The Voronovskaya type theorem for Poisson integrals of functions of two variables

Grażyna Krech

The Voronovskaya type theorem for Poisson integrals of functions of two variables

Abstract. The aim of this paper is the study the Voronovskaya type theorem for Poisson integrals of functions of two variables for Hermite and Laguerre expansions.

We also present some boundary value problems related to these integrals.

2000 Mathematics Subject Classification: 41A35, 41A36.

Key words and phrases: Poisson integral, Hermite expansions, Laguerre expansions, Voronovskaya theorem, boundary value problems.

1. Introduction. Muckenhoupt in [3], Toczek and Wachnicki in [4] considered the Poisson integrals for Hermite and Laguerre expansions in the space L ^p ( R; e ^−z

) and L ^p ( R + ; z ^α e ^−z ), respectively (1 ¬ p ¬ ∞, α > −1, R + = [0, ∞)). Some appro- ximation properties of the Poisson integrals for Hermite and Laguerre expansions in two dimensional space were studied in [1]. The aim of this work is to obtain the Voronovskaya type theorem for these operators in two dimensional space.

Let us consider the Poisson integral U of a function f ∈ L ^p ( R ⁺ ; z ^α e ^−z ), 1 ¬ p ¬ ∞, α > −1, defined by

U (f ; r, y 1 , y 2 ) = Z ∞

0

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 ) f(z 1 , z 2 ) (z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 ,

where

K(r, y, z) = X ∞ n=0

r ⁿ n!

Γ(n + α + 1) L ^α _n (y)L ^α _n (z)

= (ryz) ⁻

1 − r exp

 −r(y + z) 1 − r

 I α

2(ryz)

1 − r

!

, 0 < r < 1,

L ^α _n is the nth Laguerre polynomial and I α is the modified Bessel function (see [2]) I α (s) =

X ∞ n=0

s ^α+2n

2 ^α+2n n!Γ(α + n + 1) . The Poisson integral W is defined by

W (f ; r, y 1 , y 2 ) = Z ∞

−∞

Z ∞

−∞

P (r, y 1 , z 1 )P (r, y 2 , z 2 )f(z 1 , z 2 )e ^−z

^−z

dz 1 dz 2 ,

where

P (r, x, z) = X ∞ n=0

r ⁿ H n (x)H n (z)

√ π 2 ⁿ n!

= 1

p π(1 − r ² ) exp

 −r ² x ² + 2rxz − r ² z ² 1 − r ²

 ,

H n is the nth Hermite polynomial and f ∈ L ^p 

R ² ; e ^−z

^−z



, 0 < r < 1.

The norm in L ^p X ² ; w(z 1 , z 2 )
of a function f is given by

kfk ^p =

 

 

 



Z _∞

−∞

Z ∞

−∞ |f(t ¹ , t 2 )| ^p w(t 1 , t 2 ) dt 1 dt 2



, 1 ¬ p < ∞, sup

(t

,t

)∈R ess |f(t

¹ , t 2 )|, p = ∞,

where X = R ⁺ and w(z 1 , z 2 ) = (z 1 z 2 ) ^α e ^−z

^−z

, or X = R and w(z ¹ , z 2 ) = e ^−z

^−z

, respectively.

In this section we shall give some properties of the above operators, which we shall apply to the proofs of the main theorems.

Let ϕ n,y (z) = (z − y) ⁿ , n ∈ N = {1, 2, ...}, y, z ∈ X, where X = R ⁺ or X = R, respectively.

Krech [1] obtained the following result.

Lemma 1.1 ([1]) For every (y 1 , y 2 ) ∈ R ² + it follows U (1; r, y 1 , y 2 ) = 1,

U (ϕ 1,y

; r, y 1 , y 2 ) = (1 − r)(1 + α − y ⁱ ), U (ϕ 2,y

; r, y 1 , y 2 ) = (1 − r) 

y _i ² (1 − r) + 2(α + 2)ry ⁱ − 2(α + 1)y ⁱ + (α + 2)(α + 1)(1 − r)} ,

U (ϕ 4,y

; r, y 1 , y 2 ) = (1 − r) ² 

y ⁴ _i (r − 1) ² − 4α(r − 1) ² y _i ³ − 4(4r ² − 5r + 1) + 6(α + 4)(α + 3)r ² y ² _i − 12(α + 3)(α + 2)ry ² i

+ 6(α + 2)(α + 1)y _i ² − 12(α + 3)(α + 2)(r − 1)y ⁱ

+ (α + 4)(α + 3)(α + 2)(α + 1)(r − 1) ²

for 0 < r < 1, i = 1, 2.

Lemma 1.2 ([1]) For every (y 1 , y 2 ) ∈ R ² it follows W (1; r, y 1 , y 2 ) = 1,

W (ϕ 1,y

; r, y 1 , y 2 ) = −y ⁱ (1 − r), W (ϕ 2,y

; r, y 1 , y 2 ) = (1 − r)



y _i ² (1 − r) + 1 2 (r + 1)

 , W (ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2 ) = (1 − r) ² 

(r − 1) ² (y ₁ ⁴ + y ₂ ⁴ )

−3(r ² − 1)(y 1 ² + y ₂ ² ) + 3

2 (r + 1) ²



for 0 < r < 1, i = 1, 2.

Applying Lemma 1.1 and Lemma 1.2 it is easy to prove the following two lemmas.

Lemma 1.3 Let (y 1 , y 2 ) ∈ R ² + be a fixed point. Then

r lim →1

1

1 − r U (ϕ 1,y

; r, y 1 , y 2 ) = 1 + α − y ⁱ ,

r→1 lim

1

1 − r U (ϕ 1,y

ϕ 1,y

; r, y ₁ , y 2 ) = 0,

r→1 lim

1

1 − r U (ϕ 2,y

; r, y 1 , y 2 ) = 2y i ,

r lim →1

1

(1 − r) ² U (ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2 ) = 12(y ² ₁ + y ₂ ² ) (1)

for i = 1, 2.

Lemma 1.4 Let (y 1 , y 2 ) ∈ R ² be a fixed point. Then

r lim →1

1

1 − r W (ϕ 1,y

; r, y 1 , y 2 ) = −y ⁱ ,

r lim →1

1

1 − r W (ϕ 1,y

φ 1,y

; r, y 1 , y 2 ) = 0,

r→1 lim

1

1 − r W (ϕ 2,y

; r, y ₁ , y 2 ) = 1,

r→1 lim

1

(1 − r) ² W (ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2 ) = 6

for i = 1, 2.

Lemma 1.5 Let a, b, δ > 0 and α ­ − ¹ 2 . If q ­ 1, then

r lim →1

Z ∞

y

+δ

Z ∞

0

(K(r, y 1 , z 1 )K(r, y 2 , z 2 )) ^q (z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 = 0

uniformly with respect to (y 1 , y 2 ), on the set [0, a] × [0, b].

Proof Let q ­ 1 and δ > 0. Observe that Z ∞

y

+δ

Z ∞

0

(K(r, y 1 , z 1 )K(r, y 2 , z 2 )) ^q (z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2

= Z ∞

y

+δ

(K(r, y 2 , z 2 )) ^q z ₂ ^α e ^−z



 Z ∞

0

(K(r, y 1 , z 1 )) ^q z ₁ ^α e ^−z

dz 1



 dz 2

for 0 < r < 1, (y ₁ , y 2 ) ∈ R ² + . Using the inequality (see [5])

|I ^α (z)| ¬ 1 Γ(α + 1)

 z 2

 α

e ^z , z ­ 0, α ­ − 1 2 , we get

K(r, y 1 , z 1 ) ¬ M ¹ (1 − r) ^−(α+1) e ^y

exp − r 1 − r

 √ z 1 −

√ y 1

√ r

 ² ! (2) ,

where M 1 > 0, α ­ − ¹ 2 , 0 < r < 1. Let 0 ¬ y ¹ ¬ a. Then Z ∞

0

(K(r, y 1 , z 1 )) ^q z ₁ ^α e ^−z

dz 1

¬ M ² (1 − r) ^−q(α+1) e ^qa Z ∞

0

exp − qr 1 − r

 √ z 1 −

√ y 1

√ r

 2 !

z ₁ ^α e ^−z

dz 1

¬ M ³ (1 − r) ^−q(α+1) Z ∞

0

z ₁ ^α e ^−z

dz 1 = M 3 Γ(α + 1) (1 − r) ^−q(α+1) , where M 2 , M 3 denote some positive numbers. Observe that

z ₂ ^α+

e ^−z

¬

 α + 1

2

 α+

e ^−α−

for α > − 1 2 and

Z ∞

y

+δ

(K(r, y 2 , z 2 )) ^q z ^α ₂ e ^−z

dz 2

¬ M 4 (1 − r) ^−q(α+1) Z ∞

y

+δ

exp − qr 1 − r

 √ z 2 −

√ y 2

√ r

 2 !

z ₂ ⁻

dz 2 ,

where M 4 > 0, α ­ − ¹ 2 . Making the substitution

t =

 qr 1 − r





√ z 2 −

√ y 2

√ r



we have Z ∞

y

+δ

exp − qr 1 − r

 √ z 2 −

√ y 2

√ r

 ² !

z ₂ ⁻

dz 2 = 2

 1 − r qr



Z ^∞

d

e ^−t

dt,

where

d 1 =

 qr 1 − r



p

y 2 + δ − √ y 2

√ r

 . If 0 ¬ y 2 ¬ b, then

d 1 ­

 qr 1 − r





 b(r − 1) + rδ

√ r p

r(b + δ) + √ b 



 for 0 < r < 1

and b(r − 1) + rδ ­ ¹ 2 δ for 1 − γ < r < 1, where γ is a sufficiently small positive number. Therefore

d 1 ­

 qr 1 − r



· 1

√ r p

r(b + δ) + √ b  · δ

2 ­ M 5 (1 − r) ⁻

= d ₂ ,

Z ∞

d

e ^−t

dt ¬ Z ∞

d

e ^−t

dt

for y ₂ ¬ b, 1 − γ < r < 1, M ⁵ > 0. Because Z ∞

y

+δ

Z ∞

0

(K(r, y ₁ , z 1 )K(r, y ₂ , z 2 )) ^q (z ₁ z 2 ) ^α e ^−z

^−z

dz 1 dz 2

¬ M ⁶ (1 − r) ^−q(α+1) (1 − r) ^−q(α+1) (1 − r)

Z ∞

d

e ^−t

dt,

r→1 lim

M 5 (1 − r) ⁻

= ∞ and lim

u →∞ u ^s Z ∞

u

e ^−t

dt = 0 for s ∈ R, so

r lim →1

Z ∞

y

+δ

Z ∞

0

(K(r, y 1 , z 1 )K(r, y 2 , z 2 )) ^q (z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 = 0

uniformly on the set [0, a] × [0, b]. 

In a similar fashion we can prove following lemmas.

Lemma 1.6 Let a, b, δ > 0, α ­ − ¹ 2 and δ < b. If q ­ 1, then

r→1 lim

y Z

−δ

0

Z ∞

0

(K(r, y ₁ , z 1 )K(r, y ₂ , z 2 )) ^q (z ₁ z 2 ) ^α e ^−z

^−z

dz 1 dz 2 = 0

uniformly with respect to (y 1 , y 2 ), on the set [0, a] × [b, ∞).

Lemma 1.7 Let a, b, δ > 0. If q ­ 1, then

r→1 lim

Z

X

Z

R (P (r, y 1 , z 1 )P (r, y 2 , z 2 )) ^q e ^−z

^−z

dz 1 dz 2 = 0 uniformly with respect to (y 1 , y 2 ), on the set [−a, a] × [−b, b], where X = (y 2 + δ, ∞) or X = (−∞, y ² − δ).

Now, we present the proof of the following lemma.

Lemma 1.8 Let a, b, δ > 0 and α ­ − ¹ 2 . Then

r→1 lim

sup

z

∈R

z

∈(y

+δ,∞)

K(r, y 1 , z 1 ) K(r, y 2 , z 2 ) = 0

uniformly with respect to (y 1 , y 2 ), on the set [0, a] × [0, b].

Proof Let 0 ¬ y ¹ ¬ a, 0 ¬ y ² ¬ b and δ > 0. From (2) we get K(r, y 1 , z 1 ) K(r, y 2 , z 2 )

¬ M ¹ (1 − r) ^−2(α+1) e ^y

^+y

exp − r 1 − r

 √ z 1 −

√ y 1

√ r

 2 !

× exp − r 1 − r

 √ z 2 −

√ y 2

√ r

 2 !

¬ M 1 (1 − r) ^−2(α+1) e ^a+b exp − r 1 − r

 √ z 2 −

√ y 2

√ r

 2 ! ,

where M ₁ > 0, α ­ − ¹ 2 , 0 < r < 1, y ₁ ¬ a, y 2 ¬ b. Observe that if z 2 ­ y 2 + δ, then

√ z 2 −

√ y 2

√ r ­ y 2 (r − 1) + rδ

√ r √

ry 2 + rδ + √y ₂
­ b(r − 1) + rδ

√ r √

rb + rδ + √ b 

­

1 2 δ

√ r √

rb + rδ + √

b  ­ M ²

for y 2 ¬ b, 1 − γ < r < 1, where γ is a sufficiently small positive number, and M 2 > 0. Therefore

 √ z 2 −

√ y 2

√ r

 2

­ M 2 ²

and

K(r, y 1 , z 1 ) K(r, y ₂ , z 2 ) ¬ M 3 (1 − r) ^−2(α+1) exp



− M 4

1 − r



for y 1 ¬ a, y ² ¬ b, 1 − γ < r < 1, α ­ − ¹ 2 , where M 3 , M 4 are some positive numbers. Because

r lim →1

M 3

(1 − r) ^2(α+1) exp



− M 4

1 − r



= 0, so

r→1 lim

sup

z

∈R

z

∈(y

+δ,∞)

K(r, y 1 , z 1 ) K(r, y ₂ , z 2 ) = 0

uniformly on the set [0, a] × [0, b]. 

In a similar way we get the following lemmas.

Lemma 1.9 Let a, b, δ > 0, α ­ − ¹ 2 and δ < b. Then

r lim →1

sup

z

∈R

z

∈[0,y

−δ)

K(r, y 1 , z 1 ) K(r, y 2 , z 2 ) = 0

uniformly with respect to (y 1 , y 2 ), on the set [0, a] × [b, ∞).

Lemma 1.10 Let a, b, δ > 0. Then

r lim →1

sup

z

∈R z

∈X

P (r, y 1 , z 1 ) P (r, y 2 , z 2 ) = 0

uniformly with respect to (y 1 , y 2 ), on the set [−a, a] × [−b, b], where X = (y ² + δ, ∞) or X = (−∞, y ² − δ).

2. The Voronovskaya type theorem. In this section we give the asymptotic formulas for operator U and W . To prove the main theorems we need the following lemmas.

Theorem 2.1 Let f = f 1 + f ₂ , where f 1 ∈ L ¹ R ² + ; (z ₁ z 2 ) ^α e ^−z

^−z

and f 2 ∈ L ^∞ R ² +

, α ­ − ¹ 2 . Let y ∈ R ² + . If f is continuous at y, then

(3) lim

(r,y)→(1

,y) U (f ; r, y) = f (y).

Proof Let y = (y ₁ , y ₂ ) be a fixed point in R ² + . Let ε > 0 and let M > 0 be a number such that

sup

(z

,z

)∈R

ess |f ³ (z 1 , z 2 ))| ¬ M,

f 3 (z ₁ , z 2 ) = f ₂ (z ₁ , z 2 ) − f(y 1 , y ₂ ) and kf 1 k 1 ¬ M.

Let y ₁ > 0 and y ₂ > 0. By the continuity of f at (y ₁ , y ₂ ) there exists δ > 0 such that

(4) |f(z ¹ , z 2 ) − f(y 1 , y ₂ )| < ε

5 for |z ¹ − y 1 | < δ and |z ² − y 2 | < δ.

Observe that

|U(f; r, y ¹ , y 2 ) − f(y 1 , y ₂ )|

¬ Z ∞

0

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )|f(z ¹ , z 2 ) − f(y 1 , y ₂ )|(z ¹ z 2 ) ^α e ^−z

^−z

dz 1 dz 2

=



  Z ∞

0 y Z

−δ

0

+

y Z

−δ

0

Z

|z

−y

|<δ

+ Z

|z

−y

|<δ

Z

|z

−y

|<δ

+ Z ∞

y

+δ

Z

|z

−y

|<δ

+ Z ∞

0

Z ∞

y

+δ



 

K(r, y 1 , z 1 ) K(r, y ₂ , z 2 )|f(z 1 , z 2 ) − f(y 1 , y ₂ )|(z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2

= J 1 + J 2 + J 3 + J 4 + J 5 . From (4) we get

J 3 ¬ ε 5

Z

|z

−y

|<δ

Z

|z

−y

|<δ

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )(z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2

¬ ε 5

Z ∞

0

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )(z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 = ε 5

for every (y ₁ , y 2 ) ∈ R ² + and 0 < r < 1.

If z ₂ > y ₂ + δ and |y 2 − y 2 | < ¹ ₂ δ, then z 2 > y 2 + ¹ ₂ δ. Therefore

J 4 ¬ Z ∞

y

+

δ

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )|f ¹ (z 1 , z 2 )|(z ¹ z 2 ) ^α e ^−z

^−z

dz 1 dz 2

+ M Z ∞

y

+

δ

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )(z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 .

Observe that Z ∞

y

+

δ

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )|f ¹ (z 1 , z 2 )|(z ¹ z 2 ) ^α e ^−z

^−z

dz 1 dz 2

¬ kf ¹ k ¹ sup

z

∈R

z

∈(y

+

δ, ∞)

K(r, y 1 , z 1 ) K(r, y 2 , z 2 ).

By Lemma 1.8 and Lemma 1.5 there exists η > 0 such that sup

z

∈R

z

∈(y

+

δ, ∞)

K(r, y 1 , z 1 ) K(r, y 2 , z 2 ) < ε 10M ,

Z ∞

y

+

δ

Z ∞

0

K(r, y 1 , z 1 ) K(r, y 2 , z 2 )(z 1 z 2 ) ^α e ^−z

^−z

dz 1 dz 2 < ε 10M

for 1 − η < r < 1 and |y ² − y 2 | < ¹ 2 δ. Therefore J 4 ¬ ^ε 5 for 1 − η < r < 1,

|y ² − y 2 | < ¹ 2 δ.

In a similar fashion we obtain J i ¬ ε

5 , i ∈ {1, 2, 5}

for 1 − η < r < 1, |y ¹ − y 1 | < ¹ 2 δ and |y ² − y 2 | < ¹ 2 δ. Thus

|U(f; r, y 1 , y 2 ) − f(y 1 , y ₂ )| ¬ J 1 + J ₂ + J ₃ + J ₄ + J ₅ < ε

for 1 − min(δ, η) < r < 1, |y 1 − y 1 | < ¹ ₂ δ and |y 2 − y 2 | < ¹ ₂ δ, so we have (3) for y ₁ > 0 and y ₂ > 0.

Analogously we can get (3) for (y ₁ , y ₂ ) = (0, 0) and in the case when exactly one of the coordinates of the point (y ₁ , y ₂ ) is equal to zero. _

From Theorem 2.1 we have

Corollary 2.2 If f = f 1 +f ₂ , where f 1 ∈ L ¹ R ² + ; (z ₁ z 2 ) ^α e ^−z

^−z

, f 2 ∈ L ^∞ R ² +
, α ­ − ¹ 2 and f ∈ C(R ² + ), then

r→1 lim

U (f ; r, y 1 , y 2 ) = f(y ₁ , y 2 ) uniformly on every compact subset in R ² + .

For W we obtain the similar result.

Theorem 2.3 Let f = f 1 + f 2 , where f 1 ∈ L ¹ 

R ² ; e ^−z

^−z



, f 2 ∈ L ^∞ R ²
. If f is continuous at the point y = (y 1 , y 2 ) ∈ R ² , then

(r,y)→(1 lim

,y) W (f ; r, y) = f (y), y = (y 1 , y 2 ).

Now, we can prove the Voronovskaya type theorem.

Theorem 2.4 Let f ∈ C(R ² + ) ∩ L ^p R ² + ; (z ₁ z 2 ) ^α e ^−z

^−z

, α ­ − ¹ 2 , 1 ¬ p ¬ ∞ and (y 1 , y 2 ) ∈ R ² + . If f is of the class C ¹ in a certain neighbourhood of a point (y 1 , y 2 ) and f ⁰⁰ (y 1 , y 2 ) exists, then for every (y 1 , y 2 ) ∈ R ² + we have

r

lim

1

1 − r (U(f; r, y

, y

) − f(y

, y

)) = (1 + α − y

) ∂f(y

, y

)

∂y

+ y

∂

f(y

, y

)

∂y

+ (1 + α − y

) ∂f(y

, y

)

∂y

+ y

∂

f(y

, y

)

∂y

. Proof Let (y 1 , y 2 ) be a fixed point in R ² + . By Taylor’s formula and properties of U we get

1

1 − r U(f; r, y

, y

) − f(y

, y

)

= 1

1 − r



U (ϕ

; r, y

, y

) · ∂f(y

, y

)

∂y

+ U(ϕ

; r, y

, y

) · ∂f(y

, y

)

∂y



+ 1 2 ·

1 1 − r



U (ϕ

; r, y

, y

) · ∂

f (y

, y

)

∂y

+ 2U(ϕ

ϕ

; r, y

, y

) · ∂

f(y

, y

)

∂y

∂y

+ U(ϕ

; r, y

, y

) · ∂

f(y

, y

)

∂y



+ 1

1 − r U ψ

p ϕ

+ ϕ

; r, y

, y

, (5)

where the function ψ y

,y

is continuous in R ² + and

(z

,z

lim )→(y

,y

) ψ y

,y

(z 1 , z 2 ) = 0.

Using the H¨older inequality we obtain 1

1 − r

 U ψ y

,y

p ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2


¬  U ψ _y ²

_,y

; r, y 1 , y 2


    1

(1 − r) ² U (ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2 )    

12

.

Moreover, the function ψ _y ²

_,y

satisfying the assumptions of Corollary 2.2. Hence

r lim →1

U ψ _y ²

_,y

; r, y 1 , y 2
= ψ ² _y

_,y

(y 1 , y 2 ) = 0.

From (1) we obtain

r lim →1

1

1 − r U ψ y

,y

p

ϕ 4,y

+ ϕ 4,y

; r, y 1 , y 2
= 0.

(6)

From (5), (6) and Lemma 1.3 we get the assertion. _

Corollary 2.5 Let (y 1 , y 2 ) ∈ R ² + and let f be as in Theorem 2.4. Then

|U(f; r, y 1 , y 2 ) − f(y 1 , y 2 )| = O(1 − r) as r → 1 ⁻ . Similarly we can prove the following theorem for the operator W . Theorem 2.6 Let f ∈ C(R ² ) ∩ L ^p 

R ² ; e ^−z

^−z



, 1 ¬ p ¬ ∞ and (y ¹ , y 2 ) ∈ R ² . If f is of the class C ¹ in a certain neighboorhood of (y 1 , y 2 ) and f ⁰⁰ (y 1 , y 2 ) exists, then for every (y 1 , y 2 ) ∈ R ² we have

r→1 lim

1

1 − r (W (f; r, y ₁ , y 2 ) − f(y 1 , y 2 )) = −y 1 ∂f (y 1 , y 2 )

∂y 1 − y 2 ∂f (y 1 , y 2 )

∂y 2

+ 1 2

 ∂ ² f (y 1 , y 2 )

∂y ₂ ¹ + ∂ ² f (y 1 , y 2 )

∂y ₂ ²



and

|W (f; r, y ¹ , y 2 ) − f(y ¹ , y 2 )| = O(1 − r) as r → 1 ⁻ .

3. Boundary value problems. Now, we indicate some boundary value pro- blems related to the operators U and W . The following theorems are a consequence of Theorem 2.1 and Theorem 2.3, and the results presented in [4].

Theorem 3.1 Let y = (y ₁ , y ₂ ) ∈ R ² + . If f is as in Theorem 2.1, then U is of the class C ^∞ in the set (0, 1) × R ² + and U is a solution of the problem

−r ∂

∂r u(r, y 1 , y 2 ) =



(1 + α − y ¹ ) ∂

∂y 1 + (1 + α − y ² ) ∂

∂y 2



u(r, y 1 , y 2 )

+

 y 1 ∂ ²

∂y ² ₁ + y 2 ∂ ²

∂y ² ₂



u(r, y 1 , y 2 ),

(r,y)→(1 lim

,y) u(r, y 1 , y 2 ) = f(y ₁ , y ₂ ), y = (y 1 , y 2 ).

Theorem 3.2 Let y = (y 1 , y ₂ ) ∈ R ² . If f is as in Theorem 2.3, then W is of the class C ^∞ on (0, 1) × R ² and W is a solution of the boundary problem

2r ∂

∂r u(r, y 1 , y 2 ) = 2

 y 1 ∂

∂y 1 + y 2 ∂

∂y 2



u(r, y 1 , y 2 ) − ∆u(r, y ¹ , y 2 ),

(r,y)→(1 lim

,y) u(r, y 1 , y 2 ) = f(y ₁ , y ₂ ), y = (y 1 , y 2 ).

References

[1] G. Krech, The Poisson integrals of functions of two variables for Hermite and Laguerre expan- sions, Scientific Issues, Jan Długosz University in Cz¸estochowa, Mathematics XVI (2011), 47- 54.

[2] W. Magnus, F. Oberhettinger, Formelen und S¨atze f¨ur die speziellen Funktionen der mathe- matischen Physik, Springer-Verlag, Berlin 1943.

[3] B. Muckenhoupt, Poisson integrals for Hermite and Laguerre expansions, Trans. Amer. Math.

Soc. 139 (1969), 231-242.

[4] G. Toczek, E. Wachnicki, On the Rate of Convergence and the Voronovskaya Theorem for Poisson Integrals for Hermite and Laguerre Expansions, J. Approx. Theory 116 (2002), 113- 125.

[5] E. Wachnicki, On a Gauss-Weierstrass generalized integral, Akad. Ped. Kraków, Rocznik Nauk.-Dydakt. 204 Prace Matematyczne 17 (2000), 251-263.

Grażyna Krech

Pedagogical University, Institute of Mathematics Podchora¸żych 2, PL-30-084 Kraków, Poland E-mail: gkrech@up.krakow.pl

(Received: 12.07.2012)