In this section we will show how one can generalize the proof of existence of the Dirichlet problem on the unit disk, given by Theorem 2 in Section 3, to the case in which we have, as the boundary condition, non-integrable real functions defined on boundaries given by differentiable curves on the plane. We will be able to do this if the non-integrable real functions, despite being non-integrable over the whole curves , are however locally integrable almost everywhere on those curves, and if, in addition to this, the non-integrable hard singularities of the inner analytic functions involved have finite degrees of hardness. The definition of the concept of local integrability almost everywhere is similar to that given for the unit circle by Definition 2, in Section 3. In our case here the precise definition of local integrability almost everywhere is as follows.
The proof will follow the general lines of the one given for integrable real functions in Section 5, with the difference that, since the real functions are assumed to be non-integrable on , but locally integrable almost everywhere there, instead of showing that the corresponding functions on the unit circle are integrable there, we will show that they are locally integrable almost everywhere there. In addition to this, instead of using the result for integrable real function on the unit circle, which was given by Theorem 1 in Section 2, we will use the corresponding result for non-integrable real functions which are locally integrable almost everywhere on the unit circle, which was given by Theorem 2 in Section 3. Since that result depends also on the non-integrable hard singularities of the real functions having finite degrees of hardness, we will also show that the hypothesis that the functions have this property implies that the corresponding functions have the same property as well. In order to do this we will use the technique of piecewise integration which was introduced and employed in [#!CAoRFIV!#], where it played a crucial role.
We will start by showing the following preliminary fact about a real function defined from a real function which is locally integrable almost everywhere on , and which has non-integrable hard singularities at the finite set of points , for .
Since is integrable on we have that
(37) |
exists and is finite. If we now consider the integral
(38) |
and transform variables from to , recalling that , we get
The absolute value of the derivative shown exists and is finite on , given that
(40) |
where is analytic on and therefore differentiable there. We also have that is integrable on . It follows that, since the integrand in the right-hand side of Equation (39) is the product of a limited real function with an integrable real function, and therefore is itself an integrable real function, the integral in Equation (39) exists and is finite, thus implying that is integrable on the closed interval . This establishes Lemma 3.
As an immediate consequence of this preliminary result, under the conditions that we have here, the hypothesis that is locally integrable almost everywhere on , with the exclusion of the finite set of points , implies that is locally integrable almost everywhere on , with the exclusion of the corresponding finite set of points .
We must now discuss the issue of the degrees of hardness of the non-integrable hard singularities of the function on . Since by hypothesis has non-integrable hard singularities at the points , it clearly follows that also has hard singularities at the corresponding points , which may be non-integrable ones. In order to discuss their degrees of hardness we will use the technique of piecewise integration, that is, we will consider sectional integrals of on closed intervals contained within a neighborhood of the point where it has a single isolated hard singularity. Let us show the following preliminary fact about a real function defined from a real function which is locally integrable almost everywhere on , and which has non-integrable hard singularities with finite degrees of hardness at the finite set of points .
