Linear Elliptic PDE’s
Elliptic partial differential equations frequently arise out of conservation statements of
the form
∫∂B F⃗ ⋅ ⃗n dσ = ∫B S dx
∀B contained in bounded open set U ⊂ R n .
⃗ , S denote respectively, the flux density field and source density field associated with
Here F
a distribution of some quantity throughout a bounded open set U in R n , and B denotes an
arbitrary ball inside U with ⃗
n the unit outward normal to ∂B, the boundary of B. The
statement above then asserts that for any ball B contained in U, the flow of material through
the boundary of B is exactly balanced by the internal production as presribed by the source
term.
Using the divergence theorem, we can convert this expression to
∫B div F⃗ − S dx = 0
∀B ⊂⊂ U,
which, since B is arbitrary, implies
⃗ x − Sx = 0
div F
∀x ∈ U.
If there is now some constituitive law which asserts that
⃗ = −Kx ∇ux + uxV
⃗ x
F
diffusion
and if,
∀x ∈ U,
convection
Sx = −cx ux + fx
∀x ∈ U,

leakage source term
then
⃗ x − −cx ux + fx = 0
div −Kx ∇ux + uxV
i.e.,
⃗ ⋅ ∇ux + cx ux = fx
−div Kx ∇ux + V
∀x ∈ U,
∀x ∈ U
This last equation is a typical elliptic partial differential equation. The terms which appear in
the constituitive laws for flux and source density can frequently be given physical
interpretations as diffusive or convective transport terms (in the flux law) and loss due to
leakage (in the source equation). Then the equation can be written as
Lux = fx
∀x ∈ U,
where L can be expressed variously as
⃗ ⋅ ∇ux + cx ux
Lux = −div Kx ∇ux + V
n
n
i,j=1
i=1
= − ∑ ∂ j K ij x ∂ i ux + ∑ v i x∂ i ux + cxux
⃗  K⃗∂ux + V
⃗ ⃗∂ux + cxux.
= −∂
We place the following assumption on the coefficient K = Kx,
1
i) if K is a scalar, then
Kx ≥ k 0
n
x∈Ū
ii) if K is a matrix (tensor), then ∑ K ij xz i z j ≥ k 0 | ⃗z | 2
∀z⃗ ∈ R n , x ∈ Ū.
i,j=1
This is the assumption of uniform ellipticity for the operator L.
In addition to the elliptic equation, which is satisfied throughout U, it is usual to impose
certain conditions on the solution values on the boundary of the domain. These boundary
conditions are chosen so as to cause the resulting boundary value problem to have a
unique solution. At each point on the boundary of U we will assume that one (and only one)
of the following conditions holds
ux = gx
∇u ⋅ ⃗
nx = gx
x ∈ ∂U
Dirichlet condition
x ∈ ∂U
Neumann condition
∇u ⋅ ⃗
nx + px ux = gx
x ∈ ∂U
Robin condition
Here gx is a given function defined on ∂U.
Consider the so called Dirichlet boundary value problem,
Lux = fx
ux = 0
x∈U
x ∈ ∂U.
Then u = ux is said to be a classical solution of the BVP if u ∈ C 2 U ∩ CŪ. Clearly it
is necessary that the given data function f satisfy f ∈ CU if a classical solution is to have
any possibility of existing. There are several inconvenient aspects to classical solutions:
●
●
●
a classical solution may fail to exist.
(e.g. if ∂U is not sufficiently smooth or if f ∉ CU )
approximating a classical solution may be difficult
even if a classical solution exists, it may be easier to prove
the existence of a weak solution and subsequently prove the
weak solution is actually classical.
