Benchmark solution
=====================

The reference solution is analytical, and obtained as follows.

Next, we will note :math:`u={u}_{x}`. The equation in the relative coordinate system of :math:`{P}_{2}` is written as:

(2-1) :math:`m\ddot{u}+c\dot{u}+ku=-m\ddot{{u}_{e}}(t)`

The load is applied using the AFFE_CHAR_SEISME command. The associated multiplier function FONC_MULT defines training acceleration as:

(2-2) :math:`\ddot{{u}_{e}}(t)={A}_{e}\mathrm{sin}({\omega }_{e}t)`

The problem to be solved is therefore written as:

(2-3) :math:`m\ddot{u}+c\dot{u}+ku=-m{A}_{e}\mathrm{sin}({\omega }_{e}t)`

Or, by defining the reduced quantities :math:`{\omega }_{0}^{2}=\frac{k}{m}` and :math:`\xi =\frac{c}{2m{\omega }_{0}}`:

(2-4) :math:`\ddot{u}+2{\omega }_{0}\xi \dot{u}+{\omega }_{0}^{2}u=-{A}_{e}\mathrm{sin}({\omega }_{e}t)`

with the initial conditions:

(2-5) :math:`u(0)=0`

(2-6) :math:`\dot{u}(0)=0`

We are looking for a particular solution to () in the form:

(2-7) :math:`\stackrel{~}{u}(t)=D\mathrm{sin}({\omega }_{e}t+\varphi )`

We ask:

(2-8) :math:`\stackrel{~}{U}(t)=D{e}^{i({\omega }_{e}t+\varphi )}`

We then have :math:`\stackrel{~}{u}(t)=\mathrm{\Im }(\stackrel{~}{U}(t))` (imaginary part).

Let the complex differential equation be:

(2-9) :math:`\ddot{\stackrel{~}{U}}+2{\omega }_{0}\xi \dot{\stackrel{~}{U}}+{\omega }_{0}^{2}\stackrel{~}{U}=-{A}_{e}{e}^{i({\omega }_{e}t)}`

If :math:`\stackrel{~}{U}` is a solution of (), then is a solution of (). By injecting the expression () into (), we find:

(2-10) :math:`(-{\omega }_{\mathit{ex}}^{2}+2i{\omega }_{0}{\omega }_{e}\xi +{\omega }_{0}^{2})B{e}^{i\varphi }=-{A}_{e}`

from where we can determine the parameters :math:`D` and :math:`\varphi` of the expression ():

(2-11) :math: `D=\ left|\ frac {{A} _ {e}} {{\ omega} _ {0} ^ {2} - {\ omega} _ {e} ^ {2} +2i {\ omega} +2i {\ omega} _ {\ omega} _ {0} {\ omega} _ {\ omega} _ {e}} {\ sqrt} {\ left ({\ omega} _ {0} ^ {2} - {\ omega} _ {e} ^ {2}\ right)} ^ {2}} + {\ left (2 {\ omega} _ {0} {\ omega} _ {0} {\ omega}} {\ omega}} _ {e}\ omega} _ {e}\ xi\ right)} ^ {2}}}} `

(2-12) :math:`\varphi =\mathit{arg}\left(-\frac{{A}_{e}}{{\omega }_{0}^{2}-{\omega }_{e}^{2}+2i{\omega }_{0}{\omega }_{e}\xi }\right)=\mathit{arg}\left({\omega }_{e}^{2}-{\omega }_{0}^{2}+2i{\omega }_{0}{\omega }_{e}\xi \right)=\mathit{atan}2\left(\frac{2{\omega }_{0}{\omega }_{e}\xi }{{\omega }_{e}^{2}-{\omega }_{0}^{2}}\right)`

The general solution of the problem is then classically written as the sum of the particular solution and the homogeneous solution of a differential equation of order 2 with constant coefficients (see for example [:ref:`1 <RefNumPara__1150_2287396154>`]:

(2-13) :math:`u(t)={e}^{-\xi {\omega }_{0}t}\left(A\mathrm{cos}({\omega }_{\xi }t)+B\mathrm{sin}({\omega }_{\xi }t)\right)+D\mathrm{sin}({\omega }_{e}t)`

with:

(2-14) :math:`{\omega }_{\xi }={\omega }_{0}\sqrt{1-{\xi }^{2}}`

The initial conditions () and () give respectively:

:math:`A+D\mathrm{sin}(\varphi )=0`

:math:`{\omega }_{0}\left(-\xi A+B\sqrt{1-{\xi }^{2}}\right)+{\omega }_{e}D\mathrm{cos}(\varphi )=0`

From which we can deduce:

(2-15) :math:`A=-D\mathrm{sin}(\varphi )`

(2-16) :math:`B=-\frac{{\omega }_{e}D\mathrm{cos}\varphi }{{\omega }_{0}\sqrt{1-{\xi }^{2}}}+\frac{\xi D\mathrm{sin}(\varphi )}{\sqrt{1-{\xi }^{2}}}=-\frac{D}{\sqrt{1-{\xi }^{2}}}\left(\frac{{\omega }_{e}}{{\omega }_{0}}\mathrm{cos}(\varphi )+\xi \mathrm{sin}(\varphi )\right)`

The analytical solution is therefore written as () where :math:`A`, :math:`B`,,, :math:`D`, :math:`{\omega }_{\xi }`, and :math:`\varphi` are given by (), (), (), (), () and () respectively.


Benchmark results
----------------------

For this test case, an excitation pulse of the harmonic signal :math:`{\omega }_{e}=2{s}^{-1}`, :math:`\xi =\mathrm{0,05}` and :math:`{A}_{e}=1m\mathrm{.}{s}^{-2}` is chosen. Given the parameters of the problem, the natural frequency is :math:`{\omega }_{0}=1{s}^{-1}`.

We test the values of the relative displacement :math:`{u}_{x}` at points :math:`{P}_{2}` and :math:`{P}_{3}` to :math:`t=10s`:


.. csv-table::

    "", "Move to :math:`t=10s`"
    "Bow", ":math:`\mathit{DX}`"
    ":math:`N2` (dot :math:`{P}_{2}`)", "0.538736"
    ":math:`N3` (dot :math:`{P}_{3}`)", "0.538736"



Uncertainty about the solution
---------------------------

Clarification on the integration of time in DYNA_NON_LINE