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# scipy.integrate.solve_ivp

scipy.integrate.solve_ivp(fun, t_span, y0, method='RK45', t_eval=None, dense_output=False, events=None, vectorized=False, args=None, **options)

  Solve an initial value problem for a system of ODEs.

  This function numerically integrates a system of ordinary differential equations given an initial value:

      dy / dt = f(t, y)
      y(t0) = y0

  Here t is a 1-D independent variable (time), y(t) is an N-D vector-valued function (state), and an N-D vector-valued function f(t, y) determines the differential equations. The goal is to find y(t) approximately satisfying the differential equations, given an initial value y(t0)=y0.

  Some of the solvers support integration in the complex domain, but note that for stiff ODE solvers, the right-hand side must be complex-differentiable (satisfy Cauchy-Riemann equations [11]). To solve a problem in the complex domain, pass y0 with a complex data type. Another option always available is to rewrite your problem for real and imaginary parts separately.

  Parameters:
    fun callable

      Right-hand side of the system: the time derivative of the state `y` at time `t`. The calling signature is `fun(t, y)`, where `t` is a scalar and `y` is an ndarray with `len(y) = len(y0)`. Additional arguments need to be passed if `args` is used (see documentation of `args` argument). `fun` must return an array of the same shape as `y`. See vectorized for more information.

    t_span 2-member sequence

      Interval of integration (t0, tf). The solver starts with t=t0 and integrates until it reaches t=tf. Both t0 and tf must be floats or values interpretable by the float conversion function.

    y0 array_like, shape (n,)

      Initial state. For problems in the complex domain, pass y0 with a complex data type (even if the initial value is purely real).

    method string or `OdeSolver`, optional

      Integration method to use:

      > - ‘RK45’ (default): Explicit Runge-Kutta method of order 5(4) [1]. The error is controlled assuming accuracy of the fourth-order method, but steps are taken using the fifth-order accurate formula (local extrapolation is done). A quartic interpolation polynomial is used for the dense output [2]. Can be applied in the complex domain.

      > - ‘RK23’: Explicit Runge-Kutta method of order 3(2) [3]. The error is controlled assuming accuracy of the second-order method, but steps are taken using the third-order accurate formula (local extrapolation is done). A cubic Hermite polynomial is used for the dense output. Can be applied in the complex domain.

      > - ‘DOP853’: Explicit Runge-Kutta method of order 8 [13]. Python implementation of the “DOP853” algorithm originally written in Fortran [14]. A 7-th order interpolation polynomial accurate to 7-th order is used for the dense output. Can be applied in the complex domain.

      > - ‘Radau’: Implicit Runge-Kutta method of the Radau IIA family of order 5 [4]. The error is controlled with a third-order accurate embedded formula. A cubic polynomial which satisfies the collocation conditions is used for the dense output.

      > - ‘BDF’: Implicit multi-step variable-order (1 to 5) method based on a backward differentiation formula for the derivative approximation [5]. The implementation follows the one described in [6]. A quasi-constant step scheme is used and accuracy is enhanced using the NDF modification. Can be applied in the complex domain.

      > - ‘LSODA’: Adams/BDF method with automatic stiffness detection and switching [7], [8]. This is a wrapper of the Fortran solver from ODEPACK.

      Explicit Runge-Kutta methods (‘RK23’, ‘RK45’, ‘DOP853’) should be used for non-stiff problems and implicit methods (‘Radau’, ‘BDF’) for stiff problems [9]. Among Runge-Kutta methods, ‘DOP853’ is recommended for solving with high precision (low values of rtol and atol).

      If not sure, first try to run ‘RK45’. If it makes unusually many iterations, diverges, or fails, your problem is likely to be stiff and you should use ‘Radau’ or ‘BDF’. ‘LSODA’ can also be a good universal choice, but it might be somewhat less convenient to work with as it wraps old Fortran code.

      You can also pass an arbitrary class derived from `OdeSolver` which implements the solver.

    t_eval array_like or None, optional

      Times at which to store the computed solution, must be sorted and lie within t_span. If None (default), use points selected by the solver.

    dense_output bool, optional

      Whether to compute a continuous solution. Default is False.

    events callable, or list of callables, optional

      Events to track. If None (default), no events will be tracked. Each event occurs at the zeros of a continuous function of time and state. Each function must have the signature `event(t, y)` where additional argument have to be passed if `args` is used (see documentation of `args` argument). Each function must return a float. The solver will find an accurate value of t at which `event(t, y(t)) = 0` using a root-finding algorithm. By default, all zeros will be found. The solver looks for a sign change over each step, so if multiple zero crossings occur within one step, events may be missed. Additionally each event function might have the following attributes:

      > terminal: bool or int, optional

      >   When boolean, whether to terminate integration if this event occurs. When integral, termination occurs after the specified the number of occurences of this event. Implicitly False if not assigned.

