Numerical method
=================

The numerical method implemented in*code*\ _\ *aster* was proposed by Nigam and Jennings [:ref:`bib2 <bib2>`], which is more effective than the Newmark method (see [:ref:`Bib3 <Bib3>`]) in the case of subcritical damping that corresponds to our initial seismic problem [:ref:`R4.05.03 <R4.05.03>`].

We are therefore led to solve the differential equation:


.. math::
 :label: eq-9

    \ ddot {q}\ left (t\ right) +2\ mathrm {\ right) +2\ mathrm {\ xi}\ mathrm {\ omega}\ left (t\ right) + {\ mathrm {\ omega}}} ^ {2} q\ left (t\ right) =\ mathrm {\ alpha}\ left (t\ right)


with zero initial conditions and subcritical damping, whose solution is written as:


.. math::
 :label: eq-10

    q\ left (t\ right) =-\ frac {1} {{\ mathrm {\ omega}} _ {d}} {\ int} _ {0} ^ {t} {e} ^ {-\ mathrm {\ xi}\ mathrm {\ xi}\ xi}\ {\ xi}\\\\\\\\\\\}\ left\ mathrm {\ omega}\ left (t-\ mathrm {\ tau}\ right)}\ text {sin}\ e} ^ {-\ mathrm {\ xi}}\ left [{\ mathrm {\ mathrm {\ xi}}\ left m {\ omega}} _ {d}\ left (t-\ mathrm {\ tau}\ right)\ right]\ mathrm {\ alpha}\ left (\ mathrm {\ tau}\ right) d\ mathrm {\ tau}\ right) d\ mathrm {\ tau}


With :math:`{\mathrm{\omega }}_{d}=\mathrm{\omega }\sqrt{1-{\mathrm{\xi }}^{2}}`. Assuming that :math:`\alpha (t)` varies linearly within each :math:`\Delta (t)` interval, we can write:


.. math::
 :label: eq-11

    \ mathrm {\ alpha}\ left (\ mathrm {\ tau}\ right) =\ mathrm {\ alpha}\ left (t-\ mathrm {\ Delta} t\ right) +\ frac {\ mathrm {\ tau}} {\ mathrm {\ tau}}} {\ mathrm {\ tau}\ right}} {\ mathrm {\ delta} t\ right) +\ frac {\ mathrm {\ delta} t\ right) +\ frac {\ mathrm {\ tau}}} {\ mathrm {\ tau}}} {\ mathrm {\ tau}}} {\ mathrm {\ tau}}} {\ mathrm {\ delta} t\ right) -\ mathrm m {\ alpha}\ left (t-\ mathrm {\ Delta} t\ right)\ right]\ right]\ text {for}\ mathrm {\ tau}\ in\ left [\ mathrm {0,}\ mathrm {\ Delta} t\ right]


**Note:** the :math:`\mathrm{\Delta }t` time step is not necessarily constant.


.. image:: images/100009B0000010F300000B7993BB5AED2AA0213E.svg
    :width: 218
    :height: 148


.. _RefImage_100009B0000010F300000B7993BB5AED2AA0213E.svg:

Hence the equation to be solved (expressed in the new variable :math:`\tau`):


.. math::
 :label: eq-12

    \ ddot {q}\ left (\ mathrm {\ tau}\ right) +2\ mathrm {\ tau}\ right) +2\ mathrm {\ xi}\ left (\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) + {\ mathrm {\ tau}\ right) tau}\ text {for}\ mathrm {\ tau}\ in\ left [\ mathrm {0,}\ mathrm {\ Delta} t\ right]


Where :math:`a=\alpha (t-\Delta t )` and :math:`b=\left[\alpha (t)-\alpha (t-\Delta t )\right]/\Delta t`. With the initial conditions:


.. math::
 :label: eq-13

    \ {\ begin {array} {c} q\ left (0\ right) =q\ left (t-\ mathrm {\ Delta} t\ right)\\\ dot {q}\ left (0\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right) =\ dot {q}\ right)


The solution to this equation is the superposition of a particular solution:


.. math::
 :label: eq-14

    {q} _ {p} (t) =-\ frac {a} {{\ omega} {{\ omega} ^ {2}} +\ frac {2\ xib} {{\ omega} ^ {3}}} -\ frac {3}}} -\ frac {b} {b} {b} {b} {b} {b} {b} {b} {b} {b} {b} {b} {{b}} { 


and solutions to the homogeneous problem:


.. math::
 :label: eq-15

    {q} _ {h} (t) = {e} ^ {-\ xi\ omega\ tau}\ left [{C} _ {1}\ text {.} \ text {cos} ({\ omega} _ {d}\ tau) + {C} _ {2}\ text {.} \ text {sin} ({\ omega} _ {d}\ tau)\ right]


As a result:


.. math::
 :label: eq-16

    q (\ tau) = {e} ^ {-\ xi\ omega\ tau}\ left [{C} _ {1}\ text {.} \ text {cos} ({\ omega} _ {d}\ tau) + {C} _ {2}\ text {.} \ text {sin} ({\ omega} _ {d}\ tau)\ tau)\ right] -\ frac {a} {{\ omega} ^ {2}} +2\ frac {\ xib} {{\ omega} {\ omega} ^ {3}}} -\ frac {b\ text {.} \ tau} {{w} ^ {2}}


By deriving :math:`q` (with respect to :math:`\mathrm{\tau }`) we have:


.. math::
 :label: eq-17

    \ begin {array} {c}\ dot {q}\ left (\ mathrm {\ tau}\ right) =\ left (-\ mathrm {\ xi}\ mathrm {\ omega}\ right) {e} ^ {\ right) {e} ^ {-\ mathrm {\ xi}\ left ({C} _ {1}\ text {cos} {\ mathrm {\ omega}} _ {d}\ mathrm {\ tau} + {C} _ {2}\ text {sin} {\ mathrm {\ omega}} _ {d}\ mathrm {\ tau}\ right) +\\ {tau}\ right) +\\ {e}\ right) +\\ {e} ^ {-\ mathrm {\ xi}\ mathrm {\ omega}} _ {d}\ mathrm {\ omega}} _ {d}\ mathrm {\ omega}}\ mathrm {\ tau}\ right) +\\ {e}\ right) +\\ {e} ^ {-\ mathrm {\ xi}\ mathrm {\ omega}} _ {d}\ mathrm {\ omega}}}\ left (- {C} _ {1} {\ mathrm {\ omega}}} _ {d}\ text {sin} {\ mathrm {\ omega}} _ {d}\ mathrm {\ tau} + {\ mathrm {\ tau}} + {\ C} _ {2} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} {2} {\ mathrm {\ omega}} mathrm {\ tau}\ right) -\ frac {b} {{\ mathrm {\ omega}} ^ {2}}\ end {array}}


The coefficients :math:`{C}_{1}` and :math:`{C}_{2}` are then determined by the initial conditions at the beginning of the interval (i.e. for :math:`\tau =0`):


.. math::
 :label: eq-18

    {C} _ {1} =q\ left (t-\ mathrm {\ Delta} t\ right) +\ frac {a} {{\ mathrm {\ omega}} ^ {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}} -\ frac {2}}


And:


.. math::
 :label: eq-19

    {C} _ {2} =\ frac {1} {{\ mathrm {\ omega}}} _ {d}}\ left [\ dot {q}\ left (t-\ mathrm {\ Delta} t\ right) +\ mathrm {\ delta} t\ right) +\ mathrm {\ xi}\ right) +\ frac {\ mathrm {\ xi} a} {\ mathrm {\ omega}}} -\ frac {2 {\ mathrm {\ xi}} ^ {2} -1} {{\ mathrm {\ omega}}} ^ {2}} b\ right]


By bringing :math:`{C}_{1}` and :math:`{C}_{2}` into the expression for :math:`q` and :math:`\dot{q}` we get matrix equality for :math:`\tau =\Delta t`:

.. math::
: label: eq-20

    \ left\ {\ begin {array} {} q (t)\\ dot {q} (t)\\\ dot {q} (t)\ end {array}\ right\} =A (\ xi,\ omega,\ Delta t)\ left\ {\ begin {array} {} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {array} {\ omega,\ Delta t)\ left\ {\ begin {array} {}\ alpha (t-\ Delta t)\\ alpha (t)\\ alpha (t)\ end {array}\ array}\ right\}