MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 15 pot

17.1 Variation of paths 17.1 PRESENCE OF ONE DEPENDENT AND ONE INDEPENDENT VARIABLE 17.1.1 Euler Equation A majority of the variational problems encountered in physics and engineering

PROBLEMS 513

16.2 Show that the Fourier transform of

is given as

16.3

ing second-order inhomogeneous differential equation:

Using the Laplace transform technique, find the solution of the follow-

with the boundary conditions

y(0) = 2 and y’(0) = -1

16.4 Solve the following system of differential equations:

2z(t) - y(t) - y ’ ( t ) = 4(1 - exp(-t)) 2z’(t) + y ( t ) = 2(1 +3exp(-2t)) with the boundary conditions

Solve the heat transfer equation

where k is the thermal conductivity, c is the heat capacity, and p is the density

16.10

the end a t the origin fixed The shape of the string at t = 0 is given as

Let a semi-infinite string be extended along the positive x-axis with

Y(X, 0) = f(xL

Hint: First show that

16.12 Show that the Fourier sine transform of

xe-ax

is given as

16.13 Establish the result

16.14 Use the convolution theorem to show that

k?’ { (s2 fh2,.} = f cos ht + -sin 2h 1 bt

16.15

of differentia1 equations: Use Laplace transforms t o find the solution of the following system

- -wy1 dYl

516 INTEGRAL TRANSFORMS

with the boundary conditions yl(0) = CO, yz(0) = y3(0) = 0

16.16 Laguerre polynomials satisfy

zLK + (1 - t)L:, + nL,(z) = 0

Show that L'{L,(az)} = (s - s > 0

17

VARIATIONAL

ANALYSIS

Variational analysis is basically the study of changes We are often interested

in finding how a system reacts to small changes in its parameters It is for this reason that variational analysis hes found a wide range of applications not just

in physics and engineering but also in finance and economics In applications

we frequently encounter caees where a physical property is represented by an intwal, the extremum of which is desired Compared to ordinary calculus, where we deal with functions of numbers, these integrals are functions of some unknown function and its derivatives; thus, they are called functionals

Search for the extremum of a function yields the point a t which the function is extremum In the caee of functionals, variational analysis gives us a differential equation, which is to be solved for the extremizing function

After Newton’s formulation of mechanics mane developed a new a p

p m c h , where the equations of motion are obtained from a variational integral called action This new formulation makes applications of Newton’s theory

to many particles and continuous systems poesible Today in making the

transition to quantum mechanics and to quantum field theories w a n

formulation is a must

Geodesics are the shortest paths between two points in a given geametry and constitute one of the main applications of variational analysis In Ein- stein’s theory of gravitation geodesics play a central role tu the paths of freely

moving particles in curved spacetime Variational techniques also form the mathematical basis for the finite elements method, which is a powerful tool for

solving complex boundary value problems, and stability analysis Variational analysis and the Rayleigh-Ritz method ale0 allows us to find approximate eijpnvalues and eigenfunctions of a Sturm-Liouville system

517

518 VARIATIONAL ANALYSIS

I

I

fig 17.1 Variation of paths

17.1 PRESENCE OF ONE DEPENDENT AND ONE INDEPENDENT VARIABLE

17.1.1 Euler Equation

A majority of the variational problems encountered in physics and engineering are expressed in terms of an integral:

(17.1) where y(x) is the desired function and f is a known function depending on

y, its derivative with respect to x, that is yx, and x Because the unknown of this problem is a function, J is called a functional and we write it as

J [Y (XI1 (17.2)

Usually the purpose of these problems is to find a function, which is a path in the xy-plane between the points ( 2 1 , y1) and (zz, yz), which makes the func- tional J [y (z)] an extremum In Figure 17.1 we have shown two potentially possible paths; actually, there are infinitely many such paths The difference

of these paths from the desired path is called the variation of y, and we show

it as Sy Because Sy depends on position, we use ~ ( z ) for its position depen- dence and use a scalar parameter a as a measure of its magnitude Paths close to the desired path can now be parametrized in terms of LY as

Y(X, a ) = y(x, 0 ) + ar](x) + O(O2), (17.3)

PRESENCE OF ONE DEPENDENT AND ONE INDEPENDENT VARIABLE 519

where y(z, a = 0) is the desired path, which extremizes the functional J[Y(X)]

We can now express Sy as

SY = Y(Z, a ) - Y(Z,O) = arl(z) (17.4) and write J as a function of a as

Now the derivative of J with respect to cy is

Using equation (17.3) we can write

and

a Y x ( z , a ) - d ? 7 ( 4

aa &

Substituting these in Equation (17.8) we obtain

Integrating the second term by parts gives

520 VARlAT/ONAL ANALYSIS

Because the variation ~ ( x ) is arbitrary, the only way to satisfy this equation

is by setting the expression inside the brackets to zero, that is,

To find another version of the Euler equation we write the total derivative of

the function f (y, y,, x) as

Another Form of the Euler Equation

This can also be written as

This is another version of the Euler equation, which is extremely useful when

f (y, yx, x) does not depend on the independent variable x explicitly In such cases we can immediately write the first integral as

8.f aYx

f - yx- = constant, (17.19) which reduces the problem to the solution of a first-order differential equation

17.1.3

Example 17.1 Shortest path between two points: To find the shortest

Applications of the Euler Equation

path between two points in two dimensions we

ds = [(dx)' + ( d ~ ) ~ ] '

write the line element as

(17.20)

PRESENCE OF ONE DEPENDENT AND ONE INDEPENDENT VARIABLE 521

The distance between two points is now given as a functional of the path and in terms of the integral

(17.21)

= l: [I +YE]' dx (17.22)

To find the shortest path we must solve the Euler equation for

Because f(y, yz, x) does not depend on the independent variable ex- plicitly, we use the second form of the Euler equation [Eq (17.19)] to write

(17.23)

where c is a constant This is a first-order differential equation for y(x), and its solution can easily be found as

y = ax + b ( 17.24) This is the equation of a straight line, where the integration constants

a and b are to be determined from the coordinates of the end points The shortest paths between two points in a given geometry are called geodesics Geodesics in spacetime play a crucial role in Einstein's theory

of gravitation as the paths of free particles

Example 17.2 Shape of a soap film between two'rings: Let us find the shape of a soap film between two rings separated by a distance of 2x0 Rings pass through the points (x1,yl) and (x2,yz) as shown in Figure 17.2 Ignoring gravitation, the shape of the film is a surface of revolu- tion; thus it is sufficient to find the equation of a curve, y(x), between two points ( 2 1 , y1) and (x2, yz) Because the energy of a soap film is proportional to its surface area, y(x) should be the one that makes the area a minimum

We write the infinitesimal area element of the soap film as

522 VARIATIONAL ANALYSIS

fig 17.2 Soap film between two circles

Since f(y, yz, z) is given as

This leads us t o the first-order differential equation

PRESENCE OF MORE THAN ONE DEPENDENT VARIABLE 523

Integration constants cland c2 are to be determined so that y(x) passes through the points (x1, y1) and ( x ~ , y ~ ) Symmetry of the problem gives c2 = 0 For two rings with unit radius and xo = 1/2 we obtain

1 = c1 cash (&) (17.34)

as the equation to be solved for c1 This equation has two solutions One of these is c1 = 0.2350, which is known as the “deep curve,’’ and the other one is c1 = 0.8483, which is known as the YIat curve.” To find the correct shape we have to check which one of these makes the area,

and hence the energy, a minimum Using Equations (17.27) and (17.29)

we write the surface area as

Substituting the solution (17.33) into Equation (17.35) we get

to break In fact, the transcendental equation

17.2 PRESENCE OF MORE T H A N ONE DEPENDENT VARIABLE

In the variational integral if the function f depends on more than one depen- dent variable

and one independent variable x, then the functional J is written as

r“2

524 VARlATlONAL ANALYSIS

where yix = i3y,/6'x, i = 1,2, , n We can now write small deviations from

the desired paths, yi (x,O), which makes the functional J an extremum as

yi (2, a ) = y, ( x , ~ ) + avi(x) + o(a2), i = 1,2, , n, (17.41)

where a is again a small parameter and the functions qi(x) are independent

of each other We again take the variation at the end points as zero:

Because the variations ~ { ( z ) are independent, this equation can only be sat-

isfied if all the coefficients of Q ~ ( x ) vanish simultaneously, that is,

17.3 PRESENCE OF MORE THAN ONE INDEPENDENT VARIABLE

Sometimes the unknown function a and f in the functional J depend on more than one independent variable For example, in threedimensional problems

J may be given as

(17.46)

where us denotes the partial derivative with respect to E We now have to find

a function u(z, y, z ) such that J is an extremum We again let u(z, y, z, dy = 0)

be the function that extremizes J and write the variation about this function

as

PRESENCE OF MORE THAN ONE INDEPENDENT VARIABLE 525

where Q(Z, y, z) is a differentiable function We take the derivative of Equation (17.46) with respect to (Y and set a = 0, that is,

(17.48)

af

a.U,

We then integrate terms like -Q* by parts and use the fact that variation

a t the end points are zero to write

~ ( x , y, z)dxdydz = 0 (17.49)

a af a af

au axau, ayau, azau,

Because the variation ~ ( x , y, z ) is arbitrary, the expression inside the paren- theses must be zero; thus yielding

= 0

af a af a af a af

This is the Euler equation for one dependent and three independent variables

Example 17.3 Laplace equation: In electrostatics energy density is given

Since

f is given as

J = / J S, (a@)' dxdydz (17.54)

(17.55)

526 VARIATIONAL ANALYSIS

Writing the Euler equation [Eq (17.50)] for this f , we obtain

-2 (@XZ + @yy + @zz) = 0, (17.57) which is the Laplace equation

Now, the p,y, and the r functions that extremize J = j j s fdxdydz will be

obtained from the solutions of the following system of three Euler equations:

PRESENCE OF HIGHER-ORDER DERIVATIVES 527

Fig 17.3 Deformation of an elastic beam

17.5 PRESENCE OF HIGHER-ORDER DERIVATIVES

Sometimes in engineering problems we encounter functionals given as

(17.66)

where y(") stands for the nth order derivative, the independent variable x

takes values in the closed interval [a,b], and the dependent variable y ( z ) satisfies the boundary conditions

(n- 1)

(17.67)

Using the same method that we have used for the other cases, we can show that the Euler equation that y(z) satisfies is

d.1 = Yo, !/'(a) = Y&, , y ( n - l ) ( a ) = yyo ,

d b ) = Y1, Y'(b) = Yi, , y(n-l)(b) = &-')

This equation is also known as the Euler-Poisson equation

Example 17.4 Deformation of an elastic beam: Let us consider a ho- mogeneous elastic beam supported from its end points at ( 4 1 , O ) and ( 0 , l I ) as shown in Figure 17.3 Let us find the shape of the centerline

of this beam

From elasticity theory the potential energy E of the b a n i is given as

528 VARIATIONAL ANALYSIS

where p and p are parameters that characterize the physical properties

of the beam Assuming that the deformation is small, we can take

1 +y’2 M 1 (17.70) Now the energy becomes

11 -11

E = / [ ; ~ ( y ” ) ~ 3- py] dx (17.71) For stable equilibrium the energy of the beam must be a minimum Thus we have to minimize the energy integral with the conditions

y(Z1) = ~ ( 4 1 ) = 0 and y’(Z1) = ~ ’ ( 4 1 ) = 0 (17.72) Using

(17.73)

1

F ( z , Y, Y’, Y”) = ZP (Y/”>2 + m,

we write the Euler-Poisson equation as

py(4) + p = 0 (17.74) Solution of the Euler-Poisson equation is easily obtained as

y = ax3 + px2 + yx + s - -x P 4 (17.75) Using the boundary conditions given in Equation (17.72) we can deter- mine a , 0, y, 6 and find y(x) as

24P

y = - P [-z4 + 21;x2 - it]

For the cases where there are m dependent variables we can generalize the

(17.76) 24P

variational problem in Equation (17.66) as

and the Euler-Poisson equations become

ISOPERIMETRIC PROBLEMS AND THE PRESENCE OF CONSTRAINTS 529

17.6 ISOPERIMETRIC PROBLEMS AND THE PRESENCE OF CONSTRAINTS

In some applications we search for a function that not only extremizes a given functional

but also keeps another functional

we parametrize the possible paths in terms of two parameters €1 and ~2 as

y(z, €1 ~ 2 ) These paths also have the following properties:

i) For all values of €1 and ~2 they satisfy the boundary conditions

Y ( Z A , E l , E 2 ) = Y A and Y ( ~ B , E I , E 2 ) = Y B - (17.83) ii) y(z, 0,O) = ~ ( z ) is the desired path

iii) y(z, E I , €2) has continuous derivatives with respect to all variables to

We now substitute these paths into Equations (17.80) and (17.81) to get second order

two integrals depending on two parameters E I and € 2 as

(17.85)

While we are extremizing I ( E I , E ~ ) with respect to € 1 and €2, we are also going to ensure that J ( E I , E ~ ) takes a fixed value; thus € 1 and ~2 cannot be independent Using Lagrange undetermined multiplier X we form

K ( E 1 , E z ) = I ( € ] , E 2 ) + 1.2) (17.86) The condition for K ( E I , €2) to be an extremum is now written as

(17.87)

Solutions of this differential equation contain two integration constants and

a Lagrange undetermined multiplier A The two integration constants come

from the boundary conditions [Eq (17.82)], and X comes from the constraint that fixes the value of J , thus completing the solution of the problem Another way to reach this conclusion is to consider the variation of the two functionals (17.80) and (17.81) as

and

We now require that for all Sy that makes SJ = 0, S I should also vanish This

is possible if and only if

ISOPERIMETRIC PROBLEMS AND THE PRESENCE OF CONSTRAINTS 531

are constants independent of X , that is,

(g) / (2) = -X(const.)

This is naturally equivalent t o extremizing the functional

with respect to arbitrary variations 6y

where h is given by Equation (17.94)

Example 17.5 Isoperimetric problems: Let us find the maximum area

that can be enclosed by a closed curve of fixed perimeter L on a plane

We can define a curve on a plane in terms of a parameter t by giving its

z(t) and y ( t ) functions Now the enclosed area is

(17.97)

1 t B

A = LA (zy’ - x’y) dt, and the fixed perimeter condition is expressed as

(17.98)

where the prime denotes differentiation with respect to the independent variable t , and x and y are the two dependent variables Our only constraint is given by Equation (17.98); thus we have a single Lagrange undetermined multiplier and the h function is written as

(17.99)

1

h = 5 (zy’ - x’y) +Ad-

532 VARIATIONAL ANALYSIS

Writing the Euler equation for x(t) we get

2 dt 2 dx- -y’ - -( y + x = 0,

The first integral of this system of equations [Eqs (17.100) and (17.101))

can easily be obtained as

Because the circumference is L, we determine X as

L

A = -

Example 17.6 Shape of a freely hanging wire with fixed length: We

now find the shape of a wire with length L and fixed at both ends at

( x ~ , y ~ ) and ( x ~ , y ~ ) The potential energy of the wire is

y d s = p g 11 9J-d~ (17.108)