In deciding how to define a weak solution, consider the special case that Lu = −∇ 2 u. Then
there are the following possibilities for definitions of a weak solution to the Dirichlet BVP:
i) an ultra regular weak solution: u ∈ H 2 U ∩ H 10 U such that
∫U ∇ 2 ux + fx vx dx = 0 ∀v ∈ H 0 U
ii) a weak solution: u ∈ H 10 U such that
Bu, v = ∫ ∇u ⋅ ∇v − fv dx = 0 ∀v ∈ H 10 U
U
iii) an ultra weak solution: u ∈ H 0 U such that
∫U ux∇ 2 vx + fx dx = 0 ∀v ∈ H 2 U ∩ H 10 U
2
In case i) it is difficult to construct a family of approximate solution spaces V N that tend to
H 2 U ∩ H 10 U as N tends to infinity. Similarly, in case iii) it is difficult to construct a
sequence of ”test function” spaces W N which tend to H 2 U ∩ H 10 U as N tends to infinity.
On the other hand, case ii) is a compromise in which we take V N = W N and these spaces
are finite dimensional subspaces of CU. Moreover, the approximate problem contains a
(symmetric) positive definite matrix approximating the infinite dimensional operator, L.
Weak Formulation
Consider the following partial differential operator acting on functions defined on a bounded
open set U ⊂ R n
n
n
i,j=1
i=1
Lux = − ∑ ∂ j K ij x∂ i ux + ∑ b i x∂ i ux + cxux,
x∈U
where we suppose
i)
K ij ∈ L ∞ U
and
|K ij x| ≤ k 1 for x ∈ U
n
∑ K ij xz i z j ≥ k 0 | ⃗z | 2
also
∀z⃗ ∈ R n , x ∈ Ū
i,j=1
ii) b i ∈ L ∞ U
and
|b i x| ≤ b 1
for x ∈ U
iii) c ∈ L ∞ U
and
c 0 ≤ cx ≤ c 1
for x ∈ U.
Now define, for u, v ∈ H 10 U,
n
n
i,j=1
i=1
Bu, v = ∫  ∑ K ij x ∂ i ux ∂ j vx + vx∑ b i x∂ i ux + cxuxdx
U
Then, for f ∈ H 0 U, we define u to be a weak solution of the Dirichlet boundary value
problem,
Lux = fx
ux = 0
x ∈ U,
x∈Γ
if u ∈ H 10 U satisfies
Bu, v = f, v 0 = ∫ fxvxdx
U
∀v ∈ H 10 U.
Definition of H −1 U
If f, v are both elements of H 0 U, then the integral ∫ fv exists since
U
∫U fxvxdx ≤ ‖f‖ 0 ‖v‖ 0 < ∞.
More generally, suppose v ∈ H 10 U and that f is of the form
n
fx = g 0 x + ∑ ∂ i g i x where g i ∈ H 0 U for i = 0, . . . , n.
j=1
3
Then
n
∫U fxvxdx = ∫U g 0 x + ∑ ∂ i g i x vxdx =
j=1
n
=
n
∑ ∫∂U vg i dS + ∫U
vg 0 − ∑ g i ∂ i v dx
j=1
j=1
n
=∫
U
vg 0 − ∑ g i ∂ i v dx
∀v ∈ H 10 U
i=1
and
n
∫U fxvxdx ≤ ‖g 0 ‖ 0 ‖v‖ 0 + ∑‖g i ‖ 0 ‖∂ i v‖ 0 < ∞
∀v ∈ H 10 U
i=1
If we define
n
−1
H U =
f ∈ D U : fx = g 0 x + ∑ ∂ i g i x where g i ∈ H 0 U, 0 ≤ i ≤ n
′
j=1
n
then
∫U fxvxdx ≤
∑‖g i ‖ 20
1/2
‖v‖ 1
i=0
Evidently, if v ∈ H 10 U then f, v 0 is finite for all f ∈ H −1 U and we can can consider weak
solutions to the Dirichlet problem under the weaker assumption that f ∈ H −1 U.
