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
 | |
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:
