Functional fixed points (ie fixed point of mapping from function space C[0,1] to itself)
I am looking for some tips or guidance as to what machinery in mathematica can help me get at this problem numerically. I am looking for fixed points of a mapping, but the objects in question are themselves functions. Hence I am looking for a fixed point function.
The setup (simplified version):
suppose we restrict our search to continuous functions $f: [0,1]rightarrow [0,1]$. $p$ is a known parameter. I am looking for a fixed point (function) such that, for all $xin [0,1]$, $f(x)$ solves
$$f(x) = frac{x^p}{x^p + int_0^1 f(x) x^p , dx }.$$
It's not as simple as finding lots of fixed points for each $x$ in isolation, as the value of the expression at a single $x$ depends on the entire function $f$. Any help to try and solve this type of thing numerically would be much appreciated.
numerical-integration parametric-functions numerical-value fixed-points
asked 3 hours ago
user434180
262
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user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
I am looking for some tips or guidance as to what machinery in mathematica can help me get at this problem numerically. I am looking for fixed points of a mapping, but the objects in question are themselves functions. Hence I am looking for a fixed point function.
The setup (simplified version):
suppose we restrict our search to continuous functions $f: [0,1]rightarrow [0,1]$. $p$ is a known parameter. I am looking for a fixed point (function) such that, for all $xin [0,1]$, $f(x)$ solves
$$f(x) = frac{x^p}{x^p + int_0^1 f(x) x^p , dx }.$$
It's not as simple as finding lots of fixed points for each $x$ in isolation, as the value of the expression at a single $x$ depends on the entire function $f$. Any help to try and solve this type of thing numerically would be much appreciated.
numerical-integration parametric-functions numerical-value fixed-points
asked 3 hours ago
user434180
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensionalEigenvalueproblem which can be solved withEigensystemorNDEigensystem. -useInterpolationas function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps theVariationalMethodspackage can help with that.
– Thies Heidecke
2 hours ago
1
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved byDSolve, numerically byNDSolveor iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.
– Thies Heidecke
2 hours ago
5
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago
add a comment |
I am looking for some tips or guidance as to what machinery in mathematica can help me get at this problem numerically. I am looking for fixed points of a mapping, but the objects in question are themselves functions. Hence I am looking for a fixed point function.
The setup (simplified version):
suppose we restrict our search to continuous functions $f: [0,1]rightarrow [0,1]$. $p$ is a known parameter. I am looking for a fixed point (function) such that, for all $xin [0,1]$, $f(x)$ solves
$$f(x) = frac{x^p}{x^p + int_0^1 f(x) x^p , dx }.$$
It's not as simple as finding lots of fixed points for each $x$ in isolation, as the value of the expression at a single $x$ depends on the entire function $f$. Any help to try and solve this type of thing numerically would be much appreciated.
numerical-integration parametric-functions numerical-value fixed-points
asked 3 hours ago
user434180
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am looking for some tips or guidance as to what machinery in mathematica can help me get at this problem numerically. I am looking for fixed points of a mapping, but the objects in question are themselves functions. Hence I am looking for a fixed point function.
The setup (simplified version):
suppose we restrict our search to continuous functions $f: [0,1]rightarrow [0,1]$. $p$ is a known parameter. I am looking for a fixed point (function) such that, for all $xin [0,1]$, $f(x)$ solves
$$f(x) = frac{x^p}{x^p + int_0^1 f(x) x^p , dx }.$$
It's not as simple as finding lots of fixed points for each $x$ in isolation, as the value of the expression at a single $x$ depends on the entire function $f$. Any help to try and solve this type of thing numerically would be much appreciated.
numerical-integration parametric-functions numerical-value fixed-points
numerical-integration parametric-functions numerical-value fixed-points
asked 3 hours ago
user434180
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user434180
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
asked 3 hours ago
user434180
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user434180
262
asked 3 hours ago
user434180
262
262
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user434180 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensionalEigenvalueproblem which can be solved withEigensystemorNDEigensystem. -useInterpolationas function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps theVariationalMethodspackage can help with that.
– Thies Heidecke
2 hours ago
1
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved byDSolve, numerically byNDSolveor iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.
