Vector-valued ParametricNDSolve, solving for a combinationVector ParametricNDSolve and FindRoot interactionIs it possible to give the closed-form of the stiffness matrix of triangular prism element?Frequency domain Maxwell equations with PML boundary conditionsJacobian of ParametricNDSolve and FindRoot for the Three Body ProblemGiven a vector field construct a corresponding stream functionNumerically solving two coupled nonlinear PDEs
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Vector-valued ParametricNDSolve, solving for a combination
Vector ParametricNDSolve and FindRoot interactionIs it possible to give the closed-form of the stiffness matrix of triangular prism element?Frequency domain Maxwell equations with PML boundary conditionsJacobian of ParametricNDSolve and FindRoot for the Three Body ProblemGiven a vector field construct a corresponding stream functionNumerically solving two coupled nonlinear PDEs
$begingroup$
I am integrating two vector differential equations using ParametricNDSolve, one for $mathbfY$ and one for $mathbfZ$, and then I'm interested in a combination of the two of them. ParametricNDSolve allows for specifying the output to just be a function of the two, for instance $mathbfY cdot mathbfZ$, but the combination I want is (for the case of sixth order vector equations):
$$psi = y_1 z_6 -y_2 z_5 + y_3 z_4 + y_4 z_3 - y_5 z_2 +y_6 z_1$$
Is there a way to get ParametricNDSolve to give me this combination directly?
Example code:
A = 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, q, 0, -Sin[x], 0;
Q[x_, q_] = 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, -Sin[x],0, 0, 1, 0,
0, 0, 0, 0, 1, 0, -q, 0, 0, -Sin[x], 0, 1, 0, -q, 0, 0, 0, 0;
n = Length[Q[x, q]];
cA = With[code = N@A, Compile[x, _Real, q, _Real, code]];
F[x_?NumericQ, q_?NumericQ, Y_?VectorQ] :=
(Q[x, q] - DiagonalMatrix[ConstantArray[1, n]]).Y;
xa = -4; xb = 4; xm = 0;
k = -((-xb + xm)/(xa - xm)); c = -((-xa xm + xb xm)/(xa - xm));
Y0 = 0, 0, 0, 0, -1, 0;
Z0 = 0, 1, 0, 0, 0, 0;
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c , q, Z[x]], Z[xa] == Z0, Y, Z, x, xa, xm, q];
I'm interested in finding where $psi=0$, which I can do with this construction, but is there a more direct way to get it out of the ParametricNDSolveValue?
psi[q_?NumericQ] := (Y[1] Z[6] - Y[2] Z[5] + Y[3] Z[4] + Y[4] Z[3] -
Y[5] Z[2] + Y[6] Z[1]) /. Y[a_] :> YZsol[q][[1]][xm][[a]],
Z[a_] :> YZsol[q][[2]][xm][[a]];
FindRoot[psi[q], q, 3]
Edit:
Carl's answer does exactly what I need, but I see a strange timing effect, where after I use FindRoot it takes 6 times longer to do the same calculation:
Clear[YZsol2]; YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0, Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
(* 0.259752, Null *)
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* 1.69719, Null *)
differential-equations
asked May 31 at 12:22
KraZugKraZug
3,71421131
$endgroup$
add a comment |
$begingroup$
I am integrating two vector differential equations using ParametricNDSolve, one for $mathbfY$ and one for $mathbfZ$, and then I'm interested in a combination of the two of them. ParametricNDSolve allows for specifying the output to just be a function of the two, for instance $mathbfY cdot mathbfZ$, but the combination I want is (for the case of sixth order vector equations):
$$psi = y_1 z_6 -y_2 z_5 + y_3 z_4 + y_4 z_3 - y_5 z_2 +y_6 z_1$$
Is there a way to get ParametricNDSolve to give me this combination directly?
