Euler Lagrange Equation - Calculus of Variations, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Euler-Lagrange Equations
5.6 Fundamental lemma in the Calculus of Variations
Let f(x) be continuous in [a,b] and let ?(x) be an arbitrary function on [a,b] such that
?,?
'
,?
''
are continuous and ?(a) = ?(b) = 0. If
Z
b
a
f(x)?(x)dx = 0
for all such ?(x), then f(x)= 0 on [a,b].
Proof: Suppose to the contrary that f(x) > 0 at x = ?
0
, say.
Then there is a neighbourhood N = (?
1
,?
2
), with ?
1
< ?
0
<?
2
, in which f(x)> 0. Let
?(x) =

(x-?
1
)
4
(x-?
2
)
4
for x?N
0 elsewhere

.
Then
Z
b
a
f(x)?(x)dx> 0
contradicting the hypothesis.
Theorem 5.1: The Euler-Lagrange Equations
For the functional (5.3)
I[y] =
Z
b
a
F(x,y,y
'
)dx
(where a, b, y(a),y(b) are given, F is twice continuously di?erentiable w.r.t. its arguments
and y
''
(x) is continuous), the extremal function y =y
0
(x) satis?es the equation
d
dx

?F
?y
'

-
?F
?y
= 0. (5.4)
Page 2


Euler-Lagrange Equations
5.6 Fundamental lemma in the Calculus of Variations
Let f(x) be continuous in [a,b] and let ?(x) be an arbitrary function on [a,b] such that
?,?
'
,?
''
are continuous and ?(a) = ?(b) = 0. If
Z
b
a
f(x)?(x)dx = 0
for all such ?(x), then f(x)= 0 on [a,b].
Proof: Suppose to the contrary that f(x) > 0 at x = ?
0
, say.
Then there is a neighbourhood N = (?
1
,?
2
), with ?
1
< ?
0
<?
2
, in which f(x)> 0. Let
?(x) =

(x-?
1
)
4
(x-?
2
)
4
for x?N
0 elsewhere

.
Then
Z
b
a
f(x)?(x)dx> 0
contradicting the hypothesis.
Theorem 5.1: The Euler-Lagrange Equations
For the functional (5.3)
I[y] =
Z
b
a
F(x,y,y
'
)dx
(where a, b, y(a),y(b) are given, F is twice continuously di?erentiable w.r.t. its arguments
and y
''
(x) is continuous), the extremal function y =y
0
(x) satis?es the equation
d
dx

?F
?y
'

-
?F
?y
= 0. (5.4)
This iscalled the Euler-LagrangeEquation andisanecessary, butnotsu?cient, condition
for an extremal function.
(Every extremal function y
0
(x) satis?es the Euler-Lagrange equation; not every function
f(x) which satis?es the Euler-Lagrange equation is an extremal function.)
Proof: (NOT EXAMINABLE) Let y = y(x) be a variable function and let y =y
0
(x)
beanextremal function forthefunctionalI[y], i.e.I[y
0
]takesanextreme value(maximum
or minimum).
Let ?(x) be as in lemma 5.6 and de?ne the function
y(x) =y
0
(x)+e?(x)
where e > 0 is a parameter, and write
y = y
0
+dy.
Then dy =e?(x) is called the variation of y =y
0
(x).
For e su?ciently small, y lies in an arbitrarily small neighbourhood N
h
(y
0
) of y
0
(x).
Now, the integral I[y] =I[y
0
+e?] is a function, F(e) say, of e.
Let
dI = I[y
0
+e?]-I[y
0
] = F(e)-F(0)
then
dI =
Z
b
a
{ F(x,y
0
+e?,y
'
0
+e?
'
)-F(x,y
0
,y
'
0
)} dx.
Expanding F(e) in a Maclaurin series (that is, a Taylor series about the origin) with
respect to e to ?rst order gives
F(e) = F(0)+F
'
(0)e
and so to ?rst order in e
dI = F
'
(0)e =e
Z
b
a

?F
?y
dy
de
+
?F
?y
'
dy
'
de

y=y
0
dx
and so
F
'
(0) =
Z
b
a

?F
?y
?+
?F
?y
'
?
'

y=y
0
dx.
Now, since y = y
0
(x)+e?(x) where y
0
(x) is the extremal function, it follows that dI = 0
for all ?(x) with y? N
h
(y
0
).
For suppose not, then replace ?(x) by-?(x) and the sign of dI changes, contradicting the
fact that y
0
(x) is the extremal function!
Hence I[y] takes a stationary value for y = y
0
(x).
Thus
Z
b
a

?
?F
?y
+?
'
?F
?y
'

dx = 0.
Integrating the second integral by parts with
u =
?F
?y
'
, v
'
=?
'
,
Page 3


Euler-Lagrange Equations
5.6 Fundamental lemma in the Calculus of Variations
Let f(x) be continuous in [a,b] and let ?(x) be an arbitrary function on [a,b] such that
?,?
'
,?
''
are continuous and ?(a) = ?(b) = 0. If
Z
b
a
f(x)?(x)dx = 0
for all such ?(x), then f(x)= 0 on [a,b].
Proof: Suppose to the contrary that f(x) > 0 at x = ?
0
, say.
Then there is a neighbourhood N = (?
1
,?
2
), with ?
1
< ?
0
<?
2
, in which f(x)> 0. Let
?(x) =

(x-?
1
)
4
(x-?
2
)
4
for x?N
0 elsewhere

.
Then
Z
b
a
f(x)?(x)dx> 0
contradicting the hypothesis.
Theorem 5.1: The Euler-Lagrange Equations
For the functional (5.3)
I[y] =
Z
b
a
F(x,y,y
'
)dx
(where a, b, y(a),y(b) are given, F is twice continuously di?erentiable w.r.t. its arguments
and y
''
(x) is continuous), the extremal function y =y
0
(x) satis?es the equation
d
dx

?F
?y
'

-
?F
?y
= 0. (5.4)
This iscalled the Euler-LagrangeEquation andisanecessary, butnotsu?cient, condition
for an extremal function.
(Every extremal function y
0
(x) satis?es the Euler-Lagrange equation; not every function
f(x) which satis?es the Euler-Lagrange equation is an extremal function.)
Proof: (NOT EXAMINABLE) Let y = y(x) be a variable function and let y =y
0
(x)
beanextremal function forthefunctionalI[y], i.e.I[y
0
]takesanextreme value(maximum
or minimum).
Let ?(x) be as in lemma 5.6 and de?ne the function
y(x) =y
0
(x)+e?(x)
where e > 0 is a parameter, and write
y = y
0
+dy.
Then dy =e?(x) is called the variation of y =y
0
(x).
For e su?ciently small, y lies in an arbitrarily small neighbourhood N
h
(y
0
) of y
0
(x).
Now, the integral I[y] =I[y
0
+e?] is a function, F(e) say, of e.
Let
dI = I[y
0
+e?]-I[y
0
] = F(e)-F(0)
then
dI =
Z
b
a
{ F(x,y
0
+e?,y
'
0
+e?
'
)-F(x,y
0
,y
'
0
)} dx.
Expanding F(e) in a Maclaurin series (that is, a Taylor series about the origin) with
respect to e to ?rst order gives
F(e) = F(0)+F
'
(0)e
and so to ?rst order in e
dI = F
'
(0)e =e
Z
b
a

?F
?y
dy
de
+
?F
?y
'
dy
'
de

y=y
0
dx
and so
F
'
(0) =
Z
b
a

?F
?y
?+
?F
?y
'
?
'

y=y
0
dx.
Now, since y = y
0
(x)+e?(x) where y
0
(x) is the extremal function, it follows that dI = 0
for all ?(x) with y? N
h
(y
0
).
For suppose not, then replace ?(x) by-?(x) and the sign of dI changes, contradicting the
fact that y
0
(x) is the extremal function!
Hence I[y] takes a stationary value for y = y
0
(x).
Thus
Z
b
a

?
?F
?y
+?
'
?F
?y
'

dx = 0.
Integrating the second integral by parts with
u =
?F
?y
'
, v
'
=?
'
,
so that
u
'
=
d
dx

?F
?y
'

, v = ?,
gives

?
?F
?y
'

b
a
+
Z
b
a
?

?F
?y
-
d
dx

?F
?y
'

dx = 0.
Now, ?(a) = ?(b) = 0, so
Z
b
a
?

?F
?y
-
d
dx

?F
?y
'

dx = 0.
Hence, since ?(x) is arbitrary, and using the Fundamental Lemma of the Calculus of
Variations, we obtain

?F
?y
-
d
dx

?F
?y
'

= 0.
End of proof.
Note: For any such f(x,y,y
'
)
df
dx
=
?f
?x
+
?f
?y
dy
dx
+
?f
?y
'
dy
'
dx
=
?f
?x
+y
'
?f
?y
+y
''
?f
?y
'
.
so putting f = ?F/?y
'
gives
y
''
?
2
F
?y
'2
+y
'
?
2
F
?y
'
?y
+
?
2
F
?x?y
'
-
?F
?y
= 0.
This is a second order ode! But it could well be nonlinear and very di?cult to solve.
In this section we will use the Euler-Lagrange equation to derive extremal functions.
Because this is only a necessary condition for ?nding extremal functions, once we have
found the solutions to the equation we then have to check which one is the extremal
function. However for all problems that we will consider there will be only one unique
solution which therefore must be the extremal function.
Example: Find the extremal function for the functional
I[y] =
Z
p/2
0
[(y
'
)
2
-y
2
]dx
with y(0)= 0, and y(p/2)= 1.
Here
F(x,y,y
'
) = [(y
'
)
2
-y
2
],
and so
?F
?y
=-2y and
?F
?y
'
= 2y
'
.
The Euler-Lagrange equation (5.4) therefore yields
d
dx
(2y
'
)-(-2y) = 0,
Page 4


Euler-Lagrange Equations
5.6 Fundamental lemma in the Calculus of Variations
Let f(x) be continuous in [a,b] and let ?(x) be an arbitrary function on [a,b] such that
?,?
'
,?
''
are continuous and ?(a) = ?(b) = 0. If
Z
b
a
f(x)?(x)dx = 0
for all such ?(x), then f(x)= 0 on [a,b].
Proof: Suppose to the contrary that f(x) > 0 at x = ?
0
, say.
Then there is a neighbourhood N = (?
1
,?
2
), with ?
1
< ?
0
<?
2
, in which f(x)> 0. Let
?(x) =

(x-?
1
)
4
(x-?
2
)
4
for x?N
0 elsewhere

.
Then
Z
b
a
f(x)?(x)dx> 0
contradicting the hypothesis.
Theorem 5.1: The Euler-Lagrange Equations
For the functional (5.3)
I[y] =
Z
b
a
F(x,y,y
'
)dx
(where a, b, y(a),y(b) are given, F is twice continuously di?erentiable w.r.t. its arguments
and y
''
(x) is continuous), the extremal function y =y
0
(x) satis?es the equation
d
dx

?F
?y
'

-
?F
?y
= 0. (5.4)
This iscalled the Euler-LagrangeEquation andisanecessary, butnotsu?cient, condition
for an extremal function.
(Every extremal function y
0
(x) satis?es the Euler-Lagrange equation; not every function
f(x) which satis?es the Euler-Lagrange equation is an extremal function.)
Proof: (NOT EXAMINABLE) Let y = y(x) be a variable function and let y =y
0
(x)
beanextremal function forthefunctionalI[y], i.e.I[y
0
]takesanextreme value(maximum
or minimum).
Let ?(x) be as in lemma 5.6 and de?ne the function
y(x) =y
0
(x)+e?(x)
where e > 0 is a parameter, and write
y = y
0
+dy.
Then dy =e?(x) is called the variation of y =y
0
(x).
For e su?ciently small, y lies in an arbitrarily small neighbourhood N
h
(y
0
) of y
0
(x).
Now, the integral I[y] =I[y
0
+e?] is a function, F(e) say, of e.
Let
dI = I[y
0
+e?]-I[y
0
] = F(e)-F(0)
then
dI =
Z
b
a
{ F(x,y
0
+e?,y
'
0
+e?
'
)-F(x,y
0
,y
'
0
)} dx.
Expanding F(e) in a Maclaurin series (that is, a Taylor series about the origin) with
respect to e to ?rst order gives
F(e) = F(0)+F
'
(0)e
and so to ?rst order in e
dI = F
'
(0)e =e
Z
b
a