– Thies Heidecke
2 hours ago
5
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago
add a comment |
1
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensionalEigenvalueproblem which can be solved withEigensystemorNDEigensystem. -useInterpolationas function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps theVariationalMethodspackage can help with that.
– Thies Heidecke
2 hours ago
1
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved byDSolve, numerically byNDSolveor iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.
– Thies Heidecke
2 hours ago
5
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago
1
1
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensional
Eigenvalue problem which can be solved with Eigensystem or NDEigensystem. -use Interpolation as function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps the VariationalMethods package can help with that.– Thies Heidecke
2 hours ago
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensional
Eigenvalue problem which can be solved with Eigensystem or NDEigensystem. -use Interpolation as function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps the VariationalMethods package can help with that.– Thies Heidecke
2 hours ago
1
1
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved by
DSolve, numerically by NDSolve or iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.– Thies Heidecke
2 hours ago
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved by
DSolve, numerically by NDSolve or iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.– Thies Heidecke
2 hours ago
5
5
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
I believe this is @Lucas suggestions in the comment.
ClearAll[p, c]
p = 2;
f[x_] := x^p/(x^p + c)
Assuming[c [Element] Reals, !(
*SubsuperscriptBox[([Integral]), (0), (1)](f[x]
*SuperscriptBox[(x), (p)] [DifferentialD]x)) == c]
ConditionalExpression[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, c > 0 || c < -1]
c = c /. FindRoot[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, {c, 1}]
0.227879
fixedPoints = NSolve[f[x] == x, x]
{{x -> 0.64873}, {x -> 0.35127}, {x -> 0.}}
Plot[{f[x], x}, {x, 0, 1}, AspectRatio -> 1, Frame -> True,
GridLines -> {Flatten@Values@fixedPoints,
Flatten@Values@fixedPoints}]
answered 1 hour ago
Okkes Dulgerci
4,0251816
FixedPointappears to be faster thanNSolvefor this.
– Alan
9 mins ago
add a comment |
This uses a discretization by piecewise-linear functions.
n = 1000;
x = Subdivide[0., 1., n - 1];
p = 2;
(* quadrature weights for trapezoidal rule *)
ω = ConstantArray[1./(n - 1), n];
ω[[1]] = ω[[n]] = 0.5/(n - 1);
Applying fixed point iteration; I use Dot and ω to compute the integral:
data = FixedPointList[
f [Function] x^p/(x^p + (x^p f).ω),
ConstantArray[0.5, n]
];
Checking the $L^infty$ error:
Max[Abs[step[data[[-1]]] - data[[-1]]]]
2.22045*10^-16
Plotting the iterates:
ListLinePlot[
data,
PlotLegends -> Automatic
]
answered 2 hours ago
Henrik Schumacher
48.9k467139
add a comment |
supplement
The list of fixpoint-functions can be obtained strictly numerical (variable p) using Nintegrate:
int[c_?NumericQ, p_?NumericQ] :=NIntegrate[x^(2 p)/(x^p + c), {x, 0, 1}, Method -> "LocalAdaptive" ]
f[Infinity] =Table[ x^ p /(x^p + c) /.NMinimize[{1, c == int[c, p]}, c][[2]] , {p, 1, 5}]
(*{x/(0.323829 + x), x^2/(0.227879 + x^2), x^3/(0.1782 + x^3)
, x^4/(0.147254 + x^4), x^5/(0.125923 + x^5)}*)
Plot[f[Infinity], {x, 0, 1}, PlotRange -> {0, 1}]
answered 6 mins ago
Ulrich Neumann
7,345515
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I believe this is @Lucas suggestions in the comment.
ClearAll[p, c]
p = 2;
f[x_] := x^p/(x^p + c)
Assuming[c [Element] Reals, !(
*SubsuperscriptBox[([Integral]), (0), (1)](f[x]
*SuperscriptBox[(x), (p)] [DifferentialD]x)) == c]
ConditionalExpression[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, c > 0 || c < -1]
c = c /. FindRoot[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, {c, 1}]
0.227879
fixedPoints = NSolve[f[x] == x, x]
{{x -> 0.64873}, {x -> 0.35127}, {x -> 0.}}
Plot[{f[x], x}, {x, 0, 1}, AspectRatio -> 1, Frame -> True,
GridLines -> {Flatten@Values@fixedPoints,
Flatten@Values@fixedPoints}]
answered 1 hour ago
Okkes Dulgerci
4,0251816
FixedPointappears to be faster thanNSolvefor this.
