I read an introduction to Landau equation: $$\frac{\mathrm{d} \vert A\rvert^2}{\mathrm{d}t}=2\sigma \lvert A \rvert^2 - \ell \lvert A \rvert^4.$$ And I encountered a problem on the solution of the simple model equation.
By dividing $-\lvert A \rvert ^4$, I have $$-\frac{1}{\vert A \rvert^4}\frac{\mathrm{d} \vert A \rvert ^2}{\mathrm{d} t}+2 \sigma \vert A \rvert ^{-2} = \ell$$
Can somebody can tell me how this form can be rewritten as
$$\frac{\mathrm{d} \lvert A \rvert ^{-2}}{\mathrm{d}t}+2 \sigma \lvert A \rvert ^{-2} = \ell,$$ as shown in the following excerpt of the textbook. Thank you in advance!
Landau described the instability by the equation \begin{equation} \tag{49.3} \mathrm{d} |A|^2 / \mathrm{d} t = 2 \sigma |A|^2 - \ell |A|^4 \end{equation} for the amplitude $|A|$ of the dominant mode, where $\ell$ is some constant, now called the Landau constant. Also equation $(49.3)$ is called the Landau equation, although it is equivalent to the logistic equation in the theory of population growth. (We shall see that it is more properly regarded as a truncation of a system of ordinary differential equations whose other terms are often, but not always, negligible in hydrodynamic stability.) Of course if $\ell$ were zero equation $(49.3)$ would reduce to the equation given by the linear theory. The second term on the right-hand side of equation $(49.3)$ is due to the nonlinearity and may moderate or accelerate the exponential growth of the linear disturbance according to the signs of $\sigma$ and $\ell$.
Rewriting the Landau equation as a linear equation in $|A|^{-2}$, namely $$ \frac{\mathrm{d}|A|^{-2}}{\mathrm{d}t} + 2 \sigma |A|^{-2} = \ell, $$ we find the explicit general solution $$ |A|^{-2} = \ell/2\sigma + (A_0^{-2} - \ell/2\sigma) e^{-2 \sigma t}. $$
(Original image here.)
Use the chain rule to compute $$ \frac{d}{dt} |A|^{-2} = \frac{d}{dt} (|A|^2)^{-1} = -1(|A|^2)^{-2} \frac{d}{dt} |A|^2 = -|A|^{-4} \frac{d}{dt} |A^2| = l - 2 \sigma |A|^{-2}. $$ Then we use standard ODE to solve the equation $$ \frac{d}{dt} z + 2\sigma z = l $$ as in the picture and then plug in $z = |A|^{-2}$.