3.1 One representation of real analytic functions by Dunkl-harmonic functions
Definition 3.1
[21]
An open connected set \(\Omega\subset{\mathbf{R}^{m}}\) is a star domain with center 0 if any \(\underline{x}\in{ \Omega}\) and \(0\leq{t}\leq{1}\) imply that \(t\underline{x}\in{\Omega}\). The set is denoted by \(\Omega^{\star}\).
Definition 3.2
Let \(\Omega^{\star}\) be a star domain in \(\mathbf{R}^{m}\) with center 0. Then the generalized Euler operator on domain \({ \Omega^{\star}}\) is defined by
$$\mathbf{E}_{t}= t\mathbf{I} + \mathbf{E} =t\mathbf{I} + {\sum _{i=1}^{m}{{ x}_{i}} \partial_{{ x}_{i}}}, $$
where t is a real number, I is the identity operator, and E is the Euler operator.
Now we can see the most important intertwining relations concerning the operators \(x^{2}\), \(\Delta_{h}\), \(E_{\mu}\).
Lemma 3.3
[20]
The operators
$$E:=\frac{x^{2}}{2},\qquad F:=\frac{-\Delta_{h}}{2},\qquad H:= \mathbf{E}_{\mu} $$
generate the lie algebra
$$[H,E]=2E, \qquad [H,F]=-2F,\qquad [E,F]=H, $$
where
\(\mu=\frac{m}{2}+\gamma\)
and the Lie bracket
\([x,y]\)
is the commutator
\([x,y]=xy-yx\).
Lemma 3.4
Let
\(\Omega^{\star}\)
be a star domain in
\(\mathbf{R}^{m}\)
with center 0. If
\(f(x)\in{C^{2}(\Omega^{\star})}\otimes R_{0,m}\)
and
\(\mu\geq0\), then
$$ \Delta_{h} \bigl(x^{2s}f(x) \bigr)=x^{2s}\Delta_{h} f(x)+4sx^{2s-2} \mathbf{E}_{ {\mu}+s-1}f(x). $$
(1)
Proof
By Lemma 3.3 and the definition of \(\mathbf{E} _{t}\), we have
$$\begin{aligned} {\Delta_{h} \bigl[x^{2s}f(x) \bigr]} =& { \Delta_{h} x^{2} \bigl[x^{2s-2}f(x) \bigr]} \\ =& { \bigl(x^{2}\Delta_{h} +4\mathbf{E}_{\mu} \bigr) \bigl[x^{2s-2}f(x) \bigr]} \\ =& {x^{2}\Delta_{h} x^{2s-2}f(x)+ 4 \mathbf{E}_{\mu}x^{2s-2}f(x)} \\ =& {x^{2} \bigl(x^{2}\Delta_{h} +4 \mathbf{E}_{\mu} \bigr)x^{2s-4}f(x)+4 \bigl({ x}^{2} \mathbf{E}_{\mu}+2{ x}^{2} \bigr)x^{2s-4}f(x)=\cdots} \\ =& {x^{2s}\Delta_{h} f(x)+4sx^{2s-2} \mathbf{E}_{\mu+s-1}f(x)}. \end{aligned}$$
Thus, we finish the proof. □
Lemma 3.5
Let
\(g(x)\in{C^{1}(\Omega^{\star})}\otimes R_{0,m}\). Then
$$ (\mathbf{E}+l+1) \int_{0}^{1}\alpha^{l}g(\alpha x)\,d \alpha=g(x) $$
(2)
and
$$ (\mathbf{E}+l+1) \int_{0}^{1}\frac{(1-\alpha)^{q}}{q!}\alpha^{l}g( \alpha x)\,d\alpha = \int_{0}^{1}\frac{(1-\alpha)^{q-1}}{(q-1)!}\alpha ^{l+1}g(\alpha x)\,d\alpha $$
(3)
for
\(q\in\mathbf{N}\)
and
\(l\geq{0}\).
Proof
The proof can be referred to in the literature [22]. □
In this paper, we assume the following infinite series converges absolutely and uniformly in \(\Omega^{*}\).