Since the real functions must diverge to infinity at non-integrable hard singular points, the fact that has an isolated hard singularity on is immediate. Since these singularities are all isolated from each other, there is on a neighborhood of the point within which there are no other non-integrable hard singularities of . Since the conformal mapping is continuous, it follows that there is on a neighborhood of the corresponding point within which there are no other hard singularities of . Given that the point corresponds to the angle , let the closed interval contain the point and be contained in this neighborhood, so that we have
(41) |
where the sole hard singularity of which is contained within this interval is the one at the point . Let us now consider a pair of closed intervals contained within this neighborhood, one to the left and one to the right of the point , so that we have
(42) |
where and are two sufficiently small positive real numbers, so that we also have
(43) |
Let us now consider sectional primitives of the real function on these two intervals. Since the singularity of at may not be integrable, we cannot integrate across the singularity, but we may integrate within these two lateral closed intervals, thus defining two sectional primitives of , one to the left and another one to the right of ,
where is the notation for a primitive of with respect to , where and are two arbitrary reference points, one within each of the two lateral closed intervals, and where we have
If we change variables from to on the two sectional integrals in Equation (44), we get
where and are the reference points on corresponding respectively to and . Since the derivative which appears in these integrals is finite everywhere on , and therefore limited, due to the fact that the inverse conformal transformation is analytic and hence differentiable on the curve , it follows that there are two real numbers and such that
(46) |
everywhere on . Note that, since the conformal transformation, besides being continuous and differentiable, is also invertible on , the derivative above cannot change sign and thus cannot be zero. Therefore, the two bounds and may be chose to have the same sign. By exchanging the derivative by these extreme values we can obtain upper and lower bounds for the sectional integrals, and therefore we get for the sectional primitives in Equation (46),
We now recognize the integrals that appear in these expressions as the sectional primitives of the function , so that we get
These expressions are true so long as and , as well as the corresponding quantities and on , are not zero, but since the singularities at are non-integrable hard ones, we cannot take the limit in which these quantities tend to zero. Note that, since and have the same sign, if the sectional primitives of the function diverge to infinity in this limit, then so do the sectional primitives of the function . Therefore, we may conclude that the hard singularity of is also a non-integrable one. Since we may take further sectional integrals of these expressions, without affecting the inequalities, it is immediately apparent that, after a total of successive piecewise integrations, we get for the sectional primitives
Since the non-integrable hard singularity of at the point which corresponds to has a finite degree of hardness, according to the definition of the degrees of hardness, which was given in [#!CAoRFI!#] and discussed in detail for the case of real functions in [#!CAoRFIV!#], there is a value of such that the limit in which and can be taken for the sectional primitives and , thus implying that the piecewise primitive of is an integrable real function on the whole interval that corresponds to , with a borderline hard singularity, with degree of hardness zero, at the point . It follows from the inequalities, therefore, that the corresponding limit in which and can be taken for the functions and , thus implying that the piecewise primitive of is also an integrable real function on the whole interval , with a borderline hard singularity, with degree of hardness zero, at . Therefore, the non-integrable hard singularity of at has a finite degree of hardness, to wit the same degree of hardness of the corresponding non-integrable hard singularity of at . This establishes Lemma 4.
We have therefore established that, so long as is locally integrable almost everywhere on , and so long as its non-integrable hard singularities have finite degrees of hardness, these same two facts are true for on . Since we thus see that the necessary properties of the real functions are preserved by the conformal transformation, we are therefore in a position to use the result of Theorem 2 in Section 3 in order to extend the existence theorem of the Dirichlet problem to non-integrable real functions which are, however, integrable almost everywhere on , still for the case of differentiable curves.
In this section, using once again the results from the previous sections, we will establish the following theorem.
Similarly to what was done in the two previous cases, in Sections 5 and 6, the proof consists of using the conformal transformation between the closed unit disk and the union of the curve with its interior, which according to the analysis in Section 4 always exists, to map the given boundary condition on onto a corresponding boundary condition on the unit circle, then using the proof of existence established before by Theorem 2 in Section 3 for the closed unit disk to establish the existence of the solution of the corresponding Dirichlet problem on that disk, and finally using once more the conformal transformation to map the resulting solution back to and its interior, thus obtaining the solution of the original Dirichlet problem. The list of conditions on the real functions is now the following.
The rest of the proof is identical to that of the two previous cases, in Sections 5 and 6. Therefore, once again we may conclude that, due to the existence theorem of the Dirichlet problem on the unit disk of the plane , which in this case was established by Theorem 2 in Section 3, we know that there is an inner analytic function such that its real part is harmonic within the open unit disk and satisfies almost everywhere at the boundary . Just as before, we get on the plane the complex function which is analytic in the interior of the curve . Therefore, the real part of is harmonic and thus satisfies
(47) |
in the interior of , while we also have that
(48) |
almost everywhere on . This establishes the existence, by construction, of the solution of the Dirichlet problem on the plane, under our current hypotheses. This completes the proof of Theorem 5.
In this way we have generalized the proof of existence of the Dirichlet problem from the unit circle to all differentiable simple closed curves with finite total lengths on the plane, for boundary conditions given by non-integrable real functions which are locally integrable almost everywhere and have at most a finite set of hard singular points.