Example 1: (Flow Problem) A 500 liter container initially contains 10 kg of salt. A brine mixture of 100 grams of salt per liter is entering the container at 6 liter per minute. The well-mixed contents are being discharged from the tank at the rate of 6 liters per minute.

Express the amount of salt in the container as a function of time.

Salt is coming at the rate: 6*(0.1)=0.6 kg/min
d yin /dt =0.6 ; d yout/dt = 6x/500

dx/dt = 0.6 -6x/500 ; x(0)=10

Suppose that the rate of discharge is reduced to 5 liters per minute.

To check the answer:

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Example 2: Consider a bead of mass m on a rotating hoop of radius R.

The kinetic energy reads

The potential energy is where θ is the angle measured from the bottom of the hoop with respect to the direction of the rotation axis and α is the angle formed between the rotation axis and the vertical direction. If there exist a friction, which could be modeled by the force which leads to the generalized force Inserting these expressions into the Lagrangian \( L = \mbox{K} - \Pi , \) and applying the Euler--Lagrange equations, we obtain the equation of motion: If the tilt angle α is small, the above equation can be linearized in α by natural substitution: If rotation is along vertical axis (α = 0), we get a homogeneous differential equation The equilibrium positions θ = θ^*, associated to the condition that θ and its velocity are both zeroes, are obtained by solving either This gives three physical stationary points: subject to condition Therefore, if the angular velocity of the hoop ω is greater than a critical angular velocity ω_c, we observe the third critical point; otherwise thare are only two of them. This situation indicates that we observe a supercritical pitchfork bifurcation about the stationary value ω_c.

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Example 3:

Example: Move to chapter 4, application4 Consider the Lane--Emden type equation with an exponential nonlinearity of the solution with respect to the variable base

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Example 4: Consider the initial value problem for undamped Duffing equation:

We can rewrite the Duffing equation in the operator form: Physically, the resilience of the oscillator is directly proportional to N[y]. When k ≥ 0 and ε < 0, there is one unique equilibrium point y = 0 where the nonlinear term vanishes. However, when k ≤ 0 and ε > 0, there are two equilibrium points: y = 0 and \( y = \sqrt{|k/\varepsilon |} . \) From the physical points of view, the sum of the kinetic and potential energy of the oscillator keeps the same, therefore it is clear that the oscillation motion is periodic, no matter k is positive or negative. Thus, from physical points of view, it is easy to know that y(t) is periodic, even if we do not directly solve the Duffing equation.    ■

Example 5: (Flow Problem) Our example is based on Gladkov's article. Although the problem of brachistochrone, which was stated by Johann Bernoulli in 1696, has been discussed previously in section on Reduction higher order ODEs, we reconsider this problem when an object slides under the Coulomb friction force.

We assume that an object is spherical form is rolling over a curve without slip under gravity in vacuum and the sphere is decelerated only by the force of interaction with the body according to the Coulomb law. A schematic of the forces and geometric statement of the problem are given in the following figure.

The equation of motion can always be written as Newton’s second law:

where m is the sphere mass, F_fr is the Coulomb friction force, N is the reaction force, and g is the acceleration of gravity.

The acceleration in curvilinear coordinates is

where τ − n is the moving basis shown in the figure, R is the radius of curvature of the motion path, τ is the unit vector of the tangent directed along the velocity, and n is the normal vector. From here on, a dot above a variable means time differentiation, and it is assumed that v = v(t).

Since

where obtuse angle α(t) belongs to the closed segment [π/2, π]. Eq.(5,1) can be written as Having projected this equation to the basis τ–n, we arrive at the following system of equations: It follows from the lower equation that the reaction force is When considering brachistochrone, the following condition should be satisfied (as was demonstrated in the article by A. A. Barsuk and F. Paladi, International Journal of Non-Linear Mechanics, 148, 43 (2023).): (Note that, if \( \displaystyle \quad v = g\, \cos\alpha , \quad \) the trajectory is a parabola.) Therefore, according to \( \displaystyle \quad N = m \left( \frac{v^2}{R} - g\,\cos\alpha \right) , \quad \) the reaction force is Hence, the friction force becomes This allows us to rewrite the system of equation (5.2) as Since we arrive at the following system instead of Eq.(5.3), which is much simpler: Having divided the upper equation by the lower, we obtain After elementary integration, we find that where C ₁ is the integration constant. Since angle α is obtuse, we have Afterwards, using the equations we obtain upon integration that According to Eq.(5.4), we have we find that A solution to this equation can be written as where it is taken into account that initially α(0) = π/2. Hence, Substituting into solution (5.5), we get It follows from the law of conversation of energy that After integration, we arrive at the final analytical solution of the problem of brachistochrone with the Coulomb friction force: It follows from this solution that, in the ideal-case limit (where the friction coefficient μ → 0 ), we obtain the parametric equation of classical brachistochrone The fact that this solution describes specifically brachistochrone can easily be verified by calculation of the time of ball rolling from the upper point (where α = π/2) to the lower one (where α = π).For the classical brachistochrone, this time is Taking into account expression \( \displaystyle \quad v = - C_1 \cos\alpha \, e^{-2\mu\alpha}, \quad \) according to which we find directly the desired rolling time: which coincides with (5.6), as was to be proved.    ■

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