      > direction: float, optional

      >   Direction of a zero crossing. If direction is positive, event will only trigger when going from negative to positive, and vice versa if direction is negative. If 0, then either direction will trigger event. Implicitly 0 if not assigned.

      You can assign attributes like `event.terminal = True` to any function in Python.

    vectorized bool, optional

      Whether fun can be called in a vectorized fashion. Default is False.

      If `vectorized` is False, fun will always be called with `y` of shape `(n,)`, where `n = len(y0)`.

      If `vectorized` is True, fun may be called with `y` of shape `(n, k)`, where `k` is an integer. In this case, fun must behave such that `fun(t, y)[:, i] == fun(t, y[:, i])` (i.e. each column of the returned array is the time derivative of the state corresponding with a column of `y`).

      Setting `vectorized=True` allows for faster finite difference approximation of the Jacobian by methods ‘Radau’ and ‘BDF’, but will result in slower execution for other methods and for ‘Radau’ and ‘BDF’ in some circumstances (e.g. small `len(y0)`).

    args tuple, optional

      Additional arguments to pass to the user-defined functions. If given, the additional arguments are passed to all user-defined functions. So if, for example, fun has the signature `fun(t, y, a, b, c)`, then jac (if given) and any event functions must have the same signature, and args must be a tuple of length 3.

    **options

      Options passed to a chosen solver. All options available for already implemented solvers are listed below.

    first_step float or None, optional

      Initial step size. Default is None which means that the algorithm should choose.

    max_step float, optional

      Maximum allowed step size. Default is np.inf, i.e., the step size is not bounded and determined solely by the solver.

    rtol, atol float or array_like, optional

      Relative and absolute tolerances. The solver keeps the local error estimates less than `atol + rtol * abs(y)`. Here rtol controls a relative accuracy (number of correct digits), while atol controls absolute accuracy (number of correct decimal places). To achieve the desired rtol, set atol to be smaller than the smallest value that can be expected from `rtol * abs(y)` so that rtol dominates the allowable error. If atol is larger than `rtol * abs(y)` the number of correct digits is not guaranteed. Conversely, to achieve the desired atol set rtol such that `rtol * abs(y)` is always smaller than atol. If components of y have different scales, it might be beneficial to set different atol values for different components by passing array_like with shape (n,) for atol. Default values are 1e-3 for rtol and 1e-6 for atol.

    jac array_like, sparse_matrix, callable or None, optional

      Jacobian matrix of the right-hand side of the system with respect to y, required by the ‘Radau’, ‘BDF’ and ‘LSODA’ method. The Jacobian matrix has shape (n, n) and its element (i, j) is equal to `d f_i / d y_j`. There are three ways to define the Jacobian:

      > - If array_like or sparse_matrix, the Jacobian is assumed to be constant. Not supported by ‘LSODA’.

      > - If callable, the Jacobian is assumed to depend on both t and y; it will be called as `jac(t, y)`, as necessary. Additional arguments have to be passed if `args` is used (see documentation of `args` argument). For ‘Radau’ and ‘BDF’ methods, the return value might be a sparse matrix.

      > - If None (default), the Jacobian will be approximated by finite differences.

      It is generally recommended to provide the Jacobian rather than relying on a finite-difference approximation.

    jac_sparsity array_like, sparse matrix or None, optional

      Defines a sparsity structure of the Jacobian matrix for a finite- difference approximation. Its shape must be (n, n). This argument is ignored if jac is not None. If the Jacobian has only few non-zero elements in each row, providing the sparsity structure will greatly speed up the computations [10]. A zero entry means that a corresponding element in the Jacobian is always zero. If None (default), the Jacobian is assumed to be dense. Not supported by ‘LSODA’, see lband and uband instead.

    lband, uband int or None, optional

      Parameters defining the bandwidth of the Jacobian for the ‘LSODA’ method, i.e., `jac[i, j] != 0 only for i - lband <= j <= i + uband`. Default is None. Setting these requires your jac routine to return the Jacobian in the packed format: the returned array must have `n` columns and `uband + lband + 1` rows in which Jacobian diagonals are written. Specifically `jac_packed[uband + i - j , j] = jac[i, j]`. The same format is used in `scipy.linalg.solve_banded` (check for an illustration). These parameters can be also used with `jac=None` to reduce the number of Jacobian elements estimated by finite differences.

    min_step float, optional

      The minimum allowed step size for ‘LSODA’ method. By default min_step is zero.