– Alan
9 mins ago
add a comment |
I believe this is @Lucas suggestions in the comment.
ClearAll[p, c]
p = 2;
f[x_] := x^p/(x^p + c)
Assuming[c [Element] Reals, !(
*SubsuperscriptBox[([Integral]), (0), (1)](f[x]
*SuperscriptBox[(x), (p)] [DifferentialD]x)) == c]
ConditionalExpression[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, c > 0 || c < -1]
c = c /. FindRoot[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, {c, 1}]
0.227879
fixedPoints = NSolve[f[x] == x, x]
{{x -> 0.64873}, {x -> 0.35127}, {x -> 0.}}
Plot[{f[x], x}, {x, 0, 1}, AspectRatio -> 1, Frame -> True,
GridLines -> {Flatten@Values@fixedPoints,
Flatten@Values@fixedPoints}]
answered 1 hour ago
Okkes Dulgerci
4,0251816
FixedPointappears to be faster thanNSolvefor this.
– Alan
9 mins ago
add a comment |
I believe this is @Lucas suggestions in the comment.
ClearAll[p, c]
p = 2;
f[x_] := x^p/(x^p + c)
Assuming[c [Element] Reals, !(
*SubsuperscriptBox[([Integral]), (0), (1)](f[x]
*SuperscriptBox[(x), (p)] [DifferentialD]x)) == c]
ConditionalExpression[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, c > 0 || c < -1]
c = c /. FindRoot[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, {c, 1}]
0.227879
fixedPoints = NSolve[f[x] == x, x]
{{x -> 0.64873}, {x -> 0.35127}, {x -> 0.}}
Plot[{f[x], x}, {x, 0, 1}, AspectRatio -> 1, Frame -> True,
GridLines -> {Flatten@Values@fixedPoints,
Flatten@Values@fixedPoints}]
answered 1 hour ago
Okkes Dulgerci
4,0251816
I believe this is @Lucas suggestions in the comment.
ClearAll[p, c]
p = 2;
f[x_] := x^p/(x^p + c)
Assuming[c [Element] Reals, !(
*SubsuperscriptBox[([Integral]), (0), (1)](f[x]
*SuperscriptBox[(x), (p)] [DifferentialD]x)) == c]
ConditionalExpression[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, c > 0 || c < -1]
c = c /. FindRoot[1/3 - c + c^(3/2) ArcTan[1/Sqrt[c]] == c, {c, 1}]
0.227879
fixedPoints = NSolve[f[x] == x, x]
{{x -> 0.64873}, {x -> 0.35127}, {x -> 0.}}
Plot[{f[x], x}, {x, 0, 1}, AspectRatio -> 1, Frame -> True,
GridLines -> {Flatten@Values@fixedPoints,
Flatten@Values@fixedPoints}]
answered 1 hour ago
Okkes Dulgerci
4,0251816
edited 29 mins ago
answered 1 hour ago
Okkes Dulgerci
4,0251816
answered 1 hour ago
Okkes Dulgerci
4,0251816
answered 1 hour ago
Okkes Dulgerci
4,0251816
4,0251816
FixedPointappears to be faster thanNSolvefor this.
– Alan
9 mins ago
add a comment |
FixedPointappears to be faster thanNSolvefor this.
– Alan
9 mins ago
FixedPoint appears to be faster than NSolve for this.– Alan
9 mins ago
FixedPoint appears to be faster than NSolve for this.– Alan
9 mins ago
add a comment |
This uses a discretization by piecewise-linear functions.
n = 1000;
x = Subdivide[0., 1., n - 1];
p = 2;
(* quadrature weights for trapezoidal rule *)
ω = ConstantArray[1./(n - 1), n];
ω[[1]] = ω[[n]] = 0.5/(n - 1);
Applying fixed point iteration; I use Dot and ω to compute the integral:
data = FixedPointList[
f [Function] x^p/(x^p + (x^p f).ω),
ConstantArray[0.5, n]
];
Checking the $L^infty$ error:
Max[Abs[step[data[[-1]]] - data[[-1]]]]
2.22045*10^-16
Plotting the iterates:
ListLinePlot[
data,
PlotLegends -> Automatic
]
answered 2 hours ago
Henrik Schumacher
48.9k467139
add a comment |
This uses a discretization by piecewise-linear functions.
