A REPRESENTATION FORMULA FOR RADIALLY WEIGHTED BIHARMONIC FUNCTIONS IN THE UNIT DISC

Abstract Let w : D → (0,∞) be a radial continuous weight function. We consider the weighted biharmonic equation ∆w∆u = 0 in D with Dirichlet boundary conditions u=f0 and ∂nu=f1 on T=∂D. Under some extra conditions on the weight function w, we establish existence and uniqueness of a distributional solution u of this biharmonic Dirichlet problem. Furthermore, we give a representation formula for the solution u. The key to our analysis is a series representation of Almansi type.


Introduction
Denote by D the unit disc and let ∆ = ∂ 2 /∂x 2 + ∂ 2 /∂y 2 be the Laplacian in the complex plane.Let w : D → (0, ∞) be a radial continuous weight function.In this paper we study the weighted biharmonic equation (0.1) ∆w −1 ∆u = 0 in D.
We interpret equation (0.1) in the distributional sense.Let u ∈ D (D); here D (Ω) denotes the space of distributions in Ω.We call u a solution of (0.1) if ∆u = wh in D (D) for some harmonic function h in D. We shall see that every solution u of (0.1) is of type C 2 (D) (see Remark 1.2).A solution u of (0.1) is sometimes called a w-biharmonic function.
Of particular interest is the weighted biharmonic Dirichlet problem: (0.2) Here T = ∂D is the unit circle, ∂ n denotes differentiation in the inward normal direction and the boundary datas f j (j = 0, 1) are distributions on T.
Let us make a few comments on the interpretation of (0.2).The first equation in (0.2) is interpreted in the above distributional sense of (0.1).For 0 ≤ r < 1 we consider the function u r defined by (0.3) u r (e iθ ) = u(re iθ ), e iθ ∈ T.
The middle boundary condition in (0.2) is interpreted that lim r→1 u r = f 0 in D (T); here D (T) denotes the space of distributions on T. Similarly, we interpret the last boundary condition in (0.2) to mean that where f 0 is as above.The purpose of this paper is to study the questions of existence, uniqueness, and representation of distributional solutions of the Dirichlet problem (0.2). Let us now describe our results.
Assume that the radial weight function w : D → (0, ∞) is area integrable, that is, 1 0 w(t) dt < ∞.We can then show uniqueness of a distributional solution of the Dirichlet problem (0.2) (see Theorem 2.1).Assume also that the moment condition (0.4) (c, N > 0) is satisfied.We can then show existence of a distributional solution u of the Dirichlet problem (0.2) for arbitrary distributional boundary data f j ∈ D (T) (j = 0, 1).Furthermore, we show that the solution u of the Dirichlet problem (0.2) admits the representation (0.5) u(z) = (F w,r * f 0 )(e iθ ) + (H w,r * f 1 )(e iθ ), z = re iθ ∈ D, in terms of two functions F w and H w in D; here F w,r = (F w ) r and H w,r = (H w ) r as in (0.3), and by * we denote convolution (see Theorem 3.2).
The kernel functions F w and H w appearing in (0.5) are given by the formulas r |k| e ikθ w(t) dt ds and r |k| e ikθ w(t) dt ds for z = re iθ ∈ D (see Proposition 4.1).We also mention that the two functions F w and H w solve the Dirichlet boundary value problems in the above distributional sense; here δ e iθ denotes the unit Dirac mass at e iθ ∈ T.
The moment condition (0.4) is satisfied provided the weight function w has enough mass near the boundary T (see the end of Section 3).In particular, this moment condition is satisfied by all the so-called standard weights w = w α defined by where α > −1.In the context of these weights the case α = 0 is often referred to as the (classical) unweighted case.The standard weight w = w 1 has attracted special attention in recent papers (see [7], [8], [15]).
The above representation formula (0.5) generalizes earlier work by Abkar and Hedenmalm [1] in the unweighted case w = w 0 to our weighted context.
We consider in some more detail the above standard weight w = w 1 .In this case we show that the corresponding kernels F 1 = F w1 and H 1 = H w1 are given by the formulas Using the first of these formulas we verify that the function F 1 is not positive in D.
The motivation for our study of (weighted) biharmonic boundary value problems of the above type comes from Bergman space theory where such techniques have shown to be an important tool (see [2], [3], [6]).A recent achievement in this direction of research is the proof of positivity of certain weighted biharmonic Green functions (see [5], [14]).
The key to our analysis of the Dirichlet problem (0.2) is a series expansion of w-biharmonic functions which generalizes the classical Almansi representation of biharmonic functions.This generalized Almansi type representation is then used to prove existence and uniqueness of solutions of the Dirichlet problem (0.2).
This paper is organized as follows.In Section 1 we study an Almansi type representation of w-biharmonic functions.In Section 2 we use this Almansi representation to show uniqueness of solutions of the Dirichlet problem (0.2).In Section 3 we consider the problem of representation of a w-biharmonic function in terms of the two associated kernels F w and H w .
In Section 4 we give some formulas for the functions F w and H w , and some related work is mentioned.
The author thanks Håkan Hedenmalm for useful discussions.

Almansi representation
Our first task is to generalize the well-known Almansi representation of biharmonic functions.For this we need some preparation.Let w : D → (0, ∞) be a radial continuous weight function.We consider the associated functions w k defined by We shall also need the functions v k defined by v k (r) = r k w k (r).
In the following lemmas we establish some basic properties of these functions.
Lemma 1.1.Let the w k 's be as above.Then the function w k is of type C 2 [0, 1) and satisfies the ordinary differential equation initial value problem w k (r) + (2k + 1)w k (r)/r = w(r), 0 < r < 1, Furthermore, we have the estimates w/(4k) and 0 ≤ w k (r) ≤ r sup for 0 < r < 1 and k ≥ 1.In particular, the function w k radially extends to a function in C 2 (D).
Proof: We assume that k ≥ 1.The case k = 0 can be handled similarly.The first estimate of w k is immediate from the definition.By a change of variables we have that By differentiation we see that Similarly we check that (1.2) holds also for k = 0.An easy estimation now yields the estimate for w k .By another differentiation we see that w k satisfies (1.1).Since the compatibility condition w k (0) = 0 is satisfied, the function w k radially extends to a function in C 2 (D).
For easy reference later we shall derive one more estimate of the w k 's.Note that by (1.2) the function w k is increasing.We have that Thus the w k 's are uniformly bounded if 1 0 (1 − t)w(t) dt < ∞.We now consider the v k 's.
Lemma 1.2.Let v k (r) = w k (r)r k for 0 ≤ r < 1 and k ≥ 0. Then the function v k satisfies the ordinary differential equation Proof: This is a straightforward verification using Lemma 1.1.
We shall consider series expansions involving the functions w k .
Lemma 1.3.Let {c k } ∞ k=−∞ be a sequence of complex numbers such that lim sup Then the series We also have that Proof: Recall that by Lemma 1.1 the radial function and thus the same is true for the function z → w |k| (r)r |k| e ikθ .Using the estimates in Lemma 1.1 it is straightforward to check that for |α| ≤ 2; here ∂ = (∂/∂x, ∂/∂y) and standard multi-index notation is used.The convergence of the series expansion of u now follows.
We now compute ∆u.Recall that in polar coordinates (r, θ) the Laplacian takes the form which concludes the proof of the lemma.
Remark 1.1.We remark that in the context of the above lemma the term by term differentiated series converges absolutely and uniformly in every smaller disc tD (0 < t < 1).
The following theorem generalizes the well-known Almansi representation of biharmonic functions.
Theorem 1.1.Let w : D → (0, ∞) be a radial continuous weight function, and let the w k 's be as above.Then a function u in D satisfies the weighted biharmonic equation if and only if u has a convergent series expansion of the form Furthermore, the series (1.5) is convergent in the topology of C 2 (D).
Proof: By Lemma 1.3, we know that the series (1.5) converges in C 2 (D) and that u so defined satisfies (1.4).
Let now u be a solution of (1.4).We proceed to show that u has a series expansion of the form (1.5).We know that ∆u = wh in D (D), where Remark 1.2.Let the weight w be as above.Note that by Theorem 1.1 every w-biharmonic function is in C 2 (D).
We inform the reader that a different generalization of the Almansi representation to weights of the form w(z) = |z| 2α for z ∈ D, where α > −1, has previously been given by Hedenmalm in [4, Lemma 3.1].

Uniqueness of the Dirichlet problem
The purpose of this section is to show uniqueness of solutions of the Dirichlet problem (0.2).We shall need some asymptotic properties of the functions w k .Proposition 2.1.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable in the sense that 1 0 w(t) dt < ∞.Let the functions w k (k ≥ 0) be as above and consider the sequence of moments {A k } ∞ k=−1 defined by Then as r → 1 we have the asymptotic expansions: Proof: We first compute the expansion of w 0 .By a change of variables we see that We first consider the second integral in (2.1).A computation shows that Substituting this expansion into (2.1)we see that which is the above expansion of w 0 .We now turn to the expansion of w k .Similarly as above we see by a change of variables that We consider the second integral in (2.2).By computation we have that Substituting this expansion into (2.2) we see that which is the expansion of w k .
Using the result of Proposition 2.1 we can compute asymptotic formulas for the coefficients in (1.5).We obtain that We can now give a general uniqueness result for equation (0.1).
Theorem 2.1.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable in the sense that 1 0 w(t) dt < ∞.Let u be a solution of (0.1) and assume that lim We remark that Theorem 2.1 gives uniqueness of solutions of the Dirichlet problem (0.2).In fact, the assumption on u in the theorem can be phrased as u = ∂ n u = 0 on T in the distributional sense.
In the unweighted case w = w 0 the result of Theorem 2.1 is known (see [11,Proposition 1.1]).

A solution of the Dirichlet problem
We now turn to the problem of representation of w-biharmonic functions in terms of boundary values.First we discuss existence of distributional boundary values.Proposition 3.1.Let w : D → (0, ∞) be a radial continuous weight function such that 1 0 (1 − t)w(t) dt < ∞.Let u be given by (1.5).Then u satisfies for some positive constants C and N the growth conditions if and only if the coefficients a k and b k are of polynomial growth, that is, for some (different) constants C and N .
Proof: We first recall that the w k 's are uniformly bounded (see the paragraph after Lemma 1.1).
Assume now that u satisfies (3.1).Recall that ∆u = wh, where h = b k r |k| e ikθ (see Lemma 1.3).By the second estimate in (3.1) we see that the harmonic function h is of tempered growth in D which means that the b k 's are of polynomial growth.Since the w k 's are uniformly bounded, the sum u 1 = b k w |k| (r)r |k| e ikθ is easily seen to be of tempered growth in D. By the first estimate in (3.1) we see that the harmonic function h 1 = a k r |k| e ikθ is of tempered growth in D which means that the a k 's also are of polynomial growth.
Assume now that the coefficients a k and b k are of polynomial growth.Using the uniform boundedness of the w k 's it is easy to see that u is of tempered growth in D. Since ∆u = wh, where h = b k r |k| e ikθ , the second growth condition in (3.1) is also satisfied.
We now compute the boundary values.Proposition 3.2.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable in the sense that 1 0 w(t) dt < ∞.Let u be a w-biharmonic function satisfying (3.1).Then there exists the limits Furthermore, the distributions f 0 and f 1 are given by the Fourier series: respectively; here a k and b k are the coefficients in (1.5) and the A k 's are as in Proposition 2.1.
Proof: Consider the series expansion (1.5).By Proposition 3.1 we know that the coefficients a k and b k are of polynomial growth.We compute the boundary limit of u.Let ϕ ∈ C ∞ (T).We have that here •, • denotes the usual distributional pairing and φ(k) = T ϕ(e iθ )e −ikθ dθ/2π is the k-th Fourier coefficient of ϕ.By the Lebesgue dominated convergence theorem and Proposition 2.1 we have that We now compute the normal derivative of u.Note that there is an estimate where C is an absolute constant; here w k (1) = (A 0 − A k )/(2k).Indeed, by (1.2) we have that 0 ≤ w k (r) ≤ 1 0 w(t) dt = C and an integration yields (3.2).The estimate (3.2) allows us to use Lebesgue's theorem to compute the limit lim r→1 (u r − f 0 )/(1 − r), ϕ .By Proposition 2.1 we have that lim which concludes the proof.
Let w : D → (0, ∞) be a radial continuous area integrable weight function.We shall consider the functions F w and H w defined by the formulas for z = re iθ ∈ D. Note that since lim k→∞ A 1/k k = 1 the functions F w and H w are well-defined w-biharmonic functions in D (see Theorem 1.1).We also write F w,r (e iθ ) = F w (re iθ ) for e iθ ∈ T and 0 ≤ r < 1, and similarly for H w .
We can represent a fairly general w-biharmonic function in terms of the above kernels F w and H w .Theorem 3.1.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable.Let u be a w-biharmonic function satisfying (3.1).Then the function u admits the representation where the f j 's are as in Proposition 3.2.
Proof: We compute the Fourier coefficients.For k = 0 we have that Similarly we see that By uniqueness of Fourier coefficients we now conclude that u r = F w,r * f 0 + H w,r * f 1 , which yields (3.3).
Remark 3.1.Let f 0 , f 1 ∈ D (T) and let u be defined by (3.3).Then by Theorem 1.1 the function u is w-biharmonic in D.
Let w : D → (0, ∞) be a radial continuous weight function which is area integrable.Assume also that the moments {A k } of w are such that for some positive constants c and N .Then the coefficients of F w and H w are of polynomial growth.Propositions 3.1 and 3.2 now apply to show that the functions F w and H w solve the boundary value problems: in the distributional sense.Here δ e iθ denotes the unit Dirac mass at e iθ ∈ T. The function H w is sometimes called the harmonic compensator (see [5]).
We can now solve the Dirichlet problem (0.2) for arbitrary distributional boundary datas.Theorem 3.2.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable and such that for some positive constants c and N .Let f 0 , f 1 ∈ D (T) and let u be defined by (3.3).Then the function u solves the Dirichlet boundary value problem Proof: The function u defined by (3.3) is always w-biharmonic in D (see Remark 3.1).
We now turn to the boundary values of u.Let us first consider the function u 1 defined by u 1 (z) = (H w,r * f 1 )(e iθ ) for z = re iθ ∈ D. By Propositions 3.1 and 3.2 we know that Convolving with f 1 ∈ D (T) we see that Thus u 1 = 0 and ∂ n u 1 = f 1 on T in the distributional sense.
Similarly as above we see that the function u 0 defined by u 0 (z) = (F w,r * f 0 )(e iθ ) for z = re iθ ∈ D is such that u 0 = f 0 and ∂ n u 0 = 0 on T in the distributional sense.We now conclude that the function u = u 0 + u 1 is such that u = f 0 and ∂ n u = f 1 on T in the distributional sense.
The moment condition (3.4) is satisfied provided the weight function w : D → (0, ∞) has enough mass near the boundary T. Let us first consider the so-called standard weights w = w α defined by where α > −1.In this case it is known that where Γ(x) = ∞ 0 t x−1 e −t dt for x > 0 is the gamma function (see [13,Theorem 8.20]; see also [6, Section 1.1]).A computation using Stirling's formula (see [13,Section 8.22]) shows that an estimate of the form (3.4) holds in this case.
Let us now consider a general area integrable weight function w : D → (0, ∞).By partial integration we see that

Formulas for the functions F w and H w
We now give some formulas for the functions F w and H w .Proposition 4.1.Let w : D → (0, ∞) be a radial continuous weight function which is area integrable.Then the functions F w and H w admit the representations r |k| e ikθ w(t) dt ds and r |k| e ikθ w(t) dt ds for z = re iθ ∈ D, respectively, where A k = 1 0 t 2k+1 w(t) dt for k ≥ 0. Proof: We first derive the formula for H w .Let for k = 0.By Lemma 1.1 the function W |k| satisfies the differential equation and W |k| (0) = 0.By Proposition 2.1 we also have that W |k| (1) = 0.A standard argument shows that W |k| has the representation In fact, an integration of (1.2) yields the above formula.A computation now shows that r |k| e ikθ w(t) dt ds, which is the formula for H w .We now derive the formula for F w .Arguing as above we first see that A computation now shows that r |k| e ikθ w(t) dt ds, where The formulas in Proposition 4.1 make evident that Let us denote by u the right-hand side in (4.1).Using properties of H w we can easily verify that u = δ 1 and ∂ n u = 0 on T in the distributional sense.A uniqueness argument (see Theorem 2.1) then gives that u = F w which proves (4.1).
For easy reference we also record the following series expansions of the functions F w and H w : The formula for H w in Proposition 4.1 has appeared in a different context in [14,Formula (9)] (see also [5,Formula (8.5)]).Here we also want to mention that the function has the interpretation as the kernel function for the Bergman space of square area integrable with respect to the weight w harmonic functions in D; here f 2 w = D |f (z)| 2 w(z) dA(z), where dA is the normalized Lebesgue area measure in D.
Let us mention two cases where explicit formulas for the kernels F w and H w are available.We consider first the unweighted situation.In this case the functions F 0 = F w0 and H 0 = H w0 have the explicit expressions 2  and Here (z) denotes the real part of the complex number z.The function F 0 can also be written (see [1, Formula (0-5)]).Hedenmalm has pointed out to us that the equality of these two expressions for the function F 0 follows by a straightforward computation using the identity 2 (z) = 1+|z| 2 −|1−z| 2 (see formula (4.4) below).We mention in passing that for the harmonic compensator function H 0 a sharp monotonicity estimate is known (see [10]).
We next consider the weight w = w 1 .
Proposition 4.2.The functions F 1 = F w1 and H 1 = H w1 have the explicit expressions (1 − |z| 2 ) 5 |1 − z| 6  and Furthermore, the function F 1 assumes negative as well as positive values in every neighborhood of z = 1.
Proof: A straightforward computation shows that and that By Proposition 4.1 we now have that where (z) denotes the real part of the complex number z.Now using We shall next derive a similar formula for the function H 1 .By Proposition 4.1 we again have We shall next rewrite the real part expressions in (4.2) and (4.3).In fact, we have that To prove (4.4) we compute, using the formula 2 Let us now turn to the proof of (4.5).In analogy with the formula 2 A computation now shows that We now consider the function F 1 in some more detail.First, for z = x ∈ R, we have that F 1 (x) = (1 + x) 3 /(1 − x) > 0 for −1 < x < 1.We now demonstrate that F 1 also assumes negative values.For z on the level set (1 − |z| 2 )/|1 − z| 2 = c > 0 we have that which is clearly negative for z close to 1.
We remark that the formula for H 1 in (4.3) is known (see [8, Proof of Lemma 2.2]).
The last assertion of Proposition 4.2 can be interpreted saying that the full biharmonic maximum principle suggested in [5] does not hold true for the particular weight w = w 1 .Indeed, the above function F 1 is such that F 1 = δ 1 ≥ 0 and ∂ n F 1 = 0 on T while F 1 is not of constant sign in D. This fact has been pointed out earlier by Hedenmalm (private discussion); we have merely filled in the details.
As a direction for future research it is of interest to know detailed regularizing properties of the Dirichlet problem (0.2) discussed in the introduction.We plan to return to this topic in a forthcoming paper [12].In this context we want to mention the papers by Shimorin [14] and Hedenmalm, Jakobsson and Shimorin [5] where it is shown that the harmonic compensator function H w is nonnegative in D, that is, H w (z) ≥ 0 for z ∈ D, if the weight function w : D → (0, ∞) is logarithmically subharmonic in D.Here the function w is said to be logarithmically subharmonic in D if the function z → log w(z) is subharmonic there.mathematical physics and function theory, J. Math.Sci.(New York) 92(6) (1998), 4404-4411.
r→1 u r /(1 − r) = 0 in D (T), where u r (e iθ ) = u(re iθ ) for 0 ≤ r < 1 and e iθ ∈ T. Then u(z) = 0 for all z ∈ D. Proof: By assumption we have that ûr (k) = 1 2π T u r (e iθ )e −ikθ dθ = o(1 − r) as r → 1.By Theorem 1.1 we have that ûr (k) = a k + b k w |k| (r) r |k| , where a k , b k ∈ C. By the above asymptotic formulas for these quantities we conclude that