Example code:
A = 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, q, 0, -Sin[x], 0;
Q[x_, q_] = 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, -Sin[x],0, 0, 1, 0,
0, 0, 0, 0, 1, 0, -q, 0, 0, -Sin[x], 0, 1, 0, -q, 0, 0, 0, 0;
n = Length[Q[x, q]];
cA = With[code = N@A, Compile[x, _Real, q, _Real, code]];
F[x_?NumericQ, q_?NumericQ, Y_?VectorQ] :=
(Q[x, q] - DiagonalMatrix[ConstantArray[1, n]]).Y;
xa = -4; xb = 4; xm = 0;
k = -((-xb + xm)/(xa - xm)); c = -((-xa xm + xb xm)/(xa - xm));
Y0 = 0, 0, 0, 0, -1, 0;
Z0 = 0, 1, 0, 0, 0, 0;
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c , q, Z[x]], Z[xa] == Z0, Y, Z, x, xa, xm, q];
I'm interested in finding where $psi=0$, which I can do with this construction, but is there a more direct way to get it out of the ParametricNDSolveValue?
psi[q_?NumericQ] := (Y[1] Z[6] - Y[2] Z[5] + Y[3] Z[4] + Y[4] Z[3] -
Y[5] Z[2] + Y[6] Z[1]) /. Y[a_] :> YZsol[q][[1]][xm][[a]],
Z[a_] :> YZsol[q][[2]][xm][[a]];
FindRoot[psi[q], q, 3]
Edit:
Carl's answer does exactly what I need, but I see a strange timing effect, where after I use FindRoot it takes 6 times longer to do the same calculation:
Clear[YZsol2]; YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0, Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
(* 0.259752, Null *)
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* 1.69719, Null *)
differential-equations
asked May 31 at 12:22
KraZugKraZug
3,71421131
$endgroup$
add a comment |
$begingroup$
I am integrating two vector differential equations using ParametricNDSolve, one for $mathbfY$ and one for $mathbfZ$, and then I'm interested in a combination of the two of them. ParametricNDSolve allows for specifying the output to just be a function of the two, for instance $mathbfY cdot mathbfZ$, but the combination I want is (for the case of sixth order vector equations):
$$psi = y_1 z_6 -y_2 z_5 + y_3 z_4 + y_4 z_3 - y_5 z_2 +y_6 z_1$$
Is there a way to get ParametricNDSolve to give me this combination directly?
Example code:
A = 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, q, 0, -Sin[x], 0;
Q[x_, q_] = 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, -Sin[x],0, 0, 1, 0,
0, 0, 0, 0, 1, 0, -q, 0, 0, -Sin[x], 0, 1, 0, -q, 0, 0, 0, 0;
n = Length[Q[x, q]];
cA = With[code = N@A, Compile[x, _Real, q, _Real, code]];
F[x_?NumericQ, q_?NumericQ, Y_?VectorQ] :=
(Q[x, q] - DiagonalMatrix[ConstantArray[1, n]]).Y;
xa = -4; xb = 4; xm = 0;
k = -((-xb + xm)/(xa - xm)); c = -((-xa xm + xb xm)/(xa - xm));
Y0 = 0, 0, 0, 0, -1, 0;
Z0 = 0, 1, 0, 0, 0, 0;
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c , q, Z[x]], Z[xa] == Z0, Y, Z, x, xa, xm, q];
I'm interested in finding where $psi=0$, which I can do with this construction, but is there a more direct way to get it out of the ParametricNDSolveValue?
psi[q_?NumericQ] := (Y[1] Z[6] - Y[2] Z[5] + Y[3] Z[4] + Y[4] Z[3] -
Y[5] Z[2] + Y[6] Z[1]) /. Y[a_] :> YZsol[q][[1]][xm][[a]],
Z[a_] :> YZsol[q][[2]][xm][[a]];
FindRoot[psi[q], q, 3]
Edit:
Carl's answer does exactly what I need, but I see a strange timing effect, where after I use FindRoot it takes 6 times longer to do the same calculation:
Clear[YZsol2]; YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0, Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
(* 0.259752, Null *)
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* 1.69719, Null *)
differential-equations
asked May 31 at 12:22
KraZugKraZug
3,71421131
$endgroup$
I am integrating two vector differential equations using ParametricNDSolve, one for $mathbfY$ and one for $mathbfZ$, and then I'm interested in a combination of the two of them. ParametricNDSolve allows for specifying the output to just be a function of the two, for instance $mathbfY cdot mathbfZ$, but the combination I want is (for the case of sixth order vector equations):
$$psi = y_1 z_6 -y_2 z_5 + y_3 z_4 + y_4 z_3 - y_5 z_2 +y_6 z_1$$
Is there a way to get ParametricNDSolve to give me this combination directly?
