Sobolev orthogonality of polynomial solutions of second-order partial differential equations

Abstract

Given a second-order partial differential operator ${\mathscr {L}}$ with nonzero polynomial coefficients of degree at most 2, and a Sobolev bilinear form

$$\begin{aligned} (P,Q)_S\,=\,\sum _{i=0}^N\sum _{j=0}^i\left\langle {\textbf{u}}^{(i,j)}, \partial _x^{i-j}\partial _y^jP\,\,\partial _x^{i-j}\partial _y^{j}Q\right\rangle , \quad N\geqslant 0, \end{aligned}$$

where ${\textbf{u}}^{(i,j)}$, $0\leqslant j \leqslant i \leqslant N$, are linear functionals defined on the space of bivariate polynomials, we study the orthogonality of the polynomial solutions of the partial differential equation ${\mathscr {L}}[p]=\lambda _{n,m}\,p$ with respect to $(\cdot ,\cdot )_S$, where $\lambda _{n,m}$ are eigenvalue parameters depending on the total and partial degree of the solutions. We show that the linear functionals in the bilinear form must satisfy Pearson equations related to the coefficients of ${\mathscr {L}}$. Therefore, we also study solutions of the Pearson equations that can be obtained from univariate moment functionals. In fact, the involved univariate functionals must satisfy Pearson equations in one variable. Moreover, we study polynomial solutions of ${\mathscr {L}}[p]=\lambda _{n,m}\,p$ obtained from univariate sequences of polynomials satisfying second-order ordinary differential equations.

Appendix A

1.1 A.1 Proof of Lemma 3.2

Let ${\textbf{u}}$ be a bivariate moment functional and let $P\in \Pi ^2$.

For $i=j=0$, we have

$$\begin{aligned} {\mathscr {L}}[P]\,{\textbf{u}}\,=\,[a\,P_{xx}+2b\,P_{xy}+c\,P_{yy}+d\,P_x+e\,P_y]\,{\textbf{u}}, \end{aligned}$$

and

$$\begin{aligned} {\mathscr {L}}^*[P\,{\textbf{u}}]\,=\,(a\,P\,{\textbf{u}})_{xx}+2(b\,P\,{\textbf{u}})_{xy}+(c\,P\,{\textbf{u}})_{yy}-(d\,P\,{\textbf{u}})_x-(e\,P\,{\textbf{u}})_y. \end{aligned}$$

Using the product rule and then simplifying, we get

$$\begin{aligned} \begin{aligned} {\mathscr {L}}[P]{\textbf{u}}-{\mathscr {L}}^*[P{\textbf{u}}]=&-P\,\left( {\mathscr {L}}^{(0,0)}\right) ^*[{\textbf{u}}]-2P_x\,{\mathcal {M}}_1^{(0,0)}[{\textbf{u}}]-2P_y\,{\mathcal {M}}_2^{(0,0)}[{\textbf{u}}]. \end{aligned} \end{aligned}$$

(A.1)

Note that ${\mathscr {L}}^{(0,0)}\equiv {\mathscr {L}}$.

Now, let $i+j\geqslant 1$. On one hand, using Leibniz rule and the fact that ${\mathscr {L}}$ is in the extended Lyskova class, we can expand $\partial _x^i\partial _y^j{\mathscr {L}}[P]$. We get

$$\begin{aligned} \begin{aligned}&\partial _x^i\partial _y^j\bigg (\partial _x^i\partial _y^j{\mathscr {L}}[P]\,{\textbf{u}} \bigg )\\&\quad =\partial _x^i\partial _y^j\bigg [ \bigg ( {\mathscr {L}}[\partial _x^i\partial _y^jP]+i\,a_x\,\partial _x^{i+1}\partial _y^jP+2\,i\,b_x\,\partial _x^i\partial _y^{j+1}P+2\,j\,b_y\,\partial _x^{i+1}\partial _y^jP\\&\qquad +j\,c_y\partial _x^i\partial _y^{j+1}P+\frac{i(i-1)}{2}a_{xx}\partial _x^i\partial _y^jP+2ijb_{xy}\partial _x^i\partial _y^jP+\frac{j(j-1)}{2}c_{yy}\partial _x^i\partial _y^jP\\&\qquad +i\,d_x\,\partial _x^i\partial _y^jP+j\,e_y\,\partial _x^i\partial _y^jP\bigg )\,{\textbf{u}}\bigg ]. \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned} \begin{aligned}&{\mathscr {L}}^*\Big [\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]\\&\quad =\Big [a\,\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]_{xx}+2\Big [b\,\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]_{xy}+\Big [c\,\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]_{yy}\\&\qquad -\Big [d\,\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]_x-\Big [e\,\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]_y\\&\quad =\partial _x^i\partial _y^j\bigg [{\mathscr {L}}^*[\partial _x^i\partial _y^jP\,{\textbf{u}}]+i\,d_x\partial _x^i\partial _y^jP\,{\textbf{u}}+j\,e_y\partial _x^i\partial _y^jP\,{\textbf{u}} + \frac{i(i-1)}{2}a_{xx}\partial _x^i\partial _y^jP\,{\textbf{u}}\\&\qquad +2ij\,b_{xy}\partial _x^i\partial _y^jP\,{\textbf{u}}+\frac{j(j-1)}{2}c_{yy}\partial _x^i\partial _y^jP\,{\textbf{u}}-i\partial _x\left( a_x\partial _x^i\partial _y^jP\,{\textbf{u}}\right) \\&\qquad -2i\partial _y\left( b_x\partial _x^i\partial _y^jP\,{\textbf{u}}\right) -2j\partial _x\left( b_y\partial _x^i\partial _y^jP\,{\textbf{u}}\right) -j\partial _y\left( c_y\partial _x^i\partial _y^jP\,{\textbf{u}}\right) \bigg ]. \end{aligned} \end{aligned}$$

Then, using (A.1), we get

$$\begin{aligned} \begin{aligned}&\partial _x^i\partial _y^j\Big (\partial _x^i\partial _y^j{\mathscr {L}}[P]\,{\textbf{u}} \Big )-{\mathscr {L}}^*\Big [\partial _x^i\partial _y^j(\partial _x^i\partial _y^jP\,{\textbf{u}}) \Big ]\\&\quad =-\partial _x^i\partial _y^j\bigg [ \partial _x^i\partial _y^jP\,\Big ([(a\,{\textbf{u}})_x+(b\,{\textbf{u}})_y-(d+i\,a_x+2j\,b_y)\,{\textbf{u}}]_x\\&\qquad +[(b\,{\textbf{u}})_x+(c\,{\textbf{u}})_y-(e+2i\,b_x+j\,c_y)\,{\textbf{u}}]_y\Big )\\&\qquad +2\,\partial _x^{i+1}\partial _y^jP\,[(a\,{\textbf{u}})_x+(b\,{\textbf{u}})_y-(d+i\,a_x+2j\,b_y)\,{\textbf{u}}]\\&\qquad +2\,\partial _x^i\partial _y^{j+1}P\,[(b\,{\textbf{u}})_x+(c\,{\textbf{u}})_y-(e+2i\,b_x+j\,c_y)\,{\textbf{u}}]\bigg ]. \end{aligned} \end{aligned}$$

Remark A.1

Observe that (A.1) is satisfied even when ${\mathscr {L}}$ is not in the extended Lyskova class.

1.2 A.2 Proof of Corollary 4.3

From Proposition 4.2 with $\rho (s)\,=\,r_0+r_1\,s$, we have for all $P\in \Pi ^2$,

$$\begin{aligned} \begin{aligned}&\left\langle -y\,\widetilde{b}(x)\,\partial _x\,{\textbf{w}}-c(y)\,\partial _y\,{\textbf{w}}\,+\,\widetilde{e}(y)\,{\textbf{w}},P\right\rangle +(r_1-r_1)\left\langle \widetilde{d}(s)\,{\textbf{u}}_0^{(s)}, \left\langle t\,{\textbf{v}}^{(t)}, P \right\rangle \right\rangle \\&\quad =r_1\,\left\langle -a(s)\,D\,{\textbf{u}}_0^{(s)} + \left( \widetilde{b}(s)+\widetilde{d}(s)\right) \,{\textbf{u}}_0^{(s)},\left\langle t\,{\textbf{v}}^{(t)}, P\right\rangle \right\rangle \\&\qquad +r_1\left\langle (e_1-d_1)\,s\,{\textbf{u}}_0^{(s)}, \left\langle t\,{\textbf{v}}^{(t)}, P\right\rangle \right\rangle \\&\qquad + \left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)},(c_2-b_1)\,r_1\,s\,\partial _t\left( t^2\,P\right) \right\rangle \right\rangle \\&\qquad + \left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)},(c_2\,r_0-b_0\,r_1)\partial _t\left( t^2P\right) +c_1\partial _t(t\,P)+ \left[ e_0+(e_1\,r_0-d_0\,r_1)\,t\right] P\right\rangle \right\rangle \\&\qquad + \left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)},c_0\,\dfrac{1}{\rho (s)}\partial _t\,P \right\rangle \right\rangle . \end{aligned} \end{aligned}$$

Using conditions (a) and (b), we obtain

$$\begin{aligned} \begin{aligned}&\left\langle -y\,\widetilde{b}(x)\,\partial _x\,{\textbf{w}}-c(y)\,\partial _y\,{\textbf{w}}\,+\,\widetilde{e}(y)\,{\textbf{w}},P\right\rangle \\&\quad = \left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)},(c_2\,r_0-b_0\,r_1)\partial _t\left( t^2P\right) +c_1\partial _t(t\,P)+ \left[ e_0+(e_1\,r_0-d_0\,r_1)\,t\right] P\right\rangle \right\rangle ,\\&\quad = \left\langle {\textbf{u}}_0^{(s)},\left\langle -\left[ \left( b_1\,r_0-r_1\,b_0\right) t^2+c_1\,t \right] \,D\,{\textbf{v}}^{(t)}+ \left[ e_0+(e_1\,r_0-d_0\,r_1)\,t\right] \,{\textbf{v}}^{(t)}, P \right\rangle \right\rangle , \end{aligned} \end{aligned}$$

and from condition (c) we get $\left\langle -y\,\widetilde{b}(x)\,\partial _x\,{\textbf{w}}-c(y)\,\partial _y\,{\textbf{w}}\,+\,\widetilde{e}(y)\,{\textbf{w}},P\right\rangle =0$, and the promised result follows.

1.3 A.3 Proof of Corollary 4.4

From Proposition 4.2 with $\rho (s)\, =\, \sqrt{\ell _0+2\,\ell _1\,s+\ell _2\,s^2}$, we get

$$\begin{aligned}{} & {} \left\langle -y\,\widetilde{b}(x)\,\partial _x\,{\textbf{w}}-c(y)\,\partial _y\,{\textbf{w}}\,+\,\widetilde{e}(y)\,{\textbf{w}},P\right\rangle \\{} & {} \quad = \,\left\langle -a(s)\,D\,{\textbf{u}}_0^{(s)} +\left( \widetilde{b}(s)+\widetilde{d}(s)\right) \,{\textbf{u}}_0^{(s)},\left\langle t\,{\textbf{v}}^{(t)}, \rho '(s)\,P\right\rangle \right\rangle \\{} & {} \qquad +\,\left\langle {\textbf{u}}_0^{(s)}, \left\langle -(c_2-b_1)\,\ell _2\,t^2\,D\,{\textbf{v}}^{(t)}+\ell _2\,(e_1-d_1)\,t\,{\textbf{v}}^{(t)},\frac{s^2}{\rho (s)}P \right\rangle \right\rangle \\{} & {} \qquad -\,\left\langle {\textbf{u}}_0^{(s)}, \left\langle \left[ (2\,c_2-b_1)\,\ell _1-b_0\,\ell _2 \right] \,t^2\,D\,{\textbf{v}}^{(t)}+(2e_1\ell _1-d_1\ell _1-d_0\ell _2)\,t\,{\textbf{v}}^{(t)},\frac{s}{\rho (s)}P \right\rangle \right\rangle \\{} & {} \qquad +\,\left\langle {\textbf{u}}_0^{(s)}, \left\langle -\left[ (c_2\,\ell _0-b_0\,\ell _1)\,t^2+c_0 \right] \,D\,{\textbf{v}}^{(t)}+\left( e_1\,\ell _0-d_0\,\ell _1 \right) \,t\,{\textbf{v}}^{(t)}, \frac{1}{\rho (s)}P \right\rangle \right\rangle . \end{aligned}$$

The result follows from the conditions (a)–(d).

1.4 A.4 Proof of Theorem 4.6

We begin with (4.8). Using (4.7), we have

$$\begin{aligned}&\left\langle {\textbf{w}}, \partial _y\!\left( c(y)\rho (x)b\left( \frac{y}{\rho (x)} \right) \!P \right) +\gamma c'(y)\rho (x)b\left( \frac{y}{\rho (x)} \right) \!P+ \beta c(y)\partial _y\!\left( \rho (x)b\left( \frac{y}{\rho (x)} \right) \right) \!P \right\rangle \\&\quad =c_1\,\left\langle \rho (s)\,{\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, \partial _t(t\,b(t)\,P)+\left[ \gamma \,b(t)+\beta \,t\,b'(t) \right] P \right\rangle \right\rangle \\&\qquad +c_0\left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, \partial _t(b(t)\,P)+\beta \,b'(t)\,P \right\rangle \right\rangle . \end{aligned}$$

If either one of the two cases in the statement of the theorem holds, then we obtain that ${\textbf{w}}$ satisfies the Pearson equation (4.8).

Using (4.9) and (4.7), we compute as follows.

$$\begin{aligned}&\left\langle {\textbf{w}}, \partial _x\left( a(x)\,\rho (x)\,b\left( \frac{y}{\rho (x)} \right) \! P\!\right) +\alpha \,a'(x)\,\rho (x)\,b\left( \frac{y}{\rho (x)}\right) \!P+\beta \,a(x)\,b_0\,\rho '(x)\,P \right\rangle \nonumber \\&\quad = \left\langle {\textbf{u}}_0^{(s)},\left\langle b(t)\,{\textbf{v}}^{(t)},\partial _s\left( a(s)\,\rho (s)\,P \right) +\alpha \,a'(s)\,\rho (s)\,P \right\rangle \right\rangle \nonumber \\&\qquad +\left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, \beta \,b_0\,\rho '(s)\,a(s)\,P-\rho '(s)\,a(s)\,t\,b'(t)\,P\right\rangle \right\rangle \nonumber \\&\qquad +\left\langle {\textbf{u}}_0^{(s)}, \left\langle D\left[ t\,b(t)\,{\textbf{v}}^{(t)}\right] , \rho '(s)\,a(s)\,P \right\rangle \right\rangle . \end{aligned}$$

(A.2)

Now, we consider each of the conditions in the statement of the theorem. Suppose that (i) holds and that ${\textbf{u}}_0^{(s)}$ satisfies

$$\begin{aligned} a(s)\,\rho (s)\,D{\textbf{u}}_0^{(s)} = \left[ \alpha \,a'(s)\,\rho (s)+(1+\gamma +\beta )\,\rho '(s)\,a(s) \right] {\textbf{u}}_0^{(s)}, \end{aligned}$$

and ${\textbf{v}}^{(t)}$ satisfies $t\,b(t)\,D{\textbf{v}}^{(t)} = \left[ \gamma \,b(t)+\beta \,t\,b'(t) \right] {\textbf{v}}^{(t)}$ or, equivalently,

$$\begin{aligned} D\left[ t\,b(t)\,{\textbf{v}}^{(t)} \right] = \left[ (\gamma +1)\,b(t)+(\beta +1)\,t\,b'(t) \right] {\textbf{v}}^{(t)}. \end{aligned}$$

Substituting this in the last equality of (A.2), we get

$$\begin{aligned}&\left\langle {\textbf{u}}_0^{(s)},\left\langle b(t)\,{\textbf{v}}^{(t)},\partial _s\left( a(s)\,\rho (s)\,P \right) +\alpha \,a'(s)\,\rho (s)\,P \right\rangle \right\rangle \\&\qquad +\left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, \beta \,b_0\,\rho '(s)\,a(s)\,P+\rho '(s)\,a(s)\left[ (\gamma +1)\,b(t)+\beta \,t\,b'(t) \right] P \right\rangle \right\rangle \\&\quad =\left\langle {\textbf{u}}_0^{(s)}, \left\langle b(t)\,{\textbf{v}}^{(t)}, \partial _s\left( a(s)\,\rho (s)\,P\right) +\left[ \alpha \,a'(s)\,\rho (s)+(1+\beta +\gamma )\,a(s)\,\rho '(s)\right] P \right\rangle \right\rangle \\&\quad =0. \end{aligned}$$

Therefore, ${\textbf{w}}$ satisfies the first Pearson equation (4.8).

On the other hand, if (ii) holds, and ${\textbf{u}}_0^{(s)}$ satisfies

$$\begin{aligned} a(s)\,\rho (s)\,D{\textbf{u}}_0^{(s)} = \left[ \alpha \,a'(s)\,\rho (s)+(1+\beta )\,\rho '(s)\,a(s) \right] {\textbf{u}}_0^{(s)} \end{aligned}$$

and ${\textbf{v}}^{(t)}$ satisfies $b(t)\,D{\textbf{v}}^{(t)} = \beta \,b'(t)\,{\textbf{v}}^{(t)}$, or, equivalently,

$$\begin{aligned} D\left[ b(t)\,{\textbf{v}}^{(t)} \right] = (\beta +1)\,b'(t)\,{\textbf{v}}^{(t)}. \end{aligned}$$

Substituting this in the last equality of (A.2), we get

$$\begin{aligned}&\left\langle {\textbf{u}}_0^{(s)},\left\langle b(t)\,{\textbf{v}}^{(t)},\partial _s\left( a(s)\,\rho (s)\,P \right) +\alpha \,a'(s)\,\rho (s)\,P \right\rangle \right\rangle \\&\qquad +\left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, \beta \,b_0\,\rho '(s)\,a(s)\,P+\rho '(s)\,a(s)\left[ \beta \,t\,b'(t)+b(t) \right] P \right\rangle \right\rangle \\&\quad =\left\langle {\textbf{u}}_0^{(s)},\left\langle b(t)\,{\textbf{v}}^{(t)}, \partial _s\left( a(s)\,\rho (s)\,P\right) +\left[ \alpha \,a'(s)\,\rho (s)+(1+\beta )\,a(s)\,\rho '(s) \right] \right\rangle \right\rangle \\&\quad =0. \end{aligned}$$

Therefore, in this case we also conclude that ${\textbf{w}}$ satisfies (4.8).

1.5 A.5 Proof of Theorem 4.7

We start with (4.11). For every polynomial $P\in \Pi ^2$,

$$\begin{aligned}&\left\langle -\rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \,{\textbf{w}}_y + \alpha \,\partial _y\left( \rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \right) \,{\textbf{w}}, P \right\rangle \\&\quad =\left\langle \rho (s)\,{\textbf{u}}_0^{(s)}, \left\langle -b(t)\,D{\textbf{v}}^{(t)}+\alpha \,b'(t)\,{\textbf{v}}^{(t)}, P \right\rangle \right\rangle =0. \end{aligned}$$

For (4.10), we get

$$\begin{aligned}&\left\langle -\rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \,{\textbf{w}}_x + \alpha \,\partial _x\left( \rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \right) \,{\textbf{w}}, P \right\rangle \\&\quad =\left\langle {\textbf{u}}_0^{(s)},\left\langle {\textbf{v}}^{(t)}, b(t)\left[ \partial _s\left( \rho (s)^2\,P \right) +\frac{1}{2}\left( \rho (s)^2\right) '\,P\right] \right\rangle \right\rangle \\&\qquad +\left\langle {\textbf{u}}_0^{(s)},\left\langle t\,b(t)\,D{\textbf{v}}^{(t)},\frac{1}{2}\left( \rho (s)^2\right) '\,P \right\rangle \right\rangle \\&\qquad +\left\langle {\textbf{u}}_0^{(s)}, \left\langle {\textbf{v}}^{(t)}, \alpha \,\left( \rho (s)^2\right) '\left( b_0+\frac{1}{2}b_1\,t \right) \,P \right\rangle \right\rangle . \end{aligned}$$

Using $b(t)\,D{\textbf{v}}^{(t)}=\alpha \,b'(t)\,{\textbf{v}}^{(t)}$, we obtain

$$\begin{aligned}&\left\langle -\rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \,{\textbf{w}}_x + \alpha \,\partial _x\left( \rho (x)^2\,b\left( \frac{y}{\rho (x)} \right) \right) \,{\textbf{w}}, P \right\rangle \\&\quad =\left\langle {\textbf{u}}_0^{(s)},\left\langle b(t)\,{\textbf{v}}^{(t)},\partial _s\left( \rho (s)^2\,P \right) +\left( \alpha +\frac{1}{2}\right) \,\left( \rho (s)^2\right) ' \right\rangle \right\rangle =0. \end{aligned}$$