In this section, we show how obeying general -dependent boundary conditions transforms, under (2.1), to obeying various types of -dependent boundary conditions. The exact form of these boundary conditions is obtained by considering the number of zeros and poles (singularities) of the various Nevanlinna functions under discussion and these correlations are illustrated in the different graphs depicted in this section.

Lemma 3.1.

If obeys the boundary condition

then the domain of may be extended from to by forcing the condition

where

with .

Proof.

The transformed equation (2.2), for , together with (3.2) gives

Also the mapping (2.1), together with (3.1), yields

Substituting (3.5) into (3.4), we obtain

Now (2.1), with , gives

which when substituted into (3.6) and dividing through by results in

This may be rewritten as

Using (1.1), with , together with (3.1), gives

Subtracting (3.10) from (3.9) results in

Rearranging the above equation and dividing through by yields

and hence

Thus obeys the equation on the extended domain.

The remainder of this section illustrates why it is so important to distinguish between the two cases of obeying or not obeying the boundary conditions.

Theorem 3.2.

Consider obeying the boundary condition (3.1) where is a positive Nevanlinna function, that is, for . Under the mapping (2.1), obeying (3.1) transforms to obeying (3.2) as follows.

If does not obey (3.1) then obeys

If does obey (3.1) for then obeys

where , that is, are positive Nevanlinna functions.

In (A) and (B), is not possible.

Proof.

The fact that is by construction, see Lemma 3.1. We now examine the form of in Lemma 3.1. Let , , and then

But

thus

Now has the expansion

where and the 's correspond to where , that is, the singularities of (3.20).

Since is a positive Nevanlinna function it has a graph of the form shown in Figure 1.

Clearly, the gradient of at is positive for all , that is,

If does not obey (3.1), then the zeros of

are the poles of , that is, the 's and where for . It is evident, from Figure 1, that the number of 's is equal to the number of 's, thus in (3.21), .

We now examine the form of in (3.21). As it follows that . Thus

Therefore

Hence, substituting into (3.20) gives

Let

Then since , and we have that and clearly if then giving (3.14), that is,

If then we want so that we have a positive Nevanlinna function, that is

which means that either,

giving that, since ,

which is as shown in Figure 1, or,

giving that

but this means that which is not possible.

Thus, for , that is, given , the ratio must be chosen suitably to ensure that is a positive Nevanlinna function as required. Hence we obtain (3.15), that is

If obeys (3.1), for , then . Thus in Figure 1, one of the 's is equal to and since is less than the least eigenvalue of the boundary value problem (1.1), (3.1) together with a boundary condition at (specified later) it follows that , as for all .

Now

and as

Thus is a removable singularity. Alternatively,

which illustrates that the singularity at is removable.

We now have that the number of nonremovable singularities, , in (3.20) is one less than the number of 's , see Figure 1. Thus (3.21) becomes

which may be rewritten as

where , for .

We now examine the form of in (3.39). As , we have that, as before, . Thus

Hence, from (3.20),

Let

Then since , and we have that and clearly if then giving (3.16), that is,

If then we need so that we have a positive Nevanlinna function, that is

which means that either

giving that, since ,

which is as shown in Figure 1, or,

giving that

but this means that which is not possible.

Thus, for , that is, given , the ratio must be chosen suitably to ensure that is a positive Nevanlinna function as required. Hence, we obtain (3.17), that is,

In the theorem below, we increase the dependence by introducing a nonzero term in the original boundary condition. As in Theorem 3.2, the dependence of the transformed boundary condition depends on whether or not obeys the given boundary condition. In addition, to ensure that the dependence of the transformed boundary condition is given by a positive Nevanlinna function it is necessary that the transformed boundary condition is imposed at 0 and 1 as opposed to −1 and 0. Thus the interval under consideration shrinks by one unit at the initial end point. By routine calculation it can be shown that the form of the dependence of the transformed boundary condition, if imposed at −1 and 0, is neither a positive Nevalinna function nor a negative Nevanlinna function.

Theorem 3.3.

Consider obeying the boundary condition

where is a positive Nevanlinna function, that is, and for . Under the mapping (2.1), obeying (3.50) transforms to obeying the following.

