Warum bekomme ich so viele Fehler für meinen Runge Kutta Fehlberg-Solver?

Mein aktuelles Projekt ist die Neuprogrammierung eines Proteinfaltungsmodells, bei dem Tausende von ODEs in C ++ gelöst werden. Ich habe einige Stopp- und Startfortschritte gemacht, während ich den Solver schreibe, der vollständig auf der GPU ausgeführt werden soll. Ich habe es endlich integriert, aber wenn ich versuche, dC / dt = -C mit der Lösung fünfter Ordnung aus dem RKF45-Algorithmus mit einer festen Schrittgröße von h = 0,001 zu lösen, erhalte ich eine Entscheidung aus dem Taschenrechnerwert von e ^ {- t } in der Größenordnung von 10 ^ {- 4}, wenn ein globaler Fehler in der Größenordnung von 10 ^ {- 12} erwartet wird.

Liegt dies daran, dass ich keine adaptiven Schrittgrößen und keine Fehlerkontrolle mit Lösungen der vierten und fünften Ordnung verwende? Mein Gedanke war, dass ich, da ein adaptiver Algorithmus den Unterschied zwischen den Vermutungen vierter und fünfter Ordnung betrachtet und die fünfte Ordnung im Grunde genommen als "richtige Antwort" behandelt, diese Antwort nur verwenden würde, um den Integrator auszuprobieren.

BEARBEITEN:

Ich dachte das seitdem

Globaler Fehler = (Anzahl der Punkte) * (lokaler Fehler)

und

Lokaler Fehler = h ^ O.

wobei h die Schrittgröße und O die Reihenfolge ist, berechne ich

O = ln (globaler Fehler / Anzahl der Punkte) / ln h

Wenn ich eine Bestellanalyse sowohl so durchführe, wie ich es für richtig gehalten habe, als auch wie in der Antwort vorgeschlagen, erhalte ich Folgendes:

Number of Points    h       Global Error    My Way      The Answer's Way
10                  0.1     2.89E-06        6.539E+00   
20                  0.05    7.09E-08        6.495E+00   5.350E+00
40                  0.025   1.76E-07        5.216E+00   -1.315E+00
50                  0.02    1.128E-07       5.089E+00   2.003E+00
80                  0.0125  4.401E-08       4.866E+00   2.002E+00
100                 0.01    2.816E-08       4.775E+00   2.001E+00
500                 0.002   1.126E-09       4.316E+00   2.000E+00
1000                0.001   2.810E-10       4.184E+00   2.003E+00
10000               0.0001  3.000E-12       3.881E+00   1.972E+00

Trotz meines Quasi-Durchbruchs letzte Nacht, bei dem ich tatsächlich einen Programmierfehler hatte, ist die Reihenfolge meines Lösers immer noch falsch. Der oben erwähnte Fehler, der mit der Schrittgröße 0,001 verbunden ist, hat sich jedoch dramatisch von O (10 ^ -4) auf O (10 ^ -10) verringert.

2. BEARBEITEN

Hier ist ein Diagramm von LN (E) -LN (a) gegen LN (h). Ich warf mich auf eine lineare Anpassung, damit ich erwarten würde, dass mir die Reihenfolge offenbart wird. Was jetzt wirklich verwirrend ist, ist, dass ich drei Berechnungen für die Reihenfolge meiner Methode habe. Die Grafik sagt mir ~ 3, Godrics Methode sagt mir fast genau 2 und meine Methode zeigt, dass sie sich von 6-> 4 ändert, um die Schrittgröße zu erhöhen. Ich bin eher geneigt, meinen Berechnungen zu glauben, weil der globale Fehler so niedrig ist (~ 10 ^ -10) und dem am nächsten kommt, was ich für diese Schrittgröße erwarten würde (10 ^ -15 * 10 ^ 3 = 10 ^ -12) (10) ^ -3).

EDIT 3:

Mein Code. Verzeihen Sie die Noobishness.

