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Solving system of ODEs with extra parameter
Transforming ParametricFunction in expression depending on the parameterSolving a system of ODEs with the Runge-Kutta methodTips for efficiently solving large system coupled (nonlinear) ODEsHow do I pull from a data list for parameter values in a system of ODEs and then solve and plot?Solving PDEs with complicated boundary conditionsSolving Differential Algebraic Equations as BVPSolving a system of coupled Nonlinear ODEs using numeric and get `ndnum` errorIssues to modelize a system of differential equation (NDSolve)Solving PDEs over a region in different co-ordinate system?Error when solving 't Hooft-Polyakov radial equations using NDSolveHow to apply NIntegrate three times
$begingroup$
I would like to solve a $2times 2$ system of the form
$$fracddthetaT=TA,quad T(0)=Id$$
where $theta$ is real and $A$ is of the form
$$A=beginpmatrix 0 & frace^-i thetalambda\ frac136e^-ithetaleft(9lambda + 2(lambda-1)^2 (6costheta + cos2theta + 6)right) & 0endpmatrix,$$
with $lambda$ a free parameter in the unit circle.
In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
sys = T'[θ] == T[θ].A[θ];
The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried
NSol = NDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11[θ], T12[θ], T21[θ], T22[θ],
θ,
θ, 0, 2 Pi
];
which gives me the output
NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."
I have also tried
Nsol2 = ParametricNDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
which gives me as output $T_11,dots,T_22$ as ParametricFunctions depending on each other and on $lambda$.
I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.
EDIT
Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.
differential-equations numerical-integration
$endgroup$
add a comment |
$begingroup$
I would like to solve a $2times 2$ system of the form
$$fracddthetaT=TA,quad T(0)=Id$$
where $theta$ is real and $A$ is of the form
$$A=beginpmatrix 0 & frace^-i thetalambda\ frac136e^-ithetaleft(9lambda + 2(lambda-1)^2 (6costheta + cos2theta + 6)right) & 0endpmatrix,$$
with $lambda$ a free parameter in the unit circle.
In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
sys = T'[θ] == T[θ].A[θ];
The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried
NSol = NDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11[θ], T12[θ], T21[θ], T22[θ],
θ,
θ, 0, 2 Pi
];
which gives me the output
NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."
I have also tried
Nsol2 = ParametricNDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
which gives me as output $T_11,dots,T_22$ as ParametricFunctions depending on each other and on $lambda$.
I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.
EDIT
Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.
differential-equations numerical-integration
$endgroup$
add a comment |
$begingroup$
I would like to solve a $2times 2$ system of the form
$$fracddthetaT=TA,quad T(0)=Id$$
where $theta$ is real and $A$ is of the form
$$A=beginpmatrix 0 & frace^-i thetalambda\ frac136e^-ithetaleft(9lambda + 2(lambda-1)^2 (6costheta + cos2theta + 6)right) & 0endpmatrix,$$
with $lambda$ a free parameter in the unit circle.
In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
sys = T'[θ] == T[θ].A[θ];
The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried
NSol = NDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11[θ], T12[θ], T21[θ], T22[θ],
θ,
θ, 0, 2 Pi
];
which gives me the output
NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."
I have also tried
Nsol2 = ParametricNDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
which gives me as output $T_11,dots,T_22$ as ParametricFunctions depending on each other and on $lambda$.
I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.
EDIT
Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.
differential-equations numerical-integration
$endgroup$
I would like to solve a $2times 2$ system of the form
$$fracddthetaT=TA,quad T(0)=Id$$
where $theta$ is real and $A$ is of the form
$$A=beginpmatrix 0 & frace^-i thetalambda\ frac136e^-ithetaleft(9lambda + 2(lambda-1)^2 (6costheta + cos2theta + 6)right) & 0endpmatrix,$$
with $lambda$ a free parameter in the unit circle.
In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
sys = T'[θ] == T[θ].A[θ];
The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried
NSol = NDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11[θ], T12[θ], T21[θ], T22[θ],
θ,
θ, 0, 2 Pi
];
which gives me the output
NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."
I have also tried
Nsol2 = ParametricNDSolve[
sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
which gives me as output $T_11,dots,T_22$ as ParametricFunctions depending on each other and on $lambda$.
I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.