Inhomogeneous Boundary Data
To consider weak solutions for the Dirichlet boundary value problem with
inhomogeneous boundary data,
Lux = fx
ux = gx
x ∈ U,
x∈Γ
where g ∈ H 1/2 Γ, we recall that for Γ sufficiently regular, there exists a G ∈ H 1 U such
that T 0 G = g. Then w = u − G ∈ H 10 U and w satisfies,
Lwx = Lux − Gx = fx − LGx = F
wx = ux − gx = 0
x ∈ U,
x ∈ Γ.
Since G ∈ H 1 U then LG ∈ H −1 U, so F ∈ H −1 U and we are back to the weak BVP
having homogeneous data. Then the weak solution of the problem with inhomogeneous
data is obtained by finding the weak solution for,
Bw, v = F, v 0
∀v ∈ H 10 U,
and then u = w + G.
Alternatively, let G be any function such that G ∈ H 1 U and T 0 G = g ∈ H 1/2 Γ. Then
u ∈ H 1 U is a weak solution of the Dirichlet boundary value problem with inhomogeneous
boundary data, g, if u − G = w ∈ H 10 U and Bw, v = F, v 0
∀v ∈ H 10 U.
4
Existence of Weak Solutions
There are several different ways to prove existence of a weak solution to the Dirichlet
problem.
●
Symmetric Bilinear form (Poincare inequality approach)
Suppose that the bilinear form B has been shown to satisfy
Bu, v = Bv, u
|Bu, v| ≤ α|| u|| 1 ||v|| 1
Bu, u ≥ β || u|| 21
Then
defines a new inner product on H 10 U that is equivalent to
u, v 1 := Bu, v
the standard H 10 U inner product via the Poincare inequality . In this case we can write
the weak equation in the form
u, v 1 = f, v 0 = Fv
∀v ∈ H 10 U.
But F ∈ H −1 U, hence by the Riesz theorem, there exists a unique z F ∈ H 10 U such that
Fv = z F , v 1 ∀v ∈ H 10 U.
Then u = z F is the unique weak solution of the BVP.
●
Non-Symmetric Bilinear form (Lax-Milgram approach)
Now suppose that the bilinear form B is not symmetric but has been shown to satisfy
|Bu, v| ≤ α|| u|| 1 ||v|| 1
Bu, u ≥ β || u|| 21 .
It follows from the first condition that for any fixed u ∈ H 10 U,
H 10 U ∋ v  Bu, v ∈ R
defines a bounded linear functional on H 10 U. Then the Riesz theorem implies the
existence of a unique element Au ∈ H 10 U such that
Bu, v = Au, v 1
∀v ∈ H 10 U.
Note that since B is not symmetric, we are using the standard inner product on H 10 U. It
now follows from the second condition that A is an isomorphism of H 10 U onto H 10 U.
i.e.,the two conditions together imply
β ‖u‖ 1 ≤ ‖Au‖ 1 ≤ α ‖u‖ 1
Then Bu, v = Fv becomes
Bu, v = Au, v 1 = Fv = z F , v 1
∀v ∈ H 10 U,
and Au = z F has a unique solution for every z F ∈ H 10 U and this solution is the unique
weak solution of the BVP.
A slight variation on the Lax-Milgram approach is offered by the following argument.
Define a mapping T : H 10 U into H 10 U by
5
Tu = u − ρAu − z F 
where ρ is a real number, A denotes the previously defined bounded linear mapping
associated with the bilinear form Bu, v and z F is associated with F via the Riesz theorem.
Then
‖Tu 1 − Tu 2 ‖ 21 = ‖u 1 − u 2  − ρAu 1 − Au 2 ‖ 21
= ‖u 1 − u 2 ‖ 21 − 2ρBu 1 − u 2 , u 1 − u 2  + ρ 2 ‖Au 1 − u 2 ‖ 21
≤ 1 − 2ρβ + ρ 2 α 2 ‖u 1 − u 2 ‖ 21
and for
0<ρ<
2β
α2
it is clear that T is a strict contraction on H 10 U. Then T has a unique fixed point which must
satisfy Au = z F , or equivalently, Bu, v = Fv ∀v ∈ H 10 U.