  Returns:
    Bunch object with the following fields defined:
    t ndarray, shape (n_points,)

      Time points.

    y ndarray, shape (n, n_points)

      Values of the solution at t.

    sol `OdeSolution` or None

      Found solution as `OdeSolution` instance; None if dense_output was set to False.

    t_events list of ndarray or None

      Contains for each event type a list of arrays at which an event of that type event was detected. None if events was None.

    y_events list of ndarray or None

      For each value of t_events, the corresponding value of the solution. None if events was None.

    nfev int

      Number of evaluations of the right-hand side.

    njev int

      Number of evaluations of the Jacobian.

    nlu int

      Number of LU decompositions.

    status int

      Reason for algorithm termination:

      > - -1: Integration step failed.

      > - 0: The solver successfully reached the end of tspan.

      > - 1: A termination event occurred.

    message string

      Human-readable description of the termination reason.

    success bool

      True if the solver reached the interval end or a termination event occurred (`status >= 0`).

  References

  [1]

  J. R. Dormand, P. J. Prince, “A family of embedded Runge-Kutta formulae”, Journal of Computational and Applied Mathematics, Vol. 6, No. 1, pp. 19-26, 1980.

  [2]

  L. W. Shampine, “Some Practical Runge-Kutta Formulas”, Mathematics of Computation,, Vol. 46, No. 173, pp. 135-150, 1986.

  [3]

  P. Bogacki, L.F. Shampine, “A 3(2) Pair of Runge-Kutta Formulas”, Appl. Math. Lett. Vol. 2, No. 4. pp. 321-325, 1989.

  [4]

  E. Hairer, G. Wanner, “Solving Ordinary Differential Equations II: Stiff and Differential-Algebraic Problems”, Sec. IV.8.

  [5]

  Backward Differentiation Formula on Wikipedia.

  [6]

  L. F. Shampine, M. W. Reichelt, “THE MATLAB ODE SUITE”, SIAM J. SCI. COMPUTE., Vol. 18, No. 1, pp. 1-22, January 1997.

  [7]

  A. C. Hindmarsh, “ODEPACK, A Systematized Collection of ODE Solvers,” IMACS Transactions on Scientific Computation, Vol 1., pp. 55-64, 1983.

  [8]

  L. Petzold, “Automatic selection of methods for solving stiff and nonstiff systems of ordinary differential equations”, SIAM Journal on Scientific and Statistical Computing, Vol. 4, No. 1, pp. 136-148, 1983.

  [9]

  Stiff equation on Wikipedia.

  [10]

  A. Curtis, M. J. D. Powell, and J. Reid, “On the estimation of sparse Jacobian matrices”, Journal of the Institute of Mathematics and its Applications, 13, pp. 117-120, 1974.

  [11]

  Cauchy-Riemann equations on Wikipedia.

  [12]

  Lotka-Volterra equations on Wikipedia.

  [13]

  E. Hairer, S. P. Norsett G. Wanner, “Solving Ordinary Differential Equations I: Nonstiff Problems”, Sec. II.

  [14]

  Page with original Fortran code of DOP853.

  Examples

  Basic exponential decay showing automatically chosen time points.

      >>> import numpy as np
      >>> from scipy.integrate import solve_ivp
      >>> def exponential_decay(t, y): return -0.5 * y
      >>> sol = solve_ivp(exponential_decay, [0, 10], [2, 4, 8])
      >>> print(sol.t)
      [ 0.          0.11487653  1.26364188  3.06061781  4.81611105  6.57445806
        8.33328988 10.        ]
      >>> print(sol.y)
      [[2.         1.88836035 1.06327177 0.43319312 0.18017253 0.07483045
        0.03107158 0.01350781]
       [4.         3.7767207  2.12654355 0.86638624 0.36034507 0.14966091
        0.06214316 0.02701561]
       [8.         7.5534414  4.25308709 1.73277247 0.72069014 0.29932181
        0.12428631 0.05403123]]

  Specifying points where the solution is desired.

      >>> sol = solve_ivp(exponential_decay, [0, 10], [2, 4, 8],
      ...                 t_eval=[0, 1, 2, 4, 10])
      >>> print(sol.t)
      [ 0  1  2  4 10]
      >>> print(sol.y)
      [[2.         1.21305369 0.73534021 0.27066736 0.01350938]
       [4.         2.42610739 1.47068043 0.54133472 0.02701876]
       [8.         4.85221478 2.94136085 1.08266944 0.05403753]]

  Cannon fired upward with terminal event upon impact. The `terminal` and `direction` fields of an event are applied by monkey patching a function. Here `y[0]` is position and `y[1]` is velocity. The projectile starts at position 0 with velocity +10. Note that the integration never reaches t=100 because the event is terminal.