EDIT
Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.
differential-equations numerical-integration
differential-equations numerical-integration
edited Mar 30 at 19:56
Carl Woll
77.1k3101202
77.1k3101202
asked Mar 30 at 13:31
EduEdu
1476
1476
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π]
instead of t
) of JM's formulation:
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
pf = ParametricNDSolveValue[
t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
,
t[2π],
θ, 0, 2π,
λ
];
Then, pf
will return the matrix value at $theta = 2 pi$. For example:
pf[1]
pf[Exp[I Pi/3]]
1. + 3.62818*10^-9 I,
6.82646*10^-8 - 5.73536*10^-9 I, 1.70661*10^-8 - 1.43384*10^-9 I,
1. + 3.62818*10^-9 I
0.985595 + 1.17074 I, -0.425572 + 0.737112 I, -0.788363 - 1.36549 I,
0.985595 - 1.17074 I
Finding the series expansion is simple:
DecimalForm[Series[pf[λ], λ, 1, 5], 4,4] //TeXForm
$left(
beginarraycc
1.0000 & 6.2830 \
1.5710 & 1.0000 \
endarray
right)+left(
beginarraycc
0.0000+0.0000 i & -6.2830+0.0000 i \
1.5710+0.0000 i & 0.0000+0.0000 i \
endarray
right) (lambda -1)+left(
beginarraycc
0.0000-0.7869 i & 0.6691-0.0000 i \
0.8799+0.0000 i & -0.0000+0.7869 i \
endarray
right) (lambda -1)^2+left(
beginarraycc
0.0000+0.7869 i & -1.3380+0.0000 i \
0.0000+0.0000 i & 0.0000-0.7869 i \
endarray
right) (lambda -1)^3+left(
beginarraycc
-0.0152-0.4524 i & 1.8410-0.0000 i \
-0.5708-0.0000 i & -0.0152+0.4524 i \
endarray
right) (lambda -1)^4+left(
beginarraycc
0.0305+0.1179 i & -2.1780+0.0000 i \
0.5708-0.0000 i & 0.0305-0.1179 i \
endarray
right) (lambda -1)^5+Oleft((lambda -1)^6right)$
$endgroup$
add a comment |
$begingroup$
As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue
; it returns a ParametricFunction
object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction
for your 4 functions.
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] = 0, E^(-I θ)/λ, 1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0;
sys = T'[θ] == T[θ].A[θ];
Tsol = ParametricNDSolveValue[sys, T11[0] == 1, T12[0] == 0,
T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
In order to obtain the numerical values for all the solutions at θ = 2 Pi
for a given parameter, say λ = 0.1
, you may use Through
:
Through[Tsol[0.1][2. Pi]]
-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
2.80298*10^-8 I, -0.545795 - 1.00532 I
In order to make that into a function, you may use
f = λ [Function] Through[Tsol[λ][2. Pi]]
$endgroup$
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
add a comment |
$begingroup$
For this case, you don't even need to write out the components of your matrix function:
pf = ParametricNDSolveValue[t'[θ] == t[θ].0, Exp[-I θ]/λ,
Exp[-I θ] (9 λ + 2 (λ - 1)^2
(6 Cos[θ] + Cos[2 θ] + 6))/36, 0,
t[0] == IdentityMatrix[2], t, θ, 0, 2 π, λ,
Method -> "StiffnessSwitching"];
sol = pf[(3 + 4 I)/5];
ParametricPlot[ReIm[Tr[sol[θ]]], θ, 0, 2 π]
ParametricPlot[ReIm[Det[sol[θ]]], θ, 0, 2 π]
You can even make a plot where the parameter is varying:
Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], ϕ, 0, 2 π]
$endgroup$
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π]
instead of t
) of JM's formulation:
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
pf = ParametricNDSolveValue[
t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
,
t[2π],
θ, 0, 2π,
λ
];
Then, pf
will return the matrix value at $theta = 2 pi$. For example:
pf[1]
pf[Exp[I Pi/3]]
1. + 3.62818*10^-9 I,
6.82646*10^-8 - 5.73536*10^-9 I, 1.70661*10^-8 - 1.43384*10^-9 I,
1. + 3.62818*10^-9 I
0.985595 + 1.17074 I, -0.425572 + 0.737112 I, -0.788363 - 1.36549 I,
0.985595 - 1.17074 I
Finding the series expansion is simple:
DecimalForm[Series[pf[λ], λ, 1, 5], 4,4] //TeXForm
$left(
beginarraycc
1.0000 & 6.2830 \
1.5710 & 1.0000 \
endarray
right)+left(
beginarraycc
0.0000+0.0000 i & -6.2830+0.0000 i \
