(hi, 3.0)
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}
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S1.Data[0] = temp1.Data[0] * yi / math.Pow(hi, 3.0)
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//2.2 S2
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temp0 = ZeroMatrix(4, 1)
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temp1 = ZeroMatrix(4, 1)
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temp0.Data[2] = 1.0
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temp0.Data[1] = -2.0 * xi
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temp0.Data[0] = xi * xi
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for j := 3; j > 0; j-- {
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temp0.Data[j] = temp0.Data[j-1]
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temp1.Data[j-1] = -1.0 * xi_1 * temp0.Data[j-1]
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S2.Data[j] = (temp0.Data[j] + temp1.Data[j]) * mi_1 / math.Pow(hi, 2.0)
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}
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S2.Data[0] = temp1.Data[0] * mi_1 / math.Pow(hi, 2.0)
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//2.3 S3
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temp0 = ZeroMatrix(4, 1)
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temp1 = ZeroMatrix(4, 1)
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temp0.Data[2] = 1.0
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temp0.Data[1] = -2.0 * xi_1
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temp0.Data[0] = xi_1 * xi_1
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for j := 3; j > 0; j-- {
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temp0.Data[j] = temp0.Data[j-1]
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temp1.Data[j-1] = -1.0 * xi * temp0.Data[j-1]
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S3.Data[j] = (temp0.Data[j] + temp1.Data[j]) * mi / math.Pow(hi, 2.0)
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}
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S3.Data[0] = temp1.Data[0] * mi / math.Pow(hi, 2.0)
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//2.4 Si(x)
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for j := 0; j < 4; j++ {
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sol.SetMatrix(j, i-1, S0.Data[j]+S1.Data[j]+S2.Data[j]+S3.Data[j])
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}
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}
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err = true
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return sol, err
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}
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