Effective conductivity of 2D random composites
==============================================

.. note::
   This section is under constant development. Its content will change and improve over time.

Approximation of the symbolic formula
-------------------------------------

.. code:: ipython3

    from basicsums.conductivity import coefficient_B, effective_conductivity

.. code:: ipython3

    from sympy import symbols, pi, init_printing, expand, simplify
    init_printing()
    nu_s, rho_s = symbols('\u03BD \u03C1')

One can derive symbolic forms of terms of the coeeficient :math:`B_q`:

.. code:: ipython3

    (coefficient_B(1, rho_s),
     coefficient_B(2, rho_s),
     coefficient_B(3, rho_s))




.. math::

    \left ( e(2) ρ, \quad e(2,2) ρ^{2}, \quad e(2,2,2) ρ^{3} - 2 e(3,3) ρ^{2}\right )



.. code:: ipython3

    coefficient_B(4, rho_s)




.. math::

    e(2,2,2,2) ρ^{4} - 2 e(2,3,3) ρ^{3} - 2 e(3,3,2) ρ^{3} + 3 e(4,4) ρ^{2}



or sumbolic approximation of the effective conductivity series:

.. code:: ipython3

    ec3 = effective_conductivity(nu=nu_s, q=3, rho=rho_s, pi=pi)
    ec3




.. math::

    \frac{2 e(2) ν^{2} ρ^{2}}{\pi} + \frac{2 e(2,2) ν^{3} ρ^{3}}{\pi^{2}} + \frac{2 ν^{4} ρ \left(e(2,2,2) ρ^{3} - 2 e(3,3) ρ^{2}\right)}{\pi^{3}} + 2 ν ρ + 1



.. code:: ipython3

    simplify(ec3)




.. math::

    \frac{2 \pi^{2} e(2) ν^{2} ρ^{2} + 2 \pi e(2,2) ν^{3} ρ^{3} + 2 ν^{4} ρ^{3} \left(e(2,2,2) ρ - 2 e(3,3)\right) + \pi^{3} \left(2 ν ρ + 1\right)}{\pi^{3}}



.. code:: ipython3

    init_printing(pretty_print=False)

Numerical-symbolic calculations
-------------------------------

Let us compute the symbolic-numerical approximation of the effective
conductivity for a given configuration of disks. First, we import
example data and define periods for the quare cell.

.. code:: ipython3

    import numpy as np
    from basicsums.examples import load_example01
    A, r = load_example01()
    N, nu = len(A), sum(r**2)*np.pi
    
    w1 = 1
    w2 = 1j

.. code:: ipython3

    from basicsums.show_disks import show_disks
    show_disks(A, r, w1, w2)



.. image:: tutorial_effective_conductivity_files/tutorial_effective_conductivity_15_0.png


Let our approximation incorporate ``q=6`` first coefficients :math:`B_q`
and compute values of required basic sums.

.. code:: ipython3

    from basicsums.multi_indexes import sums_in_Gq_prime
    from basicsums.basic_sums import Cell, BasicSums
    
    q = 6
    cell = Cell(w1, w2, q)

.. code:: ipython3

    sums = sums_in_Gq_prime(q)
    bso = BasicSums(A, cell, r)
    results = bso.esums(sums, dict_output=True)
    results




.. parsed-literal::

    OrderedDict([((2,), (3.15438582262448-0.09180159581592136j)),
     ((2, 2), (11.757827037864901+5.82076609134674e-17j)),
     ((2, 2, 2), (43.07907709266533-1.7681300908296955j)),
     ((3, 3), (-3.557295937476818-7.450580596923828e-17j)),
     ((2, 2, 2, 2), (173.12898896402592-6.103515625e-15j)),
     ((3, 3, 2), (-10.80733726705118+1.833254246831019j)),
     ((4, 4), (7.9877213504124365-2.86102294921875e-16j)),
     ((2, 2, 2, 2, 2), (670.9285779378052-33.06492019934968j)),
     ((2, 3, 3, 2), (-50.034946200507406-2.9296875e-15j)),
     ((3, 3, 2, 2), (-42.96183376801067+6.393431885893068j)),
     ((3, 4, 3), (1.0336374840805849-1.3287290405279937j)),
     ((4, 4, 2), (25.69722017834029-1.0060002202097529j)),
     ((5, 5), (-16.913965412216037+2.44140625e-16j)),
     ((2, 2, 2, 2, 2, 2), (2801.2242234351697-2.05e-13j)),
     ((2, 3, 3, 2, 2), (-188.15069189815577+10.601837977710497j)),
     ((2, 4, 4, 2), (107.30650709107093-1.875e-15j)),
     ((3, 3, 2, 2, 2), (-155.34009851021497+41.19150453994921j)),
     ((3, 3, 3, 3), (39.59075459725217+2.5e-15j)),
     ((3, 4, 3, 2), (2.5510335185673747-4.11358081707618j)),
     ((4, 4, 2, 2), (103.60547850654284-2.0883152988130544j)),
     ((4, 5, 3), (0.9316723876832275+0.7154107026400275j)),
     ((5, 5, 2), (-54.55631093979708+1.40829410225262j)),
     ((6, 6), (26.34716019717142+1.5625e-16j))])



One can retrieve symbolic-numerical expression dependent on
:math:`\rho`:

.. code:: ipython3

    expr = effective_conductivity(nu=nu, q=q, rho=rho_s, results=results)
    simplify(expr)




.. parsed-literal::

    0.0217754043970236*ρ**7 - 1.59357393244145e-18*I*ρ**7 + 0.0470914953661529*ρ**6 - 0.00340488494383155*I*ρ**6 + 0.0891543602557013*ρ**5 - 1.98923904879937e-18*I*ρ**5 + 0.142246781888971*ρ**4 - 0.0084363108452773*I*ρ**4 + 0.249710501141987*ρ**3 + 1.09687688161722e-18*I*ρ**3 + 0.406649237832642*ρ**2 - 0.0118346489838408*I*ρ**2 + 0.9*ρ + 1.0



In order to compute a numerical value of the effective conductivity, we
apply function ``effective_conductivity`` with :math:`\rho=1`.

.. code:: ipython3

    rho = 1
    effective_conductivity(nu, q, rho, results)




.. parsed-literal::

    (2.8566277808824787-0.023675844772949668j)



One can also define function of parameter :math:`\rho`:

.. code:: ipython3

    from sympy import lambdify
    lam = lambdify((rho_s), expr, 'numpy')

and plot :math:`\lambda(\rho)`:

.. code:: ipython3

    import matplotlib.pyplot as plt
    
    fig, ax = plt.subplots()
    
    rho_array = np.linspace(0, 1, 100)
    lam_array = np.real(lam(rho_array))
    
    ax.plot(rho_array, lam_array,'-')
    ax.set_xlabel(r'contrast parameter $\rho$')
    ax.set_ylabel(r'effective conductivity $\lambda$')
    ax.grid(True)
    plt.show()



.. image:: tutorial_effective_conductivity_files/tutorial_effective_conductivity_26_0.png