n = 1000;
x = Subdivide[0., 1., n - 1];
p = 2;
(* quadrature weights for trapezoidal rule *)
ω = ConstantArray[1./(n - 1), n];
ω[[1]] = ω[[n]] = 0.5/(n - 1);
Applying fixed point iteration; I use Dot and ω to compute the integral:
data = FixedPointList[
f [Function] x^p/(x^p + (x^p f).ω),
ConstantArray[0.5, n]
];
Checking the $L^infty$ error:
Max[Abs[step[data[[-1]]] - data[[-1]]]]
2.22045*10^-16
Plotting the iterates:
ListLinePlot[
data,
PlotLegends -> Automatic
]
answered 2 hours ago
Henrik Schumacher
48.9k467139
add a comment |
This uses a discretization by piecewise-linear functions.
n = 1000;
x = Subdivide[0., 1., n - 1];
p = 2;
(* quadrature weights for trapezoidal rule *)
ω = ConstantArray[1./(n - 1), n];
ω[[1]] = ω[[n]] = 0.5/(n - 1);
Applying fixed point iteration; I use Dot and ω to compute the integral:
data = FixedPointList[
f [Function] x^p/(x^p + (x^p f).ω),
ConstantArray[0.5, n]
];
Checking the $L^infty$ error:
Max[Abs[step[data[[-1]]] - data[[-1]]]]
2.22045*10^-16
Plotting the iterates:
ListLinePlot[
data,
PlotLegends -> Automatic
]
answered 2 hours ago
Henrik Schumacher
48.9k467139
This uses a discretization by piecewise-linear functions.
n = 1000;
x = Subdivide[0., 1., n - 1];
p = 2;
(* quadrature weights for trapezoidal rule *)
ω = ConstantArray[1./(n - 1), n];
ω[[1]] = ω[[n]] = 0.5/(n - 1);
Applying fixed point iteration; I use Dot and ω to compute the integral:
data = FixedPointList[
f [Function] x^p/(x^p + (x^p f).ω),
ConstantArray[0.5, n]
];
Checking the $L^infty$ error:
Max[Abs[step[data[[-1]]] - data[[-1]]]]
2.22045*10^-16
Plotting the iterates:
ListLinePlot[
data,
PlotLegends -> Automatic
]
answered 2 hours ago
Henrik Schumacher
48.9k467139
answered 2 hours ago
Henrik Schumacher
48.9k467139
answered 2 hours ago
Henrik Schumacher
48.9k467139
answered 2 hours ago
Henrik Schumacher
48.9k467139
48.9k467139
add a comment |
add a comment |
supplement
The list of fixpoint-functions can be obtained strictly numerical (variable p) using Nintegrate:
int[c_?NumericQ, p_?NumericQ] :=NIntegrate[x^(2 p)/(x^p + c), {x, 0, 1}, Method -> "LocalAdaptive" ]
f[Infinity] =Table[ x^ p /(x^p + c) /.NMinimize[{1, c == int[c, p]}, c][[2]] , {p, 1, 5}]
(*{x/(0.323829 + x), x^2/(0.227879 + x^2), x^3/(0.1782 + x^3)
, x^4/(0.147254 + x^4), x^5/(0.125923 + x^5)}*)
Plot[f[Infinity], {x, 0, 1}, PlotRange -> {0, 1}]
answered 6 mins ago
Ulrich Neumann
7,345515
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
add a comment |
supplement
The list of fixpoint-functions can be obtained strictly numerical (variable p) using Nintegrate:
int[c_?NumericQ, p_?NumericQ] :=NIntegrate[x^(2 p)/(x^p + c), {x, 0, 1}, Method -> "LocalAdaptive" ]
f[Infinity] =Table[ x^ p /(x^p + c) /.NMinimize[{1, c == int[c, p]}, c][[2]] , {p, 1, 5}]
(*{x/(0.323829 + x), x^2/(0.227879 + x^2), x^3/(0.1782 + x^3)
, x^4/(0.147254 + x^4), x^5/(0.125923 + x^5)}*)
Plot[f[Infinity], {x, 0, 1}, PlotRange -> {0, 1}]
answered 6 mins ago
Ulrich Neumann
7,345515
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
add a comment |
supplement
The list of fixpoint-functions can be obtained strictly numerical (variable p) using Nintegrate:
int[c_?NumericQ, p_?NumericQ] :=NIntegrate[x^(2 p)/(x^p + c), {x, 0, 1}, Method -> "LocalAdaptive" ]
f[Infinity] =Table[ x^ p /(x^p + c) /.NMinimize[{1, c == int[c, p]}, c][[2]] , {p, 1, 5}]
(*{x/(0.323829 + x), x^2/(0.227879 + x^2), x^3/(0.1782 + x^3)