None of these abstract approaches gives any insight into how a solution might be
constructed. Therefore, we consider an additional approach to existence under the
assumptions of the second example above.
Galerkine Proof of Existence
We assume that the bilinear form is bounded and coercive but not necessarily
symmetric. Then we begin by defining a sequence of approximate solutions.
●
a) Approximate Solution
We let φ k  denote an ON basis for H 10 U, and for each positive integer N let
N
u N x = ∑ C k,N φ k x
k=1
where
Then
Bu N , φ j  = Fφ j 
1 ≤ j ≤ N.
⃗N = F
⃗ N,
B jk C
where
B jk  = Bφ j , φ k 
⃗ N = C 1,N , … , C N,N  
C
⃗ N = Fφ 1 , … , Fφ N   .
F
Note that
⃗  B jk  C
⃗ N = Bu N , u N  ≥ β || u N || 2 = β C
⃗N
C
N
1
2
and hence B jk  is positive definite so that a unique approximate solution exists for every N.
We now obtain estimates on the H 10 U norm of the sequence of approximate solutions
b) A-priori Estimate
1 ≤ j ≤ N,
It follows from
Bu N , φ j  = Fφ j 
that
β || u N || 21 ≤ | Fu N | ≤ C F || u N || 1
hence
6
|| u N || 1 ≤ 1 C F
β
∀N
Then the uniform (in N) bound for the norms implies that the sequence u N  contains a
weakly convergent subsequence u ν ;
i. e. ,
u ν , v 1 → u, v 1
∀v ∈ H 10 U.
We must now show that the weak limit of this subsequence is, in fact, a weak solution.
c) Passing to the (weak) Limit
Let V N = spanφ 1 , … , φ N . Then for each ν,
Bu ν , v = Fv
↓ ν→∞
Bu, v = Fv
∀v ∈ V ν
∀v ∈ ⋃ V ν ,
ν>0
where the convergence Bu ν , v → Bu, v follows from the weak convergence of u ν .
Since the φ ′ s form a basis for V = H 10 U, ⋃ V N is dense in V, and we have that
N>0
Bu, v = Fv
∀v ∈ H 10 U.
We have just shown that if u ν  is a subsequence of approximate solutions having weak
limit, u, in H 10 U then u must be a weak solution of the BVP. This can be done for any
weakly convergent subsequence of the sequence of approximate solutions, and since the
weak solution can be shown to be unique, it follows that all sub-sequences have the same
weak limit, u. But then it follows that the sequence of approximate solutions, u N , must
itself converge weakly to u. In fact, the sequence converges strongly to u, as we shall now
show.
d) Passing to the (strong) Limit
To see the strong convergence, let z N  denote a sequence with
z N ∈ V N , and ||z N − u|| 1 → 0 as N → ∞. For each N
Bu N , v = Fv
Bu, v = Fv
∀v ∈ V N
∀v ∈ H 10 U,
Bu − u N , v = 0
∀v ∈ V N .
hence
In particular, for v = z N − u N ∈ V N ,
Bu − u N , z N − u N  = Bu − u N , z N − u + Bu − u N , u − u N  = 0,
i.e.,
Then
Bu − u N , u − z N  = Bu − u N , u − u N ,
α||u − u N || 1 ||u − z N || 1 ≥ β ||u − u N || 21 .
||u − u N || 1 ≤ α ||u − z N || 1 → 0 as N → ∞,
β
which shows that u N  converges strongly to u.
To see that the weak solution is unique, suppose there are two weak solutions, u 1 , u 2 . Then
7
their difference satisfies
Bu 1 − u 2 , v = 0
for all v in H 10 U.
In particular, choosing v = u 1 − u 2 , and using the coercivity of B leads to u 1 = u 2 .
