Skip to main content

numerics - How to implement custom NIntegrate integration strategies?


How can new integration strategies algorithms be used with NIntegrate?


This is a different type of extension than the extensions with new integration rules, as described in the answer for the question "How to implement custom integration rules for use by NIntegrate?". (Integration strategies perform and guide the core integration process, using or leveraging different integration rules and/or preprocessing algorithms.)



Answer



Motivation (for a new semi-symbolic integration strategy)



Consider the following integral, which cannot be done neigther by Integrate:


Integrate[BesselJ[y, x^3], {x, 0, ∞}, {y, 0, 1}]

(* Integrate[If[Re[y] > -(1/3), Gamma[1/6 + y/2]/(3*2^(2/3)*Gamma[5/6 + y/2]),
Integrate[BesselJ[y, x^3], {x, 0, Infinity},
Assumptions -> Re[y] <= -(1/3)]], {y, 0, 1}] *)

nor NIntegrate:


NIntegrate[BesselJ[y, x^3], {x, 0, ∞}, {y, 0, 1}, 
Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> 2000}]



NIntegrate::slwcon: Numerical integration converging too slowly;


NIntegrate::eincr: The global error of the strategy GlobalAdaptive has ...



(* 0.524338 *)

Here is a plot of the integrand function over a much smaller domain:


Plot3D[BesselJ[y, x^3], {x, 0, 10}, {y, 0, 1}, PlotPoints -> {100, 10}, 
MaxRecursion -> 5, PlotRange -> All, BoxRatios -> {10, 3}]



Because of the oscillatory nature of the integrand we can see why NIntegrate has difficulties.


On the other hand, Integrate can find the value of the integral integrating over $x$:


In[14]:= Integrate[BesselJ[y, x^3], {x, 0, ∞}]

Out[14]= ConditionalExpression[Gamma[1/6 + y/2]/(3*2^(2/3)*Gamma[5/6 + y/2]),
Re[y] > -(1/3)]

but it has problems integrating over $y$:



In[15]:= Integrate[BesselJ[y, x^3], {y, 0, 1}]

Out[15]= Integrate[BesselJ[y, x^3], {y, 0, 1}]

Since Integrate can do partially the integral along one of the axes, we can just take that symbolic expression and give it to NIntegrate for integration over the other axis.


Semi-symbolic NIntegrate implementation


Here we make an integration strategy that combines Integrate and NIntegrate -- it uses Integrate over some of the integration range(s) and then NIntegrate for the rest of the range(s) with the symbolic expressions obtained by Integrate.


The following defintion is for the initialization of the integration strategy SemiSymbolic.


Clear[SemiSymbolic];
Options[SemiSymbolic] = {"AnalyticalVariables" -> {}};

SemiSymbolicProperties = Options[SemiSymbolic][[All, 1]];
SemiSymbolic /:
NIntegrate`InitializeIntegrationStrategy[SemiSymbolic, nfs_, ranges_,
strOpts_, allOpts_] :=
Module[{t, anVars},
t = NIntegrate`GetMethodOptionValues[SemiSymbolic, SemiSymbolicProperties,
strOpts];
If[t === $Failed, Return[$Failed]];
{anVars} = t;
SemiSymbolic[First /@ ranges, anVars]

];

This is the implementation of the integaration strategy SemiSymbolic:


SemiSymbolic[vars_, anVars_]["Algorithm"[regions_, opts___]] :=
Module[{ranges, anRanges, funcs, t},

ranges = Map[Flatten /@ Transpose[{vars, #@"OriginalBoundaries"}] &, regions];
ranges = Map[Flatten, ranges, {-2}];
anRanges = Map[Select[#, MemberQ[anVars, #[[1]]] &] &, ranges];
ranges = Map[Select[#, ! MemberQ[anVars, #[[1]]] &] &, ranges];

funcs = (#@"Integrand"[])@"FunctionExpression"[] & /@ regions;

t = MapThread[
Integrate[#1, Sequence @@ #2,
Assumptions -> (#[[2]] <= #[[1]] <= #[[3]] & /@ #3)] &, {funcs,
anRanges, ranges}];
Print["SemiSymbolic::Integrate's result:", t];

If[! FreeQ[t, Integrate], Return[$Failed]];


Total[MapThread[
NIntegrate[#1, Sequence @@ #2 // Evaluate,
Sequence @@ DeleteCases[opts, Method -> _] // Evaluate] &, {t, ranges}]]
];

(Note the implementation prints the intermediate result obtained by Integrate.)


Signatures


Initialization


We can see that the new rule SemiSymbolic is defined through TagSetDelayed for SemiSymbolic and NIntegrate`InitializeIntegrationStrategy. The rest of the arguments are:


nfs -- numerical function objects; several might be given depending on the integrand and ranges;



ranges -- a list of ranges for the integration variables;


strOpts -- the options given to the strategy;


allOpts -- all options given to NIntegrate.


Algorithm


StrategySymbol[strategyData___]["Algorithm"[regions_, opts___]] := ...

The algorithm can use regions objects as described in this answer of "Determining which rule NIntegrate selects automatically".


Remarks



Testing SemiSymbolic



The strategy works without (observable) problems for the motivational integral:


In[85]:= NIntegrate[BesselJ[y, x^3], {x, 0, Infinity}, {y, 0, 1}, 
Method -> {SemiSymbolic, "AnalyticalVariables" -> {x}}]

During evaluation of In[85]:= SemiSymbolic::Integrate's result:{(2^-y HypergeometricPFQ[{1/6+y/2},{7/6+y/2,1+y},-(1/4)])/((1+3 y) Gamma[1+y])}

Out[85]= 0.524448

Note the printout for the intermediate result by Integrate.


Since SemiSymbolic passes inside its body the non-method NIntegrate options it was invoked with we can also see the sampling points used by SemiSymbolic through EvaluationMonitor.



res = 
Reap@NIntegrate[BesselJ[y, x^3], {x, 0, Infinity}, {y, 0, 1},
Method -> {SemiSymbolic, "AnalyticalVariables" -> {x}},
EvaluationMonitor :> Sow[{x, y}]]

During evaluation of In[78]:= SemiSymbolic::Integrate's result:{(2^-y HypergeometricPFQ[{1/6+y/2},{7/6+y/2,1+y},-(1/4)])/((1+3 y) Gamma[1+y])}

(* {0.524448, {{{x, 0.00795732}, {x, 0.0469101}, {x, 0.122917}, {x,
0.230765}, {x, 0.360185}, {x, 0.5}, {x, 0.639815}, {x,
0.769235}, {x, 0.877083}, {x, 0.95309}, {x, 0.992043}, {x,

0.00397866}, {x, 0.023455}, {x, 0.0614583}, {x, 0.115383}, {x,
0.180092}, {x, 0.25}, {x, 0.319908}, {x, 0.384617}, {x,
0.438542}, {x, 0.476545}, {x, 0.496021}, {x, 0.503979}, {x,
0.523455}, {x, 0.561458}, {x, 0.615383}, {x, 0.680092}, {x,
0.75}, {x, 0.819908}, {x, 0.884617}, {x, 0.938542}, {x,
0.976545}, {x, 0.996021}}}} *)

ListPlot[res[[2, 1, All, 2]], Frame -> True]

enter image description here



Further tests


Below are some other tests / examples.


In[50]:= NIntegrate[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
Method -> {SemiSymbolic, "AnalyticalVariables" -> {x, y}}]

During evaluation of In[50]:= SemiSymbolic::Integrate's result:{2/3+z^2}

Out[50]= 1.

Note that the symbolic integration was done over two variables.



Let us use the same integrand but with different range boundaries for the different variables in order to evaluate better the variable correspondence in the 2D sampling points pattern.


In[66]:= res = 
Reap@NIntegrate[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 2}, {z, 0, 10},
Method -> {SemiSymbolic, "AnalyticalVariables" -> {x}},
EvaluationMonitor :> Sow[{y, z}]]

During evaluation of In[66]:= SemiSymbolic::Integrate's result:{1/3+y^2+z^2}

Out[66]= {700., {{{1., 5.}, {1.35857, 5.}, {0.641431, 5.}, {1.94868,
5.}, {0.0513167, 5.}, {1., 6.79284}, {1., 3.20716}, {1.,

9.74342}, {1., 0.256584}, {1.94868, 9.74342}, {1.94868,
0.256584}, {0.0513167, 0.256584}, {0.0513167, 9.74342}, {0.311753,
1.55876}, {0.311753, 8.44124}, {1.68825, 1.55876}, {1.68825,
8.44124}}}}

In[69]:= ListPlot[res[[2, 1]], Frame -> True]

enter image description here


Another example


This Lebesgue integration implementation, AdaptiveNumericalLebesgueIntegration.m -- discussed in detail in "Adaptive numerical Lebesgue integration by set measure estimates" -- has implementations of integration strategy (and rules) with the complete signatures for the plug-in mechanism.



Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...