\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 108, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/108\hfil Small functions and the solutions] {Relations between the small functions and the solutions of certain second-order differential equations} \author[H. Liu, Z. Mao \hfil EJDE-2007/108\hfilneg] {Huifang Liu, Zhiqiang Mao} % in alphabetical order \address{Huifang Liu \newline School of Mathematics, South China Normal University, Guangzhou 510631, China. \newline Institute of Mathematics and Informatics, Jiangxi Normal University, Nanchang 330027, China} \email{liuhuifang73@sina.com} \address{Zhiqiang Mao\newline School of Mathematics, South China Normal University, Guangzhou 510631, China. \newline Depatment of Mathematics, Jiangxi Science and Teachers college, Nanchang 330013, China} \email{maozhiqiang1@sina.com} \thanks{Submitted February 12, 2007. Published August 7, 2007.} \thanks{Supported by grant 04010360 from the Natural Science Foundation of Guangdong, China} \subjclass[2000]{34M10} \keywords{Entire function; exponent of convergence of the zero-sequence} \begin{abstract} In this paper, we investigate the relations between the small functions and the solutions, first, second derivatives, and differential polynomial of the solutions to the differential equation $$f''+A_1e^{P(z)}f'+A_0e^{Q(z)}f=0,$$ where $P(z)=a_nz^n+\dots+a_0$, $Q(z)=b_nz^n+\dots+b_0$ are polynomials with degree $n$ ($n\geq1$), $a_i$, $b_i$ ($i=0,1,\dots,n$), $a_nb_n\neq 0$ are complex constants, $A_j(z)\not \equiv 0$ ($j=0,1$) are entire functions with $\sigma(A_j)0$, there is a set $E\subset[0,2\pi)$ that has linear measure zero, such that if $\varphi\in[0,2\pi)\setminus E$, then there is a constant $R_0=R_0(\varphi)>1$, such that for all $z$ satisfying $\arg z=\varphi$ and $|z|\geq R_0$, we have $\exp{\{-r^{\sigma+\varepsilon}\}}\leq|f(z)| \leq\exp{\{r^{\sigma+\varepsilon}\}}$. \end{lemma} \begin{lemma} \label{lem2} Let $P(z)=(\alpha+i\beta)z^n+\dots$ ($\alpha,\beta$ are real, $|\alpha|+|\beta|\neq0)$ be a polynomial with degree $n\geq1$, $A(z)\not \equiv 0$ be a meromorphic function with $\sigma(A)0$, there is a set $H_1\subset[0,2\pi)$ that has linear measure zero, such that for any $\theta\in[0,2\pi)\setminus(H_1\bigcup H_2)$ and a sufficiently large $r$, we have \begin{itemize} \item[(i)] If $\delta(P,\theta)>0$, then $\exp{\{(1-\varepsilon)\delta(P,\theta)r^n\}}\leq|g(re^{i\theta})| \leq\exp{\{(1+\varepsilon)\delta(P,\theta)r^n\}};$ \item[(ii)] If $\delta(P,\theta)<0$, then $\exp{\{(1+\varepsilon)\delta(P,\theta)r^n\}}\leq|g(re^{i\theta})| \leq\exp{\{(1-\varepsilon)\delta(P,\theta)r^n\}},$ where $H_2=\{\theta\in[0,2\pi);\delta(P,\theta)=0\}$ is a finite set. \end{itemize} \end{lemma} \begin{proof} Rewrite $g(z)$ as $g(z)=we^{(\alpha+i\beta)z^n}$, where $w(z)=A(z)e^{P(z)-(\alpha+i\beta)z^n}$ is a meromorphic function with $\sigma(w)=s1$, for all $z$ satisfying $\arg z=\theta$ and $|z|\geq R$, we have $\exp{\{-r^{s+\varepsilon}\}}\leq|w(z)|\leq\exp{\{r^{s+\varepsilon}\}}$. By $|e^{(\alpha+i\beta)z^n}|=e^{{Re}(\alpha+i\beta)z^n=e^{\delta(P,\theta)r^n}}$, when $\theta\in[0,2\pi)\setminus(H_1\bigcup H_2)$ and $|z|=r>R$, we have $\exp{\{-r^{s+\varepsilon}+\delta(P,\theta)r^n\}}\leq|g(z)|\leq\exp{\{r^{s+\varepsilon}+\delta(P,\theta)r^n\}}$. So by the above inequality and $\delta(P,\theta)>0$ or $\delta(P,\theta)<0$, we complete the proof. \end{proof} \begin{lemma}[\cite{c4}] \label{lem3} Let $f(z)$ be an entire function with infinite order, $d_j(z)$ ($j=0,1,2$) be polynomials that are not all equal to zero. Then $w(z)=d_2f''+d_1f'+d_0f$ has infinite order. \end{lemma} \begin{lemma} \label{lem4} Let $a_i,b_i(i=0,1\dots n)$ be complex constants such that $a_nb_n\neq 0$ and $\arg{a_n}\neq\arg{b_n}$ or $a_n=cb_n (01$, such that for all $z$ satisfying $\arg z=\theta$ and $|z|=r\geq R$, we have $$|H_0(z)|\leq\exp{\{r^{\beta+\varepsilon}\}}. \label{e2.1}$$ By lemma \ref{lem2}, there exists a ray $\arg z=\theta\in[0,2\pi) \setminus (E_0\cup E_1\cup E_2)$, where $E_1\subset[0,2\pi)$ has linear measure zero, $E_2=\{\theta\in[0,2\pi);\delta(P,\theta)=0$ or $\delta(Q,\theta)=0$ or $\delta(P+Q,\theta)=0\}$ is a finite set, such that $\delta(P,\theta)<0$, $\delta(P+Q,\theta)<0$, $\delta(Q,\theta)>0$, and for the above given $\varepsilon$, we have, when $r$ is sufficiently large, \begin{gather} |H_{2Q}e^{2Q}|\geq\exp{\{(1-\varepsilon)2\delta(Q,\theta)r^n\}}, \label{e2.2} \\ |H_{Q}e^{Q}|\leq\exp{\{(1+\varepsilon)\delta(Q,\theta)r^n\}}, \label{e2.3} \\ |H_{P+Q}e^{P+Q}|\leq\exp{\{(1-\varepsilon)\delta(P+Q,\theta)r^n\}}<1, \label{e2.4}\\ |H_{2P}e^{2P}|\leq\exp{\{(1-\varepsilon)2\delta(P,\theta)r^n\}}<1, \label{e2.5} \\ |H_{P}e^{P}|\leq\exp{\{(1-\varepsilon)\delta(P,\theta)r^n\}}<1. \label{e2.6} \end{gather} If $\psi_2(z)+H_{2Q}e^{{2Q}}\equiv 0$, then by \eqref{e2.1}-\eqref{e2.6}, we have \begin{align*} \exp{\{(1-\varepsilon)2\delta(Q,\theta)r^n\}} &\leq|H_{2Q}e^{2Q}|\\ &\leq\exp{\{r^{\beta+\varepsilon}\}}+\exp{\{(1+\varepsilon) \delta(Q,\theta)r^n\}}+3\\ &\leq 3\exp{\{r^{\beta+\varepsilon}\}}\exp{\{(1+\varepsilon) \delta(Q,\theta)r^n\}}. \end{align*} Because $2(1-\varepsilon)-(1+\varepsilon)=1-3\varepsilon>\frac{2}{5}$, we have $\exp{\{\frac{2}{5}\delta(Q,\theta)r^n\}}\leq3\exp{\{r^{\beta+\varepsilon}\}}.$ This is a contradiction to $\beta+\varepsilon0$, and when $|z|=r$ is sufficiently large, we have \eqref{e2.1}-\eqref{e2.3} and \begin{gather} |H_{P+Q}e^{P+Q}|\leq\exp{\{(1+\varepsilon)(1+c)\delta(Q,\theta)r^n\}}, \label{e2.7}\\ |H_{2P}e^{2P}|\leq\exp{\{(1+\varepsilon)2c\delta(Q,\theta)r^n\}}, \label{e2.8}\\ |H_{P}e^{P}|\leq\exp{\{(1+\varepsilon)c\delta(Q,\theta)r^n\}}. \label{e2.9} \end{gather} If $\psi_2(z)+H_{2Q}e^{{2Q}}\equiv 0$, then by \eqref{e2.1}-\eqref{e2.3}, and \eqref{e2.7}-\eqref{e2.9}, we have \begin{aligned} \exp{\{(1-\varepsilon)2\delta(Q,\theta)r^n\}} &\leq|H_{2Q}e^{2Q}|\\ &\leq\exp{\{r^{\beta+\varepsilon}\}} +2\exp{\{(1+\varepsilon)(1+c)\delta(Q,\theta)r^n\}}\\ &\quad+ \exp{\{(1+\varepsilon)2c\delta(Q,\theta)r^n\}} +\exp{\{(1+\varepsilon)c\delta(Q,\theta)r^n\}}. \end{aligned} \label{e2.10} Because $0<\varepsilon<\min\{\frac{1-c}{5},n-\beta\}$, when $r\to+\infty$, we have \begin{gather} \frac{\exp{\{r^{\beta+\varepsilon}\}}}{\exp{\{(1-\varepsilon) 2\delta(Q,\theta)r^n\}}}\to 0, \label{e2.11}\$5pt] \frac{\exp{\{(1+\varepsilon)(1+c)\delta(Q,\theta)r^n\}}}{\exp{\{(1-\varepsilon) 2\delta(Q,\theta)r^n\}}}\to 0, \label{e2.12}\\[5pt] \frac{\exp{\{(1+\varepsilon)2c\delta(Q,\theta)r^n\}}}{\exp{\{(1-\varepsilon) 2\delta(Q,\theta)r^n\}}}\to 0, \label{e2.13}\\[5pt] \frac{\exp{\{(1+\varepsilon)c\delta(Q,\theta)r^n\}}}{\exp{\{(1-\varepsilon) 2\delta(Q,\theta)r^n\}}}\to 0. \label{e2.14} \end{gather} By \eqref{e2.10}-\eqref{e2.14}, we get a contradiction. Hence \psi_2(z)+H_{2Q}e^{{2Q}}\not\equiv 0. \smallskip Proof of (vi). By \sigma(\varphi)<\infty and \cite[p. 89]{g2} we know, for any given \varepsilon>0, there exists a set E\subset[0,2\pi) that has linear measure zero, if \theta\in[0,2\pi)\setminus E, then there exists a constant R=R(\theta)>1, such that for all z satisfying \arg z=\theta and |z|\geq R, we have \[ |\frac{\varphi^{(k)}(z)}{\varphi(z)}|\leq|z|^{k(\sigma(\varphi)-1 +\varepsilon)}\quad (k=1,2).