Then 0 = – h(?i = ^[{^1 + m2)ii9i + m2(.ih92 cos(6»i – в2)\ + m2(.ih9i92…



подпись: d

подпись: dt
+g(m1 + m2)t1 sin 91
0 = – h(?i = ^[{^1 + m2)ii9i + m2(.ih92 cos(6»i – в2)\ + m2(.ih9i92 8т(6»1 – в2)




0 = — ~ ^^2 = ^ [m2iii29i cos(6»i – 92) + 777,2^2^2] + gm2i2 sin 92 (25)

– m2ti^29i92 sin(^i – 92)


Consider the harmonic oscillator whose Lagrangian is given by

L{t, y,y) = – ^ky^.

1 p 2 1

H = – L + yLy =

Which is the sum of the kinetic and potential energy of the system. Differentiating


~7T = дУ

ДН p dp m

So Hamilton’s equations are

To solve these equations in the yp plane (the so called phase plane) we divide them to get

Dp —ky dy p/m


Pdy + kmydy = 0

After integration, we have

подпись: c is constantP2 + kmy"^ = c,

Which is a family of ellipses in the py plane. These represent trajectories that the system evolves along in the position-momentum space. Fixing an initial value at time t0 selects out the particular trajectory that the system takes.


подпись: i (y) подпись: b подпись: r(t)y‘^ + q(t)y^ подпись: dt.

1. Consider the functional

Find the Hamiltonian and write the canonical equations for the problem.

2. Give Hamilton’s equations for

Ну) = f + y2)(i +y2)dc

J a

Solve these equations and plot the solution curves in the yp plane.

3. A particle of unit mass moves along the y axis under the influence of a potential

F (y) = —^2y + ay2

Where ш and a are positive constants.

A. What is the potential energy V(y)? Determine the Lagrangian and write down the equations of motion.

B. Find the Hamiltonian H(y, p) and show it coincides with the total energy. Write down Hamilton’s equations. Is energy conserved? Is momentum conserved?


C. If the total energy if is —, and y(0) = 0, what is the initial velocity?

/ dg \ 2 Restriction of the problem since the equation ^ j = — a) deduced…

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/ dg \ 2

Restriction of the problem since the equation ^ j = — a) deduced above shows

That a particle started on a curve with the velocity v\ at the point 1 will always come to rest if it reaches the altitude y = a on the curve, and it can never rise above that altitude. For the present we shall restrict our curves to lie in the half-plane y > a.

подпись: f подпись: (34)

In our study of the shortest distance problems the arcs to be considered were taken in the form у : y[x) {xi < x < X2) with y(x) and y'{x) continuous on the interval X\ < x < X2, An admissible arc for the brachistochrone problem will always be understood to have these properties besides the additional one that it lies entirely in the half-plane y > a. The integrand F(y, y’) and its partial derivatives are:

Since our integrand in (33) is independent of x we may use the case 2 special result (21) of the Euler equations.

When the values of F and its derivative Fy’ for the brachistochrone problem are substituted from (34) this equation becomes

Integrating, we get x

X = a + b{u + sin u)

Where a is the new constant of integration. It will soon be shown that curves which satisfy the first and third of these equations are the cycloids described in the following theorem: Theorem 7 A curve down which a particle, started with the initial velocity v\ at the point 1, vjill fall in the short^est time to a second point 2 is necessarily an arc having equations of the form

X — a = b(u + sinu) , y — a = b(1 + cos u). (37)

These represent the locus of a point fixed on the circumference of a circle of radius b as the


Circle rolls on the lower side of the line у = a = у\——- Such a curve is called a cycloid.


Cycloids. The fact that (37) represent a cycloid of the kind described in the theorem is proved as follows: Let a circle of radius b begin to roll on the line y = a at the point whose co-ordinates are (a, a), as shown in Figure 14. After a turn through an angle of u radians the point of tangency is at a distance bu from (a, a) and the point which was the lowest in the circle has rotated to the point (x, y). The values of x and y may now be calculated in terms of u from the figure, and they are found to be those given by (37).

