Elliptic boundary value problem explained

In mathematics, an elliptic boundary value problem is a special kind of boundary value problem which can be thought of as the stable state of an evolution problem. For example, the Dirichlet problem for the Laplacian gives the eventual distribution of heat in a room several hours after the heating is turned on.

Differential equations describe a large class of natural phenomena, from the heat equation describing the evolution of heat in (for instance) a metal plate, to the Navier-Stokes equation describing the movement of fluids, including Einstein's equations describing the physical universe in a relativistic way. Although all these equations are boundary value problems, they are further subdivided into categories. This is necessary because each category must be analyzed using different techniques. The present article deals with the category of boundary value problems known as linear elliptic problems.

Boundary value problems and partial differential equations specify relations between two or more quantities. For instance, in the heat equation, the rate of change of temperature at a point is related to the difference of temperature between that point and the nearby points so that, over time, the heat flows from hotter points to cooler points. Boundary value problems can involve space, time and other quantities such as temperature, velocity, pressure, magnetic field, etc.

Some problems do not involve time. For instance, if one hangs a clothesline between the house and a tree, then in the absence of wind, the clothesline will not move and will adopt a gentle hanging curved shape known as the catenary.[1] This curved shape can be computed as the solution of a differential equation relating position, tension, angle and gravity, but since the shape does not change over time, there is no time variable.

Elliptic boundary value problems are a class of problems which do not involve the time variable, and instead only depend on space variables.

The main example

In two dimensions, let

x,y

be the coordinates. We will use the notation

ux,uxx

for the first and second partial derivatives of

u

with respect to

x

, and a similar notation for

y

. We will use the symbols

Dx

and

Dy

for the partial differential operators in

x

and

y

. The second partial derivatives will be denoted
2
D
x
and
2
D
y
. We also define the gradient

\nablau=(ux,uy)

, the Laplace operator

\Deltau=uxx+uyy

and the divergence

\nabla(u,v)=ux+vy

. Note from the definitions that

\Deltau=\nabla(\nablau)

.

The main example for boundary value problems is the Laplace operator,

\Deltau=fin\Omega,

u=0on\partial\Omega;

where

\Omega

is a region in the plane and

\partial\Omega

is the boundary of that region. The function

f

is known data and the solution

u

is what must be computed. This example has the same essential properties as all other elliptic boundary value problems.

The solution

u

can be interpreted as the stationary or limit distribution of heat in a metal plate shaped like

\Omega

, if this metal plate has its boundary adjacent to ice (which is kept at zero degrees, thus the Dirichlet boundary condition.) The function

f

represents the intensity of heat generation at each point in the plate (perhaps there is an electric heater resting on the metal plate, pumping heat into the plate at rate

f(x)

, which does not vary over time, but may be nonuniform in space on the metal plate.) After waiting for a long time, the temperature distribution in the metal plate will approach

u

.

Nomenclature

Let

Lu=auxx+buyy

where

a

and

b

are constants.
2
L=aD
y
is called a second order differential operator. If we formally replace the derivatives

Dx

by

x

and

Dy

by

y

, we obtain the expression

ax2+by2

.

If we set this expression equal to some constant

k

, then we obtain either an ellipse (if

a,b,k

are all the same sign) or a hyperbola (if

a

and

b

are of opposite signs.) For that reason,

L

is said to be elliptic when

ab>0

and hyperbolic if

ab<0

. Similarly, the operator

L=Dx+D

2
y
leads to a parabola, and so this

L

is said to be parabolic.

We now generalize the notion of ellipticity. While it may not be obvious that our generalization is the right one, it turns out that it does preserve most of the necessary properties for the purpose of analysis.

General linear elliptic boundary value problems of the second degree

Let

x1,...,xn

be the space variables. Let

aij(x),bi(x),c(x)

be real valued functions of

x=(x1,...,xn)

. Let

L

be a second degree linear operator. That is,
n
Lu(x)=\sum
i,j=1

(aij(x)

u
xi
)
xj

+

n
\sum
i=1

bi(x)

u
xi

(x)+c(x)u(x)

(divergence form).
n
Lu(x)=\sum
i,j=1

aij(x)

u
xixj

+

n
\sum
i=1

\tildebi(x)

u
xi

(x)+c(x)u(x)

(nondivergence form)

We have used the subscript

xi
to denote the partial derivative with respect to the space variable

xi

. The two formulae are equivalent, provided that

\tildebi(x)=bi(x)+\sumj

a
ij,xj

(x)

.

In matrix notation, we can let

a(x)

be an

n x n

matrix valued function of

x

and

b(x)

be a

n

-dimensional column vector-valued function of

x

, and then we may write

Lu=\nabla(a\nablau)+bT\nablau+cu

(divergence form).

One may assume, without loss of generality, that the matrix

a

is symmetric (that is, for all

i,j,x

,

aij(x)=aji(x)

. We make that assumption in the rest of this article.

