2. Preliminaries on exponential polynomials

We need to introduce several concepts some of which are new.

For example, $f(z)=\textrm {e}^{\pi z}+3\textrm {e}^{2\pi z}+z\textrm {e}^{3\pi z}$ and $g(z)=\textrm {e}^{4iz}+\textrm {e}^{6iz}$ are simple exponential polynomials, both of their frequencies are commensurable, and examples for common factors are $\pi ,\pi /2$ for $f$ and $i,2i$ for $g$. In particular, a common factor is not unique.

Note that it is usual to say that non-zero real numbers $a$ and $b$ are commensurable if their ratio $a/b$ is a rational number. Equivalently, there exist a real number $c$ and integers $m$ and $n$ such that $a=mc$ and $b=nc$. In definition 2.1 we are concerned with a simple exponential polynomial and a fixed $\theta \in [0, 2\pi )$, and thus all the non-zero frequencies are of the form $w_j=r_j\textrm {e}^{i\theta }$ for $r_j>0$, and the ratio of $w_j$ and $w_i$ is

\[ \frac{w_j}{w_i}=\frac{r_j}{r_i}=\frac{w_j/w}{w_i/w}. \]

This is a positive rational number for a common factor $w$. If we consider the dual exponential polynomial which are both commensurable with the same common factor, all those frequencies are commensurable in the usual sense, that is, each ratio of the frequencies can be an integer not restricted to a positive integer any more.

If $f$ is a simple exponential polynomial of order one with constant multipliers, and if its frequencies are commensurable as in Frei's case (1.2), then by the fundamental theorem of algebra, $f$ can be written as

\[ f(z)=A\prod_{j=1}^{m}\left(\textrm{e}^{wz}-\alpha_j\right), \]

where $A\neq 0$, $\alpha _j$'s are complex constants and $m$ is a positive integer. In particular, all the zeros of $f$ lie on at most $m$ lines.

We note that if the non-zero frequencies of $f$ are commensurable, then they are clearly linearly dependent over rationals (see [Reference Moreno13] for results in this direction), but not the other way around. For example, the points $w_1=1, w_2=\sqrt {2}, w_3=\sqrt {2}-1$ are linearly dependent over rationals but not commensurable.

For example, the functions $f(z)=1+z\textrm {e}^{z}+2\textrm {e}^{3z}$ and $g(z)=1-\textrm {e}^{-z}$ are strongly dual exponential polynomials, while $f(z)$ and $h(z)=g(z)+2z^{2}\textrm {e}^{-2z}$ are not. Note that if $\arg (w_j)=\theta$, then $\arg (\lambda _j)=\theta +\pi$ by duality, and moreover, if $w_j+\lambda _i\neq 0$, then precisely one of $\arg (w_j+\lambda _i)=\theta$ or $\arg (w_j+\lambda _i)=\theta +\pi$ holds for all $i,j>0$. Alternatively, strong duality of $f$ and $g$ of order $q$ can be expressed as follows: there exists a non-zero constant $w$ such that

(2.1)\begin{equation} f(z) = \sum_{j = 0}^{m} F_j(z)(\textrm{e}^{wz^{q}})^{j}\quad \textrm{and}\quad g(z) = \sum_{i = 0}^{m} G_i(z)(\textrm{e}^{{-}wz^{q}})^{i}, \end{equation}

where $F_j, G_i$ are exponential polynomials of order $\leq q-1$. Hence $f$ is a polynomial in $\textrm {e}^{wz^{q}}$ and $g$ is a polynomial in $\textrm {e}^{-wz^{q}}$, with smaller exponential polynomials as multipliers. Using the notation above, $f\in \operatorname {Exp}_{q-1}[\textrm {e}^{wz^{q}}]$ and $g\in \operatorname {Exp}_{q-1}[\textrm {e}^{-wz^{q}}]$. Differing from the situation in (1.5), some of the multipliers $F_j, G_i$ $(i, j>0)$ in (2.1) must suitably vanish identically so that either of $j-i\geq 0$ or $j-i\leq 0$ always holds for all non-vanishing multipliers $F_j, G_i$ $(i, j>0)$. This is a consequence of definition 2.2.

We may think that being strong in our duality means that the product of $f(z)-F_0(z)$ and $g(z)-G_0(z)$ becomes again a commensurable exponential polynomial with either $w$ or $-w$ as a common factor. In the case when both $F_0(z)$ and $G_0(z)$ are constant, each product of the derivatives $f^{(k)}(z)$ and $g^{(\ell )}(z)$, $k, \ell \in \mathbb {N}$, is a commensurable exponential polynomial with the same common factor as the product of $f(z)-F_0(z)$ and $g(z)-G_0(z)$.

The set $\operatorname {co}(W_f)$ is defined as the intersection of all closed convex sets containing $W_f$, and as such it is either a convex polygon or a line segment. The latter occurs when $f$ is simple, and, in particular, when $w_1, \ldots , w_m$ are commensurable. The vertices of $\operatorname {co} (W_f)$ are formed by some (possibly all) of the points $\overline {w}_0,\overline {w}_1,\ldots ,\overline {w}_m$. The circumference $C(\operatorname {co}(W_f))$ of $\operatorname {co}(W_f)$ plays an important role in describing the value distribution of $f$, see [Reference Heittokangas, Ishizaki, Tohge and Wen8, Reference Heittokangas and Wen10, Reference Steinmetz16].

Let $h$ be a quotient of two transcendental exponential polynomials, say

\[ h(z)=f(z)/g(z), \]

where $f$ is of the form (1.5) and $g$ is an exponential polynomial of the normalized form

\[ g(z)=G_0(z)+G_1(z)\textrm{e}^{w_1z^{q}}+\cdots+G_m(z)\textrm{e}^{w_mz^{q}}. \]

In these representations of $f$ and $g$ for the quotient $h$, we allow that some of the multipliers $F_j$ or $G_j$ may vanish identically, but we suppose that the matching multipliers $F_j$ and $G_j$ do not both vanish identically for any $j$.

For the quotient $h=f/g$, define the set $W_h=\{\overline {w}_0,\overline {w}_1,\ldots ,\overline {w}_m\}$. The proximity function of $h$ satisfies

(2.2)\begin{equation} m(r,h)=\big(C(\operatorname{co}(W_h))-C(\operatorname{co}(W_g))\big)\frac{r^{q}}{2\pi}+o(r^{q}), \end{equation}

see [Reference Steinmetz17, satz 1]. In particular, if $g\equiv 1$, then $W_g=\{0\}$ and $C(\operatorname {co}(W_g))=0$. This yields [Reference Steinmetz16, satz 1] as a special case, namely

(2.3)\begin{equation} T(r,f)=m(r,f)=C(\operatorname{co}(W^{0}_f))\frac{r^{q}}{2\pi}+o(r^{q}), \end{equation}

where $W^{0}_f=W_f\cup \{0\}$. The estimates (2.2) and (2.3) are consistent with the estimate

\[ m\left(r,\frac{f'}{f}\right)=o(T(r,f)), \]

known as the lemma on the logarithmic derivative, since $W_{f'/f}=W_f$ holds for any given exponential polynomial $f$ of the form (1.5). We also point out that $W_{f/f'}=W_{f'}$. This fact will be used in proving our main results in § 3.

3. The main results

Motivated by example 1.3, we improve theorem 1.6(a) under weaker assumptions on $B(z)$.

Proof. (a) Suppose that $0\leq \rho (f)\leq \rho (A)-1$. The case $\rho (f)=0$ is not possible because $f$ is transcendental. Hence $\rho (f)\geq 1$. But now

\begin{equation*} |A|\leq |f''/f'|+|B||f/f'| \end{equation*}

and the assumption $T(r,B)=o(T(r,A))$ imply

\[ T(r,A)=m(r,A)\leq m(r,B)+O(r^{\rho(A)-1})=o(T(r,A)), \]

which is a contradiction. Here we have used (2.2) for $h=f/f'$ and $g=f'$, as well as (2.3) for $A$ in place of $f$. The following two cases are also impossible by the proof of [Reference Heittokangas, Laine, Tohge and Wen9, theorem 3.6]:

1. (1) $\rho (f)=\rho (A)$ and either $F_0(z)\equiv 0$ or $F'_0(z)\not \equiv 0$.

2. (2) $\rho (f)\geq \rho (A)+1$.

Thus $\rho (f)=\rho (A)=q\geq 1$ and $f$ has the representation (1.6).

We proceed to prove that $f$ and $A(z)$ are dual exponential polynomials. Using (1.3), we find that

\[ m\left(r,\frac{Af'}{f}\right)=O(\log r)+m(r,B)=o(T(r,A))=o\left(r^{q}\right). \]

Formula (7.3) in [Reference Wen, Gundersen and Heittokangas18] should be replaced by this. Thus formula (7.7) in [Reference Wen, Gundersen and Heittokangas18] holds, and the reasoning in [Reference Wen, Gundersen and Heittokangas18] shows that $f$ and $A(z)$ are dual exponential polynomials.

To complete the proof of (a), it suffices to prove that $B(z)$ is an exponential polynomial of order $\rho (B)\leq q-1$. Since the frequencies $w_j$ of $f$ are all on one ray, we may appeal to a rotation, and suppose that $w_1,\ldots ,w_m\in \mathbb {R}_+$. By renaming the frequencies $w_j$, if necessary, we may further suppose that $0< w_1<\cdots < w_m$. Thus the dual coefficient must be of the form

(3.1)\begin{equation} A(z)=A_0(z)+\sum_{j=1}^{k} A_j(z)\textrm{e}^{-\lambda_j z^{q}}, \end{equation}

where $A_j(z)\not \equiv 0$ for all $j\in \{1,\ldots ,k\}$ and $\lambda _1,\ldots ,\lambda _k\in \mathbb {R}_+$. Renaming the frequencies $\lambda _j$, if necessary, we may suppose that $0<\lambda _1<\cdots <\lambda _k$. Write

\[ f'(z)=\sum_{j=1}^{m} G_j(z)\textrm{e}^{w_jz^{q}}\quad \textrm{and}\quad f''(z)=\sum_{j=1}^{m} H_j(z)\textrm{e}^{w_jz^{q}}, \]

where $G_j(z)=F_j'(z)+qw_jz^{q-1}F_j(z)\not \equiv 0$ and $H_j(z)=G_j'(z)+qw_jz^{q-1}G_j(z)\not \equiv 0$. Next, write $-Af'=Bf+f''$ in the form

(3.2)\begin{equation} -\left(\sum_{j=1}^{k}A_j\textrm{e}^{-\lambda_jz^{q}}\right)\left(\sum_{j=1}^{m} G_j\textrm{e}^{w_jz^{q}}\right)=cB+\sum_{j=1}^{m} (A_0G_j+BF_j+H_j)\textrm{e}^{w_jz^{q}}. \end{equation}

From (3.2) we find that $B$ is an exponential polynomial of order $\rho (B)\leq q$. In fact, from (2.3) and the assumption $T(r,B)=o(T(r,A))$, it follows that $\rho (B)\leq q-1$.

(b) We begin with some preparations. From [Reference Steinmetz16] and [Reference Steinmetz17], we have

\[ m\left(r,\sum_{j=1}^{k}A_j\textrm{e}^{-\lambda_jz^{q}}\right)=T\left(r,\sum_{j=1}^{k}A_j\textrm{e}^{-\lambda_jz^{q}}\right) =\frac{\lambda_k}{\pi}r^{q}+o(r^{q}), \]

and

\begin{eqnarray*} &m\left(r,\bigg\{cB+\sum_{j=1}^{m} (A_0G_j+BF_j+H_j)\textrm{e}^{w_jz^{q}}\bigg\}\Big/\sum_{j=1}^{m} G_j\textrm{e}^{w_jz^{q}}\right)\\ &\quad= \frac{2w_m-2(w_m-w_1)}{2\pi}r^{q}+o(r^{q})= \frac{w_1}{\pi}r^{q}+o(r^{q}). \end{eqnarray*}

Therefore, we deduce that

(3.3)\begin{equation} 0<\lambda_1<\cdots <\lambda_k=w_1<\cdots < w_m. \end{equation}

Thus from (3.2), it follows that

(3.4)\begin{equation} {-}A_kG_1=cB. \end{equation}

If $A_0G_m+BF_m+H_m\not \equiv 0$, then from [Reference Heittokangas and Wen10, theorem 2.2] and (3.3), we get

\begin{eqnarray*} N(r,0,L)&=&\frac{2(w_m-w_1)+2(\lambda_k-\lambda_1)}{2\pi}r^{q}+O(r^{q-1}+\log r)\\ &=&\frac{w_m-\lambda_1}{\pi}r^{q}+O(r^{q-1}+\log r),\\ N(r,0,R)&=&\frac{w_m}{\pi}r^{q}+O(r^{q-1}+\log r), \end{eqnarray*}

where $N(r,0,L)$ and $N(r,0,R)$ are the counting functions of zeros of the exponential polynomials on the left-hand side and on the right-hand side of (3.2), respectively. This implies $w_m=w_m-\lambda _1$, which is impossible. Thus we have

(3.5)\begin{equation} A_0G_m+BF_m+H_m\equiv 0. \end{equation}

Now (3.2) reduces to

(3.6)\begin{equation} -\left(\sum_{j=1}^{k}A_j\textrm{e}^{-\lambda_jz^{q}}\right)\left(\sum_{j=1}^{m} G_j\textrm{e}^{w_jz^{q}}\right) =cB+\sum_{j=1}^{m-1} (A_0G_j+BF_j+H_j)\textrm{e}^{w_jz^{q}}. \end{equation}

From the Borel–Nevanlinna theorem, and from $A_iG_j\not \equiv 0$ for $j\in \{1,2,\ldots ,m\}$ and $i\in \{1,2,\ldots ,k\}$, it follows that there are only two possibilities:

1. (I) For some pairs $(\,j,i)$, where $j\in \{1,2,\ldots ,m\}$ and $i\in \{1,2,\ldots ,k\}$, there exists $\ell \in \{0,1,\ldots ,m-1\}$ such that

   (3.7)\begin{equation} w_j-\lambda_i=w_\ell. \end{equation}

2. (II) For some pairs $(\,j,i)$, where $j\in \{1,2,\ldots ,m\}$ and $i\in \{1,2,\ldots ,k\}$, there exist $s\in \{1,2,\ldots ,m\}\setminus \{j\}$ and $t\in \{1,2,\ldots ,k\}\setminus \{i\}$ such that

   (3.8)\begin{equation} w_j-\lambda_i=w_s-\lambda_t. \end{equation}

After these preparations we proceed to prove that the frequencies of $f$ are commensurable if and only if the frequencies of $A(z)$ are commensurable. By appealing to (3.3) and to a change of variable as in example 1.3, we may suppose that $w_1=\lambda _k\in \mathbb {N}$. Thus we prove that $w_j\in \mathbb {N}$ for $j\in \{1,\ldots ,m\}$ if and only if $\lambda _i\in \mathbb {N}$ for $i\in \{1,\ldots ,k\}$.

1. (i) Suppose that $w_j\in \mathbb {N}$ for $j\in \{1,\ldots ,m\}$. From (3.3), we see that $w_m-\lambda _1=\max _{j,i}\{w_j-\lambda _i\}$ and $w_m-\lambda _1>w_j-\lambda _i$ for any $j\neq m$ and $i\neq 1$. Hence, from (3.7) and (3.8), there exists $p< m$ such that $w_m-\lambda _1=w_p$, which implies that $\lambda _1\in \mathbb {N}$. In addition, from (3.3), we have $w_m-\lambda _2>w_j-\lambda _i$ for any $j\neq m$ and $i>2$. Thus, from (3.7) and (3.8), there are only two possibilities: (1) there exists $p< m$ such that $w_m-\lambda _2=w_p-\lambda _1$ and (2) there exists $p< m$ such that $w_m-\lambda _2=w_p$. In both cases, it follows that $\lambda _2\in \mathbb {N}$. Repeating this argument for $k$ times gives us $\lambda _i\in \mathbb {N}$ for $i\in \{1,\ldots ,k\}$.

2. (ii) Suppose that $\lambda _i\in \mathbb {N}$ for $i\in \{1,\ldots ,k\}$. From (3.3), we have $\lambda _k=w_1$, and consequently $w_1\in \mathbb {N}$. Moreover, from (3.3), we have $w_2-\lambda _k< w_j-\lambda _i$ for any $j>1$ and $i\neq k$. Thus, from (3.7) and (3.8), there are only two possibilities: there exists $p< k$ such that either $w_2-\lambda _k=w_1-\lambda _p$ or $w_2-\lambda _k=w_1$. In both cases, we have $w_2\in \mathbb {N}$. Repeating this argument for $m$ times gives us $w_j\in \mathbb {N}$ for $i\in \{1,\ldots ,m\}$.

If the frequencies are commensurable for one of $f,A(z)$, then they are commensurable for both of $f,A(z)$ by the reasoning above. The remaining fact that $f$ and $A(z)$ are strongly dual exponential polynomials now follows by (3.3).

The assumption $\rho (Af')<\rho (f)$ in theorem 1.6(b) seems to be the only known sufficient condition for the conclusion $q=1$. However, in the case of Frei's result (1.1), we have

\[ A(z)f'(z)=\textrm{e}^{{-}z}\sum_{j=1}^{m} jC_j\textrm{e}^{jz}=\sum_{j=0}^{m-1}(\,j+1)C_{j+1}\textrm{e}^{jz}, \]

and so $\rho (Af')=\rho (f)=1$. This shows that $q=1$ may happen even if $\rho (Af')=\rho (f)$. Theorem 3.2 shows that $f$ having only one large exponential term is also a sufficient condition for $q=1$. In contrast, if $A(z)$ has only one large exponential term, then $f$ can have multiple large exponential terms as in (1.1).

Theorem 3.2 Suppose that $f(z)=F_0(z)+F_1(z)\textrm {e}^{wz^{q}}$ is a solution of (1.3), where $A(z)$ is an exponential polynomial and $B(z)$ is an entire function satisfying $T(r,B)=o(T(r,A))$. Then $q=1$, and there are constants $c,b\in \mathbb {C}\setminus \{0\}$ and a non-trivial polynomial $P(z)$ such that

\[ f(z)=c+b\textrm{e}^{wz},\ A(z)=\frac{b}{c}P(z)-w+P(z)\textrm{e}^{{-}wz}\ \text{and}\ B(z)={-}\frac{wb}{c}P(z). \]

Proof. We proceed similarly as in the proof of theorem 3.1 until (3.6), which now reduces to the form

(3.9)\begin{equation} -\left(\sum_{j=1}^{k}A_j\textrm{e}^{-\lambda_jz^{q}}\right)G_1\textrm{e}^{wz^{q}}=cB, \end{equation}

where $F_0(z)\equiv c\in \mathbb {C}\setminus \{0\}$. Hence $k=1$, and consequently $A(z)$ reduces to the form

\[ A(z)=A_0(z)+A_1(z)\textrm{e}^{{-}wz^{q}}. \]

From (3.9) and (3.5), with $k=1=m$, we find that

\[{-}A_1G_1=cB\quad \textrm{and}\quad -A_0G_1=BF_1+H_1. \]

In other words,

(3.10)\begin{equation} c^{{-}1}A_1G_1F_1=A_0G_1+H_1=A_0G_1+G_1'+qwz^{q-1}G_1. \end{equation}

Dividing both sides of (3.10) by $G_1$, we observe that at every zero of $G_1$ the right-hand side has a pole but the left-hand side does not. Thus $G_1$ has no zeros, and so we may write it in the form $G_1=\textrm {e}^{g}$, where $g(z)=a_{q-1}z^{q-1}+\cdots +a_0$ is a polynomial of degree $\leq q-1$. Since

\[ G_1=F_1'+qwz^{q-1}F_1=\textrm{e}^{g}, \]

we obtain $(F_1(z)\textrm {e}^{wz^{q}})'=\textrm {e}^{wz^{q}+g(z)}$, and consequently

(3.11)\begin{equation} F_1(z)\textrm{e}^{wz^{q}}=\int^{z} \textrm{e}^{w\zeta^{q}+a_{q-1}\zeta^{q-1}+\cdots+a_0}\, \textrm{d}\zeta. \end{equation}

Here the right-hand side is an exponential polynomial, which happens only if $q=1$.

Since $q=1$, we see from (3.11) that $F_1(z)$ reduces to a non-zero constant, say $F_1(z)\equiv b$. Thus $f(z)=c+b\textrm {e}^{wz}$, and we have $G_1(z)\equiv w b$ and $H_1(z)\equiv w^{2}b$. A substitution to (3.10) followed by a simplification gives

\[ \frac{b}{c}A_1=A_0+w. \]

There is no restriction for $A_1$ other than the fact that $A$ is an exponential polynomial. Thus we may suppose that $A_1$ is any non-trivial polynomial, say $A_1=P$. This gives us $A_0=({b}/{c})P-w$, and finally $B=-({wb}/{c})P$.

Example 1.1 shows that the coefficient $B(z)$ in (1.3) can be a polynomial. Next, we prove that this is equivalent to $A_0(z)$ in (3.1) being a polynomial, and reveal another sufficient condition for the conclusion $q=1$.

Proof. From the proof of theorem 3.1 we find that (3.5) holds, that is,

(3.12)\begin{equation} A_0G_m+BF_m+H_m=0, \end{equation}

where

\begin{eqnarray*} G_m &=& F_m'+w_mqz^{q-1}F_m,\\ H_m &=& F_m''+2w_mqz^{q-1}F_m'+\left(w_mq(q-1)z^{q-2}+w_m^{2}q^{2}z^{2q-2}\right)F_m. \end{eqnarray*}

Thus $F_m$ solves the second-order differential equation

(3.13)\begin{equation} F''_m + P(z)F'_m+Q(z)F_m=0, \end{equation}

where

\begin{eqnarray*} P(z) &=& 2w_mqz^{q-1}+A_0,\\ Q(z) &=& w_mqz^{q-1}A_0+w_mq(q-1)z^{q-2}+w_m^{2}q^{2}z^{2(q-1)}+B. \end{eqnarray*}

Suppose first that $A_0(z)$ is a polynomial. If $B(z)$ is transcendental, then it follows from (3.13) and [Reference Gundersen5, corollary 1] that $\rho (F_m)=\infty$, which is a contradiction. Hence $B(z)$ must be a polynomial. Conversely, suppose that $B(z)$ is a polynomial. Suppose on the contrary to the assertion that $A_0(z)$ is a transcendental exponential polynomial. Then there exists an open sector $S$ such that $A_0(z)$ blows up exponentially in $S$. Using [Reference Gundersen6, corollary 1] and $\rho (F_m)\leq q-1$ in (3.13), we obtain on almost every ray in $S$ that

\[ |w_mqz^{q-1}||A_0(z)| \leq O\left(|z|^{\max\{2q,\deg(B)\}}\right)+O\left(|z|^{q-2+\varepsilon}|A_0(z)|\right). \]

However, this is obviously a contradiction, and hence $A_0(z)$ is a polynomial.