      >>> def upward_cannon(t, y): return [y[1], -0.5]
      >>> def hit_ground(t, y): return y[0]
      >>> hit_ground.terminal = True
      >>> hit_ground.direction = -1
      >>> sol = solve_ivp(upward_cannon, [0, 100], [0, 10], events=hit_ground)
      >>> print(sol.t_events)
      [array([40.])]
      >>> print(sol.t)
      [0.00000000e+00 9.99900010e-05 1.09989001e-03 1.10988901e-02
       1.11088891e-01 1.11098890e+00 1.11099890e+01 4.00000000e+01]

  Use dense_output and events to find position, which is 100, at the apex of the cannonball’s trajectory. Apex is not defined as terminal, so both apex and hit_ground are found. There is no information at t=20, so the sol attribute is used to evaluate the solution. The sol attribute is returned by setting `dense_output=True`. Alternatively, the y_events attribute can be used to access the solution at the time of the event.

      >>> def apex(t, y): return y[1]
      >>> sol = solve_ivp(upward_cannon, [0, 100], [0, 10],
      ...                 events=(hit_ground, apex), dense_output=True)
      >>> print(sol.t_events)
      [array([40.]), array([20.])]
      >>> print(sol.t)
      [0.00000000e+00 9.99900010e-05 1.09989001e-03 1.10988901e-02
       1.11088891e-01 1.11098890e+00 1.11099890e+01 4.00000000e+01]
      >>> print(sol.sol(sol.t_events[1][0]))
      [100.   0.]
      >>> print(sol.y_events)
      [array([[-5.68434189e-14, -1.00000000e+01]]),
       array([[1.00000000e+02, 1.77635684e-15]])]

  As an example of a system with additional parameters, we’ll implement the Lotka-Volterra equations [12].

      >>> def lotkavolterra(t, z, a, b, c, d):
      ...     x, y = z
      ...     return [a*x - b*x*y, -c*y + d*x*y]
      ...

  We pass in the parameter values a=1.5, b=1, c=3 and d=1 with the args argument.

      >>> sol = solve_ivp(lotkavolterra, [0, 15], [10, 5], args=(1.5, 1, 3, 1),
      ...                 dense_output=True)

  Compute a dense solution and plot it.

      >>> t = np.linspace(0, 15, 300)
      >>> z = sol.sol(t)
      >>> import matplotlib.pyplot as plt
      >>> plt.plot(t, z.T)
      >>> plt.xlabel('t')
      >>> plt.legend(['x', 'y'], shadow=True)
      >>> plt.title('Lotka-Volterra System')
      >>> plt.show()

  A couple examples of using solve_ivp to solve the differential equation `y' = Ay` with complex matrix `A`.

      >>> A = np.array([[-0.25 + 0.14j, 0, 0.33 + 0.44j],
      ...               [0.25 + 0.58j, -0.2 + 0.14j, 0],
      ...               [0, 0.2 + 0.4j, -0.1 + 0.97j]])

  Solving an IVP with `A` from above and `y` as 3x1 vector:

      >>> def deriv_vec(t, y):
      ...     return A @ y
      >>> result = solve_ivp(deriv_vec, [0, 25],
      ...                    np.array([10 + 0j, 20 + 0j, 30 + 0j]),
      ...                    t_eval=np.linspace(0, 25, 101))
      >>> print(result.y[:, 0])
      [10.+0.j 20.+0.j 30.+0.j]
      >>> print(result.y[:, -1])
      [18.46291039+45.25653651j 10.01569306+36.23293216j
       -4.98662741+80.07360388j]

  Solving an IVP with `A` from above with `y` as 3x3 matrix :

      >>> def deriv_mat(t, y):
      ...     return (A @ y.reshape(3, 3)).flatten()
      >>> y0 = np.array([[2 + 0j, 3 + 0j, 4 + 0j],
      ...                [5 + 0j, 6 + 0j, 7 + 0j],
      ...                [9 + 0j, 34 + 0j, 78 + 0j]])

      >>> result = solve_ivp(deriv_mat, [0, 25], y0.flatten(),
      ...                    t_eval=np.linspace(0, 25, 101))
      >>> print(result.y[:, 0].reshape(3, 3))
      [[ 2.+0.j  3.+0.j  4.+0.j]
       [ 5.+0.j  6.+0.j  7.+0.j]
       [ 9.+0.j 34.+0.j 78.+0.j]]
      >>> print(result.y[:, -1].reshape(3, 3))
      [[  5.67451179 +12.07938445j  17.2888073  +31.03278837j
          37.83405768 +63.25138759j]
       [  3.39949503 +11.82123994j  21.32530996 +44.88668871j
          53.17531184+103.80400411j]
       [ -2.26105874 +22.19277664j -15.1255713  +70.19616341j
         -38.34616845+153.29039931j]]