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

Modify casing of marked letters

How to verify if g is a generator for p?

Your magic is very sketchy

Is there a good way to store credentials outside of a password manager?

Lay out the Carpet

Teaching indefinite integrals that require special-casing

How does residential electricity work?

How does it work when somebody invests in my business?

Using parameter substitution on a Bash array

What is the oldest known work of fiction?

Can criminal fraud exist without damages?

Transcription Beats per minute

when is out of tune ok?

Can I Retrieve Email Addresses from BCC?

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

Why are on-board computers allowed to change controls without notifying the pilots?

Greatest common substring

Can a monster with multiattack use this ability if they are missing a limb?

Mapping a list into a phase plot

Is it okay / does it make sense for another player to join a running game of Munchkin?

Was the picture area of a CRT a parallelogram (instead of a true rectangle)?

What would be the benefits of having both a state and local currencies?

Is a roofing delivery truck likely to crack my driveway slab?

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



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

Lioubotyn Sommaire Géographie | Histoire | Population | Notes et références | Liens externes | Menu de navigationlubotin.kharkov.uamodifier« Recensements et estimations de la population depuis 1897 »« Office des statistiques d'Ukraine : population au 1er janvier 2010, 2011 et 2012 »« Office des statistiques d'Ukraine : population au 1er janvier 2011, 2012 et 2013 »Informations officiellesCartes topographiquesCarte routièrem

Isabella Eugénie Boyer Biographie | Références | Menu de navigationmodifiermodifier le codeComparator to Compute the Relative Value of a U.S. Dollar Amount – 1774 to Present.

Mpande kaSenzangakhona Biographie | Références | Menu de navigationmodifierMpande kaSenzangakhonavoir la liste des auteursm