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MT5802 Calculus of variations › ~rac › MT5802 › Calculus of varia · PDF file MT5802 - Calculus of variations Introduction. Suppose y(x)is defined on the interval a,b and so

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  • MT5802 - Calculus of variations

    Introduction.

    Suppose y(x) is defined on the interval a,b

      and so defines a curve on the

    x,y( ) plane. Now suppose

    I = F(y, ′y ,x)

    a

    b

    ∫ dx (1)

    with ′y the derivative of y(x) . The value of this will depend on the choice of the function y and the basic problem of the calculus of variations is to find the form of the function which makes the value of the integral a minimum or maximum (most commonly a minimum). The sort of question which gives rise to this kind of problem is exemplified by the “Brachistochrone” problem, solved by Newton and the Bernoullis (the name comes from the Greek for “shortest time”). This considers a particle sliding down a smooth curve under the action of gravity and poses the question as to what curve minimises the time for the particle to slide between fixed points A and B.

    A

    Clearly the time will need to be found by calculating the speed at each point then integrating along the curve. Other examples arise in various areas of physics in which the basic laws can be stated in terms of variational principles. For example in optics Fresnel’s principle says that the path of a light ray between two points is such as to minimise the time of travel between the two points.

    The Euler-Lagrange equation.

    First recall the condition under which an ordinary function y(x) has an extremum. If we expand in a Taylor series

    B

  • y(x + δx) = y(x)+ δx ′y (x)+

    1 2

    δx 2 ′′y (x)+ ........

    then the condition is that the term proportional to δx must vanish, so that if the second derivative is non-zero the difference between y(x + δx)and y(x) will always have the same sign for small δx . The same principle applies to our present problem. What we do is consider a small change in the function y(x) , replacing it with

    y(x)+ η(x) . (Note that all the functions we introduce are assumed to have appropriate properties of differentiability etc, without particular comment being made.) We then produce a change in the integral, which can be expanded in powers of η . We demand that the term proportional to η vanishes. Substituting into (1) we get

    I(y + η) = F(y + η, ′y + ′η ,x)dx a

    b

    = F(y, ′y ,x)dx + ∂F ∂y

    η + ∂F ∂ ′y

    ′η 

      

      dx

    a

    b

    ∫ a

    b

    ∫ +O(η2)

    so that what we want is

    ∂F ∂y

    η + ∂F ∂ ′y

    ′η 

      

      dx

    a

    b

    ∫ = 0 . (2)

    Integrating the second term by parts gives

    ∂F ∂y

    − d dx

    ∂F ∂ ′y

     

     

      

       η(x)dx

    a

    b

    ∫ = 0 (3)

    In obtaining this we have assumed that η(a) = η(b) = 0 , ie the perturbation vanishes at the end points, leaving the end points A and B of the curve unchanged, as shown below.

    y

    x The unperturbed curve (full line) and the perturbed curve ( dotted line)

    Since this must hold for all η(x)we obtain

    A

    B

  • ∂F ∂y

    − d dx

    ∂F ∂ ′y

     

     

    = 0 . (4)

    This is the Euler-Lagrange equation, the basic equation of this theory. It is a differential equation which determines y as a function of x .

    Examples.

    (a) Find the curve which gives the shortest distance between two points on a plane.

    If the curve is y = y(x) then the element of length is

    dl = dx 2 +dy 2 = 1 + ′y 2dx

    so we want to minimise

    1 + ′y 2

    a

    b

    ∫ dx

    (where a and b are the x-coordinates of the points of interest). The integrand is independent of y so we just get

    d dx

    ∂ ∂ ′y

    1 + ′y 2 

     

     

    = 0

    giving

    ′y

    1 + ′y 2 = const , or ′y = const . As expected this just gives a straight line

    y = mx +c with the constants fixed by the positions of the end points.

    (b) Find the curve which minimises

    (y 2 + ′y 2)dx

    a

    b

    The Euler-Lagrange equation for this is

    d dx

    (2 ′y )− 2y = 2 ′′y − 2y = 0

    and if we multiply by ′y we get a first integral ′y 2 −y 2 = const. Assuming this constant

    to be positive and equal to a2 we get the solution y = a sinh(x +b) . If the constant is negative we can take it to be −a2 and get the solution y = a cosh(x +b) . In both cases b is a constant and a and b need to be found using given end points.