Coercivity and V-H Coercivity
The conditions i), ii) and iii) on K ij , b j and c are sufficient to show that operator L generates a
bounded bilinear form B; i.e., the form B satisfies,
|Bu, v| ≤ α|| u|| 1 ||v|| 1 .
These conditions do not, in general imply that the bilinear form B is coercive; i.e., they do
not imply the existence of β > 0 such that
Bu, u ≥ β || u|| 21 .
However, they do imply that B is V − H coercive for V = H 10 U and H = H 0 U; i. e. , they
imply the existence of β, λ > 0 such that
Bu, u ≥ β || u|| 21 − λ||u|| 20 .
To see this, write
n
n
i,j=1
i=1
Bu, v = ∫  ∑ K ij x ∂ i ux ∂ j vx + vx∑ b i x∂ i ux + cxuxdx
U
and note that conditions i), ii) and iii) imply
(a) ⃗z  Kx ⃗z ≥ k 0 | ⃗z | 2
∀z⃗ ∈ R n , x ∈ U
(b) || K ij || ∞ ≤ k 1 , || b j || ∞ ≤ b 1 , c 0 ≤ cx ≤ c 1 .
Then
n
n
i,j=1
i=1
|Bu, v| ≤ |∫U  ∑ K ij x ∂ i ux ∂ j vxdx |+|∫U vx ∑ b i x∂ i uxdx|+|∫U cxuxdx|
≤ k 1 ∫ |∇u ⋅ ∇v| dx + b 1 ∫ |v ∇u| dx + || c|| ∞ ∫ |uv| dx
U
U
U
≤ C||∇u|| 0 ||∇v|| 0 + ||∇u|| 0 ||v|| 0 + ||u|| 0 ||v|| 0  ≤ α ||u|| 1 ||v|| 1 .
Similarly,
n
n
i,j=1
i=1
|Bu, u| = ∫U  ∑ K ij x ∂ i ux ∂ j ux + ux∑ b i x∂ i ux + cxuxdx
≥ k 0 ∫ |∇u ⋅ ∇u| dx + c 0 ∫ | u| 2 dx − b 1 ∫ | u | | ∇u| dx.
U
U
U
Now
∫U | u | | ∇u| dx ≤ ∫U   | ∇u| 2 + b 0 | u| 2  dx =  ||∇u|| 20 + b 0 || u|| 20
4
4
b1
b1
and
|Bu, u| ≥ k 0 − ||∇u|| 20 + c 0 −
b 20
4
|| u|| 20 ≥ β || u|| 21 − λ||u|| 20 .
Under these conditions we then are able to prove the Fredholm alternative theorem for the
8
ellliptic boundary value problem. When the space V is chosen to be H 1 U instead of H 10 U
then one obtains the Neumann boundary value problem. By choosing V to be a space lying
between H 1 U and H 10 U, one obtains various mixed boundary value problems. (See the
notes #8)
The Fredholm Alternative Theorem
What happens in the case that B is V − H coercive but not coercive (i.e., V elliptic)? In this
case we can gain insight from looking back at the finite dimensional situation.
Suppose A is an n × n real matrix. Then
R n = R A ⊕ NA = NA ⊕ R A
dim N A = n − rankA = dim N A 
dim R A = n − dim N A  = rankA
and we have the following alternatives for the problem Ax = b :
a) rankA = n
there is a unique solution for every b ∈ R n
b) rankA = m < n
there is no solution unless b  N A 
in which case there is an n − m parameter family of (nonunique) solutions
This is the so called Fredholm alternative stated for a system of linear algebraic equations.
We will find a similar result for elliptic boundary value problems.