Finally, suppose that the multipliers of $f$ and of $A(z)$ are constants. From (3.4), we find that $B(z)=Cz^{q-1}$ for some constant $C\in \mathbb {C}\setminus \{0\}$. Since $F_m(z)$ is a non-zero constant function, it follows that the coefficient $Q(z)$ in (3.13) vanishes identically. But this is not possible because $A_0(z)$ is a constant function, unless $q=1$.

Open problem 1. Under the assumptions of theorem 3.1, is it always true that $q=1$ and $B(z)$ is a polynomial?

This problem is fragile in the sense that the desired conclusion is not valid if a minor modification in the assumptions of theorem 3.1 is performed. For example, the differential equation

\[ f''-\bigl(qwz^{q-1} + z^{{-}1}\textrm{e}^{{-}wz^{q}}\bigr)f'-q(q-1)wz^{q-2}f=0 \]

possesses an exponential polynomial solution $f(z)=\textrm {e}^{wz^{q}}-1/(q-1)$ for any $q\geq 2$. Moreover, the function $f(z)=\textrm {e}^{z^{2}}+1$ satisfies the differential equations

(3.14)\begin{equation} \begin{split} & f''+\left(\frac{\textrm{e}^{{-}z^{2}}-1}{2z}-2z\right)f'-f = 0,\\ & f''-\frac{\textrm{e}^{{-}z^{2}}(z-1)+4z^{2}+z+1}{2z}f'+(z-1)f = 0.\end{split} \end{equation}

The transcendental coefficients in (3.14) are entire exponential polynomials with rational multipliers because $z=0$ is a removable singularity for both.

4. Duality for higher order functions

Next we construct examples of differential equations of order $n\geq 2$ having an exponential polynomial solution $f$ of order $\rho (f)=n-1$ which is dual with one of the coefficients.

Example 4.1 If $H$ is an arbitrary entire function, then $f(z)=\textrm {e}^{z^{2}}+1$ solves

(4.1)\begin{equation} \begin{split} & f'''+\left(1+\textrm{e}^{{-}z^{2}}\right)Hf''-(6+4z^{2})f'-(2+4z^{2})Hf=0,\\ & f'''-2zf''+(H-4+H\textrm{e}^{{-}z^{2}})f'-2zHf = 0. \end{split} \end{equation}

A particularly interesting case is when $H$ is either a polynomial or an exponential polynomial of order one. Thus either of the two possible coefficients can be dual with $f$. Examples of second-order dual solutions for third-order equations can be found in [Reference Wen, Gundersen and Heittokangas18] but for polynomial coefficients only.

We can use the relation $zf''(z)=(2z^{2}+1)f'$ to see that, in addition to (4.1), the function $f(z)=\textrm {e}^{z^{2}}+1$ satisfies the equations

\begin{eqnarray*} f'''(z)-2zf'(z)-4f'(z)&=&0,\\ (1+\textrm{e}^{{-}z^{2}})f'(z)-2zf(z)&=&0,\\ (1+\textrm{e}^{{-}z^{2}})f''(z)-(2+4z^{2})f(z)&=&0. \end{eqnarray*}

Example 4.2 If $H$ is an arbitrary entire function, then $f(z)=\textrm {e}^{z^{3}}+1$ solves

\begin{eqnarray*} f^{(4)}+\left(1+\textrm{e}^{{-}z^{3}}\right)Hf'''-9z^{4}f''-30\left(2+3z^{3}\right)f'-\left(6+54z^{3}+27z^{6}\right)Hf &=& 0,\\ f^{(4)}-3z^{2}f'''+\left(H-18z+H\textrm{e}^{{-}z^{3}}\right)f''-18f'-H\left(6z+9z^{4}\right)f &=& 0,\\ f^{(4)}-3z^{2}f'''-27zf''+\left(H+27z^{3}+H\textrm{e}^{{-}z^{3}}\right)f'-3z^{2}Hf &=& 0. \end{eqnarray*}

A particularly interesting case is when $H$ is an exponential polynomial of order at most two. Thus all three of the possible coefficients can be dual with $f$. Previous examples of third-order dual solutions do not seem to be known.

