# Preface

This section remind the reader about a classical approach for derivation of differential equations from classical and quantum mechanics.

Introduction to Linear Algebra with Mathematica

A set of generalized coordinates q1, q2, …, qn completely describes the positions of all particles in a mechanical system. In a system with d degrees of freedom and k constraints, n = d − k independent generalized coordinates are needed to completely specify all the positions. The generalized coordinates may have units of length, or angle, or perhaps something totally different. In the theory of small oscillations, the normal coordinates are conventionally chosen to have units of (mass) ½ ×(length). However, once a choice of generalized coordinate is made, with a concomitant set of units, the units of the conjugate momentum and force are determined.

# Euler--Lagrange Equations

Lagrangian mechanics is a reformulation of classical mechanics that expresses the equations of motion in terms of a scalar quantity, called the Lagrangian (that has units of energy). In Lagrangian mechanics, the evolution of a physical system is described by the solutions to the Euler--Lagrange equations for the action of the system. The Lagrangian formulation, in contrast to Newtonian one, is independent of the coordinates in use. The Euler--Lagrange equation was first discovered in the middle of 1750s by Leonhard Euler (1707--1783) from Berlin (Prussia) and the young Italian mathematician from Turin Giuseppe Lodovico Lagrangia (1736--1813) while they worked together on the tautochrone problem.

We consider a system of n particles that is described by at most 3n coordinates $$\left\{ x_1 , x_2 , \ldots , x_{3n} \right\} .$$ Not all of these coordinates are independent---there may exist some constraints. These constraints are easiest to handle when they are in the form $$f(x_1 , x_2 , \ldots , x_{3n} , t ) =0,$$ in which case they are called holonomic. If there are k holonomic constraints, then k of the 3n coordinates can be eliminated, leaving 3n-k independent variables $$\left\{ q_1 , q_2 , \ldots , q_{3n-k} \right\} ,$$ known as the generalized coordinates of the motion. Then the original coordinates will be functions $$x_i = x_i \left( q_1 , q_2 , \ldots , q_{3n-k} , t \right) , \quad i = 1, 2, \ldots , 3n ,$$ of the generalized coordinates. The generalized coordinates could be distances, angles, or other quantities relating to the description of the motion. The number of generalized coordinates is the number of independent degrees of freedom.

We derive the Euler–Lagrange equations from d’Alembert’s Principle. Suppose that the system is described by generalized coordinates q. The virtual displacement of the i-th particle δri is consistent with the constraints acting on the system. To first order in δ's, we have

$\sum_i \left[ m_i \ddot{\bf r}_i - {\bf F}_i \right] \cdot \delta {\bf r}_i =0 ,$
which is d’Alembert’s Principle, is valid for the virtual displacement δri.

Note that d’Alembert’s Principle is expressed as a scalar equation involving what is termed virtual work Fi \cdot \deltari, which is the work that would be done on the mass mi in the virtual displacement δri. This virtual work is equal to a corresponding virtual change in the kinetic energy of the mass mi, which is $$m_i \ddot{\bf r}_i \cdot \delta{\bf r}_i .$$ This formulation includes, in principle, work by dissipative forces as well. Work and energy then replace forces in a formulation of mechanics based on d’Alembert’s Principle.

The above equations in Cartesian coordinates can be written as

$\sum_{i,k} \left[ m_i \left( \ddot{x}_i \frac{\partial x_i}{\partial q_k} + \ddot{y}_i \frac{\partial y_i}{\partial q_k} + \ddot{z}_i \frac{\partial z_i}{\partial q_k} \right) - \left( F_{xi} \frac{\partial x_i}{\partial q_k} + F_{yi} \frac{\partial y_i}{\partial q_k} + F_{zi} \frac{\partial z_i}{\partial q_k} \right) \right] \delta q_k =0 .$
This equation is of the form
$\sum_k \alpha_k \delta q_k =0 .$
Since the generalized coordinates qk are independent of one another, the δqk are arbitrary. Therefore, the above equation can only be valid if each αk is independently zero. That is
$\sum_{i} \left[ m_i \left( \ddot{x}_i \frac{\partial x_i}{\partial q_k} + \ddot{y}_i \frac{\partial y_i}{\partial q_k} + \ddot{z}_i \frac{\partial z_i}{\partial q_k} \right) = \sum_i \left( F_{xi} \frac{\partial x_i}{\partial q_k} + F_{yi} \frac{\partial y_i}{\partial q_k} + F_{zi} \frac{\partial z_i}{\partial q_k} \right) \right] \delta q_k$
for each component k. This Equation must then be valid if the system obeys Newton’s laws.

