ECE 451 LECTURE 10

1

Preface

In the last lecture, we introduced examples of solving the Euler-Lagrange equations for a scalar function argument with various boundary conditions. In this lecture, we treat the case of Euler-Lagrange equations involving a vector argument. The solution of these equations is one of our primary objectives going forward. We present examples of solving the multiple-function Euler-Lagrange equation.

Euler-Lagrange Equations (cont’d)

We now wish to generalize our discussion to include functionals that may contain several independent functions and their derivatives, i.e.

ft

t

tdtttgJ0

),)(,)(()( xxx (10.1)

Where, we have the boundary conditions 0)0( xx and fft xx )( . Proceeding as we did in the scalar

case, we obtain the Euler-Lagrange equations, i.e.

0xxxx

),)(,)((),)(,)(( ttt

x

g

td

dttt

x

g

(10.2)

Let’s take a simple example as follows:

Find the extremal x of tdtxtxtxtxJ 2/

0

2122

21 )()()(4)()(

x which satisfies

the boundary conditions TT 01)4/(,10)0( xx .

First, we form the Euler-Lagrange equations, i.e. we use equation (10.2) and we obtain

0)()(8

0)()(2

12

21

txtx

txtx

(10.3)

Note that equations (10.3) are linear ODE’s with constant coefficients, and so they are readily solved using classical techniques. Doing so, we can write one familiar form of the solution as follows, i.e.

tctcecectx

tctcecectx

tt

tt

sin2

1cos

2

1

2

1

2

1)(

sincos)(

432

22

12

432

22

11

(10.4)

ECE 451 LECTURE 10

2

We now apply the boundary conditions to equation (10.4) in order to solve for the four arbitrary constants, i.e.

0)2/(

1)2/(

1)0(

0)0(

2

1

2

1

x

x

x

x

Thus, the arbitrary constants are given by:

2

1

1

2

1

2

1

4

3

2/2/

2/

1

2/2/

2/

1

c

cee

ec

ee

ec

(10.5)

As before, let’s treat the most general case, i.e. both ft and )( ftx are free. Again, the Euler-Lagrange

equation, i.e. equation (10.2) must be satisfied, and the boundary conditions at the final time are specified by the following expression:

0)(),)(,)((),)(,)((

),)(,)((

T

T

ffffffff

ffff

tttttg

tttg

tttg

xxxx

xx

xxxx

(10.6)

Now, let’s consider some examples of free end conditions. For example,

Find an extremal of the functional tdtxtxtxtxJft

1

2221

21 )()()()()( x with

boundary conditions T2/31)( 0x Tfree2)4/( x .

As always, we start with the Euler-Lagrange equation, and we obtain

ECE 451 LECTURE 10

3

0)()(2

0)()(2

12

21

txtx

txtx

(10.7)

Eliminating )(2 tx from equation (10.7) above, we get

0)(4)( 11 txtx

This equation has solution

tctctx 2sin2cos)( 211 (10.8)

Substituting equation (10.8) into equation (10.7), we obtain

tctctx 2sin22cos2)( 212

Integrating the above equation twice, we get

4321

2 2sin2

2cos2

)( ctctc

tc

tx (10.9)

Now, we must find the values of the unknown constants in equations (10.8) and (10.9). Returning to equation (10.6), we have that

0)())4/(,)4/((2

ftxx

g xx

But, we know that )( ftx is arbitrary, and thus

0))4/(,)4/((2

xx x

g (10.10)

Using equation (10.10), we have that

02)4/(2)4/())4/(,)4/(( 3212

cxxx

g

xx

And so, we have that 03 c . We also have that

ECE 451 LECTURE 10

4

2)2()0()4/(

2)0()0(2

)1(22

3)0(

)0()1(1)0(

2213

44321

2

1211

cccx

ccccc

x

cccx

Thus, the extremal curve is given by

22sin2cos2

12sin22cos

)(

)()(

2

1

tt

tt

tx

txtx (10.11)

EXERCISE

Consider the following function:

dtxxxJ 2

0

2 42/1)(

1) Find the extremal )(* tx given the boundary values 10)2(*,1)0(* xx .

2) Find the extremal )(* tx and 0ft , if ft is free, and if 1)0(* x , and 9)(* ftx .