ParametricNDSolveValue
ParametricNDSolveValue
ParametricNDSolveValue[eqns,expr,{x,xmin,xmax},pars]
给出由常微分方程 eqns 的数值解决定的函数的 expr 值，其中自变量 x 位于从 xmin 到 xmax 的范围内，参数为 pars.
ParametricNDSolveValue[eqns,expr,{x,xmin,xmax},{y,ymin,ymax},pars]
在矩形区域上求解偏微分方程 eqns.
ParametricNDSolveValue[eqns,expr,{x,y}∈Ω,pars]
在区域 Ω 上求解偏微分方程 eqns.
ParametricNDSolveValue[eqns,expr,{t,tmin,tmax},{x,y}∈Ω,pars]
在区域 Ω 上求解时间相关偏微分方程 eqns.
更多信息和选项
- ParametricNDSolveValue 给出具有 ParametricFunction 对象的结果.
- {pspec1,pspec2,…} 的参数 pars 的规范可用于指定范围.
- pspeci 的可能形式为：
-
p p 值为 Reals 或 Complexes Element[p,Reals] p 值为 Reals Element[p,Complexes] p 值为 Complexes Element[p,{v1,…}] p 具有离散值 {v1,…} {p,pmin,pmax} p 值为 
- 通常 expr 会通过微分方程组的解间接与参数相关，但是可能显式与参数相关.
- 由此得出关于参数的 ParametricFunction 对象的导数会尽可能地组合使用符号和数值灵敏度方法进行计算.
- ParametricNDSolveValue 接受与 NDSolve 同样的选项和设置.
- NDSolve 和 ParametricNDSolveValue 通常通过多个不同的阶段求解微分方程组. 当 Method->{s1->m1,s2->m2,…} 时，阶段 si 由方法 mi 处理. 基于所要解决的问题，实际使用的阶段和顺序是由 NDSolve 决定的.
- 可能的解阶段与 NDSolve 是一样的，但还包括：
-
"ParametricCaching" 缓存计算的解 "ParametricSensitivity" 关于参数导数的计算
所有选项的列表
范例
打开所有单元 关闭所有单元基本范例 (3)
pfun = ParametricNDSolveValue[{y'[t] == a y[t], y[0] == 1}, y, {t, 0, 10}, {a}]y1 = pfun[1]
y1[10]Plot[Evaluate[Table[pfun[a][t], {a, -1, 1, .1}]], {t, 0, 1}, PlotRange -> All]
pfun = ParametricNDSolveValue[{f''[t] + a f[t] == 0, f[0] == 1, f'[0] == 0}, f[10], {t, 0, 10}, {a}]Plot[pfun[a], {a, 0, 2}]使用 FindRoot 函数求根：
FindRoot[pfun[a], {a, 1}]pfun = ParametricNDSolveValue[{f''[t] + a f[t] == 0, f[0] == b, f'[0] == 0}, f, {t, 0, 20}, {a, b}];Plot[Evaluate[(pfun[a, b][t] + {0, .1, -.1}D[pfun[a, b][t], a]) /. {a -> 1, b -> 1}], {t, 0, 20}, Filling -> {2 -> {3}}]Plot[Evaluate[(pfun[a, b][t] + {0, .1, -.1}D[pfun[a, b][t], b]) /. {a -> 1, b -> 1}], {t, 0, 20}, Filling -> {2 -> {3}}]范围 (6)
参数相关 (4)
ParametricNDSolveValue 返回一个 ParametricFunction 对象：
pfun = ParametricNDSolveValue[{y''[t] + y[t] == 3a Sin[y[t]], y[0] == y'[0] == 1}, y, {t, 0, 5}, {a}]
if0 = pfun[0]
Plot[if0[t], {t, 0, 5}]
Plot[Evaluate@Table[pfun[a][t], {a, -.5, .5, .05}], {t, 0, 5}, PlotRange -> All]pfun = ParametricNDSolveValue[{y''[t] + 2y[t] == Sin[.4t] + Cos[y[t]], y[0] == a, y'[0] == b}, y, {t, 0, 10}, {a, b}]
Plot[Evaluate@Table[pfun[a, 1][t], {a, -2, 2, .5}], {t, 0, 10}, PlotRange -> All]
Plot[Evaluate@Table[pfun[1, b][t], {b, -2, 2, .5}], {t, 0, 10}, PlotRange -> All]
eqns = {y''[t] + y[t] == 3a Sin[y[t]], y[0] == y'[0] == 1};
pfun = ParametricNDSolveValue[eqns, y[1] + y[2], {t, 0, 5}, {a}];Plot[pfun[a], {a, -2, 2}]
pfun = ParametricNDSolveValue[eqns, Underoverscript[∑, i = 1, 10]0.5 y[0.5 i], {t, 0, 5}, {a}];Plot[pfun[a], {a, -2, 2}]
pfun = ParametricNDSolveValue[eqns, Subsuperscript[∫, 0, 5]y[s]ⅆs, {t, 0, 5}, {a}];Plot[pfun[a], {a, -2, 2}]
pfun = ParametricNDSolveValue[eqns, y'[3], {t, 0, 5}, {a}];Plot[pfun[a], {a, -2, 2}]
pfun = ParametricNDSolveValue[eqns, y[a], {t, 0, 5}, {a}];Plot[pfun[a], {a, 0, 5}]pfun = ParametricNDSolveValue[{u''[x] == a u[x], DirichletCondition[u[x] == 1, x == 0], DirichletCondition[u[x] == b, x == 1]}, u, {x}∈Line[{{0}, {1}}], {a, b}]
Plot[Evaluate@Table[pfun[a, 1 / 2][t], {a, -2, 2, .5}], {t, 0, 10}, PlotRange -> All]参数灵敏度 (2)
pfun = ParametricNDSolveValue[{y''[t] + y[t] == 0, y[0] == a, y'[0] == 0}, y, {t, 0, 10}, {a}];
Show[Plot[Evaluate@Table[pfun[a][t], {a, 1, 3, .2}], {t, 0, 10}, PlotStyle -> Thin], Plot[pfun[2][t], {t, 0, 10}]]
Plot[pfun'[2][t], {t, 0, 10}]
Plot[Evaluate[pfun[2][t] + {-1, 0, 1}pfun'[2][t]], {t, 0, 10}, PlotStyle -> {Gray, {Blue}, Gray}, Filling -> {1 -> {3}}]pfun = ParametricNDSolveValue[{y''[t] + 2y[t] == Sin[t] + Cos[y[t]], y[0] == a, y'[0] == b}, y, {t, 0, 10}, {a, b}];ifun = pfun[1, 1]
ifuna = D[pfun[a, 1], a] /. a -> 1;Plot[Evaluate[ifun[t] + {-.5, 0, .5}ifuna[t]], {t, 0, 10}, PlotStyle -> {Gray, {Blue}, Gray}, Filling -> {1 -> {3}}]
ifunb = D[pfun[1, b], b] /. b -> 1;Plot[Evaluate[ifun[t] + {-.5, 0, .5}ifunb[t]], {t, 0, 10}, PlotStyle -> {Gray, {Blue}, Gray}, Filling -> {1 -> {3}}]推广和延伸 (2)
求各种 WorkingPrecision 值的 
且 
的解，并绘制误差：
err = ParametricNDSolveValue[{x'[t] == x[t], x[0] == 1}, Abs[x[1] - E], {t, 0, 1}, {wp}, Method -> "Extrapolation", WorkingPrecision -> wp ];ListLogPlot[Table[{wp, err[wp]}, {wp, 16, 60}]]考虑求出 NDSolveValue 无法直接求解的高度非线性问题的解. 设置边界条件、区域和依赖于参数 
的方程：
eqn[p_] := Inactive[Div][((x ^ 2 + 3) * D[u[x], x] ^ (p - 1)) Inactive[Grad][u[x], {x}], {x}] - Sin[x]
bc = DirichletCondition[u[x] == 0, True];
Ω = Line[{{0}, {1}}];NDSolveValue 无法求解：
NDSolveValue[{eqn[3] == 0, bc}, u, Element[{x}, Ω]]
Clear[seeding]
seeding[x_ ? NumericQ] := 0基于参数 
创建 ParametricNDSolveValue 函数：
psol = ParametricNDSolveValue[{eqn[p] == 0, bc}, u, Element[{x}, Ω], {p}, InitialSeeding -> {u[x] == seeding[x]}]
s1 = psol[1]
Clear[seeding]
seeding[x_ ? NumericQ] := s1[x]
s2 = psol[3]Plot[s2[x], {x, 0, 1}]选项 (2)
Method (2)
ParametricCaching (1)
pfun = ParametricNDSolveValue[{x''[t] + a x[t] == 0, x[0] == b, x'[0] == 0}, x, {t, 0, 1000}, {a, b}, MaxSteps -> ∞, Method -> {"ParametricCaching" -> None}];Do[Print[Module[{m = MemoryInUse[]}, AbsoluteTiming[{ByteCount[pfun[2, 2]], MemoryInUse[] - m}]]], {3}]
pfun = ParametricNDSolveValue[{x''[t] + a x[t] == 0, x[0] == b, x'[0] == 0}, x, {t, 0, 1000}, {a, b}, MaxSteps -> ∞];Do[Print[Module[{m = MemoryInUse[]}, AbsoluteTiming[{ByteCount[pfun[2, 2]], MemoryInUse[] - m}]]], {3}]
ParametricSensitivity (1)
pfun = ParametricNDSolveValue[{x'[t] == a x[t], x[0] == b}, x[1], {t, 0, 1}, {a, b}, Method -> {"ParametricSensitivity" -> None}];Plot3D[pfun[a, b], {a, 1, 2}, {b, 0, 1}]dfun = Derivative[1, 0][pfun];
dfun[1, 1]
应用 (15)
参数扫描 (8)
pfun = ParametricNDSolveValue[{y''[t] == -9.8bounce[t], y[0] == height, y'[0] == 0, bounce[0] == 1, WhenEvent[y[t] == 0, If[Abs[y'[t]] > 10 ^ -6, y'[t] -> -0.7 y'[t], {bounce[t], y'[t]} -> {0, 0}]]}, y[t], {t, 0, 4}, {height}, DiscreteVariables -> bounce];Plot[Evaluate[Table[pfun[height], {height, 1, 5, .2}]], {t, 0, 4}, PlotRange -> All]
pfun = ParametricNDSolveValue[{x''[t] - x'[t] + x[t] == 0, x[0] == 1, x'[0] == s}, x[10], {t, 0, 10}, {s}];
root = FindRoot[pfun[s] == 0, {s, 6}]NDSolve[{x''[t] - x'[t] + x[t] == 0, x[0] == 1, x'[0] == s} /. root, x, {t, 0, 10}];Plot[x[t] /. %, {t, 0, 10}]psol = ParametricNDSolveValue[{x''[t] - x'[t] + x[t] == 0, x[0] == 1, x'[0] == s}, x, {t, 0, 10}, {s}];Show[Plot[Evaluate@Table[Tooltip[psol[s][t], s], {s, 1, 1.8, 0.1}], {t, 0, 10}, PlotStyle -> Lighter[Gray, 0.5]],
Plot[psol[s][t] /. root, {t, 0, 10}, PlotStyle -> Orange]]求边界值问题 
, 
, 
的多个解. 首先考虑 
的多个可能的值：
psol = ParametricNDSolveValue[{y''[t] + Sin[y[t]] == 0, y[0] == 0, y'[0] == s}, y, {t, 0, 10}, {s}];Plot[Evaluate@Table[Tooltip[psol[s][t], s], {s, 0.3, 1.9, 0.1}], {t, 0, 10}]
pfun = ParametricNDSolveValue[{y''[t] + Sin[y[t]] == 0, y[0] == 0, y'[0] == s}, y[10], {t, 0, 10}, {s}];Plot[pfun[s], {s, 0, 2.1}]val = Table[FindRoot[pfun[s], {s, s0}], {s0, {0.9, 1.9, 2}}]Plot[Evaluate[psol[s][t] /. val], {t, 0, 10}]求典型谐波振荡器 
，其中 
的所有特征值 
和特征函数 
. 通过探索可能的参数值开始：
psol = ParametricNDSolveValue[{y''[x] == -λ^2 y[x], y[0] == 0, y'[0] == λ}, y, {x, 0, 1}, {λ}];Plot[Evaluate@Table[Tooltip[psol[λ][x], λ], {λ, π / 2, 11π / 2, π / 2}], {x, 0, 1}]
pfun = ParametricNDSolveValue[{y''[x] == -λ^2 y[x], y[0] == 0, y'[0] == λ}, y[1], {x, 0, 1}, {λ}];Plot[pfun[λ], {λ, π / 2, 11π / 2}]sol = Table[FindRoot[pfun[λ] == 0, {λ, λ0}], {λ0, {3, 6, 9, 12, 16}}]//QuietPlot[Evaluate[psol[λ][x] /. sol], {x, 0, 1}]求量子谐振子 
，
，![lim_(TemplateBox[{x}, Abs]->infty) y(x)=0](Files/ParametricNDSolveValue.zh/78.png "lim_(TemplateBox[{x}, Abs]->infty) y(x)=0"){kind=small}
的前三个特征函数. 开始于绘制 
的解，其中 
：
pfun = ParametricNDSolveValue[{-y''[x] + x ^ 2 y[x] == λ y[x], y[0] == 0, y'[0] == 1}, y, {x, -15, 15}, {λ}];Plot[Evaluate@Table[pfun[λ][x], {λ, 0, 13, 1}], {x, -5, 5}, PlotRange -> {-2, 2}]
Plot[pfun[λ][5]Exp[-5], {λ, 0, 13}, PlotRange -> {-7, 7}]Map[FindRoot[pfun[λ][5], {λ, #}]&, {2.8, 6.5, 10}]
Show[Plot[Evaluate@Table[pfun[λ][x], {λ, 0, 13, 1}], {x, -5, 5}, PlotRange -> {-1, 1}, PlotStyle -> Lighter[Gray, .5]], Plot[{pfun[3][x], pfun[7][x], pfun[11][x]}, {x, -5, 5}]]ψ[n_, x_] := (HermiteH[n, x] E^-(x^2/2)/Sqrt[2^n n! Sqrt[π]])Plot[{ψ[1, x], -ψ[3, x], ψ[5, x]}, {x, -5, 5}, PlotRange -> {-1, 1}]求 
，
的解具有最小弧长度时，
的值，其中 
到 
. 开始绘制 
的值为 0 到 1 的解：
pfun = ParametricNDSolveValue[{y''[t] + y[t] == 3a Sin[y[t]], y[0] == y'[0] == 1, α'[t] == Sqrt[1 + y[t] ^ 2], α[0] == 0}, {y[t]}, {t, 0, 10}, {a}];
alfun = ParametricNDSolveValue[{y''[t] + y[t] == 3a Sin[y[t]], y[0] == y'[0] == 1, α'[t] == Sqrt[1 + y[t] ^ 2], α[0] == 0}, α[10], {t, 0, 10}, {a}];Plot[Evaluate@Table[Tooltip[pfun[a], a], {a, 0, 1, .1}], {t, 0, 10}]
Plot[alfun[a], {a, 0, 1}, MaxRecursion -> 0, PlotPoints -> 100]
Show[Plot[Evaluate@Table[pfun[a], {a, 0, 1, .1}], {t, 0, 10}, PlotRange -> All, PlotPoints -> 500, PlotStyle -> Lighter[Gray, 0.3]], Plot[pfun[0], {t, 0, 10}, PlotStyle -> Thick]]
lmin = FindMinimum[alfun[a], {a, 0.725}, AccuracyGoal -> 7]Show[Plot[Evaluate@Table[pfun[a], {a, .65, .8, .005}], {t, 0, 10}, PlotRange -> All, PlotPoints -> 500, PlotStyle -> Lighter[Gray, 0.3]], Plot[pfun[a] /. lmin[[2]], {t, 0, 10}, PlotStyle -> Thick]]求解径向形式的 Klein-Gordon 方程，势能形式为 
：
sol = ParametricNDSolveValue[{Derivative[2][R][r] - ((1 - (UnitConvert[Quantity[1, "FineStructureConstant"]] E^-r Z + ϵ)^2) R[r]) == 0, R[0] == 0, R'[0] == 1}, R, {r, 0, 50}, {ϵ, Z}]
ContourPlot[sol[ϵ, Z][50] == 0, {Z, 50, 800}, {ϵ, -1, 1}, FrameLabel -> Automatic, PlotPoints -> 100]
Subscript[v, 1][t_] := (Sign[t - .01] + 1) / 2;components = {r iR[t] == Subscript[v, 2][t] - Subscript[v, 3][t], l iL'[t] == Subscript[v, 1][t] - Subscript[v, 2][t], iC[t] == c Subscript[v, 3]'[t]};
connections = {iV[t] - iL[t] == 0, iL[t] - iR[t] == 0, iR[t] - iC[t] == 0};
ic = {iR[0] == 0, iL[0] == 0, iC[0] == 0, Subscript[v, 2][0] == 0, Subscript[v, 3][0] == 0};pfun = ParametricNDSolveValue[{components, connections, ic}, Subscript[v, 2], {t, 0, .03}, {r, l, c}];step = Plot[Subscript[v, 1][t], {t, 0, .03}, Filling -> {1 -> 0}, Ticks -> {Range[0.01, 0.03, 0.01], Automatic}]Show[step, Plot[Evaluate@Table[pfun[r, 10^-2, 10^-4][t], {r, {5, 10, 20, 40}}], {t, 0, .03}], PlotRange -> All]Show[step, Plot[Evaluate@Table[pfun[10, 10^-2l, 10^-4][t], {l, {.25, 1, 2}}], {t, 0, .03}], PlotRange -> All]Show[step, Plot[Evaluate@Table[pfun[10, 10^-2, 10^-4c][t], {c, {.25, 1, 2}}], {t, 0, .03}], PlotRange -> All]参数灵敏度 (5)
pfun = ParametricNDSolveValue[{y''[t] + y[t] == a Sin[2t ] + Cos[2y[t]], y[0] == 1, y'[0] == 0}, y, {t, 0, 10}, {a}];Show[Plot[Evaluate@Table[pfun[a][t], {a, -.6, .6, .1}], {t, 0, 5}, PlotStyle -> Lighter[Gray, 0.3], PlotRange -> All], Plot[pfun[0][t], {t, 0, 5}]]Plot[Evaluate[pfun[0][t] + {-.5, 0, .5}pfun'[0][t]], {t, 0, 5}, PlotStyle -> {Gray, {Blue}, Gray}, Filling -> {1 -> {3}}]pfun = ParametricNDSolveValue[{θ''[t] + (9.8 / 10)Sin[θ[t]] == -(a / 10)ω ^ 2 Sin[ω t]Sin[θ[t]], θ[0] == 3.14158, θ'[0] == 0.01}, θ, {t, 0, 5}, {ω, a}];
if = pfun[50, 2];
ifa = D[pfun[ω, a], a] /. {ω -> 50, a -> 2};Plot[Evaluate[if[t] + {-.1, 0, .1}ifa[t]], {t, 0, 5}, PlotStyle -> {Gray, Blue, Gray}, Filling -> {1 -> {3}}]ndf = ParametricNDSolveValue[{Derivative[1][x][t] == -3 (x[t] - y[t]), Derivative[1][y][t] == -x[t] z[t] + a x[t] - y[t], Derivative[1][z][t] == x[t] y[t] - z[t], x[0] == z[0] == 0, y[0] == 1}, x[T], {t, 0, T}, {a, T}, MaxSteps -> ∞];Plot[Evaluate[RealExponent[D[ndf[a, 8], a]]], {a, 26, 27}, PlotRange -> All]
pfun = ParametricNDSolveValue[{D[u[t, x], t] == c^2 D[u[t, x], x, x], u[0, x] == Exp[-(a x) ^ 2], u[t, -10] == u[t, 10] == 0}, u, {t, 0, 5}, {x, -10, 10}, {a, c}];
ifun = pfun[1, 1];
ifda = D[pfun[a, 1], a] /. {a -> 1};
ifdc = D[pfun[1, c], c] /. {c -> 1};Panel@Grid[Join[{{"", "a", "c"}}, Table[{Row[{"t=", τ}], Plot[Evaluate[(ifun[τ, x] + .5{0, 1, -1}ifda[τ, x])], {x, -10, 10}, Filling -> {2 -> {3}}, PlotRange -> All], Plot[Evaluate[(ifun[τ, x] + .5{0, 1, -1}ifdc[τ, x])], {x, -10, 10}, Filling -> {2 -> {3}}, PlotRange -> All]}, {τ, {1, 3, 5}}]]]ColorSensitivityPlot3D[f_, fsense_, opts___] :=
Block[{fs = fsense}, Plot3D[f, {t, 0, 5}, {x, -10, 10}, Mesh -> None, ColorFunction -> Function[{t, x}, ColorData["LakeColors"][fs]], PlotPoints -> 60, ColorFunctionScaling -> False, opts]]{ColorSensitivityPlot3D[pfun[1, 1][t, x], 1.5Abs[ifda[t, x]], PlotLabel -> a, PlotRange -> All], ColorSensitivityPlot3D[pfun[1, 1][t, x], Abs[ifdc[t, x]], PlotLabel -> c, PlotRange -> All]}
pfun = ParametricNDSolveValue[{D[u[t, x], t, t] == c ^ 2 D[u[t, x], x, x], u[0, x] == Exp[-(a x) ^ 2], Derivative[1, 0][u][0, x] == 0, u[t, -10] == u[t, 10]}, u, {t, 0, 5}, {x, -10, 10}, {a, c}];
ifun = pfun[1, 1];
ifda = D[pfun[a, 1], a] /. {a -> 1};
ifdc = D[pfun[1, c], c] /. {c -> 1};Panel@Grid[Join[{{"", "a", "c"}}, Table[{StringJoin["t = ", ToString[τ]], Plot[Evaluate[(pfun[a, c][τ, x] + .5{0, 1, -1}D[pfun[a, c][τ, x], a]) /. {a -> 1, c -> 1}], {x, -10, 10}, Filling -> {2 -> {3}}, PlotRange -> All], Plot[Evaluate[(pfun[a, c][τ, x] + .5{0, 1, -1}D[pfun[a, c][τ, x], c]) /. {a -> 1, c -> 1}], {x, -10, 10}, Filling -> {2 -> {3}}, PlotRange -> All]}, {τ, {1, 3, 5}}]]]ColorSensitivityPlot3D[f_, fsense_, opts___] :=
Block[{fs = fsense}, Plot3D[f, {t, 0, 5}, {x, -10, 10}, Mesh -> None, ColorFunction -> Function[{t, x}, ColorData["LakeColors"][fs]], PlotPoints -> 60, ColorFunctionScaling -> False, opts]]{ColorSensitivityPlot3D[pfun[1, 1][t, x], Abs[ifda[t, x]], PlotLabel -> a, PlotRange -> All], ColorSensitivityPlot3D[pfun[1, 1][t, x], .25Abs[ifdc[t, x]], PlotLabel -> c, PlotRange -> All]}参数拟合 (2)
{sol, samples} = NDSolve[{y''[t] == -9.8, y[0] == 1, y'[0] == 40, WhenEvent[Mod[t, .5] == 0, Sow[{t, y[t]}]]}, y, {t, 0, 10}]//Reap;
noisysamples = Map[({#[[1]], #[[2]] + RandomReal[{-10, 10}]})&, First@samples];ListPlot[noisysamples]pfun = ParametricNDSolveValue[{y''[t] == -w y[t], y[0] == a, y'[0] == b}, y, {t, 0, 10}, {w, a, b}];fit = FindFit[noisysamples, pfun[w, a, b][t], {{w, .1}, {a, 1}, {b, 1}}, t]Plot[pfun[w, a, b][t] /. fit, {t, 0, 10}, Epilog -> {Point@noisysamples}]pfun = ParametricNDSolveValue[{y''[t] == -g, y[0] == p, y'[0] == v}, y, {t, 0, 10}, {g, p, v}];fit = FindFit[noisysamples, pfun[g, p, v][t], {{g, 5}, {p, .5}, {v, .5}}, t]Plot[pfun[g, p, v][t] /. fit, {t, 0, 10}, Epilog -> {Point@noisysamples}]time = Quantity[{0, 25, 34, 42, 54, 63, 80, 89, 103, 115, 131, 145, 158, 175, 195, 213, 225, 250, 274, 298, 315, 335, 353, 387, 411, 440, 475, 492, 520, 530, 550, 560, 570, 585, 600, 610, 620, 630, 640, 647, 660, 670, 680, 690, 700, 710, 720, 730, 740, 750, 760, 770, 778, 790, 800, 810, 820, 830, 840, 850, 860, 870, 880, 890, 900, 910, 920, 930, 940, 950, 960, 970, 980, 990, 1000, 1010, 1020, 1030, 1040, 1050, 1060, 1070, 1080, 1090, 1100, 1110, 1120, 1130, 1140, 1150, 1160, 1170, 1180, 1190, 1200, 1210, 1220, 1230, 1240, 1250, 1260, 1270, 1280, 1300, 1340, 1356, 1380, 1418, 1725, 1740, 1825, 1890, 1955, 2022, 2077, 2160, 2244, 2320, 2408, 2473, 2613}, "Seconds"];
temp = Quantity[{210, 204, 200, 198, 196, 194, 192, 190, 188, 186, 184, 182, 180, 178, 176, 174, 172, 170, 168, 166, 164, 162, 160, 158, 156, 154, 152, 150, 150, 148, 148, 147, 146, 145, 144, 144, 144, 143, 143, 142, 142, 141, 141, 140, 140, 139.5, 139, 139, 138, 138, 137, 137, 136, 136, 136, 135, 135, 134, 134, 133, 133, 133, 132.5, 132, 132, 131, 131, 131, 130, 130, 129.5, 129, 129, 128.5, 128, 128, 128, 127.5, 127, 127, 126, 126, 126, 125.5, 125, 125, 124.5, 124, 124, 123.5, 123, 123, 122.5, 122, 122, 122, 121.5, 121, 121, 121, 120.5, 120.5, 120, 120, 119, 118, 117, 116, 111, 110.5, 109, 108, 107, 106, 105, 104, 103, 102, 101, 100, 99}, "DegreesFahrenheit"];tempK = UnitConvert[temp, "Kelvins"];
Subscript[tempK, 0] = QuantityMagnitude[First[tempK]];
Subscript[tempK, ∞] = QuantityMagnitude[UnitConvert[Quantity[79, "DegreesFahrenheit"], "Kelvins"]];
tend = QuantityMagnitude[Last[time]];BoltzmannConstant = Quantity[1.3806504`*^-23 , "Joules" / "Kelvins"];