Because the Cartesian coordinates are functions of the generalized coordinates and the time, the time derivative of the coordinate xi is

$\dot{x}_i = \frac{{\text d}x_i}{{\text d}t} = \sum_k \frac{\partial x_i}{\partial q_k} \,\dot{q}_k + \frac{\partial x_i}{\partial t} .$
If we now take the partial derivative of of the above relation with respect to $$\dot{q}_k ,$$ we obtain
$\frac{\partial \dot{x}_i}{\partial \dot{q}_k} = \frac{\partial x_i}{\partial q_k} .$
This equation is often called cancellation of the dots because it appears as though we have simply canceled the dots (time derivatives) in $$\partial \dot{x}_i /\partial \dot{q}_k$$ to obtain $$\partial x_i/\partial q_k .$$ This means mathematically that the Cartesian coordinate xi depends only on q and the time t and is independent of the velocities $$\dot{q}_k$$

With this in hand, we have

$\ddot{x}_i \frac{\partial x_i}{\partial q_k} = \ddot{x}_i \frac{\partial \dot{x}_i}{\partial \dot{q}_k} .$
Now
$\ddot{x}_i \frac{\partial \dot{x}_i}{\partial \dot{q}_k} = \frac{\text d}{{\text d}t} \left( \dot{x}_i \frac{\partial \dot{x}_i}{\partial \dot{q}_k} \right) - \dot{x}_i \frac{\text d}{{\text d}t} \left( \frac{\partial x_i}{\partial q_k} \right)$
We calculate next
$\frac{\text d}{{\text d}t} \left( \frac{\partial x_i}{\partial q_k} \right) = \sum_j \frac{\partial^2 x_i}{\partial q_j \,\partial q_k} \, \dot{q}_j + \frac{\partial^2 x_i}{\partial t\,\partial q_k} .$
Also we have
$\frac{\partial \dot{x}_i}{\partial q_k} = \sum_j \frac{\partial^2 x_i}{\partial q_j \,\partial q_k}\, \dot{q}_j + \frac{\partial^2 x_i}{\partial t\,\partial q_k} ,$
since the order of partial differentiation is immaterial. Form the above equations, we see that
$\frac{\text d}{{\text d}t} \left( \frac{\partial x_i}{\partial q_k} \right) = \frac{\partial \dot{x}_i}{\partial q_k} .$
This allows us to conclude that
$\ddot{x}_i \frac{\partial x_i}{\partial q_k} = \frac{\text d}{{\text d}t} \left( \dot{x}_i \frac{\partial \dot{x}_i}{\partial \dot{q}_k} \right) - dot{x}_i \frac{\partial \dot{x}_i}{\partial q_k} = \left[ \frac{\text d}{{\text d}t} \,\frac{\partial}{\partial \dot{q}_k} - \frac{\partial}{\partial q_k} \right] \left( \frac{1}{2}\,\dot{x}_i^2 \right) .$
Using the latter, we have
$\sum_i m_i \left( \ddot{x}_i \frac{\partial x_i}{\partial q_k} + \ddot{y}_i \frac{\partial y_i}{\partial q_k} + \ddot{z}_i \frac{\partial z_i}{\partial q_k} \right) = \left[ \frac{\text d}{{\text d}t} \,\frac{\partial}{\partial \dot{q}_k} - \frac{\partial}{\partial q_k} \right] \sum_i \frac{1}{2} \left( \dot{x}_i^2 + \dot{y}_i^2 + \dot{z}_i^2 \right) .$
We recognize the last term as the kinetic energy , which, in keeping with the notation of Lagrange and Hamilton, we shall designate as
$\mbox{K} = \frac{1}{2} \, \sum_i m_i \left( \dot{x}_i^2 + \dot{y}_i^2 + \dot{z}_i^2 \right) .$
According to Newton's law,
$\left[ \frac{\text d}{{\text d}t} \,\frac{\partial}{\partial \dot{q}_k} - \frac{\partial}{\partial q_k} \right] \mbox{K} = \sum_i \left( F_{xi} \frac{\partial x_i}{\partial q_k} + F_{yi} \frac{\partial y_i}{\partial q_k} + F_{zi} \frac{\partial z_i}{\partial q_k} \right)$
We now recall that the forces remaining are those arising from external fields. In modern notation these forces are equal to the negative gradient of a scalar potential Π, which is a function only of spatial coordinates. That is
$F_{xi} = - \frac{\partial \Pi}{\partial x_i} , \quad F_{yi} = - \frac{\partial \Pi}{\partial y_i} , \quad F_{zi} = - \frac{\partial \Pi}{\partial z_i} .$
Therefore, the right hand side becomes
$\sum_i \left( \frac{\partial \Pi}{\partial x_i} \, \frac{\partial x_i}{\partial q_k} + \frac{\partial \Pi}{\partial y_i} \, \frac{\partial y_i}{\partial q_k} + \frac{\partial \Pi}{\partial z_i} \, \frac{\partial z_i}{\partial q_k} \right) = \frac{\partial \Pi}{\partial _k}$
using the chain rule. This allows us to simplify
$\left[ \frac{\text d}{{\text d}t} \,\frac{\partial}{\partial \dot{q}_k} - \frac{\partial}{\partial q_k} \right] \mbox{K} = - \frac{\partial \Pi}{\partial q_k} .$
Since the potential Π depends only on the coordinates and not on the velocities, the last equation may be written as
$\left[ \frac{\partial}{\partial q_k} - \frac{\text d}{{\text d}t} \,\frac{\partial}{\partial \dot{q}_k} \right] \left( \mbox{K} - \Pi \right) = 0 .$
These are the Euler–Lagrange equations. The combination K − Π is called the Lagrangian, which is a scalar function of the generalized coordinates q, the time derivatives of the generalized coordinates $$\dot{\bf q} ,$$ and possibly the time t.
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In terms of generalized coordinates qi, the equations of motion follow from 3n-k equations