Example code:
A = 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, q, 0, -Sin[x], 0;
Q[x_, q_] = 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, -Sin[x],0, 0, 1, 0,
0, 0, 0, 0, 1, 0, -q, 0, 0, -Sin[x], 0, 1, 0, -q, 0, 0, 0, 0;
n = Length[Q[x, q]];
cA = With[code = N@A, Compile[x, _Real, q, _Real, code]];
F[x_?NumericQ, q_?NumericQ, Y_?VectorQ] :=
(Q[x, q] - DiagonalMatrix[ConstantArray[1, n]]).Y;
xa = -4; xb = 4; xm = 0;
k = -((-xb + xm)/(xa - xm)); c = -((-xa xm + xb xm)/(xa - xm));
Y0 = 0, 0, 0, 0, -1, 0;
Z0 = 0, 1, 0, 0, 0, 0;
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c , q, Z[x]], Z[xa] == Z0, Y, Z, x, xa, xm, q];
I'm interested in finding where $psi=0$, which I can do with this construction, but is there a more direct way to get it out of the ParametricNDSolveValue?
psi[q_?NumericQ] := (Y[1] Z[6] - Y[2] Z[5] + Y[3] Z[4] + Y[4] Z[3] -
Y[5] Z[2] + Y[6] Z[1]) /. Y[a_] :> YZsol[q][[1]][xm][[a]],
Z[a_] :> YZsol[q][[2]][xm][[a]];
FindRoot[psi[q], q, 3]
Edit:
Carl's answer does exactly what I need, but I see a strange timing effect, where after I use FindRoot it takes 6 times longer to do the same calculation:
Clear[YZsol2]; YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0, Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
(* 0.259752, Null *)
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* 1.69719, Null *)
differential-equations
differential-equations
asked May 31 at 12:22
KraZugKraZug
3,71421131
asked May 31 at 12:22
KraZugKraZug
3,71421131
edited May 31 at 19:42
KraZug
asked May 31 at 12:22
KraZugKraZug
3,71421131
asked May 31 at 12:22
KraZugKraZug
3,71421131
asked May 31 at 12:22
KraZugKraZug
3,71421131
3,71421131
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define a quadratic form matrix:
form =
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, -1, 0,
0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0,
0, -1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0
;
Then:
YZsol2 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
Y[xm] . form . Z[xm],
x,xa,xm,
q
];
will produce the desired function. For example:
YZsol2 /@ Range[0, 4]
psi /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Another possibility is to include your desired output as an additional equation (constraint):
YZsol3 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
U[x] == Y[x] . form . Z[x]
,
U[xm],
x,xa,xm,
q
];
Check:
YZsol3 /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Note that there is a bug as of June 2019 with using this inside a FindRoot, which makes it much slower unless we add Method -> "ParametricSensitivity" -> None as detailed in Vector ParametricNDSolve and FindRoot interaction).
edited Jun 7 at 7:20
KraZug
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
$endgroup$
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on theNDSolve.
$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow withFindRoot.
$endgroup$
– KraZug
Jun 7 at 7:35
add a comment |
$begingroup$
A hybrid of yours and @CarlWoll's approach seems to perform well after FindRoot.
Clear[YZsol, psi2]
form = Reverse@DiagonalMatrix@PadRight[, 6, 1, -1, 1];
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y, Z, x, xa,
xm, q];
psi2[q_?NumericQ] := (Y.form.Z) /. Y :> YZsol[q][[1]][xm],
Z :> YZsol[q][[2]][xm];
FindRoot[psi2[q], q, 3]
AbsoluteTiming[psi2 /@ Range[20]]
Clear[YZsol2];
YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* q -> 3.55393 *)
(* 0.279002, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 0.337883, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 2.20664, -148.785, -6240.83, -7841.05, 12815., 63393.6,
134180., 197976., 214748.,
139054., -71335.6, -446840., -998421., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22467*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
$endgroup$
$begingroup$
Er… what's the difference betweenYZsol2andYZsol3?
$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, ifFindRootis called first, the correspondingParametricFunctionwill be slower?