1.5710+0.0000 i & 0.0000+0.0000 i \
endarray
right) (lambda -1)+left(
beginarraycc
0.0000-0.7869 i & 0.6691-0.0000 i \
0.8799+0.0000 i & -0.0000+0.7869 i \
endarray
right) (lambda -1)^2+left(
beginarraycc
0.0000+0.7869 i & -1.3380+0.0000 i \
0.0000+0.0000 i & 0.0000-0.7869 i \
endarray
right) (lambda -1)^3+left(
beginarraycc
-0.0152-0.4524 i & 1.8410-0.0000 i \
-0.5708-0.0000 i & -0.0152+0.4524 i \
endarray
right) (lambda -1)^4+left(
beginarraycc
0.0305+0.1179 i & -2.1780+0.0000 i \
0.5708-0.0000 i & 0.0305-0.1179 i \
endarray
right) (lambda -1)^5+Oleft((lambda -1)^6right)$
$endgroup$
add a comment |
$begingroup$
Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π]
instead of t
) of JM's formulation:
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
pf = ParametricNDSolveValue[
t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
,
t[2π],
θ, 0, 2π,
λ
];
Then, pf
will return the matrix value at $theta = 2 pi$. For example:
pf[1]
pf[Exp[I Pi/3]]
1. + 3.62818*10^-9 I,
6.82646*10^-8 - 5.73536*10^-9 I, 1.70661*10^-8 - 1.43384*10^-9 I,
1. + 3.62818*10^-9 I
0.985595 + 1.17074 I, -0.425572 + 0.737112 I, -0.788363 - 1.36549 I,
0.985595 - 1.17074 I
Finding the series expansion is simple:
DecimalForm[Series[pf[λ], λ, 1, 5], 4,4] //TeXForm
$left(
beginarraycc
1.0000 & 6.2830 \
1.5710 & 1.0000 \
endarray
right)+left(
beginarraycc
0.0000+0.0000 i & -6.2830+0.0000 i \
1.5710+0.0000 i & 0.0000+0.0000 i \
endarray
right) (lambda -1)+left(
beginarraycc
0.0000-0.7869 i & 0.6691-0.0000 i \
0.8799+0.0000 i & -0.0000+0.7869 i \
endarray
right) (lambda -1)^2+left(
beginarraycc
0.0000+0.7869 i & -1.3380+0.0000 i \
0.0000+0.0000 i & 0.0000-0.7869 i \
endarray
right) (lambda -1)^3+left(
beginarraycc
-0.0152-0.4524 i & 1.8410-0.0000 i \
-0.5708-0.0000 i & -0.0152+0.4524 i \
endarray
right) (lambda -1)^4+left(
beginarraycc
0.0305+0.1179 i & -2.1780+0.0000 i \
0.5708-0.0000 i & 0.0305-0.1179 i \
endarray
right) (lambda -1)^5+Oleft((lambda -1)^6right)$
$endgroup$
add a comment |
$begingroup$
Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π]
instead of t
) of JM's formulation:
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
pf = ParametricNDSolveValue[
t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
,
t[2π],
θ, 0, 2π,
λ
];
Then, pf
will return the matrix value at $theta = 2 pi$. For example:
pf[1]
pf[Exp[I Pi/3]]
1. + 3.62818*10^-9 I,
6.82646*10^-8 - 5.73536*10^-9 I, 1.70661*10^-8 - 1.43384*10^-9 I,
1. + 3.62818*10^-9 I
0.985595 + 1.17074 I, -0.425572 + 0.737112 I, -0.788363 - 1.36549 I,
0.985595 - 1.17074 I
Finding the series expansion is simple:
DecimalForm[Series[pf[λ], λ, 1, 5], 4,4] //TeXForm
$left(
beginarraycc
1.0000 & 6.2830 \
1.5710 & 1.0000 \
endarray
right)+left(
beginarraycc
0.0000+0.0000 i & -6.2830+0.0000 i \
1.5710+0.0000 i & 0.0000+0.0000 i \
endarray
right) (lambda -1)+left(
beginarraycc
0.0000-0.7869 i & 0.6691-0.0000 i \
0.8799+0.0000 i & -0.0000+0.7869 i \
endarray
right) (lambda -1)^2+left(
beginarraycc
0.0000+0.7869 i & -1.3380+0.0000 i \
0.0000+0.0000 i & 0.0000-0.7869 i \
endarray
right) (lambda -1)^3+left(
beginarraycc
-0.0152-0.4524 i & 1.8410-0.0000 i \
-0.5708-0.0000 i & -0.0152+0.4524 i \
endarray
right) (lambda -1)^4+left(
beginarraycc
0.0305+0.1179 i & -2.1780+0.0000 i \
0.5708-0.0000 i & 0.0305-0.1179 i \
endarray
right) (lambda -1)^5+Oleft((lambda -1)^6right)$
$endgroup$
Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π]
instead of t
) of JM's formulation:
A[θ_] =
0, E^(-I θ)/λ,
1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0
;
pf = ParametricNDSolveValue[
t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
,
t[2π],
θ, 0, 2π,
λ
];
Then, pf
will return the matrix value at $theta = 2 pi$. For example:
pf[1]
pf[Exp[I Pi/3]]
1. + 3.62818*10^-9 I,
6.82646*10^-8 - 5.73536*10^-9 I, 1.70661*10^-8 - 1.43384*10^-9 I,