подпись: figure 14: cycloidподпись: xподпись: y(a, a)

6R = Cr5r + eer59 + e\r cos 95Л Where r = |R| is distance from origin to particle…

подпись: (22)6R = Cr5r + eer59 + e\r cos 95Л

Where r = |R| is distance from origin to particle and er, ee, C\ are the unit direction vectors (see Figure 25) that R changes due to respective changes in the spherical coordinates (r, 9, Л). Then by (22)

подпись: (23)F • 5R = F • er6r + F • eer69 + F • C\r cos 95Л = F • er6r

Where the last equality follows since F is along er according to our assumption. Now using (15), (23) and second part of (13) results in

подпись: (24)DV dV dV

F • 6r6r = F -5R = -5V = -[^5r + —59 + —^A]

Dr d9 ДЛ

And since 5r, 59, 5Л are independent, then this gives





Figure 25: Unit vectors Cr, ee, and e\


I. e. V is only a functions of r (actually we know V =———– where /и is a constant)


U = U (r)




Now for our particle in the xy plane in spherical coordinates, the velocity, of our particle at “to” has components along er ,ee, e\ respectively of




R, 0, rX


The second value being due to the velocity vector being in P and ee being perpendicular to P. Then the kinetic energy T is


[ V\l^ +y’ dx. J Xi The problem of determining the form of the soap film…

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подпись: i[ V\l^ +y’ dx.

J Xi

The problem of determining the form of the soap film surface between the two circles is analytically that of finding in the class of arcs y = y(x) whose ends are at the points 1 and 2 one which minimizes the last-written integral I.

As a third example of problems of the calculus of variations consider the problem of the brachistochrone (shortest time) i. e. of determining a path down which a particle will fall from one given point to another in the shortest time. Let the y-axis for convenience be taken vertically downward, as in Figure 6, the two fixed points being 1 and 2.

подпись: 0



The initial velocity v\ at the point 1 is supposed to be given. Later we shall see that for an arc dehned by an equation of the form у = y{x) the time of descent from 1 to 2 is

подпись: i

times the value of the integral


Where g is the gravitational constant and a has the constant value a = у\—– The problem


Of the brachistochrone is then to hnd, among the arcs у : y{x) which pass through two points 1 and 2, one which minimizes the integral I.

As a last example, consider the boundary value problem

where utV = {vtC‘^[0,1],v(0) = 0,v(1) = 0} . The function r(x) can be viewed as force per unit mass.

подпись: within the class of arcs which are continuously differentiable and also satisfy the endpoint conditions
y(xi) = yi y(x2) = у2 (4)
where j/i, j/2 are constants. in the previous three problems f was respectively f = -\/l + y' ^ ^ / \/l + y' ^
f = yjl + y' ^ , f = — and yi, уг were the у coordinates associated with the points

1 and 2.

It should be noted that in (3) the symbols x, y, y’ denote free variables and are not directly related to arcs. For example, we can differentiate with respect to these variables to get in the case of our last example

Thus we solve for y(x) on xi < x < x2 by solving the second…

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Thus we solve for y(x) on xi < x < x2 by solving the second order differential equation (12) with initial conditions

N(xi) = 0 n (x-]) = 1.

For example, suppose we wish to find the minimum of







Y(0) = 0

Y(1) = 1.

The function odeinput. m supplies the user with the boundary conditions, a guess for the initial slope, tolerance for convergence. All the derivatives of f required in (4) are supplied in rhs2f. m.

Function [fy1y1,fy1y, fy, fy1x, t0,tf, y1,y2,rhs2,sg, tol] = odeinput % Defines the problem for solving the ode:

% (f_{y’y’} )y" + (f_{y’y})y’ = f_y – f_{y’x}

% t0 – start time

% tf – end time

% y1 – left hand side boundary value

% y2 – right hand side boundary value % sg – initial guess for the slope % tol – tolerance e. g. 1e-4

= 1;

подпись: to
= 0;

= 1;

= 1;

= 1e-4;

%rhs2f. m


Function [rhs2]=rhs2f(t, x)



% t is the time

% x is the solution vector (y, y’)


% fy1fy1 – fy’y’ (2nd partial wrt y’ y’)

% fy1y – fy’y (2nd partial wrt y’ y)

% fy – fy (1st partial wrt y)

% fy1x – fy’x (2nd partial wrt y’ x)


Fy1y1 = 2; fy1y = 0; fy = 2*x(1);

Fy1x = 0;


The main program is odel. m which uses a modified version of ode23 from matlab. This modified version is called ode23m. m. Since we have to solve a second order ordinary differential equation, we have to transform it to a system of first order to be able to use ode23. To solve the n equation, the ode23 is used without any modifications. We also need the right hand side of the 2 equations to be solved (one for y and one for n). These are called odef. m and feta. m, respectively. All these programs (except the original ode23.m) are given here

% ode1.m

% This program requires an edited version of ode23 called ode23m. m % Also required is odef. m, feta. m & odeinput. m % All changes to a problem should ONLY be entered in odeinput. m

[fy1y1,fy1y, fy, fy1x, t0,tf, y1,y2,rhs2,sg, tol] = odeinput;