# The “Energy transformation” of a second order differential equation

After asking a question regarding the math myself, as well as finding some others, I am still puzzled by the opening sections of Jordan, D. and Smith, P. (2011). Nonlinear ordinary differential equations. We start off with the equation of motion for a simple pendulum, where $$x$$ denotes the angular displacement of the pendulum: $$\ddot{x} + \omega^2 \sin{x} = 0.$$ Here, $$\ddot{x} = \mathop{\mathrm{d}}^2x/\mathop{\mathrm{d}}t^2$$, and similarily, $$\dot{x}= \mathop{\mathrm{d}}x / \mathop{\mathrm{d}}{t}$$. Then, the first element that confuses me is the statement $$\ddot{x} = \frac{\mathop{\mathrm{d}}}{\mathop{\mathrm{d}}t} \frac{\mathop{\mathrm{d}}x}{\mathop{\mathrm{d}}t} = \frac{\mathop{\mathrm{d}}\dot{x}}{\mathop{\mathrm{d}}t} = \frac{\mathop{\mathrm{d}}\dot{x}}{\mathop{\mathrm{d}}x} \frac{\mathop{\mathrm{d}x}}{\mathop{\mathrm{d}}t} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2} \dot{x}^2 \right).$$ According to the authors, this is known as the energy transformation. I haven't been able to find any other literature supporting this, so my first question is this: Is energy transformation a general term for ODEs, or is it a special name for this operation on the pendulum equation?

To continue, I understand the equation given that one may write $$\dot{x} = \dot{x}(x)$$, but that doesn't seem right to me, as the angular velocity may be both increasing and decreasing at any given $$x$$, depending upon if the pendulum is currently moving "up" or "down". I wouldn't know how to evaluate $$\mathop{\mathrm{d}}\dot{x} / \mathop{\mathrm{d}}x$$ by itself. My second question is therefore: Is there a more rigorous way to arrive at the same result?

Simply multiply the equation with $$\dot x$$ and integrate for the time to get $$\int (\dot x\ddot x+ω^2\sin(x)\dot x)dt=\frac12\dot x^2 + ω^2(1-\cos x) = E(x,\dot x)=const.$$ The second part can be substituted as $$\int \sin x\,dx$$. Forcing the same substitution on the first term instead of integrating directly as product $$\int v\dot v\,dt$$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context $$\ddot x =\frac{d}{dt}\frac{\partial(\frac12\dot x^2)}{\partial \dot x}.$$ as part of the Euler-Lagrange equations $$0=\frac{d}{dt}\frac{∂L}{∂\dot x}-\frac{∂L}{∂x}.$$
• Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $\int \left( \dot{x} \ddot{x} + \omega^2 \sin{(x)} \dot{x} \right) dt = \frac{1}{2} \dot{x}^2 - \omega^2 \cos{x}$? – Erik André Jan 11 at 19:59
• There is a free integration constant. I choose to have the energy at rest to be zero, $E=\frac12\dot x^2+\frac{ω^2}2(2\sin\frac x2)^2.$ Your choice gives a minimality of terms. – LutzL Jan 11 at 20:50