Suppose the bilinear form B satisfies
∃ positive β, λ such that
|Bu, u| + λ||u|| 20 ≥ β || u|| 21
∀u ∈ H 10 U
Then B is not coercive but the modified form B μ u, v = Bu, v + μu, v 0 is coercive for
μ ≥ λ. This implies that
∀f ∈ H −1 U there exists a unique u ∈ H 10 U such that
B μ u, v = ⟨f, v⟩
∀v ∈ H 10 U
This is equivalent to the statement
L μ = L + μI : H 10 U → H −1 U
is an isomorphism
Denote the unique weak solution of the boundary value problem with B μ by u = L −1
μ f and
recall that we want to solve Lu = f, not L μ u = f. If we write
L μ u = Lu + μu = f + μu := g,
then
−1
u = L −1
μ g = L μ f + μu,
Write
K = μL −1
μ ,
and
so our equation can be written
where
K : H 0 U  H 10 U
or
−1
u − μL −1
μ u = L μ f.
F = L −1
μ f,
I − Ku = F,
is bounded.
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To see this write
B μ u, u = g, u 0
which leads to
β‖u‖ 21 ≤ |B μ u, u| ≤ ‖g‖ 0 ‖u‖ 0 ≤ ‖g‖ 0 ‖u‖ 1 ;
‖u‖ 1 = ‖L −1
μ g‖ 1 ≤
i.e.,
‖g‖ 0
;
β
But then the definition of K leads to the estimate
μ
‖Kg‖ 1 = ‖μL −1
‖g‖ 0
μ g‖ 1 ≤
β
and since K : H 0 U  H 10 U is bounded, and the embedding i : H 10 U  H 0 U is
compact, it follows that
K : H 0 U  H 0 U
is compact.
Now we list for convenient reference, several equivalent problems:
u is a weak solution for
Lux = fx
u=0
x∈U
x∈Γ
if and only if
a
u ∈ H 10 U, satisfies
Bu, v = f, v 0
u ∈ H 10 U, satisfies
B μ u, v = f, v 0 + μu, v 0
∀v ∈ H 10 U
if and only if
b
∀v ∈ H 10 U
if and only if
−1
u = L −1
μ f + μL μ u,
c
if and only if
d
I − Ku = L −1
μ f,
The problems a through d are all equivalent formulations and u ∈ H 10 U solves one if
and only if it solves all the others. However, the operator I − K is an example of what is
called a Fredholm operator on H 0 U. This means
H 0 U = RngI − K ⊕ NI − K ∗  = NI − K ⊕ RngI − K ∗ 
dim NI − K = dim NI − K ∗  < ∞
RngI − K is closed and RngI − K = NI − K ∗  
RngI − K ∗  is closed and RngI − K ∗  = NI − K 
The analogy with the statements at the beginning of the section are obvious. We can now
state without proof the Fredholm alternative theorem for a compact operator on a Hilbert
space.
Fredholm Alternative Theorem- Consider the equation I − Ku = F ∈ H, for K a compact
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operator on Hilbert space, H. Then exactly one of the following alternatives must hold:
i NI − K = 0 : in this case ∀F ∈ H there exists a unique u ∈ H satisfying
I − Ku = F
ii NI − K = spanz 1 , … , z p  and NI − K ∗  = spanw 1 , … , w p  for a positive
integer p. In this case I − Ku = F has no solution unless
F, w j  H = 0, 1 ≤ j ≤ p.
Then
I − Ku 0 + C 1 z 1 + ⋯ + C p z p  = F
for all constants, C j and any u 0 ∈ H such that I − Ku 0 = F.
Here, the operator K ∗ on H is defined as follows. For v ∈ H, fixed we can define a linear
functional on H by
λu =: Ku, v H
∀u ∈ H.
Then the Riesz theorem implies the existence of an element, denoted by K ∗ v to indicate its
dependence on v, such that
λu =: u, K ∗ v H
i.e.,
∀u ∈ H.
Ku, v H = u, K ∗ v H
∀u, v ∈ H.
is easily seen to be bounded and linear and it can be shown
The operator K ∗ : H  H
to be compact of K is compact. The operator is called the adjoint operator for K.