$\frac{\text d}{{\text d} t} \left( \frac{\partial {\mbox K}}{\partial \dot{q}_i} \right) - \frac{\partial {\mbox K}}{\partial q_i} = Q_i , \qquad i = 1,2, \ldots , 3n-k ,$
where K is the kinetic energy and Qi are the generalized forces. If Fi is the i-th component of the force, then the generalized force Qi is defined by
\begin{equation} \label{EqL.1} Q_i = \sum_{j=1}^{3n} F_j \, \frac{\partial x_j (q_1 , \ldots , q_{3n-k} , t)}{\partial q_i} , \qquad i = 1,2, \ldots , 3n-k. \end{equation}
If some of the forces are conservative, then they can be expressed in term of the potential Π, where $$F = - \nabla \, \Pi .$$ There are many sources of potential energy, of which we idicate the following:
• Gravitational Energy. This is simply m g h, where m is the mass, h is the height, and g is gravity.
• Spring energy. This is the energy held by compressing/elongating a spring or extending it.The simplest (Hooke's law) spring has an energy, EH = ½K(ℓ - ℓ0)², where K is a constant and ℓ0 is the rest length of the spring.
• Inverse distance energy. Many phenomena have energy that depends of the inverse of the distance, Ei = K/r, where K is a constant and r is a distance.
Chemical and electrical energy are other types that can be important in applications, but almost all of the systems we study here rely on the above three examples. Note that in all physical systems, there is friction.

Expressing the conservative forces by a potential Π and nonconservative forces by the generalized forces Qi, the equation of motion follow from Euler--Lagrange's equations