As in the previous example, we can use the relations $zf''(z)=(3z^{3}+2)f'(z)$ and $zf'''(z)=(3z^{3}+1)f''(z)+9z^{2}f'(z)$ to see that $f(z)=\textrm {e}^{z^{3}}+1$ satisfies the equations

\begin{eqnarray*} f^{(4)}(z)-3z^{2}f'''(z)-18zf''(z)-18f'(z)&=&0,\\ (1+\textrm{e}^{{-}z^{3}})f'(z)-3z^{2}f(z)&=&0,\\ (1+\textrm{e}^{{-}z^{3}})f''(z)-(6z+9z^{4})f(z)&=&0,\\ (1+\textrm{e}^{{-}z^{3}})f'''(z)-(6+54z^{3}+27z^{6})f(z)&=&0. \end{eqnarray*}

In light of open problem 1 and the examples just discussed, it is natural to pose our second open problem.

Open problem 2. If a solution and the dominant coefficient are dual exponential polynomials of order $q$, then is the differential equation in question of order at least $q+1$?

For the fragility of this problem, recall equations (3.14) satisfied by $f(z)=\textrm {e}^{z^{2}}+1$. Moreover, the function $f(z)=\textrm {e}^{z^{3}}+1$ satisfies the third-order equation

\begin{equation*} f'''+\left(\frac{\textrm{e}^{{-}z^{3}}-1}{2z}\right)f''-3z(3z^{3}+5)f'-\frac{3}{2}(3z^{3}+2)f=0 \end{equation*}

with entire coefficients.

As the first initial step to knowing more about open problem 2, we make a summary of the fundamental ideas in constructing examples 4.1 and 4.2.

Proof. First, let us prove (i) by induction on $j$. Of course, by taking their logarithmic derivatives, we have $(1+\textrm {e}^{-z^{q}})f'(z)=qz^{q-1}f(z)$ immediately, that is, the case when $j=0$ follows with $P_{0,0}(z)=qz^{q-1}$. Assume (i) is true for each $j=0, 1, \ldots , n$. Then

\begin{eqnarray*} (1+\textrm{e}^{{-}z^{q}})f^{(n+2)}(z) &=& qz^{q-1}\textrm{e}^{{-}z^{q}}f^{(n+1)}+\sum_{k=0}^{n} \bigl\{ P_{n,k}'(z)f^{(k)}(z)+P_{n,k}(z)f^{(k+1)}(z)\bigr\}\\ &=& qz^{q-1}(1+\textrm{e}^{{-}z^{q}})f^{(n+1)}(z)+\bigl\{P_{n,n}(z)-qz^{q-1}\bigr\}f^{(n+1)}(z)\\ & & +\sum_{k=1}^{n} \bigl\{ P_{n,k}'(z)f^{(k)}(z)+P_{n,k-1}(z)\bigr\} f^{(k)}(z)+P_{n,0}'(z)f(z) \\ &=& qz^{q-1} \sum_{k=0}^{n} P_{n,k}(z)f^{(k)}(z) + \bigl\{P_{n,n}(z)-qz^{q-1}\bigr\}f^{(n+1)}(z)\\ & & +\sum_{k=1}^{n} \bigl\{ P_{n,k}'(z)f^{(k)}(z)+P_{n,k-1}(z)\bigr\} f^{(k)}(z)+P_{n,0}'(z)f(z) \\ &=& \bigl\{P_{n,n}(z)-qz^{q-1}\bigr\}f^{(n+1)}(z) \\ & & +\sum_{k=1}^{n} \bigl\{ P_{n,k}'(z)f^{(k)}(z)+qz^{q-1}P_{n,k}(z)+P_{n,k-1}(z)\bigr\}f^{(k)}(z)\\ & & +\bigl\{P_{n,0}'(z)+qz^{q-1}P_{n,0}(z)\bigr\}f(z), \end{eqnarray*}

which is the one to be proved.

Second, let us calculate the $q$-th order derivative of the product $f'(z)\textrm {e}^{-z^{q}}=qz^{q-1}$. The Leibniz rule gives

\[ \sum_{\ell=0}^{q} \binom{q}{\ell} f^{(\ell+1)}(z)(\textrm{e}^{{-}z^{q}})^{(q-\ell)}\equiv 0. \]

Denoting $Q_{\ell +1}(z)=-\binom{q}{\ell }(\textrm {e}^{-z^{q}})^{(q-\ell )}\textrm {e}^{z^{q}}$ for $0\leq \ell \leq q-1$, we have

\[ f^{(q+1)}(z)=\sum_{\ell=0}^{q-1}Q_{\ell+1}(z)f^{(\ell+1)}(z)=\sum_{\ell=1}^{q}Q_{\ell}(z)f^{(\ell)}(z), \]

as desired.