, x^4/(0.147254 + x^4), x^5/(0.125923 + x^5)}*)
Plot[f[Infinity], {x, 0, 1}, PlotRange -> {0, 1}]
answered 6 mins ago
Ulrich Neumann
7,345515
supplement
The list of fixpoint-functions can be obtained strictly numerical (variable p) using Nintegrate:
int[c_?NumericQ, p_?NumericQ] :=NIntegrate[x^(2 p)/(x^p + c), {x, 0, 1}, Method -> "LocalAdaptive" ]
f[Infinity] =Table[ x^ p /(x^p + c) /.NMinimize[{1, c == int[c, p]}, c][[2]] , {p, 1, 5}]
(*{x/(0.323829 + x), x^2/(0.227879 + x^2), x^3/(0.1782 + x^3)
, x^4/(0.147254 + x^4), x^5/(0.125923 + x^5)}*)
Plot[f[Infinity], {x, 0, 1}, PlotRange -> {0, 1}]
answered 6 mins ago
Ulrich Neumann
7,345515
answered 6 mins ago
Ulrich Neumann
7,345515
answered 6 mins ago
Ulrich Neumann
7,345515
answered 6 mins ago
Ulrich Neumann
7,345515
7,345515
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
add a comment |
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
This is a very helpful addition. I was about to comment on Henrik's answer that it requires the integral to be analytically solvable by mathematica. Although the one I posted in the question is solvable, the actual ones I care about are generally not. So this is very useful. Thanks!
– user434180
4 mins ago
add a comment |
user434180 is a new contributor. Be nice, and check out our Code of Conduct.
user434180 is a new contributor. Be nice, and check out our Code of Conduct.
user434180 is a new contributor. Be nice, and check out our Code of Conduct.
user434180 is a new contributor. Be nice, and check out our Code of Conduct.
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1
I can think of multiple ways to go about this: -discretize your function by representing it by a vector and solve the discretized problem as an approximation. Then this might result in a finite dimensional
Eigenvalueproblem which can be solved withEigensystemorNDEigensystem. -useInterpolationas function representation and sample and reinterpolate after each iteration. - use a variational approach to find the fixed point, perhaps theVariationalMethodspackage can help with that.– Thies Heidecke
2 hours ago
1
- perhaps the problem can be stated as an ordinary differential equation and either be directly solved by
DSolve, numerically byNDSolveor iteratively by a Picard iteration. I'm not sure if all of those methods can successfully tackle your problem, but all those alleys could be explored in Mathematica. Hope this gives you some ideas! When you try something and need further help, update your question with Mathematica code and specific questions, so that people can help you with the details.– Thies Heidecke
2 hours ago
5
The following should work: replace the integral with $c$. Solve for $f(x)$. Compute the integral as a function of $c$. Set the integral equal to $c$ and solve for $c$.
– Lukas Lang
2 hours ago
Thank you very much for these suggestions. I have tried the ODE approach. I don't think that it can be transformed into an ODE as I don't see a way to remove the integral. The ratio means it can't be transformed into a Fredholm equation, for which code already exists to numerically solve. I will try discretizing in line with the great answer below. And I am a little lost by your suggestion, Lukas, although I will try work through it as well.
– user434180
2 hours ago
@LukasLang Great idea, this seems like the most simple and best approach!
– Thies Heidecke
1 hour ago