$\frac{\text d}{{\text d} t} \left( \frac{\partial {\cal L}}{\partial \dot{q}_i} \right) - \frac{\partial {\cal L}}{\partial q_i} = Q_i , \qquad i = 1,2, \ldots , 3n-k .$
The Lagrangian $${\cal L} = \mbox{K} - \Pi$$ describes the conservative forces. The force field F(x) is said to be conservative if
${\bf F} ({\bf x}) = - \nabla \Pi ({\bf x})$
for a smooth potential function $$\Pi\,:\,\mathbb{R}^n \to \mathbb{R} ,$$ where ∇ denotes the gradient with respect to x. Equivalently, the force field is conservative if the work done by F on the particle as it moves from point A to point B,
$\int_{\Gamma (A,B)} \, {\bf F} \cdot {\text d} {\bf x}$
is independent of the path Γ(A,B) between the two endpoints. When all forces are conservatives, Euler--Lagrange's equations are reduced to
$\frac{\text d}{{\text d} t} \left( \frac{\partial {\cal L}}{\partial \dot{q}_i} \right) - \frac{\partial {\cal L}}{\partial q_i} = 0 , \qquad i = 1,2, \ldots , 3n-k .$
Given a Lagrangian, $${\cal L} ,$$ which is a function of the location in space and the velocity, we define the action:
$S({\bf x}, a,b) = \int_{t =a}^{t=b} {\cal L}\left( {\bf x} , \dot{\bf x} \right){\text d}t ,$
which is a real-valued function, defined on a space of trajectories $$\left\{ {\bf x}: [a,b] \to \mathbb{R}^n \right\} .$$ A scalar-valued function of functions, such as the action, is often called a functional. The principle of stationary action (also called Hamilton’s principle) states that, for fixed initial and final positions x(a) and x(b), the trajectory of the particle x(t) is a stationary point of the action. Then for fixed initial and final positions, the path of least action (stationary action) will be the path along which Newton's laws are obeyed. So for Newtonian mechanics, the Lagrangian is chosen to be $${\cal L} = \mbox{K} - \Pi .$$ To explain what this means in more detail, suppose that $$\left\{ {\bf h}: [a,b] \to \mathbb{R}^n \right\}$$ is a trajectory with h(a) = h(b) = 0. The directional (or Gâteaux) derivative of S at x(t) in the direction h(t) is
${\text d}S({\bf x})\, {\bf h} = \left. \frac{\text d}{{\text d}t}\,S \left( {\bf x} + \epsilon {\bf h} \right) \right\vert_{\epsilon =0} .$

Friction is not conservative, and so it doesn't fit neatly into the scheme we've outlined so far. In this case, one needs to take into account the non-conservative generalized forces.

Example: Consider an elastic spring attached to mass m; its Lagrangian is

${\cal L} = \frac{m}{2} \, \dot{x}^2 - \frac{k}{2}\, x^2 ,$
where k is a spring constant, because its kinetic energy is $$\mbox{K} = \frac{m}{2} \, \dot{x}^2$$ and the potential energy is $$\Pi = \frac{k}{2}\, x^2 .$$ The position x(t) of a one-dimensional oscillator moving in a potential $$\Pi \,:\, \mathbb{R} \to \mathbb{R}$$ satisfies the ordinary differential equation
$m\,\ddot{x} + \Pi' (x) =0 ,$
where prime denotes the derivative with respect to x. The solutions lie on the curves in the $$( x, \dot{x})$$ phase plane given by
$\frac{m}{2}\, \dot{x} + \Pi (x) = E \qquad \mbox{the total energy is a constant}.$
The equilibrium solutions are the critical points of the potential Π. Local minima of Π correspond to stable equilibria, while other critical points correspond to unstable equilibria. For example, the quadratic potential $$\Pi = \frac{k}{2}\, x^2$$ gives the linear simple harmonic oscillator, $$\ddot{x} + \omega^2 x =0 ,$$ with frequency $$\omega = \sqrt{k/m} .$$ Its solution curves in the phase plane are ellipses, and the origin is a stable equilibrium.

Example: The simplest version of the Atwood machine consists of two masses, m1 and m2, suspended on either side of a massless, frictionless pulley of radius R by a massless rope of length $$\ell .$$ Picking an origin in space at the center of the pulley, we can characterize the configuration of the system by, say, the vertical position of m1, which we specify via the vertical distance s from the origin. So we choose s as the generalized coordinate for our system---Atwood's machine. The location of the other mass is now determined; its vertical distance from the origin is $$\ell -\pi R -s .$$ Evidently, if the velocity of m1 is $$\dot{s} = {\text d}s/{\text d}t ,$$ then the velocity of m2 is $$-\dot{s} .$$ The kinetic energy of the system (the two masses) is then

$\mbox{K} = \frac{m_1}{2}\,\dot{s}^2 + \frac{m_2}{2}\left( -\dot{s} \right)^2 = \frac{1}{2} \left( m_1 + m_2 \right) \dot{s}^2 .$
The potential energy
$\Pi = - m_1 gs -m_2 g \left( \ell -\pi R -s \right) = - \left( m_1 - m_2 \right) gs + \mbox{constant} .$
We will suppress the additive constant since only the derivative of Π will be needed to make the Euler--Lagrange equation. Then the Lagrangian becomes
${\cal L} \left( s, \dot{s} \right) = \mbox{K} - \Pi = \frac{1}{2} \left( m_1 + m_2 \right) \dot{s}^2 + \left( m_1 - m_2 \right) gs .$
From the Euler--Lagrange equations we derive the equation of motion for the Atwood machine
$\dot{s} = \frac{m_1 - m_2}{m_1 + m_2} \, g .$
The Atwood machine (or Atwood's machine) was invented in 1784 by the English mathematician George Atwood.