We say that the operator

L

is elliptic if, for some constant

\alpha>0

, any of the following equivalent conditions hold:

λmin(a(x))>\alpha   \forallx

(see eigenvalue).

uTa(x)u>\alphauTu   \forallu\inRn

.
n
\sum
i,j=1

aijuiuj>\alpha

n
\sum
i=1
2
u
i

   \forallu\inRn

.

An elliptic boundary value problem is then a system of equations like

Lu=fin\Omega

(the PDE) and

u=0on\partial\Omega

(the boundary value).

This particular example is the Dirichlet problem. The Neumann problem is

Lu=fin\Omega

and

u\nu=gon\partial\Omega

where

u\nu

is the derivative of

u

in the direction of the outwards pointing normal of

\partial\Omega

. In general, if

B

is any trace operator, one can construct the boundary value problem

Lu=fin\Omega

and

Bu=gon\partial\Omega

.

In the rest of this article, we assume that

L

is elliptic and that the boundary condition is the Dirichlet condition

u=0on\partial\Omega

.

Sobolev spaces

The analysis of elliptic boundary value problems requires some fairly sophisticated tools of functional analysis. We require the space

H1(\Omega)

, the Sobolev space of "once-differentiable" functions on

\Omega

, such that both the function

u

and its partial derivatives
u
xi
,

i=1,...,n

are all square integrable. There is a subtlety here in that the partial derivatives must be defined "in the weak sense" (see the article on Sobolev spaces for details.) The space

H1

is a Hilbert space, which accounts for much of the ease with which these problems are analyzed.

The discussion in details of Sobolev spaces is beyond the scope of this article, but we will quote required results as they arise.

Unless otherwise noted, all derivatives in this article are to be interpreted in the weak, Sobolev sense. We use the term "strong derivative" to refer to the classical derivative of calculus. We also specify that the spaces

Ck

,

k=0,1,...

consist of functions that are

k

times strongly differentiable, and that the

k

th derivative is continuous.

Weak or variational formulation

The first step to cast the boundary value problem as in the language of Sobolev spaces is to rephrase it in its weak form. Consider the Laplace problem

\Deltau=f

. Multiply each side of the equation by a "test function"

\varphi

and integrate by parts using Green's theorem to obtain

-\int\Omega\nablau\nabla\varphi+\int\partialu\nu\varphi=\int\Omegaf\varphi

.

We will be solving the Dirichlet problem, so that

u=0on\partial\Omega

. For technical reasons, it is useful to assume that

\varphi

is taken from the same space of functions as

u

is so we also assume that

\varphi=0on\partial\Omega

. This gets rid of the

\int\partial

term, yielding

A(u,\varphi)=F(\varphi)

(*)

where

A(u,\varphi)=\int\Omega\nablau\nabla\varphi

and

F(\varphi)=-\int\Omegaf\varphi

.

If

L

is a general elliptic operator, the same reasoning leads to the bilinear form

A(u,\varphi)=\int\Omega\nablauTa\nabla\varphi-\int\OmegabT\nablau\varphi-\int\Omegacu\varphi

.

We do not discuss the Neumann problem but note that it is analyzed in a similar way.

Continuous and coercive bilinear forms

The map

A(u,\varphi)

is defined on the Sobolev space
1
H
0\subset

H1

of functions which are once differentiable and zero on the boundary

\partial\Omega

, provided we impose some conditions on

a,b,c

and

\Omega

. There are many possible choices, but for the purpose of this article, we will assume that

aij(x)

is continuously differentiable on

\bar\Omega

for

i,j=1,...,n,

bi(x)

is continuous on

\bar\Omega

for

i=1,...,n,

c(x)

is continuous on

\bar\Omega

and

\Omega

is bounded.

The reader may verify that the map

A(u,\varphi)

is furthermore bilinear and continuous, and that the map

F(\varphi)

is linear in

\varphi

, and continuous if (for instance)

f

is square integrable.

We say that the map

A

is coercive if there is an

\alpha>0

for all

u,\varphi\in

1(\Omega)
H
0
,

A(u,\varphi)\geq\alpha\int\Omega\nablau\nabla\varphi.

This is trivially true for the Laplacian (with

\alpha=1

) and is also true for an elliptic operator if we assume

b=0

and

c\leq0

. (Recall that

uTau>\alphauTu

when

L

is elliptic.)

Existence and uniqueness of the weak solution

One may show, via the Lax–Milgram lemma, that whenever

A(u,\varphi)

is coercive and

F(\varphi)

is continuous, then there exists a unique solution

u\in

1(\Omega)
H
0
to the weak problem (*).

If further

A(u,\varphi)

is symmetric (i.e.,

b=0

), one can show the same result using the Riesz representation theorem instead.

This relies on the fact that

A(u,\varphi)

forms an inner product on
1(\Omega)
H
0
, which itself depends on Poincaré's inequality.

