Table of Contents

First Order IVP Solver

Routine Name: firstOrderIVPSolver

Author: Philip Nelson

Language: C++

Description

This method is used to solve the simple first order initial value problem

\[u` = \lambda u\]

with \(u(0)=P_0\). This ordinary differential equations has the soltion

\[u(t)=\alpha e^{\lambda t}\]

Input

firstOrderIVPSolver(const T& lambda, const T& alpha) requires:

• const T& l - \(\lambda\)
• const T& a - \(\alpha\)

Output

This method returns a function that can be evaluated at any time \(t\) to obtain the exact solution to the analytic equation (including roundoff error).

Code

template <typename T>
auto firstOrderIVPSolver(const T& l, const T& a)
{
  return [=](const T& t) { return a * std::exp(l * t); };
}

Example

int main()
{

  auto solveIVP = firstOrderIVPSolver(-1.5, 7.3);

  for(auto t = 0; t < 10; ++t)
  {
    std::cout << "t = " << t << " -> " << solveIVP(t)
      << "\tt = " << t+10 << " -> " << solveIVP(t+10) << '\n';
  }

  return EXIT_SUCCESS;
}

Result

t = 0 -> 7.3		t = 10 -> 2.23309e-06
t = 1 -> 1.62885	t = 11 -> 4.98269e-07
t = 2 -> 0.363446	t = 12 -> 1.11179e-07
t = 3 -> 0.0810957	t = 13 -> 2.48074e-08
t = 4 -> 0.0180949	t = 14 -> 5.53527e-09
t = 5 -> 0.00403752	t = 15 -> 1.23509e-09
t = 6 -> 0.000900892	t = 16 -> 2.75585e-10
t = 7 -> 0.000201016	t = 17 -> 6.14913e-11
t = 8 -> 4.48528e-05	t = 18 -> 1.37206e-11
t = 9 -> 1.0008e-05	t = 19 -> 3.06147e-12

Last Modification date: 3 April 2018