1. + 3.62818*10^-9 I
0.985595 + 1.17074 I, -0.425572 + 0.737112 I, -0.788363 - 1.36549 I,
0.985595 - 1.17074 I
Finding the series expansion is simple:
DecimalForm[Series[pf[λ], λ, 1, 5], 4,4] //TeXForm
$left(
beginarraycc
1.0000 & 6.2830 \
1.5710 & 1.0000 \
endarray
right)+left(
beginarraycc
0.0000+0.0000 i & -6.2830+0.0000 i \
1.5710+0.0000 i & 0.0000+0.0000 i \
endarray
right) (lambda -1)+left(
beginarraycc
0.0000-0.7869 i & 0.6691-0.0000 i \
0.8799+0.0000 i & -0.0000+0.7869 i \
endarray
right) (lambda -1)^2+left(
beginarraycc
0.0000+0.7869 i & -1.3380+0.0000 i \
0.0000+0.0000 i & 0.0000-0.7869 i \
endarray
right) (lambda -1)^3+left(
beginarraycc
-0.0152-0.4524 i & 1.8410-0.0000 i \
-0.5708-0.0000 i & -0.0152+0.4524 i \
endarray
right) (lambda -1)^4+left(
beginarraycc
0.0305+0.1179 i & -2.1780+0.0000 i \
0.5708-0.0000 i & 0.0305-0.1179 i \
endarray
right) (lambda -1)^5+Oleft((lambda -1)^6right)$
answered Mar 30 at 18:58
Carl WollCarl Woll
77.1k3101202
77.1k3101202
add a comment |
add a comment |
$begingroup$
As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue
; it returns a ParametricFunction
object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction
for your 4 functions.
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] = 0, E^(-I θ)/λ, 1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0;
sys = T'[θ] == T[θ].A[θ];
Tsol = ParametricNDSolveValue[sys, T11[0] == 1, T12[0] == 0,
T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
In order to obtain the numerical values for all the solutions at θ = 2 Pi
for a given parameter, say λ = 0.1
, you may use Through
:
Through[Tsol[0.1][2. Pi]]
-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
2.80298*10^-8 I, -0.545795 - 1.00532 I
In order to make that into a function, you may use
f = λ [Function] Through[Tsol[λ][2. Pi]]
$endgroup$
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
add a comment |
$begingroup$
As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue
; it returns a ParametricFunction
object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction
for your 4 functions.
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] = 0, E^(-I θ)/λ, 1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0;
sys = T'[θ] == T[θ].A[θ];
Tsol = ParametricNDSolveValue[sys, T11[0] == 1, T12[0] == 0,
T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
In order to obtain the numerical values for all the solutions at θ = 2 Pi
for a given parameter, say λ = 0.1
, you may use Through
:
Through[Tsol[0.1][2. Pi]]
-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
2.80298*10^-8 I, -0.545795 - 1.00532 I
In order to make that into a function, you may use
f = λ [Function] Through[Tsol[λ][2. Pi]]
$endgroup$
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
add a comment |
$begingroup$
As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue
; it returns a ParametricFunction
object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction
for your 4 functions.
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] = 0, E^(-I θ)/λ, 1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0;
sys = T'[θ] == T[θ].A[θ];
Tsol = ParametricNDSolveValue[sys, T11[0] == 1, T12[0] == 0,
T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
In order to obtain the numerical values for all the solutions at θ = 2 Pi
for a given parameter, say λ = 0.1
, you may use Through
:
Through[Tsol[0.1][2. Pi]]
-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
2.80298*10^-8 I, -0.545795 - 1.00532 I
In order to make that into a function, you may use
f = λ [Function] Through[Tsol[λ][2. Pi]]
$endgroup$
As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue
; it returns a ParametricFunction
object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction
for your 4 functions.