Theorem 3.6
Let
\(G(x)\in C^{\infty}(\Omega^{*})\otimes R_{0,m}\). Then
$$ G(x)=f_{0}+\sum_{s=1}^{\infty} \frac{x^{2s}}{4^{s}s!(s-1)!} \int_{0} ^{1} (1-\alpha)^{s-1} \alpha^{{\mu}-1}f_{s}(\alpha x)\,d\alpha, $$
(4)
where
\(\Delta_{h} f_{s}(x)=0 \)
and
$$ f_{s}(x)=\Delta_{h}^{s}G(x)+\sum _{l=1}^{\infty} \frac{(-1)^{l}x^{2l}}{4^{l}l!} \int_{0}^{1}\frac{(1-\alpha)^{l-1} \alpha^{l-1}}{(l-1)!}\alpha^{{\mu}-1} \Delta_{h}^{s+l}G(\alpha x)\,d\alpha. $$
(5)
Proof
First we prove that the functions \(f_{s}({x})\) satisfy (5). Substituting \(f_{s}({x})\) into the right-hand side of the identity (5), we have
$$\begin{aligned}& f_{0}+\sum_{s=1}^{\infty} \frac{{x}^{2s}}{4^{s}s!(s-1)!} \int_{0}^{1} (1-\alpha)^{s-1} \alpha^{{\mu}-1}f_{s}(\alpha{x})\,d\alpha \\& \quad = G({x})+\sum_{s=1}^{\infty} \frac{(-1)^{s}{x}^{2s}}{4^{s}s!} \int_{0} ^{1}\frac{(1-\alpha)^{s-1}\alpha^{s-1}}{(s-1)!}\alpha^{{\mu}-1} \Delta_{h}^{s}G(\alpha{x})\,d\alpha \\& \qquad {} +\sum_{s=1}^{\infty} \frac{{x}^{2s}}{4^{s}s!(s-1)!} \int_{0}^{1}(1- \alpha)^{s-1} \alpha^{{\mu}-1}\Delta_{h}^{s}G(\alpha{x})\,d\alpha \\& \qquad {} +\sum_{s=1}^{\infty} \frac{{x}^{2s}}{4^{s}s!(s-1)!} \int_{0}^{1}(1- \alpha)^{s-1} \alpha^{{\mu}-1} \\& \qquad {}\times \sum_{l=1}^{\infty} \frac{(-1)^{l}( \alpha{x})^{2l}}{4^{l}l!} \int_{0}^{1}\frac{(1-\beta)^{l-1}\beta ^{l-1}}{(l-1)!}\beta^{\frac{\mu}{2}-1} \Delta_{h}^{l+s}G(\alpha \beta{x})\,d\beta \,d\alpha. \end{aligned}$$
(6)
Denote by \(A_{1}({x})\) the fourth term on the right side of equation (6). Then
$$\begin{aligned} {A_{1}({x})} =& {\sum_{s=1}^{\infty} \sum_{l=1}^{\infty} \frac{(-1)^{l}{x}^{2s+2l}}{4^{l+s}s!l!} \int_{0}^{1}\frac{(1-\alpha)^{s-1} \alpha^{2l+{\mu}-1}}{(s-1)!} \int_{0}^{1}\frac{(1-\beta)^{l-1} \beta^{l+{\mu}-2}}{(l-1)!}\Delta_{h}^{l+s}G( \alpha\beta{x})\,d\beta\, d \alpha} \\ =& {\sum_{s=1}^{\infty}\sum _{l=1}^{\infty} \frac{(-1)^{l}{x}^{2s+2l}}{4^{l+s}s!l!} \int_{0}^{1}\frac{\alpha^{2}(1- \alpha)^{s-1}}{(s-1)!} \int_{0}^{1} \frac{(\alpha-\alpha\beta)^{l-1}( \alpha\beta)^{l+{\mu}-2}}{(l-1)!}\Delta_{h}^{l+s}G( \alpha\beta {x})\,d\beta \,d\alpha}. \end{aligned}$$
