changes.mady.by.user Arthur Aslanyan (Nafta College)
Saved on Feb 08, 2019
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Image Modified
Q^{\downarrow}_{AQ}= B \cdot \int_0^t W_{eD}(t - \tau) \dot p d\tau
q^{\downarrow}_{AQ}(t)= \frac{dQ^{\downarrow}_{AQ}}{dt}
W_{eD}(t)= \int_0^{t} \frac{\partial p_1}{\partial r_D} \bigg|_{r_D = 1} dt_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}{\partial r_D} \bigg|_{(t_D, r_D=r_a/r_e)} = 0