Example 4.4 We may apply the two identities in lemma 4.3 to construct differential equations of arbitrary order. Given any entire function $H$, we have the identity

\[ f^{(q+1)}(z)-\sum_{\ell=1}^{q} Q_{\ell}(z)f^{(\ell)}(z)=H(z)\left( (1+\textrm{e}^{{-}z^{q}})f^{(\,j)}(z)-\sum_{k=0}^{j-1} P_{j-1,k}(z)f^{(k)}(z) \right), \]

that is, $f(z)=\textrm {e}^{z^{q}}+1$ solves

\begin{eqnarray*} f^{(q+1)}(z) &-&\sum_{\ell=j+1}^{q} Q_{\ell}(z)f^{(\ell)}(z)- \left( (1+\textrm{e}^{{-}z^{q}})H(z) +Q_j(z) \right) f^{(\,j)}(z)\\ &+&\sum_{\ell=1}^{j-1} \bigl(P_{j-1,\ell}(z) H(z)-Q_{\ell}(z)\bigr)f^{(\ell)}(z) +P_{j-1,0}(z)H(z)f(z)=0, \end{eqnarray*}

where $1\leq j \leq q$, the sum $\sum _{\ell =1}^{j-1}$ is empty if $j=1$ and the sum $\sum _{\ell =j+1}^{q}$ is empty if $j=q$.

5. Multiple duality

The possibility that a solution $f$ would be dual to more than one coefficient has not been studied rigorously. In this case there would be at least two equally strong dominant coefficients, or, in the case of (1.3), both coefficients $A(z),B(z)$ would be equally strong. For example, $f(z)=\textrm {e}^{-z}$ solves

\[ f''+\textrm{e}^{z}f'+(\textrm{e}^{z}-1)f=0 \]

and is dual to both coefficients. Obviously the coefficients are not dual to each other. More examples can be produced from example 1.1. Note that $f(z)=\textrm {e}^{z}$ solves (1.3) if $A(z)=-B(z)-1$. Hence $f$ is not necessarily dual with either of $A(z),B(z)$.

If $H$ is any entire function, then $f(z)=\textrm {e}^{z^{q}}$ solves

\[ f''+\left(H(z)-qz^{q-1}\right)f'-\left(q(q-1)z^{q-2}+qz^{q-1}H(z)\right)f=0. \]

This example is from [Reference Gundersen5]. Note that $f(z)=\textrm {e}^{z^{q}}$ satisfies both $f'(z)-qz^{q-1}f(z)=0$ and $f''(z)-qz^{q-1}f'(z)-q(q-1)z^{q-2}f(z)=0$.

Recall [Reference Heittokangas, Laine, Tohge and Wen9, theorem 2.1], according to which there cannot be even one ray on which $B(z)$ would be stronger than $A(z)$ in the sense of the Phragmén–Lindelöf indicator, for otherwise all solutions of (1.3) are of infinite order. This happens, for example, when $A(z)$ and $B(z)$ are dual to each other.

Example 5.1 One may observe the necessity of the assumption on the duality of $f$ and $A(z)$ as well as that on the dominance of $A(z)$ over $B(z)$ by the following example: the function $f(z)=\bigl (\textrm {e}^{z}+\textrm {e}^{-z}\bigr )\textrm {e}^{z^{q}}$, $q\in \mathbb {N}$, satisfies

\begin{eqnarray*} f''&+&\bigl\{H\textrm{e}^{z}+H\textrm{e}^{{-}z}-2qz^{q-1}\bigr\}f' \\ &-&\bigl\{(qz^{q-1}+1)H\textrm{e}^{z}+(qz^{q-1}-1)H\textrm{e}^{{-}z}-q^{2}z^{2(q-1)}+q(q-1)z^{q-2}+1 \bigr\}f=0 \end{eqnarray*}

for any entire function $H$. When $q=1$, this becomes

\[ f''+\bigl\{H\textrm{e}^{z}+H\textrm{e}^{{-}z}-2\bigr\}f'-2H\textrm{e}^{z}f=0 \]

with $f(z)=\textrm {e}^{2z}+1$. Thus we may use it in order to observe the duality of $A(z)$ and $B(z)$ by several choices of $H$ such as $H(z)=\textrm {e}^{nz}$ for $n\in \mathbb {Z}$ or $H(z)=\textrm {e}^{iz}$.

Here we note that $f(z)=F(z)\textrm {e}^{z^{q}}$ satisfies ${f'}/{f}={F'}/{F}+qz^{q-1}$ and

\[ \frac{f''}{f}=\frac{F''}{F}+2qz^{q-1}\frac{f'}{f}+\bigl(q(q-1)z^{q-2}-q^{2}z^{2(q-1)}\bigr) \]

so that there is no large freedom to choose the function $F$. For example, taking an Airy function as $F$, we cannot have our desired equation $f''+A(z)f'+B(z)f=0$ with the exponential polynomial coefficients $A(z)$ and $B(z)$.