Runge—Khutta—Fehlberg initial value problem solver.

• Solves a system of ordinary differential equations.
• The algorithm is described in [9] and [10].

Types

• using value_type = T

Methods

• verbose() const -> bool
• verbose(bool rhs) -> void
• tolerance(T rhs) -> void
• tolerance() const -> T
• max_step(T rhs) -> void
• max_step() const -> T
• min_step(T rhs) -> void
• min_step() const -> T
• template <class Function, int N>
operator()(Function f, T t0, T t1, const blitz::TinyVector< T, N > & x0) -> blitz::TinyVector< T, N >

  Solve system of ordinary differential equations with right hand sides f.

  f

  right hand sides of the equations

  
  

  t0

  initial time instant

  
  

  t1

  final time instant

  
  

  x0

  initial value at $t=t_0$

  
  

  • The system of equations is written as $\dot{\vec{x}} = f\left(\vec{x}\right).$
  • The solution is computed for interval $(t_0,t_1]$.
  • The algorithm is described in [9].

  Date

  2018-07-13

  Author

  Ivan Gankevich

  Return

  $\vec{x}$ at $t=t_1$

• template <class Function, int N>
solve(Function f, T t0, T t1, const blitz::TinyVector< T, N > & x0) -> blitz::TinyVector< T, N >

  Solve system of ordinary differential equations with right hand sides f.

  f

  right hand sides of the equations

  
  

  t0

  initial time instant

  
  

  t1

  final time instant

  
  

  x0

  initial value at $t=t_0$

  
  

  • The system of equations is written as $\dot{\vec{x}} = f\left(\vec{x}\right).$
  • The solution is computed for interval $(t_0,t_1]$.
  • The algorithm is described in [9].
  Calculate coefficients: $$\begin{aligned} k_1 &= h f\left(t_i, x_i\right) \\ k_2 &= h f\left( t_i + \frac{1}{4}h, x_i + \frac{1}{4}k_1 \right) \\ k_3 &= h f\left( t_i + \frac{3}{8}h, x_i + \frac{3}{32}k_1 + \frac{9}{32}k_2 \right) \\ k_4 &= h f\left( t_i + \frac{12}{13}h, x_i + \frac{1932}{2197}k_1 - \frac{7200}{2197}k_2 + \frac{7296}{2197}k_3 \right) \\ k_5 &= h f\left( t_i + h, x_i + \frac{439}{216}k_1 - 8 k_2 + \frac{3680}{513}k_3 - \frac{845}{4104}k_4 \right) \\ k_6 &= h f\left( t_i + \frac{1}{2}h, x_i - \frac{8}{27}k_1 + 2 k_2 - \frac{3544}{2565}k_3 + \frac{1859}{4104}k_4 - \frac{11}{40}k_5 \right) \end{aligned}$$ Calculate $x_{i+1}$ using a Runge—Kutta method of order 4. N.B. $k_2$ is not used here. $$x_{i+1} = x_i + \frac{25}{216}k_1 + \frac{1408}{2565}k_3 + \frac{2197}{4101}k_4 - \frac{1}{5}k_5$$ Calculate $x_{i+1}$ (denoted here as $z_{i+1}$) using a Runge—Kutta method of order 5. N.B. $k_2$ is not used here. $$z_{i+1} = z_i + \frac{16}{135}k_1 + \frac{6656}{12825}k_3 + \frac{28561}{56430}k_4 - \frac{9}{50}k_5 + \frac{2}{55}k_6$$ Calculate optimal step size $sh$. $$s = \left( \frac{\textrm{tol} \, h} {2\,\textrm{max}\left|x_{i+1} - z_{i+1}\right|} \right)^{1/4}$$ Clamp step size $h$ to $[h_{\textrm{min}},h_{\textrm{max}}]$. Clamp time instant to $[t_{\textrm{min}},t_{\textrm{max}}]$.

  Date

  2018-07-13

  Author

  Ivan Gankevich

  Return

  $\vec{x}$ at $t=t_1$

• ~RKF45()
• operator=(RKF45 &&) -> RKF45 &
• RKF45(RKF45 &&)
• operator=(const RKF45 &) -> RKF45 &
• RKF45(const RKF45 &)
• RKF45()