# Variational Formulation of Boundary Value Problems I

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Variational Formulation of Boundary Value Problems I
```Week 4
Variational Formulation of Boundary Value
Problems I
1
Introduction
The importance of the variational formulations is based on the following Characterization
Theorem:
Let V be a linear space, a(v, v) : V × V → R a symmetric positive bilinear form and
f : V → R a linear functional. Then, the quantity
1
I(v) = a(v, v) − (f, v)
2
attains a minimum over V at u if and only if
a(u, v) = (f, v)
for all v ∈ V ; there is at most one solution of the above.
The problem of finding the function u which makes I(u) ≤ I(v) for any v ∈ V is known
as the functional minimization problem while that of determining the function u such that
a(u, v) = (f, v) for any v ∈ V is known as the variational problem.
Moreover, using the calculus of variations one can show that the function u which solves
the above two problems is the solution of a particular differential equation called Euler’s
equation with the form
Lu = f
where L is a linear differential operator.
As a specific example consider the problem of finding the function u(x) satisfying the
differential equation
−
d2 u
= f (x)
dx2
1
subject to the boundary conditions
u(0) = 0
u(1) = 0
This is Euler’s equation associated with the problem of determining the function u(x)
which minimizes the functional
1
I(v) = (u0 , v 0 ) − (f, v)
2
R
R
where (u0 , v 0 ) = 01 u0 v 0 dx and (f, v) = 01 f vdx.
Furthermore, the same function u(x) which minimizes the functional (and solves the
original differential equation subject to the stated boundary conditions) is also the solution
of the variational problem
(u0 , v 0 ) = (f, v)
To better understand the connection between the differential and the variational statements of problems it is useful to review the foundations of the calculus of variations.
2
Extrema of Functions
Let u(x) be a function of a single independent variables x, u(x). Let ∆u(x) = u(x+h)−u(x)
where h is a small number. The necessary and sufficient condition for u to have a relative
minimum at x0 is
∆u(x0 ) ≥ 0
I.e. any small change whatsoever of x increases the value of u.
Alternatively, the function has a relative maximum if
∆u(x0 ) ≤ 0
I.e. any small change whatsoever of x decreases the value of u.
Consider now a function of two independent variables u(x, y). Let ∆u(x, y) = u(x+h, y +
k) − u(x, y) where h and k are small numbers. . The condition for u to posses a relative
minimum at (x0 , y0 ) is
∆u(x0 , y0 ) ≥ 0
Alternatively, the function has a relative maximum if
∆u(x0 , y0 ) ≤ 0
2
Since when h and k are sufficiently small the sign of ∆u(x, y) is the same as the sign of
hux (x, y) + kuy (x, y) regardless of the signs of h and k, necessary conditions for a relative
extremum at (x0 , y0 ) are
ux = uy = 0
Consider now that this is indeed the case for a given u(x, y) and let the last term in the
Taylor series with N = 3, h2 uxx (x, y) + 2hkuxy (x, y) + k 2 uyy (x, y) = h2 A + 2hkB + k 2 C.
Now the sign of ∆u(x, y) must be the same as that of the quadratic h2 A + 2hkB + k 2 C.
Whenever the discriminant B 2 − AC > 0, there will be two distinct values of the ratio k/h
for which h2 A + 2hkB + k 2 C = 0 and the sign of the quadratic will change when crossing
the distinct values if k/h.
Therefore, a necessary condition for u to have a relative extremum at (x0 , y0 ) is
B 2 − AC ≤ 0
or alternatively
δ = uxx uyy − u2xy ≥ 0
Whether the extremum is maximum or minimum depends on the signs of uxx and uyy which
must necessarily be equal to each other for δ to be ≥ 0. Specifically, if ux = uy = 0 at a
point (x0 , y0 ) then u has
a) a relative maximum if uxx < 0 and uxx uyy > u2xy there,
b) a relative minimum if uxx > 0 and uxx uyy > u2xy there,
c) a saddle point if uxx uyy < u2xy there.
If uxx uyy = u2xy more research is needed.
Example. Consider the function u(x, y) = 1 − x2 − y 2 at (x, y) = (0, 0). Here
ux = −2x = 0
and
uy = −2y = 0
Also
uxx = −2 = uyy < 0
and since uxy = 0,
uxx uyy = 4 > u2xy
the point (0, 0) is a relative maximum.
3
3
Extrema of Functionals: Calculus of Variations
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
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
0
I (0) =
Z
b
a
[
d ∂F
∂F
− ( 0 )]ηdx = 0
∂u
dx ∂u
4
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
0
( 0) −
= Fu u 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.
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 1√
Z 1
02
1+u =
F (u0 )dx
I=
0
0
The Euler equation is this case becomes simply u00 = 0 and the required function minimizing
I is u(x) = c1 x + c2 = x
4
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
0
Z
I (0) =
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
d2 ∂F
∂F
− ( 0 ) + 2 ( 00 )− = 0
∂u
dx ∂u
dx ∂u
5
where the function u satisfyies the boundary conditions
u(a) = α1
u0 (a) = α2
u(b) = β1
u0 (b) = β2
5
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.
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=
6
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
6
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
7
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
8
Boundary Conditions
Let u(x) be a function of the single independent variable x. The necessary condition for I
to attain a minimum is
Z b
d ∂F
dI
∂F
∂F
|=0 = [
− ( 0 )]ηdx + ( 0 )η)|ba
d
dx ∂u
∂u
a ∂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:
7
• 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.
8
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