T[θ_] = T11[θ], T12[θ], T21[θ], T22[θ];
A[θ_] = 0, E^(-I θ)/λ, 1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0;
sys = T'[θ] == T[θ].A[θ];
Tsol = ParametricNDSolveValue[sys, T11[0] == 1, T12[0] == 0,
T21[0] == 0, T22[0] == 1,
T11, T12, T21, T22,
θ, 0, 2 Pi,
λ
];
In order to obtain the numerical values for all the solutions at θ = 2 Pi
for a given parameter, say λ = 0.1
, you may use Through
:
Through[Tsol[0.1][2. Pi]]
-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
2.80298*10^-8 I, -0.545795 - 1.00532 I
In order to make that into a function, you may use
f = λ [Function] Through[Tsol[λ][2. Pi]]
edited Mar 30 at 14:25
answered Mar 30 at 13:47
Henrik SchumacherHenrik Schumacher
61.5k585171
61.5k585171
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
add a comment |
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
$begingroup$
Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
$endgroup$
– Edu
Mar 30 at 14:17
add a comment |
$begingroup$
For this case, you don't even need to write out the components of your matrix function:
pf = ParametricNDSolveValue[t'[θ] == t[θ].0, Exp[-I θ]/λ,
Exp[-I θ] (9 λ + 2 (λ - 1)^2
(6 Cos[θ] + Cos[2 θ] + 6))/36, 0,
t[0] == IdentityMatrix[2], t, θ, 0, 2 π, λ,
Method -> "StiffnessSwitching"];
sol = pf[(3 + 4 I)/5];
ParametricPlot[ReIm[Tr[sol[θ]]], θ, 0, 2 π]
ParametricPlot[ReIm[Det[sol[θ]]], θ, 0, 2 π]
You can even make a plot where the parameter is varying:
Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], ϕ, 0, 2 π]
$endgroup$
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
|
show 1 more comment
$begingroup$
For this case, you don't even need to write out the components of your matrix function:
pf = ParametricNDSolveValue[t'[θ] == t[θ].0, Exp[-I θ]/λ,
Exp[-I θ] (9 λ + 2 (λ - 1)^2
(6 Cos[θ] + Cos[2 θ] + 6))/36, 0,
t[0] == IdentityMatrix[2], t, θ, 0, 2 π, λ,
Method -> "StiffnessSwitching"];
sol = pf[(3 + 4 I)/5];
ParametricPlot[ReIm[Tr[sol[θ]]], θ, 0, 2 π]
ParametricPlot[ReIm[Det[sol[θ]]], θ, 0, 2 π]
You can even make a plot where the parameter is varying:
Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], ϕ, 0, 2 π]
$endgroup$
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
|
show 1 more comment
$begingroup$
For this case, you don't even need to write out the components of your matrix function:
pf = ParametricNDSolveValue[t'[θ] == t[θ].0, Exp[-I θ]/λ,
Exp[-I θ] (9 λ + 2 (λ - 1)^2
(6 Cos[θ] + Cos[2 θ] + 6))/36, 0,
t[0] == IdentityMatrix[2], t, θ, 0, 2 π, λ,
Method -> "StiffnessSwitching"];
sol = pf[(3 + 4 I)/5];
ParametricPlot[ReIm[Tr[sol[θ]]], θ, 0, 2 π]
ParametricPlot[ReIm[Det[sol[θ]]], θ, 0, 2 π]
You can even make a plot where the parameter is varying:
Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], ϕ, 0, 2 π]
$endgroup$
For this case, you don't even need to write out the components of your matrix function:
pf = ParametricNDSolveValue[t'[θ] == t[θ].0, Exp[-I θ]/λ,
Exp[-I θ] (9 λ + 2 (λ - 1)^2
(6 Cos[θ] + Cos[2 θ] + 6))/36, 0,
t[0] == IdentityMatrix[2], t, θ, 0, 2 π, λ,
Method -> "StiffnessSwitching"];
sol = pf[(3 + 4 I)/5];
ParametricPlot[ReIm[Tr[sol[θ]]], θ, 0, 2 π]
ParametricPlot[ReIm[Det[sol[θ]]], θ, 0, 2 π]
You can even make a plot where the parameter is varying:
Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], ϕ, 0, 2 π]
answered Mar 30 at 13:57
community wiki
J. M. is away
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
|
show 1 more comment
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
$endgroup$
– Edu
Mar 30 at 14:23
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
$endgroup$
– J. M. is away♦
Mar 30 at 14:24
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
$endgroup$
– Edu
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
$endgroup$
– J. M. is away♦
Mar 30 at 14:26
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
$begingroup$
For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
$endgroup$
– Edu
Mar 30 at 14:41
|
show 1 more comment
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