I'm just learning the Einstein index notation, and came across this derivation in a textbook. I couldn't follow the steps. Can someone please help me out?
The first order differential equation:
$$\bigg(- ik_i \frac{dM_i}{dt} - \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - iY_{ij}k_iM_j - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j \bigg)P = 0$$
where non-indicial $i$ (prefix in first and third term in the brackets above) is $\sqrt{-1}$ and
$$P = \exp \bigg(-ik_iM_i(t) - \frac{1}{2}k_ik_jY_{ij}(t) \bigg)$$
implies that ($\bf{I \: don't \: understand \: this \: step}$)
$$\frac{dU_{ij}}{dt} = - Y_{il}U_{lj} - Y_{jl}U_{li} + 2D_{ij}$$
given that $U_{ij} = U_{ji}$
Any help is much appreciated. Thanks.
$$(- ik_i \frac{dM_i}{dt} - \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - iY_{ij}k_iM_j - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j)P = 0$$
We know $P \ne 0$, real and imagine equal zero separately.
$$- \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j = 0 ~~~~~~~(1)$$
$$- k_i \frac{dM_i}{dt} - Y_{ij}k_iM_j = 0~~~~~~~~(2)$$
for (1), $$- \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j = 0 $$
$$ k_i k_j \frac{dU_{ij}}{dt}= - 2Y_{ij}k_iU_{jl}k_l + 2D_{ij}k_ik_j $$ $$ k_i k_j \frac{dU_{ij}}{dt} =- Y_{ij}k_iU_{jl}k_l- Y_{ij}k_iU_{lj}k_l + 2D_{ij}k_ik_j $$
then you need to exchange j, l so you can remove $k_ik_j$ $$ k_i k_j \frac{dU_{ij}}{dt} =- Y_{il}k_iU_{lj}k_j- Y_{il}k_iU_{jl}k_j + 2D_{ij}k_ik_j $$
$$ \frac{dU_{ij}}{dt} =- Y_{il}U_{lj}- Y_{il}U_{jl} + 2D_{ij} $$