$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that usingParametricNDSolveto calculate the quantity of interest would be faster and perhaps more accurate.
$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And theFindRootis slower too. It is marginally faster to evaluate individual points until you useFindRoot, then it slows down considerably
$endgroup$
– KraZug
Jun 1 at 9:13
add a comment |
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2 Answers
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$begingroup$
Define a quadratic form matrix:
form =
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, -1, 0,
0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0,
0, -1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0
;
Then:
YZsol2 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
Y[xm] . form . Z[xm],
x,xa,xm,
q
];
will produce the desired function. For example:
YZsol2 /@ Range[0, 4]
psi /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Another possibility is to include your desired output as an additional equation (constraint):
YZsol3 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
U[x] == Y[x] . form . Z[x]
,
U[xm],
x,xa,xm,
q
];
Check:
YZsol3 /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Note that there is a bug as of June 2019 with using this inside a FindRoot, which makes it much slower unless we add Method -> "ParametricSensitivity" -> None as detailed in Vector ParametricNDSolve and FindRoot interaction).
edited Jun 7 at 7:20
KraZug
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
$endgroup$
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on theNDSolve.
$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow withFindRoot.
$endgroup$
– KraZug
Jun 7 at 7:35
add a comment |
$begingroup$
Define a quadratic form matrix:
form =
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, -1, 0,
0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0,
0, -1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0
;
Then:
YZsol2 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
Y[xm] . form . Z[xm],
x,xa,xm,
q
];
will produce the desired function. For example:
YZsol2 /@ Range[0, 4]
psi /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Another possibility is to include your desired output as an additional equation (constraint):
YZsol3 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
U[x] == Y[x] . form . Z[x]
,
U[xm],
x,xa,xm,
q
];
Check:
YZsol3 /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Note that there is a bug as of June 2019 with using this inside a FindRoot, which makes it much slower unless we add Method -> "ParametricSensitivity" -> None as detailed in Vector ParametricNDSolve and FindRoot interaction).
edited Jun 7 at 7:20
KraZug
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
$endgroup$
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on theNDSolve.
$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow withFindRoot.
$endgroup$
– KraZug
Jun 7 at 7:35
add a comment |
$begingroup$
Define a quadratic form matrix:
form =
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, -1, 0,
0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0,
0, -1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0
;
Then:
YZsol2 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
Y[xm] . form . Z[xm],
x,xa,xm,
q
];
will produce the desired function. For example:
YZsol2 /@ Range[0, 4]
psi /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Another possibility is to include your desired output as an additional equation (constraint):
YZsol3 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
U[x] == Y[x] . form . Z[x]
,
U[xm],
x,xa,xm,
q
];
Check:
YZsol3 /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Note that there is a bug as of June 2019 with using this inside a FindRoot, which makes it much slower unless we add Method -> "ParametricSensitivity" -> None as detailed in Vector ParametricNDSolve and FindRoot interaction).
edited Jun 7 at 7:20
KraZug
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
$endgroup$
Define a quadratic form matrix:
form =
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, -1, 0,
0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0,
0, -1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0
;
Then:
YZsol2 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
Y[xm] . form . Z[xm],
x,xa,xm,
q
];
will produce the desired function. For example:
YZsol2 /@ Range[0, 4]
psi /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Another possibility is to include your desired output as an additional equation (constraint):
YZsol3 = ParametricNDSolveValue[
Y'[x] == F[x,q,Y[x]], Y[xa]==Y0,
Z'[x] == 1/k F[k x+c,q,Z[x]], Z[xa]==Z0,
U[x] == Y[x] . form . Z[x]
,
U[xm],
x,xa,xm,
q
];
Check:
YZsol3 /@ Range[0, 4]
-24.1853, -148.784, -6240.84, -7841.06, 12815.
Note that there is a bug as of June 2019 with using this inside a FindRoot, which makes it much slower unless we add Method -> "ParametricSensitivity" -> None as detailed in Vector ParametricNDSolve and FindRoot interaction).
edited Jun 7 at 7:20
KraZug
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
edited Jun 7 at 7:20
KraZug
3,71421131
edited Jun 7 at 7:20
KraZug
3,71421131
edited Jun 7 at 7:20
KraZug
3,71421131
3,71421131
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
answered May 31 at 14:13
Carl WollCarl Woll
84.1k3107218
84.1k3107218
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on theNDSolve.