The Adjoint Boundary Value Problem
Suppose that u is a weak solution for
Lux = fx
u=0
x∈U
x∈Γ
That is,
u ∈ H 10 U, satisfies
Bu, v = f, v 0
∀v ∈ H 10 U
Note that for φ, ψ ∈ C ∞c Ū we have
Lφ, ψ 0 = Bφ, ψ − ∫Γ ψ K∇φ ⋅ n dS = Bφ, ψ − 0
and if we extend this by continuity to u ∈ H 1 U, this becomes
Lu, ψ 0 = Bu, ψ = f, ψ 0
∀ψ ∈ C ∞c U
Evidently, if u is a weak solution of the BVP, then
Lu = f
in the sense of distributions on U.
If we integrate by parts again in order to move all of the differentiation onto ψ, we get,
Lu, ψ 0 = Bu, ψ
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n
n
i,j=1
i=1
= ∫  ∑ K ij x ∂ i ux ∂ j ψx + ψx∑ b i x∂ i ux + cxuxdx
U
= ∫ ∇ψK∇u + ψb ⋅ ∇u + cψu dx
U
∫U u−∇ ⋅ K∇ψ − ∇ ⋅ ψb + cψ dx = u, L ∗ ψ 0
where
L ∗ ψ = −∇K∇ψ − ∇ψb + cψ
n
n
i,j=1
i=1
= − ∑ ∂ i K ij x ∂ j ψx − ∑ ∂ i b i xψx + cxψx
Then
Lu, ψ 0 = Bu, ψ = u, L ∗ ψ 0
and by analogy with
Lux = fx
u is a weak solution for
u=0
if and only if
x∈Γ
Bu, v = f, v 0
u ∈ H 10 U, satisfies
x∈U
∀v ∈ H 10 U
we have
L ∗ vx = gx
v is a weak solution for
v=0
if and only if
x∈Γ
B ∗ v, u = g, u 0
v ∈ H 10 U, satisfies
x∈U
∀u ∈ H 10 U
B ∗ v, u = Bu, v.
where
We refer to the BVP as the adjoint problem to the original BVP.
Proceeding just as we did above, we have that v is a weak solution of the adjoint
problem if and only if,
a
v ∈ H 10 U, satisfies
B ∗ v, u = g, u 0
v ∈ H 10 U, satisfies
B ∗μ v, u = g, u 0 + μu, v 0
∀u ∈ H 10 U
if and only if
b
∀u ∈ H 10 U
if and only if
c
−1
−1
u = L ∗μ  f + μL ∗μ  u,
if and only if
d
Since
then
−1
I − K ∗ u = L ∗μ  f,
Lu, v 0 = Bu, v = u, L ∗ v 0
μL μ  −1 f, g
0
=
for u, v ∈ H 10 U
−1
f, μL ∗μ  g
0
for f, g ∈ H 0 U
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Kf, g 0 = f, K ∗ g 0
i.e.,
for f, g ∈ H 0 U
This proves that K ∗ is the adjoint mapping for K.
Then we can restate the Fredholm alternative theorem as it applies to the elliptic boundary
value problem as follows:
Fredholm Alternative Theorem- Consider the elliptic boundary value problem
Lux = fx
u=0
x∈U
x∈Γ
Exactly one of the following alternatives must apply to the associated weak problem:
i N = u ∈ H 10 U : Bu, v = 0, ∀v ∈ H 10 U = 0 :
N ∗ = v ∈ H 10 U : B ∗ u, v = 0, ∀u ∈ H 10 U = 0
In this case ∀f ∈ H −1 U there exists a unique weak solution, u ∈ H 10 U.
ii N = spanz 1 , … , z p  and N ∗ = spanw 1 , … , w p  for a positive integer p.
In this case there is no weak solution for the BVP unless
f, w j 0 = 0, 1 ≤ j ≤ p.
Then
Bu 0 + C 1 z 1 + ⋯ + C p z p,v  = f, v 0
for all constants, C j and any weak solution u 0 ∈ H 10 U .
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