#include <cuda.h>
#include <cuda_runtime.h>
#include <device_launch_parameters.h>
#include <stdio.h>
#include <iostream>
#include <iomanip>                      //display 2 decimal places
#include <math.h>
using namespace std;

/*
Things to do log:
1. Decide how adaptive step sizes is handled by the integrator loop.  We can no longer say how many points to integrate if that is to be determined in the kernel
itself, only what the initial step size is.
2. Figure out how I'm going to graph all this stuff.
3. On paper solve for the step size.
4. Adaptive Step Size
    a. arrMin kernel to get min of array
    b. Decide how to implement tolerances
5. Use multiples of 256 for maxlength 

Important things accomplished:
1. Can create 2D array for storing concentrations over time.
2. Can store a 1D array in 2D array.
3. Can calculate all the k-arrays given static step size.
4. Can take a step given static step size.
5. Can step through an integration from t0 to tf
6. Can accomodate fixed time-step
7. Calculated the current order of integrator using logs.
8. Learned how to profile the program.
9. Fixed the GPU timeout problem with WDDM.

Ideas:
1. Just use large fixed size array
*/
__global__ void rkf5(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*, int*, size_t, double*, double*, double*);
__global__ void calcK(double*, double*, double*);
__global__ void k1(double*, double*, double*);
__global__ void k2(double*, double*, double*);
__global__ void k3(double*, double*, double*);
__global__ void k4(double*, double*, double*);
__global__ void k5(double*, double*, double*);
__global__ void k6(double*, double*, double*);
__global__ void arrAdd(double*, double*, double*);
__global__ void arrSub(double*, double*, double*);
__global__ void arrMult(double*, double*, double*);
__global__ void arrInit(double*, double);
__global__ void arrCopy(double*, double*);
__device__ void setup(double , double*, double*, double*, double*, int*);
__device__ double flux(int, double*) ;
__device__ double knowles_flux(int, double*);
__device__ void calcStepSize(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*);
__global__ void storeConcs(double*, size_t, double*, int);
__global__ void takeFourthOrderStep(double*, double*, double*, double*, double*, double*, double*);
__global__ void takeFifthOrderStep(double*, double*, double*, double*, double*, double*, double*, double*);

//Error checking that I don't understand yet.
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
   if (code != cudaSuccess) 
   {
      fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
      if (abort) exit(code);
   }
}

//Main program.
int main(int argc, char** argv)
{
    //std::cout << std::fixed;          //display 2 decimal places
    //std::cout << std::setprecision(12);   //display 2 decimal places
    const int maxlength = 1;            //Number of discrete concentrations we are tracking.
    double concs[maxlength];            //Meant to store the current concentrations 
    double temp1[maxlength];                //Used as a bin to store products of Butcher's tableau and k values.
    double temp2[maxlength];                //Used as a bin to store products of Butcher's tableau and k values.
    double tempsum[maxlength];          //Used as a bin to store cumulative sum of tableau and k values
    double k1s[maxlength];
    double k2s[maxlength];
    double k3s[maxlength];
    double k4s[maxlength];
    double k5s[maxlength];
    double k6s[maxlength];
    const int numpoints = 40;       
    double to = 0;
    double tf = 1;
    //double dt = static_cast<double>(.5)/static_cast<double>(64);
    double dt = (tf-to)/static_cast<double>(numpoints);
    double mo = 1;
    double concStorage[maxlength][numpoints];   //Stores concs vs. time                     

    //Initialize all the arrays on the host to ensure arrays of 0's are sent to the device.
    //Also, here is where we can seed the system.
    std::cout<<dt;
    std::cout<<"\n";
    concs[0]=mo;
    std::cout<<concs[0];
    std::cout<<" ";
    for (int i=0; i<maxlength; i++)
    {
        for (int j=0; j<numpoints; j++)
            concStorage[i][j]=0;
        concs[i]=0;
        temp1[i]=0;
        temp2[i]=0;
        tempsum[i]=0;
        k1s[i]=0;
        k2s[i]=0;
        k3s[i]=0;
        k4s[i]=0;
        k5s[i]=0;
        k6s[i]=0;
        std::cout<<concs[i];
        std::cout<<" ";
    }
    concs[0]=mo;
    std::cout<<"\n";

    //Define all the pointers to device array memory addresses. These contain the on-GPU
    //addresses of all the data we're generating/using.
    double *d_concs;
    double *d_temp1;
    double *d_temp2;
    double *d_tempsum;
    double *d_k1s;
    double *d_k2s;
    double *d_k3s;
    double *d_k4s;
    double *d_k5s;
    double *d_k6s;
    double *d_dt;
    int *d_maxlength;
    int *d_numpoints;
    double *d_to;
    double *d_tf;
    double *d_concStorage;

