C++ Segmentation Fault on Differential Solver - c++

Question

I have written a program to approximate the solutions to ordinary differential equations using Adam's Method.
Running the program with gdb gives me:
Program received signal EXC_BAD_ACCESS, Could not access memory.
Reason: KERN_PROTECTION_FAILURE at address: 0x00007fff5f3ffff8
0x0000000100003977 in std::vector<double, std::allocator<double> >::push_back (this=0x100005420, __x=#0x100005310) at stl_vector.h:604
604 this->_M_impl.construct(this->_M_impl._M_finish, __x);
Clearly something is wrong with my treatment of vector.push_back, but I do not know where to begin looking. I can not think of a case where modifying a vector is illegal.
Call differentiate() to begin. Mathematics is done in step(). Adaptive time advancing across an interval with advance(). Check the chosen time step with checkdt() before running step() again.
Sorry for the huge code dump. I am sure many improvements can be made purely from the standpoint of C++, without knowledge of the mathematics:
//============================================================================
// Description : An Adam's Method Ordinary Differential Equation Approximation
// Disclaimer : Posted to StackOverflow for help on a segmentation fault
//============================================================================
#include <iostream> //IO
#include <vector> //std::vector
#include <cmath> //abs, pow, sqrt
#include <numeric> //accumulate
using namespace std;
/* Terminology:
* f(t, y) = the ordinary differential equation that will be solved
* y(t) = solution of f at point t.
* told = the previous point at which f was evaluated and solved
* tnow = the current point at which f is evaluated and solved
* tnew = the new (interesting) point at which f will be evaluated and solved
*
* Yold = the corrected solution of the differential equation at told
* Ynow = the corrected solution of the differential equation at tnow
* Ynew = the corrected solution of the differential equation at tnew
*
* fold = the value of the given differential equation at told
= f(told, Yold)
* fnow = the value of the given differential equation at tnow
= f(tnow, Ynow)
* fnew = the value of the given differential equation at tnew
= f(tnew, Ynew)
*
* Pnew = prediction for the value of Ynew
* dt = abs(tnew - tnow)
* dtold = abs(tnow - told)
*/
//Information storage
struct simTime {
double told;
double tnow;
double tnew;
double dt;
double dtold;
double tol;
double agrow;
double ashrink;
double dtmin;
double dtmax;
double endTime;
double fold;
double fnow;
double fnew;
double Yold;
double Ynow;
double Ynew;
double Pnew;
int stepsSinceRejection;
int stepsRejected;
int stepsAccepted;
} test;
//Define global variables
std::vector<double> errorIndicators(0);
std::vector<double> solutions(0);
std::vector<double> differencesDDY(0);
std::vector<double> differencesDDYSquared(0);
std::vector<double> timesTNew(0);
//Function declarations
void step(double fnow, double fold, double Ynow, double tnew, double tnow,
double dtold, double dt, double(*f)(double t, double y), double told);
void advance();
void shiftvariables();
void printvector(std::vector<double>& vec);
void differentiate(double(*f)(double t, double y), double y0, double a,
double b);
double f(double t, double y);
void checkdt();
int main() {
differentiate(f, 0, 1, 5);
cout << "Time values:" << endl;
printvector(timesTNew);
cout << "Solutions:" << endl;
printvector(solutions);
cout << "Differences between Prediction and Solution:" << endl;
printvector(differencesDDY);
return 0;
}
//Shift back all the variables to make way for the new values
void shiftvariables() {
test.tnow = test.tnew;
test.dtold = test.dt;
test.Yold = test.Ynow;
test.Ynow = test.Ynew;
test.fold = test.fnow;
test.fnow = test.fnew;
advance();
}
//Ordinary differential equation to be solved
double f(double t, double y) {
return pow(t, 2);
}
//Calculate the predicted and corrected solution at a chosen tnew
void step(double fnow, double fold, double Ynow, double tnew, double tnow,
double dtold, double dt, double(*f)(double t, double y), double told) {
//The calculation for Ynew requires integration. I first thought I would need to