    This is a fairly simple, artificial example, but it illustrates a more general point. Note that we could easily find a first integral and reduce the problem to a first order DE. The existence of a first integral like this turns out to be a general property of the Euler- Lagrange equation whenever the integral has no explicit dependence on x .

  • Under these circumstances, if we multiply the E-L equation by ′y we get

    ′y d dx

    ∂F ∂ ′y

     

     

    − ′y ∂F ∂y

     

     

    = 0

    or

    d dx

    ′y ∂F ∂ ′y

     

      − ′′y

    ∂F ∂ ′y

     

      − ′y

    ∂F ∂y

     

     

    = 0 .

    Since F does not contain x explicitly, the last two terms combine to give

    dF dx

    , the total

    derivative. So, we get the first integral

    ′y ∂F ∂ ′y

     

      −F = const (5)

    As a more interesting example we return to the brachistochrone problem mentioned earlier. Suppose two points A and B are connected by a smooth ramp along which a particle can slide, starting at rest at A. Taking A at the origin and the y direction vertically downwards, then at a point (x,y) on the curve, the particle speed is given by

    v 2 = 2gy

    (with g the acceleration due to gravity). The time to move an increment (dx,dy)along the curve is

    dt = dx 2 +dy 2 / v = 1 + ′y 2

    dx

    2gy So, the integral which we need to minimise is

    1 + ′y 2

    y a

    b

    ∫ dx

    and the first integral of the Euler-Lagrange equation as derived above (Eq. (5)) is

    ′y ′y

    y(1 + ′y 2) −

    1 + ′y 2

    y = c .

    This simplifies to (with k = −1/c )

    y(1 + ′y 2 = k

    y(1 + ′y 2) = k

    or

    ′y =

    k 2 −y y

    .

    This can be integrated by making the substitution y = k 2 sin2 θ , giving

  • dy dx

    = 2k 2 sin θ cosθ dθ dx

    = cosθ sin θ

    which has the solution

    x =

    k 2

    2 2θ − sin 2θ( ) + K .

    Putting b = k 2 / 2 and φ = 2θ we get parametric equations for the curve in the form

    x = b(φ − sinφ)+ K y = b(1− cosφ) .

    As illustrated by the diagram below, these represent a cycloid, the curve traced out by a point on the circumference of a wheel of radius b rolling along the x axis.

    y

    Since the curve passes through the origin, K = 0 . The value of b is determined by the condition that the curve passes through B.

    More than one dependent variable.

    Suppose F = F(y1, ′y1,y2, ′y2,y3, ′y3,..........) with each yi = yi(x)and again we are looking

    for an extremum of

    Fdx a

    b

    ∫ . The analysis proceeds as before, replacing each yi with

    yi + ηi . Since each ηi can be chosen independently, we must let the coefficient of each in the integrand vanish. We end up with a system of Euler-Lagrange equations

    x A

    φ

    B

  • ∂F ∂y

    i

    = d dx

    ∂F ∂ ′y

    i

    

     . (6)

    It has, of course, been assumed that the end points are fixed, as before. Example: Find the curve which minimises

    ( ′y 2 +

    0

    1

    ∫ ′z 2 + y 2)dx

    and which joins the points (0,0,0)and (1,1,1) .

    The E-L equations are

    d dx

    (2 ′y )− 2y = 0

    d dx

    (2 ′z ) = 0

    with general solutions

    y = a coshx +b sinhx z = cx +d.

    Imposing the end point conditions gives the curve

    y = cosh−1 1coshx z = x .

    Hamilton’s Principle

    Suppose a conservative dynamical system is described by coordinates q

    1 ,q

    2 ,.....,q

    n( ) and the rates of change of these are &qi

    i = 1,....,n( ) . Then the kinetic energy of the system is, in general T(q1,...,qn; &q1...., &qn)and the potential energy is V(q1,...,qn) . The Lagrangian is then defined by L = T −V and Ham

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