Denote by \(A_{2}(x)\) the integral on the above expression. Let \(t=\alpha\beta\). Then \(dt=\alpha \,d\beta\). We calculate
$$\begin{aligned} {A_{2}({x})} =& { \int_{0}^{1}\frac{\alpha(1-\alpha)^{s-1}}{(s-1)!} \int_{0}^{\alpha }\frac{(\alpha-t)^{l-1}t^{l+{\mu}-2}}{(l-1)!}\Delta_{h}^{l+s}G(t {x})\,dt\,d\alpha} \\ =& { \int_{0}^{1} \int_{0}^{\alpha} \frac{\alpha(1-\alpha)^{s-1}(\alpha -t)^{l-1}}{(s-1)!(l-1)!}t^{l+{\mu}-2} \Delta_{h}^{l+s}G(t{x})\,dt\,d \alpha} \\ =& { \int_{0}^{1} \int_{t}^{1} \frac{\alpha(1-\alpha)^{s-1}(\alpha-t)^{l-1}}{(s-1)!(l-1)!}t^{l+ {\mu}-2} \Delta_{h}^{l+s}G(t{x})\,d\alpha \,dt} \\ =& { \int_{0}^{1}\frac{t^{l+{\mu}-2}}{(s-1)!(l-1)!}\Delta_{h}^{l+s}G(t {x})\,dt \int_{t}^{1} \alpha(1-\alpha)^{s-1}( \alpha-t)^{l-1}\,d\alpha.} \end{aligned}$$
Let \(\alpha=\beta+t\). Then we have
$$A_{3}(t)= \int_{t}^{1} \alpha(1-\alpha)^{s-1}( \alpha-t)^{l-1}\,d\alpha= \int_{0}^{1-t}(\beta+t) (1-\beta-t)^{s-1} \beta^{l-1}\,d\beta. $$
Let \(\beta=\alpha(1-t)\). It follows that
$$\begin{aligned} A_{3}(t) =& { \int_{0}^{1}(\alpha-\alpha t+t) (1- \alpha)^{s-1}(1-t)^{s-1}\alpha ^{l-1}(1-t)^{l} \,d\alpha} \\ =& {(1-t)^{s+l-1} \int_{0}^{1}(\alpha-\alpha t+t) (1- \alpha)^{s-1} \alpha^{l-1}\,d\alpha.} \end{aligned}$$
We calculate
$$\begin{aligned} {A_{3}(t)} =& {(1-t)^{l+s} \int_{0}^{1}\alpha^{l}(1- \alpha)^{s-1}\,d\alpha+t(1-t)^{l+s-1} \int_{0}^{1}\alpha^{l-1}(1- \alpha)^{s-1}\,d\alpha} \\ =& {(1-t)^{l+s}B(l+1,s)+t(1-t)^{s+l-1}B(l,s),} \end{aligned}$$
where the beta functions
$$ B(l,s)= \int_{0}^{1}\alpha^{l-1}(1- \alpha)^{s-1}\,d\alpha. $$
(7)
Using the properties of beta functions and gamma functions:
$$ B(l,s)=\frac{\Gamma(l)\Gamma(s)}{\Gamma(s+l)} $$
(8)
and
$$ \Gamma(s)=(s-1)!, $$
(9)
we have
$$\begin{aligned} A_{3}(t) =& {\frac{\Gamma(l+1)\Gamma(s)}{\Gamma(l+s+1)}(1-t)^{s+l}+ \frac{ \Gamma(l)\Gamma(s)}{\Gamma(l+s)}t(1-t)^{l+s-1}} \\ =& {\frac{l!(s-1)!}{(l+s)!}(1-t)^{s+l}+\frac {(l-1)!(s-1)!}{(l+s-1)!}t(1-t)^{l+s-1}.} \end{aligned}$$
By substituting \(A_{3}(t)\) into \(A_{1}({x})\), we have
$$\begin{aligned}& \sum_{s=1}^{\infty}\sum _{l=1}^{\infty} \frac{(-1)^{l}{x}^{2s+2l}}{4^{l+s}} \int_{0}^{1} \biggl[ \frac{(1-t)^{s+l}}{s!(l-1)!(l+s)!}+ \frac{t(1-t)^{l+s-1}}{s!l!(l+s-1)!} \biggr] t^{l+{\mu}-2}\Delta_{h} ^{l+s}G(t{x})\,dt \\& \quad = {\sum_{s=1}^{\infty}\sum _{l=1}^{\infty} \frac{(-1)^{l}{x}^{2s+2l}}{4^{l+s}} \int_{0}^{1} \biggl[ \frac{t^{l-1}(1-t)^{s+l}}{s!(l-1)!(l+s)!}+ \frac{t ^{l}(1-t)^{l+s-1}}{s!l!(l+s-1)!} \biggr] t^{{\mu}-1}\Delta_{h}^{l+s}G(t {x})\,dt} \\& \quad = {\sum_{i=2}^{\infty}\sum _{l=1}^{i-1}\frac{(-1)^{l}{x}^{2i}}{4^{i}} \int_{0}^{1} \biggl[ \frac{t^{i-l-1}(1-t)^{i}}{l!(i-l-1)!i!}+ \frac{t ^{i-l}(1-t)^{i-1}}{l!(i-l)!(i-1)!} \biggr] t^{{\mu}-1}\Delta_{h}^{i}G(t {x})\,dt} \\& \quad = {\sum_{i=2}^{\infty}\frac{{x}^{2i}}{4^{i}} \int_{0}^{1} \Biggl[ \frac{(1-t)^{i}}{i!} \sum _{l=1}^{i-1}\frac{(-1)^{l}t^{i-l-1}}{l!(i-l-1)!}+ \frac{(1-t)^{i-1}}{(i-1)!}\sum_{l=1}^{i-1} \frac{(-1)^{l}t^{i-l}}{l!(i-l)!} \Biggr] t^{{\mu}-1}\Delta_{h}^{i}G(t {x})\,dt.} \end{aligned}$$
We calculate
$$\begin{aligned} \sum_{l=1}^{i-1}\frac{(-1)^{l}t^{i-l-1}}{l!(i-l-1)!} =&\sum _{s=0}^{i-1}\frac{(-1)^{l}t ^{i-l-1}}{l!(i-l-1)!}- \frac{t^{i-1}}{(i-1)!} \\ =& \frac{(t-1)^{i-1}}{(i-1)!}-\frac{t^{i-1}}{(i-1)!} \end{aligned}$$
and
$$\sum_{l=1}^{i-1}\frac{(-1)^{l}t^{i-l}}{l!(i-l)!}=\sum _{s=0}^{i-1}\frac{(-1)^{l}t ^{i-l}}{l!(i-l)!}- \frac{t^{i}}{i!}-\frac{(-1)^{i}}{i!} = \frac{(t-1)^{i}}{i!}-\frac{t^{i}}{i!}- \frac{(-1)^{i}}{i!}. $$
Thus, we have
$$\begin{aligned} A_{1}({x}) =& {\sum_{i=2}^{\infty} \frac{{x}^{2i}}{4^{i}} \int_{0}^{1} \biggl[ -\frac{(1-t)^{i-1}t ^{i-1}}{(i-1)!i!} - \frac{(-1)^{i}(1-t)^{i-1}}{(i-1)!i!} \biggr] t^{ {\mu}-1}\Delta_{h}^{i}G(t{x}) \,dt} \\ =& {\sum_{i=1}^{\infty}\frac{{x}^{2i}}{4^{i}} \int_{0}^{1} \biggl[ -\frac{(1-t)^{i-1}t ^{i-1}}{(i-1)!i!} - \frac{(-1)^{i}(1-t)^{i-1}}{(i-1)!i!} \biggr] t^{ {\mu}-1}\Delta_{h}^{i}G(t{x}) \,dt.} \end{aligned}$$
By substituting \(A_{1}(x)\) into (6), we have (5).
Next, we prove that \(\Delta_{h} f_{s}({x})=0\). By Lemma 3.4, we have
$$\begin{aligned}& {\Delta_{h} f_{s}({x})} \\& \quad = {\Delta_{h}^{s+1}G({x})+\sum _{l=1}^{\infty}\frac{(-1)^{l}}{4^{l}l!} \Delta_{h} \biggl( x^{2l} \int_{0}^{1} \frac{(1-\alpha)^{l-1}\alpha^{l-1}}{(l-1)!}\alpha^{{\mu}-1} \Delta _{h}^{s+l}G(\alpha{x})\,d\alpha \biggr) } \\& \quad = {\Delta_{h}^{s+1}G({x})+\sum _{l=1}^{\infty}\frac {x^{2l}}{4^{l}l!} \int _{0}^{1} \frac{(1-\alpha)^{l-1}\alpha^{l+1}}{(l-1)!} \alpha^{{\mu}-1} \Delta_{h}^{s+l+1}G(\alpha{x})\,d\alpha} \\& \qquad {} -\sum_{l=1}^{\infty} \frac{{x}^{2(l-1)}}{4^{l-1}(l-1)!} ( \mathbf{E}+ {\mu}+l-1 ) \int_{0}^{1} \frac{(1-\alpha)^{l-1}\alpha^{l-1}}{(l-1)!}\alpha^{{\mu}-1} \Delta _{h}^{s+l}G(\alpha{x})\,d\alpha. \end{aligned}$$
Denote by \(B_{1}({x})\) the third term of the above equality. From Lemma 3.5, we have
$$\begin{aligned} B_{1}(x) =& \sum_{l=1}^{\infty} \frac{(-1)^{l}{x}^{2(l-1)}}{4^{l-1}(l-1)!} ( \mathbf{E}+ {\mu}+l-1 ) \int_{0}^{1} \frac{(1-\alpha)^{l-1}\alpha^{l-1}}{(l-1)!} \alpha^{{\mu}-1}\Delta _{h}^{s+l}G(\alpha{x})\,d\alpha \\ =& - ( \mathbf{E}+{\mu} ) \int_{0}^{1}\alpha^{{\mu }-1}\Delta _{h}^{s+1}G(\alpha{x})\,d\alpha \\ &{} -\sum_{l=2}^{\infty}\frac{{x}^{2(l-1)}}{4^{l-1}(l-1)!} ( \mathbf{E}+ {\mu}+l-1 ) \int_{0}^{1} \frac{(1-\alpha)^{l-1}\alpha^{l-1}}{(l-1)!} \alpha^{{\mu}-1}\Delta _{h}^{s+l}G(\alpha{x})\,d\alpha \\ =& -\Delta_{h}^{s+1}G({x})-\sum _{l=2}^{\infty} \frac{(-1)^{l}x^{2(l-1)}}{4^{l-1}(l-1)!} \int_{0}^{1}\frac{(1-\alpha)^{l-2} \alpha^{l}}{(l-2)!} \alpha^{{\mu}-1} \Delta_{h}^{s+l}G(\alpha{x})\,d\alpha \\ =& -\Delta_{h}^{s+1}G({x})-\sum _{l=2}^{\infty} \frac{x^{2(l-1)}}{4^{l-1}(l-1)!} \int_{0}^{1}\frac{(1-\alpha)^{l-2} \alpha^{l}}{(l-2)!} \alpha^{{\mu}-1} \Delta_{h}^{s+l}G(\alpha{x})\,d\alpha \\ =& -\Delta_{h}^{s+1}G({x})-\sum _{l=1}^{\infty}\frac{x^{2l}}{4^{l}l!} \int_{0}^{1}\frac{(1-\alpha)^{l-1}\alpha^{l+1}}{(l-1)!} \alpha^{ {\mu}-1} \Delta_{h}^{s+l+1}G(\alpha{x})\,d\alpha, \end{aligned}$$
which implies that \(\Delta_{h} f_{s}({x})=0\). Thus, we finish the proof. □
Corollary 3.7
Let
\(P_{l}(x)\)
be a homogeneous polynomial of degree
l. Then
$$ P_{l}(x)=R_{l}(x)+x^{2}R_{l-2}(x)+ \cdots+x^{2k}R_{l-2k}(x), $$
(10)
where
\(R_{l-2k}(x)\)
are homogeneous Dunkl-harmonic polynomials and
$$ R_{l-2k}(x)=\frac{1}{4^{k}k!(l-k+\mu)\cdots(l-2k+\mu)} \sum _{s=0} ^{[l-k]}\frac{(-1)^{s}x^{2s}\Delta_{h}^{s+k} P_{l}(x)}{4^{s}s!(l-2k-s+ \mu-1)_{s}}. $$
(11)
Proof
Let \(P_{l}(x)\) be a homogeneous polynomial of degree l. By Theorem 3.6, we have
$$P_{l}(x)=R_{l}(x)+x^{2}R_{l-2}(x)+ \cdots+x^{2k}R_{l-2k}(x), $$
where
$$ R_{l-2k}(x)=\frac{f_{k}(x)}{4^{k}k!(k-1)!} \int_{0}^{1}(1-\alpha)^{k-1} \alpha^{l-2k+\mu-1}\,d\alpha = \frac{f_{k}(x)B(k, l-2k+\mu)}{4^{k}k!(k-1)!}. $$
(12)
Using equation (5), we have
$$\begin{aligned} {f_{0}(x)} =& {P_{l}(x)+\sum _{s=1}^{\infty}\frac{(-1)^{s}x^{2s}}{4^{s}s!} \int_{0} ^{1}\frac{(1-\alpha)^{s-1}\alpha^{s-1}}{(s-1)!}\alpha^{{\mu}-1} \Delta_{h}^{s}P_{l}(\alpha x)\,d\alpha} \\ =& {P_{l}(x)+\sum_{s=1}^{\infty} \frac{(-1)^{s}x^{2s}\Delta_{h}^{s} P _{l}(x)}{4^{s}s!(s-1)!} \int_{0}^{1}{(1-\alpha)^{s-1} \alpha^{l-s+ \mu-2}}\,d\alpha} \end{aligned}$$
for the case \(G(x)=P_{l}(x)\). Using equations (7) and (8), we have
$$\int_{0}^{1}{(1-\alpha)^{s-1} \alpha^{l-s+\mu-2}}\,d\alpha=B(s,l-s+ \mu-1) =\frac{(s-1)!}{(l-s+\mu-1)_{s}}, $$
where \((m)_{s} = m( m + 1)\cdots(m +s-1)\) is the Pochhammer symbol. Therefore,
$$f_{0}(x)=P_{l}(x)+\sum_{s=1}^{\infty} \frac{(-1)^{s}x^{2s}\Delta_{h} ^{s} P_{l}(x)}{4^{s}s!(l-s+\mu-1)_{s}}=\sum_{s=0}^{\infty} \frac{(-1)^{s}x ^{2s}\Delta_{h}^{s} P_{l}(x)}{4^{s}s!(l-s+\mu-1)_{s}}. $$
By equation (5), we have
$$f_{k}(x)=\sum_{s=0}^{\infty} \frac{(-1)^{s}x^{2s}\Delta_{h}^{s+k} P _{l}(x)}{4^{s}s!(l-2k-s+\mu-1)_{s}}. $$
Thus, it follows from (12) that
$$\begin{aligned} {R_{l-2k}(x)} =& {\frac{f_{k}(x)B(k, l-2k+\mu)}{4^{k}k!(k-1)!}} \\ =& {\frac{1}{4^{k}k!(l-k+\mu)\cdots(l-2k+\mu)} \sum_{s=0}^{\infty} \frac{(-1)^{s}x ^{2s}\Delta_{h}^{s+k} P_{l}(x)}{4^{s}s!(l-2k-s+\mu-1)_{s}},} \end{aligned}$$
which completes the proof. □
3.2 Solutions of the Dunkl-Poisson equation in Clifford analysis
In this section, we study the Dunkl-Poisson equation in Clifford analysis,
$$ {\Delta_{h}}G(x)=f({x}), $$
(13)
where \(f({x})\in C^{\infty}(\Omega) \otimes R_{0,m}\) is a real analytic function.