$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow withFindRoot.
$endgroup$
– KraZug
Jun 7 at 7:35
add a comment |
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on theNDSolve.
$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow withFindRoot.
$endgroup$
– KraZug
Jun 7 at 7:35
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Thanks, of course that is a way to write it as a vector/matrix equation
$endgroup$
– KraZug
May 31 at 14:31
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on the
NDSolve.$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
Huh, that works really well, except they are both much slower than the original formulation. The constraint version also fails to find the root (presumably because it is using a different internal method on the
NDSolve.$endgroup$
– KraZug
May 31 at 14:42
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow with
FindRoot.$endgroup$
– KraZug
Jun 7 at 7:35
$begingroup$
I edited your answer to include a link to another question where I show the workaround to not be slow with
FindRoot.$endgroup$
– KraZug
Jun 7 at 7:35
add a comment |
$begingroup$
A hybrid of yours and @CarlWoll's approach seems to perform well after FindRoot.
Clear[YZsol, psi2]
form = Reverse@DiagonalMatrix@PadRight[, 6, 1, -1, 1];
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y, Z, x, xa,
xm, q];
psi2[q_?NumericQ] := (Y.form.Z) /. Y :> YZsol[q][[1]][xm],
Z :> YZsol[q][[2]][xm];
FindRoot[psi2[q], q, 3]
AbsoluteTiming[psi2 /@ Range[20]]
Clear[YZsol2];
YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* q -> 3.55393 *)
(* 0.279002, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 0.337883, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 2.20664, -148.785, -6240.83, -7841.05, 12815., 63393.6,
134180., 197976., 214748.,
139054., -71335.6, -446840., -998421., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22467*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
$endgroup$
$begingroup$
Er… what's the difference betweenYZsol2andYZsol3?
$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, ifFindRootis called first, the correspondingParametricFunctionwill be slower?
$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that usingParametricNDSolveto calculate the quantity of interest would be faster and perhaps more accurate.
$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And theFindRootis slower too. It is marginally faster to evaluate individual points until you useFindRoot, then it slows down considerably
$endgroup$
– KraZug
Jun 1 at 9:13
add a comment |
$begingroup$
A hybrid of yours and @CarlWoll's approach seems to perform well after FindRoot.
Clear[YZsol, psi2]
form = Reverse@DiagonalMatrix@PadRight[, 6, 1, -1, 1];
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y, Z, x, xa,
xm, q];
psi2[q_?NumericQ] := (Y.form.Z) /. Y :> YZsol[q][[1]][xm],
Z :> YZsol[q][[2]][xm];
FindRoot[psi2[q], q, 3]
AbsoluteTiming[psi2 /@ Range[20]]
Clear[YZsol2];
YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* q -> 3.55393 *)
(* 0.279002, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 0.337883, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 2.20664, -148.785, -6240.83, -7841.05, 12815., 63393.6,
134180., 197976., 214748.,
139054., -71335.6, -446840., -998421., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22467*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
$endgroup$
$begingroup$
Er… what's the difference betweenYZsol2andYZsol3?
$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, ifFindRootis called first, the correspondingParametricFunctionwill be slower?
$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that usingParametricNDSolveto calculate the quantity of interest would be faster and perhaps more accurate.
$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And theFindRootis slower too. It is marginally faster to evaluate individual points until you useFindRoot, then it slows down considerably
$endgroup$
– KraZug
Jun 1 at 9:13
add a comment |
$begingroup$
A hybrid of yours and @CarlWoll's approach seems to perform well after FindRoot.
Clear[YZsol, psi2]
form = Reverse@DiagonalMatrix@PadRight[, 6, 1, -1, 1];
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y, Z, x, xa,
xm, q];
psi2[q_?NumericQ] := (Y.form.Z) /. Y :> YZsol[q][[1]][xm],
Z :> YZsol[q][[2]][xm];
FindRoot[psi2[q], q, 3]
AbsoluteTiming[psi2 /@ Range[20]]
Clear[YZsol2];
YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* q -> 3.55393 *)
(* 0.279002, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 0.337883, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 2.20664, -148.785, -6240.83, -7841.05, 12815., 63393.6,
134180., 197976., 214748.,
139054., -71335.6, -446840., -998421., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22467*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
$endgroup$
A hybrid of yours and @CarlWoll's approach seems to perform well after FindRoot.