Example: Consider a particle of constant mass m moving in n-dimensional space in a spatially-dependent conservative force field $${\bf F} ({\bf x}) = - \nabla \Pi ({\bf x}) .$$ Then according to Newton's second law, a trajectory satisfies the equation

$m\,\ddot{\bf x} = - \nabla \Pi ({\bf x}) ,$
where a dot denotes the derivative with respect to time t. Taking the scalar product with respect to velocity $$\dot{\bf x} ,$$ we get
$\frac{\text d}{{\text d}t} \left\{ \frac{m}{2} \left\vert \dot{\bf x} \right\vert^2 + \Pi ({\bf x}) \right\} =0.$
Hence the total energy of the particle is the sum
${\mbox E} = \mbox{K} + \Pi ({\bf x})$
of the potential energy Π and the kinetic energy $$\mbox{K} = \frac{m}{2} \left\vert \dot{\bf x} \right\vert^2 .$$

In particular, consider a particle moving in 3-dimensional space. Introducing spherical coordinates

$x= r\,\sin \theta\,\cos\phi , \quad y = r\,\sin \theta\,\sin\phi , \quad z=r\,\cos\theta ,$
we rewrite the kinetic energy as
$\mbox{K} = \frac{m}{2} \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) = \frac{m}{2} \left( \dot{r}^2 +r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \,\dot{\phi}^2 \right) .$
It is now very straightforward to compute the equations of motion:
\begin{eqnarray*} m\,\ddot{r} - mr\,\dot{\theta}^2 - mr\,\sin^2 \theta \,\dot{\phi}^2 &=& \frac{\partial \Pi}{\partial r} , \\ m\,\frac{\text d}{{\text d}t} \left( r^2 \dot{\theta} \right) - mr^2 \sin\theta\,\cos\theta \,\dot{\phi}^2 &=& \frac{\partial \Pi}{\partial \theta} , \\ m\,\frac{\text d}{{\text d}t} \left( r^2 \sin^2 \theta \,\dot{\phi}^2 \right) &=& \frac{\partial \Pi}{\partial \phi} . \end{eqnarray*}
From these formulas we can easily derive the equation of motion for the spherical pendulum, with gravity acting in the z direction
$r(t) \equiv \ell , \qquad \Pi = mg\ell \left( 1+ \cos \theta \right) ,$
where $$\ell$$ is the radius of the pendulum. Then from the Euler--Lagrange equations, it follows
\begin{eqnarray*} \ddot{\theta} + \frac{g}{\ell} \sin \theta - \sin\theta\,\cos\theta \,\dot{\phi}^2 &=& 0, \\ \frac{\text d}{{\text d}t} \left( \sin^2 {\theta} \,\dot{\phi} \right) &=& 0 \qquad \Longrightarrow \qquad \ddot{\phi} = - 2\dot{\theta} \,\dot{\phi} \,\cot\theta . \end{eqnarray*}

Example: The trajectory $${\bf x}\,:\,|a,b| \to \mathbb{R}^3$$ of a particle of mass m moving in gravitational 3-dimensional field satisfies the vector differential equation (Newton’s universal law of gravitation)

$\ddot{\bf x} = -GM\frac{\bf x}{\| {\bf x}\|^3} ,$
where G is the gravitational constant. The Lagrangian for classical gravitational field is
${\cal L} = \frac{1}{8\pi G}\left( \nabla \phi \right)^2 + \rho\,\phi ,$
where ϕ is the gravitational potential field and ρ is the mass density. This allows one to derive the corresponding Euler-Lagrange equation, which is actually the Poisson's equation for gravity:
$\nabla^2 \phi = 4\pi G \rho .$

Example: We use notation of r and θ for polar coordinates on the plane. The velocity has two component: in the radial direction it is dr/dt and in the tangential direction it is r(dθ/dt). Since the gravitational potential is $$-m\,k/{\bf r} ,$$ with k = GM, the kinetic and potential energy in polar coordinates become