$ So on the ray $\arg z=\theta\in[0,2\pi)\setminus E$, $\frac{\varphi^{(k)}(z)}{\varphi(z)}H_j(z)e^j$ ($k=1,2,j\in\Lambda_2$) keep the properties of $H_je^j$ which are defined as in \eqref{e2.1}, \eqref{e2.3}--\eqref{e2.6} or \eqref{e2.1}, \eqref{e2.3}, \eqref{e2.7}--\eqref{e2.9}. Using a similar reasoning to that in the proof of (ii), we can prove (vi). \end{proof} \begin{lemma}[\cite{c2}] \label{lem5} Suppose that $A_0,\dots,A_{k-1},F\not\equiv 0$ are finite-order meromorphic functions. If $f$ is an infinite-order meromorphic solution of the equation $f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=F,$ then $f$ satisfies $\overline{\lambda}(f)=\lambda(f)=\sigma(f)=\infty$. \end{lemma} \section{Proofs of Theorems} \begin{proof}[Proof of Theorem \ref{thm1}] Suppose that $f(z)\not \equiv 0$ is a solution of \eqref{e1.1}. First of all we prove that $\overline{\lambda}(f-\varphi)=\infty$. Set $g_0=f-\varphi$, then $\sigma(g_0)=\sigma(f)=\infty$,$\overline{\lambda}(g_0)=\overline{\lambda}(f-\varphi)$. Substituting $f=g_0+\varphi$,$f'=g'_0+\varphi'$,$f''=g''_0+\varphi''$ into equation \eqref{e1.1}, we have $$g''_0+A_1e^{P(z)}g'_0+A_0e^{Q(z)}g_0=-(\varphi''+A_1e^{P(z)}\varphi' +A_0e^{Q(z)}\varphi). \label{e3.1}$$ We remark that \eqref{e3.1} may have finite-order solution (For example when $\varphi(z)=z$, $g_0=-z$ solves the equation \eqref{e3.1}). But here we discuss only the case $\sigma(g_0)=\infty$. By $\varphi(z)$ being a finite-order entire function and Theorem \ref{thmA}, we know $\varphi''+A_1e^{P(z)}\varphi'+A_0e^{Q(z)}\varphi\not\equiv0$. Hence by lemma \ref{lem5}, we have $\overline{\lambda}(g_0)=\sigma(g_0)=\infty$, i.e. $\overline{\lambda}(f-\varphi)=\infty$. Secondly we prove $\overline{\lambda}(f'-\varphi)=\infty$. Set $g_1=f'-\varphi$, then $\sigma(g_1)=\sigma(f')=\sigma(f)=\infty$,$\overline{\lambda}(g_1)=\overline{\lambda}(f'-\varphi)$. Differentiating both sides of equation \eqref{e1.1}, we get $$f'''+A_1e^{P(z)}f''+[(A_1e^{P(z)})'+A_0e^{Q(z)}]f'+(A_0e^{Q(z)})'f=0. \label{e3.2}$$ Substituting $f=-\frac{1}{A_0e^{Q(z)}}[f''+A_1e^{P(z)}f']$ into \eqref{e3.2}, we get $$f'''+[A_1e^{P(z)}-\frac{(A_0e^{Q(z)})'}{A_0e^{Q(z)}}]f'' +[(A_1e^{P(z)})'+A_0e^{Q(z)}-\frac{(A_0e^{Q(z)})'}{A_0e^{Q(z)}}A_1e^{P(z)}]f'=0. \label{e3.3}$$ Substituting $f'=g_1+\varphi$, $f''=g'_1+\varphi'$, $f'''=g''_1+\varphi''$ into equation \eqref{e3.3}, we get $$g''_1+h_1g'_1+h_0g_1=h, \label{e3.4}$$ where $h_1=A_1e^{P(z)}-\frac{(A_0e^{Q(z)})'}{A_0e^{Q(z)}}$, \begin{gather*} h_0=(A_1e^{P(z)})'+A_0e^{Q(z)}-\frac{(A_0e^{Q(z)})'}{A_0e^{Q(z)}}A_1e^{P(z)},\\ -h=\varphi''-(\frac{A'_0}{A_0}+Q')\varphi'+[A_1\varphi'+A'_1\varphi +P'A_1\varphi-\frac{A'_0}{A_0}A_1\varphi-Q'A_1\varphi]e^P+A_0\varphi e^Q. \end{gather*} Now we prove $h\not\equiv0$. If $h\equiv0$, then $$\frac{\varphi''}{\varphi}-(\frac{A'_0}{A_0}+Q')\frac{\varphi'}{\varphi}+ [\frac{\varphi'}{\varphi}+\frac{A'_1}{A_1}+P'-\frac{A'_0}{A_0}-Q']A_1e^P+A_0 e^Q=0. \label{e3.5}$$ By $\sigma(\varphi)<\infty$, $\sigma(A_j)1$, such that for all $z$ satisfying $\arg z=\theta$ and $|z|\geq R$, we have \begin{gather} |\frac{\varphi^{(k)}(z)}{\varphi(z)}|\leq|z|^{k(\sigma(\varphi)-1 +\varepsilon)}\quad (k=1,2), \label{e3.6} \\ |\frac{A'_j(z)}{A_j(z)}|\leq|z|^{\sigma(A_j)-1+\varepsilon}\quad (j=0,1). \label{e3.7} \end{gather} Since $P(z)$, $Q(z)$ are polynomials with degree $n$, when $|z|=r$ is sufficiently large, we have $$|P'(z)|\leq r^n\quad \text{and}\quad |Q'(z)|\leq r^n . \label{e3.8}$$ So by \eqref{e3.6}-\eqref{e3.8}, there exists a positive constant $M$, such that for all $z$ satisfying $\arg z=\theta\in[0,2\pi)\setminus E_0$, we have, when $|z|=r$ is sufficiently large, \begin{gather} \big|(\frac{A'_0}{A_0}+Q')\frac{\varphi'}{\varphi}\big|\leq r^M, \label{e3.9} \\ \big|\frac{\varphi'}{\varphi}+\frac{A'_1}{A_1}+P'-\frac{A'_0}{A_0}-Q'\big|\leq r^M. \label{e3.10} \end{gather} If $\arg{a_n}\neq\arg{b_n}$, then by lemma \ref{lem2}, there exists a ray $\arg z=\theta\in[0,2\pi) \setminus (E_0\cup E_1\cup E_2)$, $E_1\subset[0,2\pi)$ having linear measure zero, $E_2=\{\theta\in[0,2\pi);\delta(P,\theta)=0$ or $\delta(Q,\theta)=0\}$ being a finite set, such that $\delta(P,\theta)<0, \delta(Q,\theta)>0$, and for the above given $\varepsilon$, we have, when $r$ is sufficiently large, \begin{gather} |A_0e^{Q}|\geq\exp{\{(1-\varepsilon)\delta(Q,\theta)r^n\}}, \label{e3.11} \\ |A_1e^{P}|\leq\exp{\{(1-\varepsilon)\delta(P,\theta)r^n\}}<1. \label{e3.12} \end{gather} So by \eqref{e3.5},\eqref{e3.6} and \eqref{e3.9}-\eqref{e3.12}, we get $\exp{\{(1-\varepsilon)\delta(Q,\theta)r^n\}}\leq|A_0e^{Q}| \leq r^{2(\sigma(\varphi)-1+\varepsilon)}+r^M+r^M.