    //Calculate all the sizes of the arrays in order to allocate the proper amount of memory on the GPU.
    size_t size_concs = sizeof(concs);
    size_t size_temp1 = sizeof(temp1);
    size_t size_temp2 = sizeof(temp2);
    size_t size_tempsum = sizeof(tempsum);
    size_t size_ks = sizeof(k1s);
    size_t size_maxlength = sizeof(maxlength);
    size_t size_numpoints = sizeof(numpoints);
    size_t size_dt = sizeof(dt);
    size_t size_to = sizeof(to);
    size_t size_tf = sizeof(tf);
    size_t h_pitch = numpoints*sizeof(double);
    size_t d_pitch;

    //Calculate the "pitch" of the 2D array.  The pitch is basically the length of a 2D array's row.  IT's larger 
    //than the actual row full of data due to hadware issues.  We thusly will use the pitch instead of the data 
    //size to traverse the array.
    gpuErrchk(cudaMallocPitch( (void**)&d_concStorage, &d_pitch, numpoints * sizeof(double), maxlength)); 

    //Allocate memory on the GPU for all the arrrays we're going to use in the integrator.
    gpuErrchk(cudaMalloc((void**)&d_concs, size_concs));
    gpuErrchk(cudaMalloc((void**)&d_temp1, size_temp1));
    gpuErrchk(cudaMalloc((void**)&d_temp2, size_temp1));
    gpuErrchk(cudaMalloc((void**)&d_tempsum, size_tempsum));
    gpuErrchk(cudaMalloc((void**)&d_k1s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k2s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k3s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k4s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k5s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k6s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_maxlength, size_maxlength));
    gpuErrchk(cudaMalloc((void**)&d_numpoints, size_numpoints));
    gpuErrchk(cudaMalloc((void**)&d_dt, size_dt));
    gpuErrchk(cudaMalloc((void**)&d_to, size_to));
    gpuErrchk(cudaMalloc((void**)&d_tf, size_tf));

    //Copy all initial values of arrays to GPU.
    gpuErrchk(cudaMemcpy2D(d_concStorage, d_pitch, concStorage, h_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_concs, &concs, size_concs, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_temp1, &temp1, size_temp1, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_temp2, &temp2, size_temp2, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_tempsum, &tempsum, size_tempsum, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k1s, &k1s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k2s, &k2s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k3s, &k3s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k4s, &k4s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k5s, &k5s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k6s, &k6s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_maxlength, &maxlength, size_maxlength, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_numpoints, &numpoints, size_numpoints, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_dt, &dt, size_dt, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_to, &to, size_to, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_tf, &tf, size_tf, cudaMemcpyHostToDevice));

    //Run the integrator.
    rkf5<<<1,1>>>(d_concs, d_concStorage, d_temp1, d_temp2, d_tempsum, d_k1s, d_k2s, d_k3s, d_k4s, d_k5s, d_k6s, d_maxlength, d_numpoints, d_pitch, d_dt, d_to, d_tf);
    gpuErrchk( cudaPeekAtLastError() );
    gpuErrchk( cudaDeviceSynchronize() );
    cudaDeviceSynchronize();

    //Copy concentrations from GPU to Host.  Almost defunct now that transferring the 2D array works.
    cudaMemcpy(concs, d_concs, size_concs, cudaMemcpyDeviceToHost);
    //Copy 2D array of concentrations vs. time from GPU to Host.
    gpuErrchk( cudaMemcpy2D(concStorage, h_pitch, d_concStorage, d_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyDeviceToHost) );   

    //Print concentrations after the integrator kernel runs.  Used to test that data was transferring to and from GPU correctly.
    std::cout << "\n";
    for (int i=0; i<maxlength; i++)
    {
        std::cout<<concs[i];
        std::cout<<" ";
    }

    double a[10];
    double b[10];
    double c[10];
    for(int i = 0; i< 10; i++)
    {
        a[i]=0;
        b[i]=0;
        c[i]=0;
    }



