Ways to speed up user implemented RK4Speed up Numerical IntegrationSpeed of convergence for NIntegrateTough Calculation, novice mathematica userNumerical integration's speedNumerical integral speedImprove the speed of Gaussian quadrature integrationSolving an unstable BVP numerically, accurately and efficientlyHow to speed up integral of results of PDE modelSolve BVP involving user defined functionUser defined ArcTan function

What is difference between behavior and behaviour

Are there any comparative studies done between Ashtavakra Gita and Buddhim?

Go Pregnant or Go Home

Failed to fetch jessie backports repository

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

How was Earth single-handedly capable of creating 3 of the 4 gods of chaos?

Will it be accepted, if there is no ''Main Character" stereotype?

How to be diplomatic in refusing to write code that breaches the privacy of our users

At which point does a character regain all their Hit Dice?

The baby cries all morning

Greatest common substring

What's the purpose of "true" in bash "if sudo true; then"

What is the oldest known work of fiction?

How can I replace every global instance of "x[2]" with "x_2"

Why Were Madagascar and New Zealand Discovered So Late?

Why does John Bercow say “unlock” after reading out the results of a vote?

What's a natural way to say that someone works somewhere (for a job)?

Why did Kant, Hegel, and Adorno leave some words and phrases in the Greek alphabet?

Curses work by shouting - How to avoid collateral damage?

What to do with wrong results in talks?

How do I keep an essay about "feeling flat" from feeling flat?

What defines a dissertation?

Efficiently merge handle parallel feature branches in SFDX

Opposite of a diet



Ways to speed up user implemented RK4


Speed up Numerical IntegrationSpeed of convergence for NIntegrateTough Calculation, novice mathematica userNumerical integration's speedNumerical integral speedImprove the speed of Gaussian quadrature integrationSolving an unstable BVP numerically, accurately and efficientlyHow to speed up integral of results of PDE modelSolve BVP involving user defined functionUser defined ArcTan function













3












$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question









$endgroup$







  • 1




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 hours ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    @HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
    $endgroup$
    – Shinaolord
    2 hours ago















3












$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question









$endgroup$







  • 1




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 hours ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    @HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
    $endgroup$
    – Shinaolord
    2 hours ago













3












3








3


1



$begingroup$


So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!










share|improve this question









$endgroup$




So, I've implemented RK4, and I'm wondering what I can do to make it more efficient? What I've got so far is below. I wish to still record all steps. I think AppendTo is doing the most damage to the time, is there a faster alternative?



rk4[f_, variables_, valtinit_, tinit_, tfinal_, nsteps_] := 
Module[table, xlist, ylist, step, k1, k2, k3, k4,
xlist = tinit;
step = N[(tfinal - tinit)/(nsteps)];
ylist = valtinit;
table = xlist, ylist;
Table[
k1 = step* f /. MapThread[Rule, variables, ylist]; (*
Equivalent to step* f/.Thread[Rule[variables,ylist]]*)
k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist];
k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist];
k4 = step*f /. MapThread[Rule, variables, k3 + ylist];
ylist += 1/6 (k1 + 2 (k2 + k3) + k4);
xlist += step;
AppendTo[table, xlist, ylist];
xlist, ylist, nsteps];
table
];


Example Input:



funclist = -x + y, x - y;
initials = 1, 2;
variables = x, y;
init = 0;
final = 200;
nstep = 20000;
approx = rk4[funclist, variables, initials, init, final, nstep]//AbsoluteTiming;



3.59932,...




I'd love some suggestions!







differential-equations numerical-integration






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









ShinaolordShinaolord

808




808







  • 1




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 hours ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    @HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
    $endgroup$
    – Shinaolord
    2 hours ago












  • 1




    $begingroup$
    AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
    $endgroup$
    – b3m2a1
    2 hours ago










  • $begingroup$
    I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    @HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
    $endgroup$
    – Shinaolord
    2 hours ago







1




1




$begingroup$
AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
$endgroup$
– b3m2a1
2 hours ago




$begingroup$
AppendTo is quadratic time complexity. Might be better to preallocate and set by index. Also it'll be much faster to not use Rule and instead code stuff up a little bit more explicitly. As a general rule, too, use vectorized operators. Those can be very fast. And if everything can be totally functional over "packed arrays" (look them up here) it'll be very quick too.
$endgroup$
– b3m2a1
2 hours ago












$begingroup$
I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
$endgroup$
– Shinaolord
2 hours ago




$begingroup$
I'll work on implementing it more explicity, this is what came to find first though. It'll require some changes to the inputs, I'll have to ponder this. And preallocating the list is a quick change that won't be an issue to do, I can't believe I forgot that's faster :(. Thanks though!
$endgroup$
– Shinaolord
2 hours ago












$begingroup$
Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
$endgroup$
– Henrik Schumacher
2 hours ago




$begingroup$
Shinaoloard, using Join[ xlist, ylist, Table[ k1 = step*f /. MapThread[Rule, variables, ylist]; k2 = step*f /. MapThread[Rule, variables, k1/2 + ylist]; k3 = step*f /. MapThread[Rule, variables, k2/2 + ylist]; k4 = step*f /. MapThread[Rule, variables, k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps ] ] as return value is already a first step. There is no point in appending if you use a Table anyways.
$endgroup$
– Henrik Schumacher
2 hours ago