c = QuantityMagnitude[(2260 18/6.02 10^23 BoltzmannConstant)];pfun = ParametricNDSolveValue[{(Derivative[1][T][t] == -k1(T[t] - Subscript[tempK, ∞]) - k2 (E^-c / T[t] - E^-c / Subscript[tempK, ∞])), T[0] == Subscript[tempK, 0]}, T, {t, 0, tend}, {k1, k2}];fitdata = N@QuantityMagnitude[Transpose[{time, tempK}]];fit = FindFit[fitdata, pfun[k1, k2][t], {{k1, 0.0001}, {k2, 10000}}, t]Show[ ListPlot[fitdata], Plot[Evaluate[pfun[k1, k2][t] /. fit], {t, 0, tend}, PlotStyle -> Red]]属性和关系 (3)
使用 NDSolveValue 求解带有参数的微分方程：
ndsvfun[a_] := NDSolveValue[{y''[t] == -a y[t] + t Sin[y[t]], y[0] == 1, y'[0] == 1}, y, {t, 0, 100}];Table[First@AbsoluteTiming[ndsvfun[1][100]], {10}]ParametricNDSolve 缓存解并接着重新使用缓存的解的值：
pfun = ParametricNDSolveValue[{y''[t] == -a y[t] + t Sin[y[t]], y[0] == 1, y'[0] == 1}, y, {t, 0, 100}, {a}];Table[First@AbsoluteTiming[pfun[1][100]], {10}]DSolve 可以解析式求解带有参数的某些微分方程：
dsol = y[x] /. First@DSolve[{y'[x] + a y[x] == 1, y[0] == 1}, y[x], x]使用 ParametricNDSolveValue 数值求解范例：
ndsol = ParametricNDSolveValue[{y'[x] + a y[x] == 1, y[0] == 1}, y, {x, 0, 5}, {a}, MaxStepSize -> 0.01];Plot[Evaluate[{D[dsol, a], D[ndsol[a][x], a]} /. a -> 1], {x, 0, 5}]使用 SystemModelParametricSimulate 模拟更大的分层系统模型：
pf = SystemModelParametricSimulate[\!\(\*GraphicsBox[«8»]\), "Inertia2.w", 40, {"Resistor1.R", "SpringDamper1.d"}]sims = {pf[0.1, 0.25], pf[0.4, 0.3]};Plot[Evaluate@Through[sims[t]], {t, 0, 40}]可能存在的问题 (1)
用 WhenEvent、Reap 和 Sow 采样的解只试用对每个参数值的第一次调用：
pfun = ParametricNDSolveValue[{y''[t] == -a Cos[ y[t]], WhenEvent[Mod[t, 1], Sow[{t, y[t]}]], y[0] == 1, y'[0] == 1}, y, {t, 0, 10}, {a}]{ifun, {points}} = pfun[10]//ReapPlot[ifun[t], {t, 0, 10}, Epilog -> {PointSize[Medium], Red, Point[points]}]{ifun, points} = pfun[10]//ReapPlot[ifun[t], {t, 0, 10}, Epilog -> {PointSize[Medium], Red, Point[points]}]
Wolfram Research (2012)，ParametricNDSolveValue，Wolfram 语言函数，https://reference.wolfram.com/language/ref/ParametricNDSolveValue.html （更新于 2014 年）.
文本
Wolfram Research (2012)，ParametricNDSolveValue，Wolfram 语言函数，https://reference.wolfram.com/language/ref/ParametricNDSolveValue.html （更新于 2014 年）.
CMS
Wolfram 语言. 2012. "ParametricNDSolveValue." Wolfram 语言与系统参考资料中心. Wolfram Research. 最新版本 2014. https://reference.wolfram.com/language/ref/ParametricNDSolveValue.html.
APA
Wolfram 语言. (2012). ParametricNDSolveValue. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/ParametricNDSolveValue.html 年