Download Calculus of Variations and Variational Problems

Week 10
Calculus of Variations and Variational
Problems
1
Extrema of Functionals
Sometimes one is interested in the behavior of functions whose arguments are also functions,
so-called functionals. Consider for instance the problem of determining the function y = u(x)
on the x − y plane which maximizes the integral
Z
I=
b
a
F (x, u, u0 )dx
subject to
u(a) = α
u(b) = β
where F (x, u, u0 ) is some given function. One can consider this as a competition among
a large number of admissible functions only one of which is the required winer. Let the
admissible set of functions v ∈ C 2 , be represented as
v(x) = u(x) + η(x)
where u(x) is the function we are after, η(a) = η(b) = 0 (so that v(a) = α and v(b) = β)
and is a parameter independent of x but which varies from function to function in the
admissible set. The quantity η(x) is called the variation of u(x). Using this, the integral
becomes
Z
I() =
b
a
F (x, u + η, u0 + η 0 )dx
Clearly, I is maximum when the variation of u is zero, i.e.
dI()
=0
d
1
when = 0. Differentiating with respect to under the integral sign and setting = 0 in
the result leads to
Z b
∂F
∂F
dI
|=0 = [
η + 0 η 0 ]dx = 0
d
∂u
a ∂u
The second term on the righ hand side above can now be integrated by parts to yield
I 0 (0) =
Z
b
a
Z b
∂F
∂F
d ∂F
ηdx + [ 0 η]ba −
(
)ηdx = 0
∂u
∂u
a dx ∂u0
Now, recalling that η(a) = η(b) = 0, the above becomes
Z
0
I (0) =
b
a
[
d ∂F
∂F
− ( 0 )]ηdx = 0
∂u
dx ∂u
Since this is true for any reasonable η(x), necessarily the coefficient in parenthesis must
vanish, i.e.
d ∂F
∂F
− ( 0) = 0
∂u
dx ∂u
This is the Euler differential equation associated with the minimization of the functional
under constraints. The above equation can also be written as
∂F
d2 u
du
d ∂F
( 0) −
= Fu0 u0 2 + Fuu0
+ (Fxu0 − Fu ) = 0
dx ∂u
∂u
dx
dx
The specific function u(x) which solves Euler’s equation is also the one which minimizes the
functional under the given constraints.
Another frequently useful related result is obtained whenever ∂F
= 0. In this case one
∂x
has Beltrami’s identity which states that Euler’s equation reduces to
F − u0
∂F
=C
∂u0
where C is a constant.
Example.
Consider the problem of determining the function y = u(x)√such that the distance from
(0, 0) to (1, 1) is least. The differential arc length is ds(x) = 1 + u02 dx and the distance
which is to be minimized is
Z
I=
0
1
√
1 + u02 =
Z
1
0
F (u0 )dx
The Euler equation is this case becomes simply u00 = 0 and the required function minimizing
I is u(x) = c1 x + c2 = x
2
2
Fourth Order Boundary Value Problems
Consider now the functional
Z
I=
b
a
F (x, u, u0 , u00 )dx
and consider admissible functions v ∈ C 4 such that v(a) = α1 , v 0 (a) = α2 , v(b) = β1 , v 0 (b) =
β2 , i.e.
v(x) = u(x) + η(x)
and proceed to minimize I with respect to as before. Finally, introduction of the stated
boundary conditions yields
I 0 (0) =
Z
b
a
[
d ∂F
∂F
d2 ∂F
− ( 0 ) + 2 ( 00 )]ηdx = 0
∂u
dx ∂u
dx ∂u
and the Euler equation associated with the original problem is
d ∂F
∂F
d2 ∂F
− ( 0 ) + 2 ( 00 )− = 0
∂u
dx ∂u
dx ∂u
where the function u satisfyies the boundary conditions
u(a) = α1
u0 (a) = α2
u(b) = β1
u0 (b) = β2
3
Second Order Boundary Value Problems with Two
Independent Variables
For a two-dimensional system, let the functional I be defined as
Z Z
I=
F (x, y, u, ux , uy )dxdy
The specific function u(x, y) which yields an extremum of I is the same which solves the
associated Euler equation
∂ ∂F
∂ ∂F
∂F
−
(
(
)−
)=0
∂u
∂x ∂ux
∂y ∂uy
Example.