$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
$endgroup$
– Shinaolord
2 hours ago




$begingroup$
@HenrikSchumacher do you think it would be faster to Pre-allocate a list of length nsteps, and append the values, or to join the values using table? I can obviously change Table to Do to remove the time it takes to make the table list, going by b3m2a1's method, or I could use Join as you have suggested. I'm thinking your method may be faster, though. I've already removed the MapThread part, I am testing the speed increase granted by that at the moment. Just curious which path you think will be faster.
$endgroup$
– Shinaolord
2 hours ago












$begingroup$
I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
2 hours ago




$begingroup$
I am currently testing the speed difference between the one in the post and rk4t2[f_, valtinit_, tinit_, tfinal_, nsteps_] := Module[test, table, xlist, ylist, step, k1, k2, k3, k4, xlist = tinit; step = N[(tfinal - tinit)/(nsteps)]; ylist = valtinit; table = xlist, ylist; test = Table[ k1 = step* f[ylist] ; k2 = step*f[k1/2 + ylist]; k3 = step*f[k2/2 + ylist]; k4 = step*f[k3 + ylist]; ylist += 1/6 (k1 + 2 (k2 + k3) + k4); xlist += step; xlist, ylist, nsteps]; Join[table, test] ];
$endgroup$
– Shinaolord
2 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vectorfield F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 2 million times with NestList and still need only 2 seconds.



nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First



2.08678







share|improve this answer









$endgroup$












  • $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 hours ago










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194002%2fways-to-speed-up-user-implemented-rk4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vectorfield F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 2 million times with NestList and still need only 2 seconds.



nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First



2.08678







share|improve this answer









$endgroup$












  • $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 hours ago















4












$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vectorfield F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 2 million times with NestList and still need only 2 seconds.



nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First



2.08678







share|improve this answer









$endgroup$












  • $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 hours ago













4












4








4





$begingroup$

Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vectorfield F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 2 million times with NestList and still need only 2 seconds.



nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First



2.08678







share|improve this answer









$endgroup$



Just to give you an impression how fast things may get when you use the right tools.



For given stepsize τ and given vectorfield F, this creates a CompiledFunction cStep that computes a single Runge-Kutta step



F = X [Function] -Indexed[X, 2], Indexed[X, 1];

τ = 0.01;
Block[YY, Y, k1, k2, k3, k4,

YY = Table[Compile`GetElement[Y, i], i, 1, 2];
k1 = τ F[YY];
k2 = τ F[0.5 k1 + YY];
k3 = τ F[0.5 k2 + YY];
k4 = τ F[k3 + YY];

cStep = With[code = YY + (k1 + 2. (k2 + k3) + k4)/6. ,
Compile[Y, _Real, 1,
code,
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
]
]
];


Now we can apply it 2 million times with NestList and still need only 2 seconds.



nsteps = 20000000;
xlist = Range[0., step nsteps, step];
Ylist = NestList[cStep, initials, nsteps]; // AbsoluteTiming // First



2.08678








share|improve this answer












share|improve this answer



share|improve this answer










answered 2 hours ago









Henrik SchumacherHenrik Schumacher

57.9k579159




57.9k579159











  • $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 hours ago
















  • $begingroup$
    Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
    $endgroup$
    – Shinaolord
    2 hours ago










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    2 hours ago










  • $begingroup$
    I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
    $endgroup$
    – Shinaolord
    2 hours ago















$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
2 hours ago




$begingroup$
Damn, you definitely know how to use Mathematica A LOT more efficiently than I do. Thanks!
$endgroup$
– Shinaolord
2 hours ago












$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours ago




$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours ago












$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
2 hours ago




$begingroup$
I'll have to play around with Compile, it definitely seems like a massive speed up if used correctly.
$endgroup$
– Shinaolord
2 hours ago

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194002%2fways-to-speed-up-user-implemented-rk4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

कुँवर स्रोत दिक्चालन सूची"कुँवर""राणा कुँवरके वंशावली"

Why is a white electrical wire connected to 2 black wires?How to wire a light fixture with 3 white wires in box?How should I wire a ceiling fan when there's only three wires in the box?Two white, two black, two ground, and red wire in ceiling box connected to switchWhy is there a white wire connected to multiple black wires in my light box?How to wire a light with two white wires and one black wireReplace light switch connected to a power outlet with dimmer - two black wires to one black and redHow to wire a light with multiple black/white/green wires from the ceiling?Ceiling box has 2 black and white wires but fan/ light only has 1 of eachWhy neutral wire connected to load wire?Switch with 2 black, 2 white, 2 ground and 1 red wire connected to ceiling light and a receptacle?

चैत्य भूमि चित्र दीर्घा सन्दर्भ बाहरी कडियाँ दिक्चालन सूची"Chaitya Bhoomi""Chaitya Bhoomi: Statue of Equality in India""Dadar Chaitya Bhoomi: Statue of Equality in India""Ambedkar memorial: Centre okays transfer of Indu Mill land"चैत्यभमि