Clear[YZsol, psi2]
form = Reverse@DiagonalMatrix@PadRight[, 6, 1, -1, 1];
YZsol = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0, Y, Z, x, xa,
xm, q];
psi2[q_?NumericQ] := (Y.form.Z) /. Y :> YZsol[q][[1]][xm],
Z :> YZsol[q][[2]][xm];
FindRoot[psi2[q], q, 3]
AbsoluteTiming[psi2 /@ Range[20]]
Clear[YZsol2];
YZsol2 = ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
AbsoluteTiming[YZsol2 /@ Range[20]]
Clear[YZsol3]; YZsol3 =
ParametricNDSolveValue[Y'[x] == F[x, q, Y[x]], Y[xa] == Y0,
Z'[x] == 1/k F[k x + c, q, Z[x]], Z[xa] == Z0,
Y[xm].form.Z[xm], x, xa, xm, q];
FindRoot[YZsol3[q], q, 3];
AbsoluteTiming[YZsol3 /@ Range[20]]
(* q -> 3.55393 *)
(* 0.279002, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 0.337883, -148.784, -6240.84, -7841.06, 12815., 63393.7,
134180., 197976., 214749.,
139054., -71335.6, -446840., -998422., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22468*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
(* 2.20664, -148.785, -6240.83, -7841.05, 12815., 63393.6,
134180., 197976., 214748.,
139054., -71335.6, -446840., -998421., -1.71155*10^6, -2.5422*10^6,
-3.41516*10^6, -4.22467*10^6, -4.83742*10^6, -5.09767*10^6,
-4.83432*10^6, -3.86961*10^6 *)
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
answered May 31 at 22:38
Tim LaskaTim Laska
1,1911210
1,1911210
$begingroup$
Er… what's the difference betweenYZsol2andYZsol3?
$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, ifFindRootis called first, the correspondingParametricFunctionwill be slower?
$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that usingParametricNDSolveto calculate the quantity of interest would be faster and perhaps more accurate.
$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And theFindRootis slower too. It is marginally faster to evaluate individual points until you useFindRoot, then it slows down considerably
$endgroup$
– KraZug
Jun 1 at 9:13
add a comment |
$begingroup$
Er… what's the difference betweenYZsol2andYZsol3?
$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, ifFindRootis called first, the correspondingParametricFunctionwill be slower?
$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that usingParametricNDSolveto calculate the quantity of interest would be faster and perhaps more accurate.
$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And theFindRootis slower too. It is marginally faster to evaluate individual points until you useFindRoot, then it slows down considerably
$endgroup$
– KraZug
Jun 1 at 9:13
$begingroup$
Er… what's the difference between
YZsol2 and YZsol3?$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
Er… what's the difference between
YZsol2 and YZsol3?$endgroup$
– xzczd
Jun 1 at 4:53
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@xzczd, there is no difference, just one has FindRoot applied first
$endgroup$
– KraZug
Jun 1 at 6:40
$begingroup$
@KraZug So, if
FindRoot is called first, the corresponding ParametricFunction will be slower?$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
@KraZug So, if
FindRoot is called first, the corresponding ParametricFunction will be slower?$endgroup$
– xzczd
Jun 1 at 9:09
$begingroup$
This is basically the same as my approach directly. I was thinking that using
ParametricNDSolve to calculate the quantity of interest would be faster and perhaps more accurate.$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
This is basically the same as my approach directly. I was thinking that using
ParametricNDSolve to calculate the quantity of interest would be faster and perhaps more accurate.$endgroup$
– KraZug
Jun 1 at 9:11
$begingroup$
@xzczd, exactly. And the
FindRoot is slower too. It is marginally faster to evaluate individual points until you use FindRoot , then it slows down considerably$endgroup$
– KraZug
Jun 1 at 9:13
$begingroup$
@xzczd, exactly. And the
FindRoot is slower too. It is marginally faster to evaluate individual points until you use FindRoot , then it slows down considerably$endgroup$
– KraZug
Jun 1 at 9:13
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