LECTURE 9. GREEN'S FUNCTION. EXAMPLES
In this chapter we consider the Laplace equation in bounded domains
, located on the plane or in space. The points
and
on the plane (or
and
in the space) belong to
and
(and
) - the distance between points
and
Let us assume that on the boundary of
is set to zero Dirichle condition.
The function is called theGreen's function of the Dirichleproblem in the
, if for any fixed point
of it, as a function of
, satisfies the following conditions:
a continuous
everywhere except at the point
, and
on the boundary of
;
a harmonic except at
point
;
in case the plane is
a harmonic function at the point
; if space remains
harmonic function at the point
.
As follows from the definition of the Green function is continuous and harmonic throughout the domain except
point at which it has a feature type
in the plane or
in space. Green's function is sometimes called the source function.
The Green's function ) (if it exists) is uniquely determined by the properties
. In addition,
in the domain
. Consider, for example, a flat area
. To prove the uniqueness of the Green's function, we assume the contrary: let
, and
- two functions, possessing properties
for a given domain
and the point
. Then
remains harmonic at any point in the area
, including point
, since the vicinity of the point
can be written
Each bracket on the right side is a function harmonic everywhere in
(see property
.), And therefore the difference
- everywhere in the harmonic function
. Also, at the boundary function
Consequently, by the maximum
principle in
.
Further, if - part of the region
, positioned outside a small neighborhood of the point
, according to the conditions
, the function
is continuous
in harmonic
, and on the border
takes non-negative values (as
at
). Therefore, on the basis of a maximum
of
with the zero value inside the area
function can not accept. That means that
everywhere in
.
Example1. In the plane, consider a circle of radius centered at the origin. We construct the Green's function in the circle. In the construction of this function, we need the concept of conjugate points. Points
and
are conjugate with respect to the circle if they lie on the same ray emanating from the center of the circle
, and the product of their distances from the center is equal to the square of the radius:
(See Figure16.).
Figure16.
Let and
Then
Since the points
and
lie on the same ray emanating from the origin, then
Consider the function
Where (See. Figure17). We verify that it is the Green's function for the circle.
By the theorem of cosines and
where
Figure17.
Using the equality , we obtain
In this way, the value of
and
expressed in terms of
and, ultimately, through
We show that the function
satisfies the items
determination. It is obvious that the function is continuous everywhere in the closed circle except at the point
(when
). At the boundary of a distance
and, hence,
Hence The function
consists of two terms. The first term, the fundamental solution of the Laplace equation and , therefore, harmonic function everywhere except
. The function
is harmonic throughout the domain
, since the point
belongs to the region, and the point
lies outside the region
, and hence,
. The harmony of this function is easily verified if we write the Laplace operator in polar coordinates with pole at
(sm. A similar formula
with a pole at the point
):
Therefore, the function in
harmonic everywhere except at the point Po, and the difference
) - harmonic and at the point
.
Similarly, construction of Green's function for a sphere of radius . It has the form
Where
Point
conjugate point
with respect to a sphere of radius
centered at
, that is
. The coordinates
are calculated according to the formulas:
Example2. Green's function can be viewed not only limited, but also for unbounded domains. As an example, we construct the Green's function for the half-plane. To do this, we define a point conjugate with respect to the line: points and
are conjugate with respect to a straight line, if they are symmetrical with respect to this line (see. Figure18).
Figure18. Figure19.
The function where
(See Figure19.), satisfies the properties in the half-plane
. In fact, on the boundary at
distance
, so
The harmony function
everywhere in the region
can be verified directly by calculating the partial derivatives:
So
Consequently, the function harmonic in the domain
everywhere except at the point
, and the difference
and the harmonic at the point
.
For the half and the Green's function has the form
Where
Examples:
Given the problem
Find Green's function.
First step: From demand-2 we see that
For we see from demand-3 that the
, while for
we see from demand-3 that the
(we leave it to the reader to fill in the in-between steps).
Summarize the results:
Second step: Now we shall determine and
Using demand-1 we get
Using demand- 4 we get
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Using Cramer's rule or by intelligent guess solve for and
and obtain that
Check that this automatically satisfies demand-5.
So our Green's function for this problem is: