changes.mady.by.user Arthur Aslanyan (Nafta College)
Saved on Nov 28, 2021
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Q^{\downarrow}_{AQ}= B \cdot \int_0^t W_{eD} \left( \frac{(t-\tau)\chi}{r_e^2}, \frac{r_a}{r_e} \right) \dot p(\tau) d\tau
W_{eD}(t) = \int_0^{t} \frac{\partial p_1}{\partial r_D} \bigg|_{r_D = 1} dt_D
q^{\downarrow}_{AQ}(t)= \frac{dQ^{\downarrow}_{AQ}}{dt}
p_1 = p_1(t_D, r_D)
\frac{\partial p_1}{\partial t_D} = \frac{\partial^2 p_1}{\partial r_D^2} + \frac{1}{r_D}\cdot \frac{\partial p_1}{\partial r_D}
p_1(t_D = 0, r_D)= 0
p_1(t_D, r_D=1) = 1
\frac{\partial p_1(t_D, r_D)}{\partial r_D} \Bigg|_{r_D=r_{aD}} = 0
or
p_1(t_D, r_D = \infty) = 0