    //Print out the concStorage array after the kernel runs.  Used to test that the 2D array transferred correctly from host to GPU and back.
    std::cout << "\n\n";
    std::cout << "Calculated Array";
    std::cout << "\n\n";
    for (int i=0; i<maxlength; i++)
    {
        for(int j=0; j<numpoints; j++)
        {
            if (j%(numpoints/10)==0)
            {
                a[j/(numpoints/10)]=concStorage[i][j];
                std::cout<<concStorage[i][j];
                std::cout<<"   ";
            }
        }
        std::cout << "\n";
    }
    std::cout << "\n";
    std::cout << "Exponential";
    std::cout << "\n\n";
    for (int i=0; i<10; i++)
    {
        b[i]=exp(-i*(tf-to)/10);
        std::cout<<exp(-i*(tf-to)/10);
        std::cout<<"   ";
    }
    std::cout << "\n\n";
    std::cout << "Error Array";
    std::cout << "\n\n";
    for (int i=0; i<10; i++)
    {
        c[i]=a[i]-b[i];
        std::cout<<c[i];
        std::cout<<"   ";
    }
    std::cout << "\n\n";
    cudaDeviceReset();  //Clean up all memory.

    return 0;
}
//Main kernel.  This is mean to be run as a master thread that calls all the other functions and thusly "runs" the integrator.
__global__ void rkf5(double* concs, double* concStorage, double* temp1, double* temp2, double* tempsum, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, int* maxlength, int* numpoints, size_t pitch, double* dt, double* to, double* tf)
{
    /*
    axy variables represent the coefficients in the Butcher's tableau where x represents the order of the step and the y value corresponds to the ky value 
    the coefficient gets multiplied by.  Have to cast them all as doubles, or the ratios evaluate as integers.
    e.g. a21 -> a21 * k1
    e.g. a31 -> a31 * k1 + a32 * k2
    */
    double a21 = static_cast<double>(.25);

    double a31 = static_cast<double>(3)/static_cast<double>(32);
    double a32 = static_cast<double>(9)/static_cast<double>(32);

    double a41 = static_cast<double>(1932)/static_cast<double>(2197);
    double a42 = static_cast<double>(-7200)/static_cast<double>(2197);
    double a43 = static_cast<double>(7296)/static_cast<double>(2197);

    double a51 = static_cast<double>(439)/static_cast<double>(216);
    double a52 = static_cast<double>(-8);
    double a53 = static_cast<double>(3680)/static_cast<double>(513);
    double a54 = static_cast<double>(-845)/static_cast<double>(4104);

    double a61 = static_cast<double>(-8)/static_cast<double>(27);
    double a62 = static_cast<double>(2);
    double a63 = static_cast<double>(-3544)/static_cast<double>(2565);
    double a64 = static_cast<double>(1859)/static_cast<double>(4104);
    double a65 = static_cast<double>(-11)/static_cast<double>(40);

    //for loop that integrates over the specified number of points. Actually, might have to make it a do-while loop for adaptive step sizes 
    for(int k = 0; k < *numpoints; k++)
    {
        if (k!=0)
        {
            arrCopy<<< 1, *maxlength >>>(concs, tempsum);
            cudaDeviceSynchronize();
        }
        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        arrInit<<< 1, *maxlength >>>(temp1, 0);
        cudaDeviceSynchronize();
        arrInit<<< 1, *maxlength >>>(temp2, 0);
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(concs, k1s, dt);         //k1 = dt * flux (concs)
        cudaDeviceSynchronize(); //Sync here because kernel continues onto next line before k1 finished

        setup(a21, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a21*k1
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a21*k1    
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k2s, dt);       //k2 = dt * flux (concs + a21*k1)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a31, temp1, tempsum, k1s, concs, maxlength);  //temp1sum = a31*k1
        setup(a32, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a31*k1 + a32*k2
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a31*k1 + a32*k2
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k3s, dt);           //k3 = dt * flux (concs + a31*k1 + a32*k2)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a41, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a41*k1
        setup(a42, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a41*k1 + a42*k2
        setup(a43, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a41*k1 + a42*k2 + a43*k3
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a41*k1 + a42*k2 + a43*k3
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k4s, dt);           //k4 = dt * flux (concs + a41*k1 + a42*k2 + a43*k3)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a51, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a51*k1
        setup(a52, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a51*k1 + a52*k2
        setup(a53, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a51*k1 + a52*k2 + a53*k3
        setup(a54, temp1, tempsum, k4s, concs, maxlength);  //tempsum = a51*k1 + a52*k2 + a53*k3 + a54*k4
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a51*k1 + a52*k2 + a53*k3 + a54*k4
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k5s, dt);           //k5 = dt * flux (concs + a51*k1 + a52*k2 + a53*k3 + a54*k4)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a61, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a61*k1
        setup(a62, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a61*k1 + a62*k2
        setup(a63, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3
        setup(a64, temp1, tempsum, k4s, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4
        setup(a65, temp1, tempsum, **k4s**, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k6s, dt);           //k6 = dt * flux (concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5)
        cudaDeviceSynchronize();

        //At this point, temp1 and tempsum are maxlength dimension arrays that are able to be used for other things.