\begin{align*} \mbox{K} &= \frac{m}{2} \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) , \\ \Pi &= -m \,\frac{k}{r} , \end{align*}
giving the Lagrangian
${\cal L} = \frac{m}{2} \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) + m \,\frac{k}{r} .$
In our case, polar coordinates are our generalized coordinates, so we calculate partial derivatives
\begin{eqnarray*} \frac{\partial {\cal L}}{\partial \theta} &=& 0 , \\ \frac{\partial {\cal L}}{\partial \dot{\theta}} &=& m\,r^2 \dot{\theta} , \\ \frac{\text d}{{\text d}t}\,\frac{\partial {\cal L}}{\partial \dot{\theta}} &=& m \left( 2r\,\dot{r} \dot{\theta} + r^2 \ddot{\theta} \right) , \\ \frac{\partial {\cal L}}{\partial r} &=& m\,r \dot{\theta}^2 - m\,\frac{k}{r^2} , \\ \frac{\partial {\cal L}}{\partial \dot{r}} &=& m\,\dot{r} , \\ \frac{\text d}{{\text d}t}\,\frac{\partial {\cal L}}{\partial \dot{r}} &=& m\,\ddot{r} . \end{eqnarray*}
Plugging the derivatives into the Euler--Lagrange equations, we obtain the equations of motion:
$\begin{split} 2r\,\dot{r} \dot{\theta} + r^2 \ddot{\theta} =0 , \\ \ddot{r} = r\dot{\theta}^2 - \frac{k}{r^2} . \end{split}$
We see from the latter that we can have a circular orbit if the "centrifugal force" ($$r\dot{\theta}^2$$ ) balances the gravitational attraction (k/r2); in that case there won't be any radial acceleration and, upon substitution v = r(dθ/dt), the equation of circular motion becomes
$\dot{\theta}^2 = v^2 / r^2 \qquad \mbox{and} \qquad v^2 = k/r .$

Example: Consider a particle rotating with constant angular velocity ω0:

$\begin{split} x &= r\,\cos \left( \theta + \omega_0 t \right) , \\ y &= r\,\sin \left( \theta + \omega_0 t \right) . \end{split}$
The velocities in the Cartesian frame are:
$\begin{split} \dot{x} &= \dot{r}\,\cos \left( \theta + \omega_0 t \right) -r \left( \dot{\theta} + \omega_0 \right) \sin \left( \theta + \omega_0 t \right) , \\ \dot{y} &= \dot{r}\,\sin \left( \theta + \omega_0 t \right) + r \left( \dot{\theta} + \omega_0 \right) \cos \left( \theta + \omega_0 t \right) . \end{split}$
The kinetic energy of the particle is given by
${\text K} = \frac{m}{2} \left( \dot{x}^2 + \dot{y}^2 \right) = \frac{m}{2} \left( \dot{r}^2 + r^2 \left( \dot{\theta} + \omega_0 \right)^2 \right) .$
Then the Lagrangian becomes
${\cal L} = \frac{m}{2} \left( \dot{r}^2 + r^2 \left( \dot{\theta}^2 + \omega_0^2 + 2\omega_0 \dot{\theta} \right) \right) - \Pi (r) .$
The derivatives we need are:
\begin{eqnarray*} \frac{\text d}{{\text d}t}\,\frac{\partial {\cal L}}{\partial \dot{\theta}} &=& m \left( 2r\,\dot{r} \left( \dot{\theta} + \omega_0 \right) + r^2 \ddot{\theta} \right) , \\ \frac{\partial {\cal L}}{\partial \theta} &=& 0 , \\ \frac{\text d}{{\text d}t}\,\frac{\partial {\cal L}}{\partial \dot{r}} &=& m \ddot{r} , \\ \frac{\partial {\cal L}}{\partial r} &=& mr \left( \dot{\theta}^2 + \omega_0^2 + 2\omega_0 \dot{\theta} \right) - \frac{\partial \Pi}{\partial r} . \end{eqnarray*}
Inserting the above equations into the Euler--Lagrange equations, and rearranging a little, we obtain the equations of motion:
$\begin{split} \ddot{\theta} &= - \frac{2}{r}\,\dot{r} \left( \dot{\theta} + \omega_0 \right) , \\ \ddot{r} &= r \left( \dot{\theta}^2 + \omega_0^2 \right) + 2r \dot{\theta} \omega_0 - \frac{1}{m} \,\frac{\partial \Pi}{\partial r} . \end{split}$
The right hand side of the latter equation, $$- \frac{2}{r}\,\dot{r} \left( \dot{\theta} + \omega_0 \right) ,$$ is the tangential Coriolis force; it depends on the radial velocity, and the sum of the angular velocity of the particle in the frame and rotation rate of the frame. The first term on the right of the former equation is the centrifugal force due to the particle's own motion and the rotation of the frame. The second term is the radial Coriolis force; it's proportional to ω0, so for a non-rotating frame, it vanishes.