**(1)** If does not obey (3.50) then obeys

**(2)** If does obey (3.50), for , then obeys

where .

Proof.

Since and are defined we do not need to extend the domain in order to impose the boundary conditions (3.51) or (3.52).

The mapping (2.1), at , together with (3.50) gives

Also (2.1), at , is

Substituting in for from (1.1), with , and using (3.50), we obtain that

From (3.53) and (3.55), it now follows that

As in Theorem 3.2, let and . Then (3.56) becomes

From Theorem 3.2, we have that so

Also, as in Theorem 3.2,

has the expansion

where corresponds to , that is, the singularities of (3.59). Now is a positive Nevanlinna function with graph given in Figure 2.

Clearly, the gradient of at is positive for all , that is,

If does not obey (3.50) then the zeros of

are the poles of , that is, the 's and where for . It is evident, from Figure 2, that the number of 's is one more than the number of 's, thus in (3.60), .

We now examine the form of in (3.60). As it follows that , thus

Hence, .

Using (3.58) we now obtain

Note that . Let

then

Now since if then , that is, but and so this is not possible. Therefore by Section 2, Nevanlinna result (II), we have that

that is, (3.51) holds.

If does obey (3.50) for then . Thus, in Figure 2, one of the 's, is equal to and since is less than the least eigenvalue of the boundary value problem (1.1), (3.50) together with a boundary condition at (specified later) it follows that , as for all .

Now (3.59) can be written as

and as

Thus is a removable singularity. Alternatively, we could substitute in for and to illustrate that the singularity at is removable, see Theorem 3.2. Hence the number of nonremovable 's is the same as the number of 's, see Figure 2. So (3.60) becomes

which may be rewritten as

where and for .

We now examine the form of in (3.70). As , we have that , thus

Hence, . So, from (3.58) with , we obtain

where, as before,

Thus, by Section 2, Nevanlinna result (II), we have that

that is, (3.52) holds.

In Theorem 3.4, we impose a boundary condition at the terminal end point and show how it is transformed according to whether or not obeys the given boundary condition.

Theorem 3.4.

Consider obeying the boundary condition at given by

where is a negative Nevanlinna function, that is, and for . Under the mapping (2.1), obeying (3.76) transforms to obeying the following.

**(I)** If does not obey (3.76) then obeys

**(II)** If does obey (3.76) then obeys

where .

Proof.

Since and are defined we do not need to extend the domain of in order to impose the boundary conditions (3.77) or (3.78).

The mapping (2.1), at , gives

From (1.1), with , we can substitute in for in the above equation to get

Using (3.76), we obtain

But obeys (1.1) at , for , so that (3.81) becomes

Also, for , (2.1) together with (3.76) yields

Therefore,

Let , then (3.84) may be rewritten as

By Section 2, Nevanlinna result (I), since is a negative Nevanlinna function it follows that is a positive Nevanlinna function, which has the form

by Section 2, Nevanlinna result (III).

As before has expansion

where , , corresponds to the singularities of (3.85), that is, where . The graph of is as shown in Figure 3.

As before, the gradient of at is positive for all , that is

If does not obey (3.76) then the zeros of

are the poles of , that is, the 's and where for . Clearly, from Figure 3, the number of 's is the same as the the number of 's, thus in (3.87), .

Next, we examine the form of in (3.87). As it follows that . Thus

Therefore, . Hence,

where , , and for , which is precisely (3.77).

If does obey (3.76) for then . Thus in Figure 3, one of the 's, is equal to and since is less than the least eigenvalue of the boundary value problem (1.1), (3.76) together with a boundary condition at −1 (as given in Theorems 3.2 or 3.3) it follows that , as for all .

Now

and as

Thus is a removable singularity. Again, alternatively, we could have substituted in for and to illustrate that the singularity at is removable, see Theorem 3.2. Hence the number of nonremovable 's is one less than the number of 's, see Figure 3.

So (3.87) becomes

which may be rewritten as

where and for .

Now as ,

So, we obtain

where , , , and for all , that is, we obtain (3.78).