3
Let u(x, y) be the function which satisfyies
∇2 u =
∂ 2u ∂2u
+
+f =0
∂x2 ∂y 2
inside a two-dimensional domain Ω and is subject to the condition u = 0 on the boundary
of the domain Γ.
The above is Euler’s equation is associated with the extremum of the functional
Z Z
I=
4
Ω
∂u
1 ∂u 2
[( ) + ( )2 − 2f u]dxdy
2 ∂x
∂y
Fourth Order Problems with Two Independent Variables
For fourth order boundary value problems consider the functional
Z Z
I=
F (x, y, u, ux , uy , uxx , uxy , uyy )dxdy
a similar procedure yields the following Euler equation
∂ ∂F
∂F
∂ ∂F
−
(
(
)−
)+
∂u
∂x ∂ux
∂y ∂uy
∂F
∂ 2 ∂F
∂2
∂ 2 ∂F
(
+ 2(
)+
) + 2(
)=0
∂x ∂uxx
∂x∂y ∂uxy
∂y ∂uyy
5
Problems with Multiple Dependent Variables
Additional functions can be readily incorporated into the formulation. Let u(x, y) and v(x, y)
be two functions of the independent variables x and y. The specific functions u and v which
lead to an extremum value of the functional
Z Z
I=
F (x, y, u, v, ux , vx , uy , vy )dxdy
are the one which satisfy Euler’s equations
∂ ∂F
∂F
∂ ∂F
−
(
(
)−
)=0
∂u
∂x ∂ux
∂y ∂uy
∂ ∂F
∂F
∂ ∂F
−
(
(
)−
)=0
∂v
∂x ∂vx
∂y ∂vy
4
6
Boundary Conditions
Let u(x) be a function of the single independent variable x. The necessary condition for I
to attain a minimum is
dI
|=0 =
d
Z
b
a
[
d ∂F
∂F
∂F
− ( 0 )]ηdx + ( 0 )η)|ba
∂u
dx ∂u
∂u
∂F
b
Consider the term ( ∂u
0 )η|a . This term can be equal to zero iff
η(a) = η(b) = 0; or
∂F
η(a) = 0; 0 (b) = 0; or
∂u
∂F
(a) = 0; η(b) = 0; or
∂u0
∂F
∂F
(a) = 0; 0 (b) = 0
0
∂u
∂u
Two kinds of boundary conditions are then obtained:
• Essential Boundary Conditions: Value of u specified at boundary.
• Natural Boundary Conditions: ∂F/∂u0 = 0 on the boundary.
Problems with essential boundary conditions are Dirichlet boundary value problems while
those with natural boundary conditions are Neumann boundary value problems.
7
Numerical Solution of Variational Problems
Since the problem of minimization of a certain functional is equivalent to that of solving a
boundary value problem. One can search for solutions of specific boundary value problems
by identifying functions that minimize an appropriate functional. For instance, it is known
that the function that solves the boundary value problem
−
d2 u
=f
dx2
with u(0) = u(1) = 0 is the same that minimizes the functional
Z
I(u) =
0
1
1 du
[ ( )2 − f u]dx
2 dx
5
Approximate solutions can be obtained for the above problems using the Rayleigh-RitzGalerkin method. In the Ritz method, one introduces a set of linearly independent coordinate
functions {φj }N
j=1 and expresses the approximation to u(x), uN as a linear combination i.e.
uN =
N
X
j=1
aj φj
where the aj ’s are parameters to be determined by computation. This expression is then
substituted into the expression for I(u) and the extremun conditions applied, i.e.
∂I(uN )
=0
∂aj
for j = 1, 2, ..., N .
The result is a system of N algebraic equations in N unknowns i.e.
Ka = F
Solution of the system yields the values of the coefficients aj , j = 1, 2, ..., N .
In the Galerkin method, one starts with the given differential equation and introduces
the approximation in terms of coordinate functions as in the Ritz method. Next, the result is
multiplied by coordinate function φ1 and the product integrated from 0 to 1; this produces
one equation. The complete set of equations is obtained by multiplying the differential
equation in turn by the coordinate functions φ2 , φ3 , ..., φN and again integrating from 0 to
1. The final result is a system of N algebraic equations in N unknowns which is identical to
the one obtained by the Ritz method, i.e.
Ka = F
As in the previous case, solution of the system yields the values of the coefficients aj , j =
1, 2, ..., N .
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