Strong solutions

We have shown that there is a

u\in

1(\Omega)
H
0
which solves the weak system, but we do not know if this

u

solves the strong system

Lu=fin\Omega,

u=0on\partial\Omega,

Even more vexing is that we are not even sure that

u

is twice differentiable, rendering the expressions
u
xixj
in

Lu

apparently meaningless. There are many ways to remedy the situation, the main one being regularity.

Regularity

A regularity theorem for a linear elliptic boundary value problem of the second order takes the form

Theorem If (some condition), then the solution

u

is in

H2(\Omega)

, the space of "twice differentiable" functions whose second derivatives are square integrable.

There is no known simple condition necessary and sufficient for the conclusion of the theorem to hold, but the following conditions are known to be sufficient:

  1. The boundary of

\Omega

is

C2

, or

\Omega

is convex.

It may be tempting to infer that if

\partial\Omega

is piecewise

C2

then

u

is indeed in

H2

, but that is unfortunately false.

Almost everywhere solutions

In the case that

u\inH2(\Omega)

then the second derivatives of

u

are defined almost everywhere, and in that case

Lu=f

almost everywhere.

Strong solutions

One may further prove that if the boundary of

\Omega\subsetRn

is a smooth manifold and

f

is infinitely differentiable in the strong sense, then

u

is also infinitely differentiable in the strong sense. In this case,

Lu=f

with the strong definition of the derivative.

The proof of this relies upon an improved regularity theorem that says that if

\partial\Omega

is

Ck

and

f\inHk-2(\Omega)

,

k\geq2

, then

u\inHk(\Omega)

, together with a Sobolev imbedding theorem saying that functions in

Hk(\Omega)

are also in

Cm(\bar\Omega)

whenever

0\leqm<k-n/2

.

Numerical solutions

While in exceptional circumstances, it is possible to solve elliptic problems explicitly, in general it is an impossible task. The natural solution is to approximate the elliptic problem with a simpler one and to solve this simpler problem on a computer.

Because of the good properties we have enumerated (as well as many we have not), there are extremely efficient numerical solvers for linear elliptic boundary value problems (see finite element method, finite difference method and spectral method for examples.)

Eigenvalues and eigensolutions

Another Sobolev imbedding theorem states that the inclusion

H1\subsetL2

is a compact linear map. Equipped with the spectral theorem for compact linear operators, one obtains the following result.

Theorem Assume that

A(u,\varphi)

is coercive, continuous and symmetric. The map

S:fu

from

L2(\Omega)

to

L2(\Omega)

is a compact linear map. It has a basis of eigenvectors

u1,u2,...\inH1(\Omega)

and matching eigenvalues

λ1,λ2,...\inR

such that

Suk=λkuk,k=1,2,...,

λk0

as

kinfty

,

λk\gneqq0  \forallk

,

\int\Omegaujuk=0

whenever

jk

and

\int\Omegaujuj=1

for all

j=1,2,....

Series solutions and the importance of eigensolutions

If one has computed the eigenvalues and eigenvectors, then one may find the "explicit" solution of

Lu=f

,
infty
u=\sum
k=1

\hatu(k)uk

via the formula

\hatu(k)=λk\hatf(k),  k=1,2,...

where

\hatf(k)=\int\Omegaf(x)uk(x)dx.

(See Fourier series.)

The series converges in

L2

. Implemented on a computer using numerical approximations, this is known as the spectral method.

An example

Consider the problem

u-uxx-uyy=f(x,y)=xy

on

(0,1) x (0,1),

u(x,0)=u(x,1)=u(0,y)=u(1,y)=0  \forall(x,y)\in(0,1) x (0,1)

(Dirichlet conditions).

The reader may verify that the eigenvectors are exactly

ujk(x,y)=\sin(\pijx)\sin(\piky)

,

j,k\inN

with eigenvalues

λjk={1\over1+\pi2j2+\pi2k2}.

The Fourier coefficients of

g(x)=x

can be looked up in a table, getting

\hatg(n)={(-1)n+1\over\pin}

. Therefore,

\hatf(j,k)={(-1)j+k+1\over\pi2jk}

yielding the solution

u(x,y)=

infty
\sum
j,k=1

{(-1)j+k+1\over\pi2jk(1+\pi2j2+\pi2k2)}\sin(\pijx)\sin(\piky).

Maximum principle

There are many variants of the maximum principle. We give a simple one.

Theorem. (Weak maximum principle.) Let

u\inC2(\Omega)\capC1(\bar\Omega)

, and assume that

c(x)=0 \forallx\in\Omega

. Say that

Lu\leq0

in

\Omega

. Then

maxxu(x)=maxxu(x)

. In other words, the maximum is attained on the boundary.

A strong maximum principle would conclude that

u(x)lneqqmaxyu(y)

for all

x\in\Omega

unless

u

is constant.

Further reading

Notes and References

  1. Swetz, Faauvel, Bekken, "Learn from the Masters", 1997, MAA, pp.128-9