How to show that energy is monotonically increasing?

Question

Particle of mass $m$ is moving on a straight line under the influence of a force generated by a potential and friction with the equation of motion: $$m \frac{d^2x}{dt^2}+2 \gamma \frac{dx}{dt} = -V'(x) $$

They say $\displaystyle \frac{d}{dt}E = \frac{d}{dt} \bigg[\frac{m}{2}\left(\frac{dx}{dt}\right)^2+V(x)\bigg] \color{blue}= -2\gamma \left(\frac{dx}{dt}\right)^2 \le 0$ so $E$ is monotically increasing.

Could someone please explain the second/blue equality? Thanks.

Answer 1

Multiplying the movement equation by $\dot x$ we have

$$ \frac{d}{dt}\left(\frac 12 m\dot x^2 + V(x)\right) = -\gamma \dot x ^2 $$

but

$$ E = \frac 12 m\dot x^2 + V(x) \;\;\mbox{the so called mechanical energy} $$

hence

$$ \frac{d}{dt}E = -\gamma \dot x ^2 \le 0 $$

the mechanical energy decays according to the work realized by the dissipative forces.

Answer 2

As an alternative to @Cesareo's answer, you can also differentiate directly using the chain rule on both terms: $$ \frac{d}{dt} \left( \frac{m}{2} \left(\frac{dx}{dt}\right)^2 + V(x) \right) = \frac{m}{2} \cdot 2\cdot\frac{dx}{dt}\cdot\frac{d^2x}{dt^2} + \frac{dV}{dx}\cdot\frac{dx}{dt} = \frac{dx}{dt}\left(m\frac{d^2x}{dt^2} + V'(x)\right), $$ and then use your equation of motion to obtain the blue equality.