        //Calculate acceptable step size before storing the concentrations.
        calcStepSize(temp1, temp2, tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s, dt, maxlength);    //temp1 = 4th Order guess, tempsum = 5th Order guess
        cudaDeviceSynchronize();

        //Store the initial conditions in the first column of the storage array.
        if (k==0)
        {
            storeConcs<<< 1, *maxlength >>>(concStorage, pitch, concs, k);  //Store this step's concentrations in 2D array
            cudaDeviceSynchronize();
        }
        //Store future concentration in next column of storage array.
        storeConcs<<< 1, *maxlength >>>(concStorage, pitch, tempsum, k+1);  //Store this step's concentrations in 2D array
        cudaDeviceSynchronize();
    }
}
//calcStepSize will take in an error tolerance, the current concentrations and the k values and calculate the resulting step size according to the following equation
//e[n+1]=y4[n+1] - y5[n+1]
__device__ void calcStepSize(double* temp1, double*temp2, double* tempsum, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, double* dt, int* maxlength)
{
    //do
    //{
        takeFourthOrderStep<<< 1, *maxlength >>>(temp1, concs, k1s, k2s, k3s, k4s, k5s);            //Store 4th order guess in temp1
        takeFifthOrderStep<<< 1, *maxlength >>>(tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s);  //Store 5th order guess in tempsum
        cudaDeviceSynchronize();
        arrSub<<< 1, *maxlength >>>(temp1, tempsum, temp2)
        arrMin<<< 1, *maxlength >>>
    //arrMult
    //}
    //while
}
//takeFourthOrderStep is going to overwrite the old temp1 array with the new array of concentrations that result from a 4th order step.  This kernel is meant to be launched 
// with as many threads as there are discrete concentrations to be tracked.
__global__ void takeFourthOrderStep(double* y4, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s)
{
    double b41 = static_cast<double>(25)/static_cast<double>(216);
    double b42 = static_cast<double>(0);
    double b43 = static_cast<double>(1408)/static_cast<double>(2565);
    double b44 = static_cast<double>(2197)/static_cast<double>(4104);
    double b45 = static_cast<double>(-1)/static_cast<double>(5);
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    y4[idx] = concs[idx] + b41 * k1s[idx] + b42 * k2s[idx] + b43 * k3s[idx] + b44 * k4s[idx] + b45 * k5s[idx];
}
//takeFifthOrderStep is going to overwrite the old array of concentrations with the new array of concentrations.  As of now, this will be the 5th order step.  Another function can be d
//defined that will take a fourth order step if that is interesting for any reason.  This kernel is meant to be launched with as many threads as there are discrete concentrations
//to be tracked.
//Store b values in register? Constants?
__global__ void takeFifthOrderStep(double* y5, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s)
{
    double b51 = static_cast<double>(16)/static_cast<double>(135);
    double b52 = static_cast<double>(0);
    double b53 = static_cast<double>(6656)/static_cast<double>(12825);
    double b54 = static_cast<double>(28561)/static_cast<double>(56430);
    double b55 = static_cast<double>(-9)/static_cast<double>(50);
    double b56 = static_cast<double>(2)/static_cast<double>(55);
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    y5[idx] = concs[idx] + b51 * k1s[idx] + b52 * k2s[idx] + b53 * k3s[idx] + b54 * k4s[idx] + b55 * k5s[idx] + b56 * k6s[idx];
}
//storeConcs takes the current array of concentrations and stores it in the cId'th column of the 2D concStorage array
//pitch = memory size of a row
__global__ void storeConcs(double* cS, size_t pitch, double* concs, int cId)
{
    int tIdx = threadIdx.x;
    //cS is basically the memory address of the first element of the flattened (1D) 2D array.
    double* row = (double*)((char*)cS + tIdx * pitch);
    row[cId] = concs[tIdx];
}
//Perhaps I can optimize by using shared memory to hold conc values.
__global__ void calcK(double* concs, double* ks, double* dt)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    ks[idx]=(*dt)*flux(idx, concs);
}
//Adds two arrays (a + b) element by element and stores the result in array c.
__global__ void arrAdd(double* a, double* b, double* c)
{                                                 
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]+b[idx];
}
//Subtracts two arrays (a - b) element by element and stores the result in array c.
__global__ void arrSub(double* a, double* b, double* c)
{                                                 
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]-b[idx];
}
//Multiplies two arrays (a * b) element by element and stores the result in array c.
__global__ void arrMult(double* a, double* b, double* c)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]*b[idx];
}
//Will find the min of errors array.
__global__ void arrMin(double* errors)
{
    //extern _shared_ double[7];
}
//Initializes an array a to double value b.
__global__ void arrInit(double* a, double b)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    a[idx]=b;
}
//Copies array b onto array a.
__global__ void arrCopy(double* a, double* b)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    a[idx]=b[idx];
}
//Placeholder function for the flux calculation.  It will take the size of the oligomer and current concentrations as inputs.
__device__ double flux(int r, double *concs) 
{
    return -concs[r];
}
//This function multiplies a tableau value by the corresponding k array and adds the result to tempsum.  Used to
//add all the a*k terms.
//e.g. setup(a21, temp1, tempsum, k1s, concs, maxlength) => tempsum = a21 * k1
__device__ void setup(double tableauValue, double *temp1, double *tempsum, double *ks, double *concs, int *maxlength) 
{
    //Sets tempsum to tabVal * k
    arrInit<<< 1, *maxlength >>>(temp1, tableauValue);      //Set [temp1] to tableau value, temp1 = a
    cudaDeviceSynchronize();
    arrMult<<< 1, *maxlength >>>(ks, temp1, temp1);         //Multiply tableau value by appropriate [k], temp1 = a*k
    cudaDeviceSynchronize();
    arrAdd<<< 1, *maxlength >>>(tempsum, temp1, tempsum);   //Move tabVal*k to [tempsum], tempsum = tempsum+temp1
    cudaDeviceSynchronize();
    //temp1 = tableauValue * kArray
    //tempsum = current sum (tableauValue * kArray)
}