?F
?y
dy
de
+
?F
?y
'
dy
'
de

y=y
0
dx
and so
F
'
(0) =
Z
b
a

?F
?y
?+
?F
?y
'
?
'

y=y
0
dx.
Now, since y = y
0
(x)+e?(x) where y
0
(x) is the extremal function, it follows that dI = 0
for all ?(x) with y? N
h
(y
0
).
For suppose not, then replace ?(x) by-?(x) and the sign of dI changes, contradicting the
fact that y
0
(x) is the extremal function!
Hence I[y] takes a stationary value for y = y
0
(x).
Thus
Z
b
a

?
?F
?y
+?
'
?F
?y
'

dx = 0.
Integrating the second integral by parts with
u =
?F
?y
'
, v
'
=?
'
,
so that
u
'
=
d
dx

?F
?y
'

, v = ?,
gives

?
?F
?y
'

b
a
+
Z
b
a
?

?F
?y
-
d
dx

?F
?y
'

dx = 0.
Now, ?(a) = ?(b) = 0, so
Z
b
a
?

?F
?y
-
d
dx

?F
?y
'

dx = 0.
Hence, since ?(x) is arbitrary, and using the Fundamental Lemma of the Calculus of
Variations, we obtain

?F
?y
-
d
dx

?F
?y
'

= 0.
End of proof.
Note: For any such f(x,y,y
'
)
df
dx
=
?f
?x
+
?f
?y
dy
dx
+
?f
?y
'
dy
'
dx
=
?f
?x
+y
'
?f
?y
+y
''
?f
?y
'
.
so putting f = ?F/?y
'
gives
y
''
?
2
F
?y
'2
+y
'
?
2
F
?y
'
?y
+
?
2
F
?x?y
'
-
?F
?y
= 0.
This is a second order ode! But it could well be nonlinear and very di?cult to solve.
In this section we will use the Euler-Lagrange equation to derive extremal functions.
Because this is only a necessary condition for ?nding extremal functions, once we have
found the solutions to the equation we then have to check which one is the extremal
function. However for all problems that we will consider there will be only one unique
solution which therefore must be the extremal function.
Example: Find the extremal function for the functional
I[y] =
Z
p/2
0
[(y
'
)
2
-y
2
]dx
with y(0)= 0, and y(p/2)= 1.
Here
F(x,y,y
'
) = [(y
'
)
2
-y
2
],
and so
?F
?y
=-2y and
?F
?y
'
= 2y
'
.
The Euler-Lagrange equation (5.4) therefore yields
d
dx
(2y
'
)-(-2y) = 0,
i.e.
y
''
+y = 0.
The general solution is
y(x) =Acosx+Bsinx.
The end condition y(0)= 0 now gives A = 0 and y(p/2)= 1 shows that B = 1; hence the
extremal function is
y
0
= sinx.
Thus I[y] is stationarised by the choice y =y
0
and takes the value
I[sinx] =
Z
p/2
0
[cos
2
x-sin
2
x]dx =
Z
p/2
0
cos(2x)dx = 0.
Note that this is a minimum, for example we could try the function y(x) = 2x/p, or
y(x) = (-1)
n
sin(2n + 1)x for any n ? Z (n 6= 0 of course). These satisfy the end
conditions but NOT the E-L equation (ODE) so they cannot yield the extreme function.
Special cases. IMPORTANT! First integrals of the Euler-Lagrange equations
ForspecialformsofF(x,y,y
'
)thesecondorderODE(5.4)maybeintegratedtogivea‘?rst
integral’ which is often simpler (and which can sometimes have a physically meaningful
interpretation).
Theorem 5.2: Let F =F(x,y
'
) (i.e. F has no explicit dependence on y) then (5.4) gives
?F
?y
'
= constant. (5.5)
Proof: We have
?F
?y
= 0
and so
d
dx

?F
?y
'

= 0,
which integrates w.r.t. x to give the result.
Theorem 5.3: Let F =F(y,y
'
) (i.e. F has no explicit dependence on x) then (5.4) gives
F -y
'
?F
?y
'
= constant. (5.6)
Proof: In general if F = F(x,y,y
'
), then
d
dx

F -y
'
?F
?y
'

=

?F
?x
+y
'
?F
?y
+y
''
?F
?y
'

-

y
''
?F
?y
'
+y
'
d
dx

?F
?y
'

=
?F
?x
+y
'

?F
?y
-
d
dx

?F
?y
'

=
?F
?x
by the Euler-Lagrange equation,
= 0, as F is independent of x in this case;
again integration gives the required result.