fixed dependencies
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96
vendor/github.com/nuknal/goNum/PDEDiffParabolicE.go
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96
vendor/github.com/nuknal/goNum/PDEDiffParabolicE.go
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// PDEDiffParabolicE
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-12-14
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版本 : 0.0.0
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------------------------------------------------------
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求解抛物型偏微分方程的差分解法(显式)
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理论:
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对于抛物型偏微分方程:
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du d^2u
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---- = A ------ + B
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dt dx^2
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u(x, 0) = p(x)
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u(0, t) = u1(t), u(L, t) = u2(t)
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0 < x < L, 0 < t < T
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则古典显式差分格式为,x分为m等份,t分为n等份
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u_(i,j+1) = lu_(i-1,j) + (1-2l)u_(i,j) + lu_(i+1,j) + B*tau
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A*tau
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l = -------
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h^2
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u_(i, 0) = p(ih), i=1,2,..,m-1
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u_(0, j) = u1(j*tau), u_(m, j) = u2(j, tau), j=0,1,...,n
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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 214-215.
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------------------------------------------------------
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输入 :
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funp, funu1, funu2 边界函数
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x0 求解范围,2x2
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A, B 常系数
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m, n 网格数量
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输出 :
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sol 解矩阵
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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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// PDEDiffParabolicE 求解抛物型偏微分方程的差分解法(显式)
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func PDEDiffParabolicE(funp, funu1, funu2 func(float64) float64, x0 Matrix,
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A, B float64, m, n int) (Matrix, bool) {
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/*
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求解抛物型偏微分方程的差分解法(显式)
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输入 :
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funp, funu1, funu2 边界函数
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x0 求解范围,2x2
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A, B 常系数
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m, n 网格数量
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输出 :
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sol 解矩阵
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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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//判断网格数量
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if (m < 1) || (n < 1) {
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panic("Error in goNum.PDEDiffParabolicE: Grid numbers error")
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}
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var err bool = false
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sol := ZeroMatrix(m+1, n+1)
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hx := (x0.GetFromMatrix(1, 0) - x0.GetFromMatrix(0, 0)) / float64(m) //x方向步长
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ht := (x0.GetFromMatrix(1, 1) - x0.GetFromMatrix(0, 1)) / float64(n) //t方向步长
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//1. 计算t第零层上的值u_(i,0) i=0,1,...,m
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for i := 0; i < m+1; i++ {
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sol.SetMatrix(i, 0, funp(x0.GetFromMatrix(0, 0)+float64(i)*hx))
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}
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//2. 计算左右边界上的节点u_(0,j)和u_(m,j) j=1,2,...,n
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for j := 1; j < n+1; j++ {
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sol.SetMatrix(0, j, funu1(x0.GetFromMatrix(0, 1)+float64(j)*ht)) //左边界
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sol.SetMatrix(m, j, funu2(x0.GetFromMatrix(0, 1)+float64(j)*ht)) //右边界
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}
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//内部节点循环求解
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l := A * ht / (hx * hx)
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//稳定性判断
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if (l <= 0) || (l > 0.5) {
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panic("Error in goNum.PDEDiffParabolicS: lambda less than or equal to zero, or greater than 0.5")
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}
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for j := 1; j < n+1; j++ { //层循环, ti
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for i := 1; i < m; i++ { //列循环, xi
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uij := l * sol.GetFromMatrix(i-1, j-1)
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uij += (1 - 2.0*l) * sol.GetFromMatrix(i, j-1)
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uij += l * sol.GetFromMatrix(i+1, j-1)
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sol.SetMatrix(i, j, uij+B*ht)
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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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