FINAL EDIT Ich habe meinen Fehler gefunden. Es war ein Tippfehler bei der Berechnung der K6. Ich setze Sternchen um den beleidigenden Begriff "k4s", der ein "k5s" sein sollte. Jeder, der einen CUDA RKF45-Integrator möchte, kann ihn gerne verwenden, da er jetzt funktioniert und in der richtigen Reihenfolge ist.

Number of Points    h       Global Error    Calculated Order
        10          0.1         3.59E-09    
        20          0.05        1.08E-10    5.06E+00
        40          0.025       3.31E-12    5.03E+00
        80          0.0125      1.03E-13    5.01E+00
        160         0.00625     2.20E-15    5.55E+00
        320         0.003125    1.20E-15    8.74E-01
        640         0.0015625   1.40E-15    -2.22E-01
        1280        0.00078125  2.00E-16    2.81E+00

Ich gehe davon aus, dass Sie die Fehlertoleranz auf einstellen 1e-12. Sie haben Recht, wenn ein adaptives Schema die aktuelle Schrittgröße akzeptiert, wird davon ausgegangen, dass das Schema 5. Ordnung in jeder Hinsicht die "richtige" Antwort war. Dies ist jedoch nur dann der Fall, wenn der aktuelle Schritt akzeptiert wird. Wenn die Differenz zwischen dem 4. und 5. Ordnung Schritte zu groß sind, akzeptiert es nicht entweder als gut genug. Um einen einzelnen Algorithmus wie diesen zu testen, müssen Sie ihn mit mehreren Zeitschritten ausführen und die Reihenfolge der Konvergenz anzeigen, um ihn zu überprüfen.

Auf der anderen Seite sollte ein Schritt 0.001mit einem Schema 5. Ordnung für Ihr Testproblem (ich nehme an, es war 1D) mehr als ausreichend sein, um den Fehler auf nahezu Maschinen-Epsilon zu bringen. Ich vermute, dass Sie irgendwo einen Tippfehler haben, der Ihren Integrator auf die 1. Ordnung reduziert. Führen Sie es, wie in Absatz 1 erwähnt, mit mehreren Zeitschritten aus und überprüfen Sie die Konvergenzreihenfolge. Hier erfahren Sie, ob der angezeigte Fehler angemessen ist oder ob irgendwo in Ihrem Integrator ein Fehler vorliegt.

BEARBEITEN

$h$$\frac{h}{2}$

$\frac{E(h)}{E(\frac{h}{2})} = 2^n$

$n=5$