// use my project 1 code to calculate the integration, but now I see in class we
// solved it analytically such that integration is not required:
//Linear prediction of Ynew using Ynow and fnow
double Pnew = Ynow + (dt * fnow) + (dt * dt / (2 * dtold)) * (fnow - fold);
test.Pnew = Pnew;
//Predict the value of f at tnew using Pnew
double fnew = f(tnew, Pnew);
test.fnew = fnew;
//Calculate the corrected solution at tnew
double interpolationFactor = fnew - (fnow + dt * (fnow - fold) / dtold);
double integration = (dt / 6) * (2 * dt + 3 * dtold) / (dt + dtold);
double Ynew = Pnew + interpolationFactor * integration;
test.Ynew = Ynew;
//Update the variables for the next round
shiftvariables();
}
//Check the previous solution and choose a new dt to continue evaluation
void advance() {
//The error indicator is the l2-norm of the prediction minus the correction
double err_ind = sqrt(
std::accumulate(differencesDDYSquared.begin(),
differencesDDYSquared.end(), 0));
errorIndicators.push_back(err_ind);
// Case where I reject the step and retry
if (err_ind > test.tol && test.dt > test.dtmin) {
++test.stepsRejected;
test.stepsSinceRejection = 0;
test.dt = test.dt * 0.5;
test.tnew = test.tnow + test.dt;
checkdt();
}
// Cases where I accept the step and move forward
else {
++test.stepsAccepted;
++test.stepsSinceRejection;
solutions.push_back(test.Ynew);
differencesDDY.push_back(abs(test.Pnew - test.Ynew));
differencesDDYSquared.push_back(pow((test.Pnew - test.Ynew), 2));
//Decrease dt
if (err_ind >= 0.75 * test.tol) {
test.dtold = test.dt;
test.dt = (test.dt * test.ashrink);
test.tnew = test.tnow + test.dt;
checkdt();
}
//Increase dt
else if (err_ind <= test.tol / 4) {
if ((test.stepsRejected != 0) && (test.stepsSinceRejection >= 2)) {
test.dt = (test.dt * test.agrow);
test.tnew = test.tnow + test.dt;
checkdt();
} else if (test.stepsRejected == 0) {
test.dt = (test.dt * test.agrow);
test.tnew = test.tnow + test.dt;
checkdt();
}
}
}
}
//Check that the dt chosen by advance is acceptable
void checkdt() {
if ((test.tnew < test.endTime) && (test.endTime - test.tnew < test.dtmin)) {
cout << "Reached endTime." << endl;
} else if (test.dt < test.dtmin) {
test.dt = test.dtmin;
test.tnew = test.tnow + test.dt;
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
} else if (test.dt > test.dtmax) {
test.dt = test.dtmax;
test.tnew = test.tnow + test.dt;
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
} else if ((test.tnew + test.dt) > test.endTime) {
test.dt = test.endTime - test.tnew;
test.tnew = test.tnow + test.dt;
checkdt();
} else if (((test.tnew + test.dt) < test.endTime)
&& ((test.tnew + 2 * test.dt) > test.endTime)) {
test.dt = (test.endTime - test.tnew) / 2;
test.tnew = test.tnow + test.dt;
checkdt();
}
//If none of the above are satisfied, then the chosen dt
// is ok and proceed with it
else {
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
}
}
//meta function to solve a differential equation, called only once
void differentiate(double(*f)(double t, double y), double y0, double a,
double b) {
//Set the starting conditions for the solving of the differential equation
test.fnow = f(a, y0);
test.endTime = b;
test.Ynow = y0;
//solutions.push_back(y0);
timesTNew.push_back(a);
//Set the constants
test.ashrink = 0.8;
test.agrow = 1.25;
test.dtmin = 0.05;
test.dtmax = 0.5;
test.tol = 0.1;
//Set fold = fnow for the first step
test.fold = test.fnow;
test.tnow = a;
test.told = a - test.dtmin;
test.dtold = abs(test.tnow - test.told);
//Create the first prediction, which will then lead to correcting it with step
advance();
}
// Takes a vector as its only parameters and prints it to stdout
void printvector(std::vector<double>& vec) {
for (vector<double>::iterator it = vec.begin(); it != vec.end(); ++it) {
cout << *it << ", ";
}
cout << "\n";
}
Thank you.

Answer 1

Since you're using recursion, could it be possible that you are running out of stack memory, thus causing the segfault? This could happen if either your app recurses too many times or if some bug causes it to recurse infinitely.
Note, as sth suggests in a comment, a debugger may help you decide whether or not this is the case.