Theorem 3.8
Let
\(f({x}) \in C^{\infty}(\Omega^{*})\otimes R_{0,m}\). A real analytic solution of equation (13) can be found in the form
$$ G({x})=\sum_{s=0}^{\infty} \frac{(-1)^{s}{x}^{2(s+1)}}{4^{s+1}(s+1)!s!} \int_{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+s-1} \Delta_{h}^{s}f(\alpha{x})\,d\alpha. $$
(14)
Proof
Let \(f({x})\in C^{\infty}(\Omega^{*})\otimes R _{0,m}\). Then it follows by Theorem 3.6 that
$$ G(x)=f_{0}+\sum_{s=1}^{\infty} \frac{x^{2s}}{4^{s}s!(s-1)!} \int_{0} ^{1} (1-\alpha)^{s-1} \alpha^{{\mu}-1}f_{s}(\alpha x)\,d\alpha, $$
(15)
where \(f_{s}(x)\) are Dunkl-harmonic in \(\Omega^{*}\) given by the relation
$$ f_{s}(x)=\Delta_{h}^{s}G(x)+\sum _{l=1}^{\infty} \frac{(-1)^{l}x^{2l}}{4^{l}l!} \int_{0}^{1}\frac{(1-\alpha)^{l-1} \alpha^{l-1}}{(l-1)!}\alpha^{{\mu}-1} \Delta_{h}^{s+l}G(\alpha x)\,d\alpha. $$
(16)
Note that \(\Delta_{h}^{s+1}G(x)=\Delta_{h}^{s}f\). Thus, we have
$$\begin{aligned} {G(x)-f_{0}(x)} =& {-\sum _{l=1}^{\infty}\frac{(-1)^{l}x^{2l}}{4^{l}l!} \int_{0}^{1}\frac{(1- \alpha)^{l-1}\alpha^{l-1}}{(l-1)!}\alpha^{{\mu}-1} \Delta_{h}^{l}G( \alpha x)\,d\alpha} \\ =& {\sum_{s=0}^{\infty}\frac{(-1)^{s}{x}^{2(s+1)}}{4^{s+1}(s+1)!s!} \int _{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+s-1} \Delta_{h}^{s}f(\alpha {x})\,d\alpha.} \end{aligned}$$
(17)
Since \(\Delta_{h}[G(x)-f_{0}(x)]=f{(x)}\), it implies that \([G(x)-f _{0}(x)]\) is a solution of the Poisson equation (13). Therefore, the right-hand of equation (17) is a solution of equation (13). □
Corollary 3.9
The solution of the Poisson equation
\(\Delta_{h}G(x)=P_{l}(x)\)
can be represented in the form
$$ G(x)=\sum_{s=0}^{[\frac{l}{2}]} \frac{(-1)^{s}{x}^{2(s+1)} \Delta_{h} ^{s}P_{l}({x})}{4^{s+1}(s+1)!(l+\mu)\cdots(l-s+\mu)}, $$
(18)
where
\([\frac{l}{2}]\)
is the integer part of
\(\frac{l}{2}\)
and
\((a,b)_{k}=a(a+b)\cdots(a+kb-b)\)
is the generalized Pochhammer symbol with the convention that
\((a,b)_{0}=1\).
Proof
Let \(P_{l}(x)\) be a homogeneous polynomial of degree l. Then we have \(\Delta_{h}^{k}P_{l}(\alpha x)=\alpha^{l-2k} \Delta_{h}^{k}P_{l}(x)\). Therefore, (14) can be transformed into
$$G(x)=\sum_{s=0}^{[\frac{l}{2}]}\frac{(-1)^{s}{x}^{2(s+1)}\Delta_{h} ^{s}P_{l}({x})}{4^{s+1}(s+1)!s!} \int_{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+s-1} \alpha^{l-2s}\,d\alpha. $$
Furthermore, we can write
$$G(x)=\sum_{s=0}^{[\frac{l}{2}]}\frac{(-1)^{s}{x}^{2(s+1)}\Delta_{h} ^{s}P_{l}({x})}{4^{s+1}(s+1)!s!} B(s+1,\mu+l-s), $$
where \(B(s,k)=\int_{0}^{1}\alpha^{s-1}(1-\alpha)^{k-1}\,d\alpha\). Then by the relation
$$B(s,k)=\frac{\Gamma(s)\Gamma(k)}{\Gamma(s+k)}, $$
we have
$$u(x)=\sum_{s=0}^{[\frac{l}{2}]}\frac{(-1)^{s}\Gamma(l-s+\mu){x}^{2(s+1)} \Delta_{h}^{s}P_{l}({x})}{4^{s+1}(s+1)!\Gamma(l+\mu+1)}. $$
Using the property \(\Gamma(s+1)= s\Gamma(s)\) of the gamma function, we find that
$$\Gamma(l+\mu+1)=(l+\mu) (l+\mu-1)\cdots(l+\mu-s)\Gamma(l+\mu-s). $$
It follows that
$$G(x)=\sum_{s=0}^{[\frac{l}{2}]}\frac{(-1)^{s}{x}^{2(s+1)} \Delta_{h} ^{s}P_{l}({x})}{4^{s+1}(s+1)!(l+\mu)\cdots(l-s+\mu)}, $$
which completes the proof. □
Corollary 3.10
Let
\(P_{l}(x)\)
be a homogeneous harmonic polynomial of degree
l. The solution of the equation
\(\Delta_{h} G(x)=x^{2k}P_{l}(x)\)
is given by
$$G(x)=x^{2k+2}P_{l}(x)\sum_{s=0}^{k} \frac{(-1)^{s}(2k-2s+2,2)_{s}(2l+2 \mu+2k-2s, 2)_{s}}{4^{s+1}(s+1)!(\mu+2k+l)\cdots(\mu+2k+l-s)}. $$
Proof
Let \(f(x)=x^{2k}P_{l}(x)\). We calculate this solution using equation (14) to obtain
$$G({x})=\sum_{s=0}^{\infty} \frac{(-1)^{s}{x}^{2(s+1)}}{4^{s+1}(s+1)!s!} \int_{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+s-1} \Delta_{h}^{s} \bigl[x^{2k}P_{l}(x) \bigr]\,d\alpha. $$
Let us derive an expression for \(\Delta_{h}^{s}[x^{2k}P_{l}(x)]\). By Lemma 3.5, we have
$$\Delta_{h} \bigl[x^{2k}P_{l}(x) \bigr]=4kx^{2k-2}(l+\mu+k-1)P_{l}(x). $$
Therefore, for \(2s\leq2k+l\), we have
$$\begin{aligned}& \Delta_{h}^{s}\bigl[x^{2k}P_{l}(x) \bigr] \\& \quad = 2k(2k-2)\cdots(2k-2s+2) (2l+2\mu+2k-2)\cdots(2l+2\mu+2k-2s)x^{2k-2s}P _{l}(x) \\& \quad = (2k-2s+2,2)_{s}(2l+2\mu+2k-2s, 2)_{s}x^{2k-2s}P_{l}(x). \end{aligned}$$
Thus, we find
$$\begin{aligned} G({x}) =&\sum_{s=0}^{k}\frac{(-1)^{s}{x}^{2(s+1)}}{4^{s+1}(s+1)!s!} \\ &{}\times \int _{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+s-1} (2k-2s+2,2)_{s}(2l+2 \mu+2k-2s, 2)_{s}(\alpha x)^{2k-2s}P_{l}(\alpha x)\,d\alpha \end{aligned}$$
and, since \(P_{l}(\alpha x)=\alpha^{l}P_{l}(x)\),
$$\begin{aligned} G({x}) =&x^{2k+2}P_{l}(x)\sum_{s=0}^{k} \frac{(-1)^{s}(2k-2s+2,2)_{s}(2l+2 \mu+2k-2s, 2)_{s}}{4^{s+1}(s+1)!s!} \\ &{}\times \int_{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+2k+l-s-1}\,d\alpha. \end{aligned}$$
The integral in this expression is evaluated as
$$\begin{aligned} \int_{0}^{1} (1-\alpha)^{s} \alpha^{{\mu}+2k+l-s-1}\,d\alpha =&B(s+1, \mu+2k+l-s) \\ =&\frac{s!}{(\mu+2k+l) \cdots(\mu+2k+l-s)}, \end{aligned}$$
where \(B(m,n)\) is the Euler beta function. Then \(G(x)\) is transformed into
$$G(x)=x^{2k+2}P_{l}(x)\sum_{s=0}^{k} \frac{(-1)^{s}(2k-2s+2,2)_{s}(2l+2 \mu+2k-2s, 2)_{s}}{4^{s+1}(s+1)!(\mu+2k+l)\cdots(\mu+2k+l-s)}. $$
Thus, we complete the proof. □