Weak solutions of the Poisson equation

namely, we prove the existence of weak solutions of the Poisson equation. ... 4 5 : 7 ; Moreover,. the inequality holds for any function in *, ;...

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Weak solutions of the Poisson equation  Now we are ready to demonstrate the usefulness of Sobolev spaces in the simplest situation, namely, we prove the existence of weak solutions of the Poisson equation. Let is a bounded open subset of equation:

and let us consider the Dirichlet problem for the Poisson ∆

, 0,

.

Since we are interested in the zero boundary condition, it is reasonable to consider as the class of test functions. Let be the classical solution of the above Dirichlet problem. Then multiplying the Poisson equation by an arbitrary test function and integrating the obtained equality by parts yields 1

· The following remarks are appropriate to mention here: (a) The latter equality may also be interpreted for any function (b) A bilinear form ,

:

,

of a larger class

.

·

in the left hand side looks like a scalar product. (c) The identity (1) can be interpreted now as equation on a linear functional 2

: and the scalar product , that resembles the Riesz representation theorem (d) If , the linear functional defined above is bounded in -norm for and moreover, we have subspace /

|

that is

/

|

,

extends (by Hahn-Banach theorem) to a bounded functional in

,

.

We return to the scalar product defined above. In fact, the bilinear form , is well-defined in the whole

,

:

·

, symmetric and positive there (notice that | |

, which implies, by the zero boundary condition, the associated norm.

0). Denote by | |

0 yields ,

Hence we obtain a weak formulation of (1): given a function , any classical solution to the Poisson equation with zero boundary condition is a solution of the following problem , where

,

,

3

is defined by (2). The following definition is then natural.

, which solves (3) is called a weak solution to the Poisson Definition. A function equation with zero boundary conditions.

In order to threat the existence of a weak solution we show that the new scalar product generates the same metric (and topological) structure in , .

Poincare’s inequality  Theorem  (Poincare’s inequality).  If    is  a  bounded  open  subset  of  constant  0 such that   | for all 

|

  then  there  exists  a 

  ,

. Moreover, the inequality holds for any function in 

 

Denote by the same letter the function obtained by extending by zero outside of . Let be large enough such that the support of is contained in the cube :| | . Notice that

1, hence integrating by parts we find 2

Applying Cauchy inequality

|

|

we obtain 1

1 4

2

4

|

|

and the desired property follows. Remark 1. The question on the optimal constant in the Poincare inequality has many relations to other problems in mathematics and mathematical physics. For instance, if one think of as a membrane then the best constant √ in the Poincare inequality is exactly the fundamental frequency sometimes is called also the fundamental tone of . If 0, is the one‐dimensional interval then / . Besides the fundamental tone, one distinguish also a series of higher “tones” see the first picture below . The second picture shows the oscillating two‐dimensional membrane corresponding to the 7th tone 0,1 0,1 . sin 2 sin 4 in the rectangle

Remark 2. The above Poincare inequality is a partial case of a more general relation the so‐called Poincare‐Friedrichs inequality inequality |

|

,

 

which can be proved by the same method. Corollary 1. If    is  a  bounded  open  subset  of  , 0such that for any    

  then  there  exists  a  constant 

 

,

,

Indeed, for any

Poincare’s inequality yields |

,

|

| |

|

1

| |

|

Corollary 2 Existence and uniqueness of the weak solution . For any ,  which solves (3).  unique function  By Corollary 1 we have

It follows that | | and

and, in the other direction,

,

| |

| ,

there is a

|

|

|

| |

,

.

are equivalent norms.

On the other hand, defined by (2) is a bounded linear functional in , . Hence, applying the Riesz representation theorem to the new scalar product , we immediately conclude that , there is a unique function such that , That is

,

,

is a weak solution to the Dirichlet problem formulated above.

Existence of weak solutions for general elliptic equations The method described above can be applied also to the Dirichlet problem for more general elliptic equations. Consider a differential operator in divergence form .

,

where , , ellipticity condition: there is a constant

satisfying the (uniform)

and 0 such that for any .

,

Starting from a smooth solution and integrating by parts, we see that the natural definition of a weak solution is the following. ,

Definition. A function zero boundary conditions if

which solves (3) is called a weak solution to

,

,

0 with

.

4

It is then convenient to restate the condition for a weak solution as follows: , where

,

,

5

· s defined as before by formula (2) and the new bilinear form is given by ,

.

,

The existence of a weak solution will follow immediately if we show that product whose associated norm is equivalent to the standard norm in , investigate under which conditions , meets these criteria.

, is a scalar . So we need to

First we notice that , is symmetric by our assumption on the coefficients using of the ellipticity condition we obtain the following lower estimate: , where

max

:

, . Let us assume that max

:

. Next, by

0 then

where is the constant in the Poincare inequality. If 0 then

,

,

, and if

.

Applying then Corollary 1 we obtain , where

and

,

are some positive constants. The latter inequality is called coercive condition.

Now, we notice that functions | | and | | are continuous in , hence they bounded there. Denoting by the common upper bound we find , for all

, and, consequently, for all

Summarizing, we conclude that inequality ,

,

,

,

.

is a symmetric bilinear form satisfying the bilateral ,

,

,

0 .

Let | | , . Then the new norm is equivalent to , and we can again apply the Riesz theorem to show that for any bounded linear functional there is a function , such that equality (5) above holds. Thus we have obtained Theorem. In the made assumptions, the weak solution of 

0 exists and is uniquely defined.