Lagrangian Modeling

Abstract

The basic laws of physics are used to model every system whether it is electrical, mechanical, hydraulic, or any other energy domain. In mechanics, Newton’s laws of motion provide key concepts to model-related physical phenomenon. The Lagrangian formulation of modeling derives from the basic work–energy principle and Newton’s laws of motion. The basic law states that the force acting on a body is directly proportional to its acceleration equated with constant mass.

$$ F=ma=m\frac{dv}{dt}=m\frac{d^2x}{d{t}^2} $$

In this equation, force is directly proportional to the derivative of velocity v or double derivative of displacement x with respect to time. If a body moves from point A to point B then the work done by a body is a dot product of force and displaced path integrated from point A to B.

$$ W={\displaystyle {\int}_A^BF\cdot dx} $$

$$ W={\displaystyle {\int}_A^Bm\frac{d^2x}{d{t}^2}\cdot dx}={\displaystyle {\int}_A^Bm\ddot{x}\cdot dx} $$

We now derive the equation as

$$ \ddot{x}dx=\frac{d\overset{.}{x}}{dt}dx=d\overset{.}{x}\frac{dx}{dt}=\overset{.}{x}d\overset{.}{x} $$

So Eq. (2.3) formulates as

$$ W={\displaystyle {\int}_A^Bm\overset{.}{x}d\overset{.}{x}}={\left.m\left(\frac{{\overset{.}{x}}^2}{2}\right)\right|}_A^B=\frac{m}{2}\left({\overset{.}{x}}_B^2-{\overset{.}{x}}_A^2\right.\Big) $$

$$ W={K}_B-{K}_A=\Delta K $$

It is important to know that here we are discussing motion in a conservative field where work done is independent of path and depends upon the difference between the initial and final value of kinetic energy K. We note that work done is represented as change in K between two points. By the law of conversation of energy and work–energy principle we know that a change in kinetic energy should also change the potential energy U, as the total energy of the system remains constant. So a change in kinetic and change in potential energies should have a net effect of zero as

$$ \Delta K+\Delta U=0 $$

So

$$ \Delta U=-W=-{\displaystyle {\int}_A^BF\cdot dx} $$

Using the fundamental theorem of calculus for antiderivatives, we express Eq. (2.8) as

$$ F=-\frac{dU}{dx} $$

This shows forces required to change the potential energy, whereas the force required to change the kinetic energy is determined from the definition of K.

$$ K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\overset{.}{x}}^2 $$

So

$$ \frac{\partial K}{\partial \overset{.}{x}}=m\overset{.}{x} $$

Differentiating Eq. (2.11) gives a force of kinetic energy

$$ F=\frac{d}{dt}\left(\frac{\partial K}{\partial \overset{.}{x}}\right)=\frac{d}{dt}\left(m\overset{.}{x}\right) = ma $$

As sum of forces equal to zero so

$$ \frac{d}{dt}\left(\frac{\partial K}{\partial \overset{.}{x}}\right)-\frac{dU}{dx}=0 $$

Now we define a Lagrangian function L as the difference of kinetic and potential energies

$$ L=K-U $$

We observe that kinetic energy K is a function of velocity $\overset{.}{x}$, and potential energy U is a function of displacement x. Accordingly the Lagrangian function L has two terms as a function of velocity $\overset{.}{x}$ and displacement x independently. As we know that

$$ \left(\frac{\partial L}{\partial \overset{.}{x}}\right)=\left(\frac{\partial K}{\partial \overset{.}{x}}\right)\ \mathrm{and}\ \frac{\partial L}{\partial x}=\frac{\partial U}{\partial x} $$

Lagrangian function L now represents Eq. (2.13) as

$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \overset{.}{x}}\right)-\frac{\partial L}{\partial x}=0 $$

This is Lagrangian formulation for a system with motion only in single axis.