$ This is absurd. If $a_n=cb_n(00$, and for the above given $\varepsilon$, when $r$ is sufficiently large, we have \eqref{e3.11} and $$|A_1e^{P}|\leq\exp{\{(1+\varepsilon)c\delta(Q,\theta)r^n\}}. \label{e3.13}$$ So by \eqref{e3.5}, \eqref{e3.6}, \eqref{e3.9}-\eqref{e3.11} and \eqref{e3.13}, we get \begin{align*} \exp{\{(1-\varepsilon)\delta(Q,\theta)r^n\}}&\leq|A_0e^{Q}|\\ &\leq r^{2(\sigma(\varphi)-1+\varepsilon)}+r^M +r^M\exp{\{(1+\varepsilon)c\delta(Q,\theta)r^n\}}\\ &\leq3\exp{\{(1+2\varepsilon)c\delta(Q,\theta)r^n\}}. \end{align*} This is a contradiction to $0<\varepsilon<\frac{1-c}{1+2c}$. From the above proof, we get $h\not\equiv0$. From $h\not\equiv0$ and lemma \ref{lem5} we get $\overline{\lambda}(g_1)=\sigma(g_1)=\infty$. Hence $\overline{\lambda}(f'-\varphi)=\infty$. Finally we prove that $\overline{\lambda}(f''-\varphi)=\infty$. Set $g_2=f''-\varphi$, then $\sigma(g_2)=\sigma(f'')=\sigma(f)=\infty$, $\overline{\lambda}(g_2)=\overline{\lambda}(f''-\varphi)$. Differentiating both sides of equation \eqref{e3.2}, we get $$f^{(4)}+A_1e^{P}f'''+[2(A_1e^{P})'+A_0e^{Q}]f''+[(A_1e^{P})''+2(A_0e^{Q})']f' +(A_0e^{Q})''f=0. \label{e3.14}$$ Substituting $f=-\frac{1}{A_0e^{Q}}[f''+A_1e^{P}f']$ into \eqref{e3.14}, we get \begin{aligned} f^{(4)}+A_1e^{P}f'''+[2(A_1e^{P})'+A_0e^{Q}-\frac{(A_0e^{Q})''}{A_0e^{Q}}]f''&\\ +[(A_1e^{P})''+2(A_0e^{Q})'-\frac{(A_0e^{Q})''}{A_0e^{Q}}A_1e^{P}]f'&=0. \end{aligned} \label{e3.15} By \eqref{e3.3} and \eqref{e3.15}, we have $$f^{(4)}+H_3f'''+H_2f''=0, \label{e3.16}$$ where \begin{gather} H_3=A_1e^{P}-\frac{\varphi_1(z)}{\varphi_2(z)}, \label{e3.17} \\ H_2=2(A_1e^{P})'+A_0e^{Q}-\frac{(A_0e^{Q})''}{A_0e^{Q}} -\frac{\varphi_1(z)}{\varphi_2(z)}[A_1e^{P}-\frac{(A_0e^{Q})'}{A_0e^{Q}}], \label{e3.18}\\ \varphi_1(z)=(A_1e^{P})''+2(A_0e^{Q})'-\frac{(A_0e^{Q})''}{A_0e^{Q}}A_1e^{P}, \label{e3.19}\\ \varphi_2(z)=(A_1e^{P})'+A_0e^{Q}-\frac{(A_0e^{Q})'}{A_0e^{Q}}A_1e^{P}, \label{e3.20} \end{gather} and $\varphi_2(z) \not \equiv 0$ by Lemma \ref{lem4} (i). Clearly, $H_3,H_2,\varphi_1(z),\varphi_2(z)$ are meromorphic functions with $\sigma(\varphi_k)\leq n(k=1,2)$, $\sigma(H_j)\leq n(j=2,3)$. Substituting $f''=g_2+\varphi$,$f'''=g'_2+\varphi'$,$f^{(4)}=g''_2+\varphi''$ into \eqref{e3.16}, $g''_2+H_3g'_2+H_2g_2=-(\varphi''+H_3\varphi'+H_2\varphi).$ If we can prove that $-(\varphi''+H_3\varphi'+H_2\varphi)\not\equiv0$, then by lemma \ref{lem5}, we get $\overline{\lambda}(g_2)=\sigma(g_2)=\infty$. Hence $\overline{\lambda}(f''-\varphi)=\infty$. Now we prove $-(\varphi''+H_3\varphi'+H_2\varphi)\not\equiv0$. Notice that \begin{gather*} (A_1e^P)'=(A_1'+A_1P')e^P,\quad (A_1e^P)''=(A_1''+2A_1'P'+A_1(P')^2+A_1P'')e^P,\\ \frac{(A_0e^Q)'}{A_0e^Q}=\frac{A_0'}{A_0}+Q',\quad \frac{(A_0e^Q)''}{A_0e^Q}=\frac{A_0''}{A_0}+2\frac{A_0'}{A_0}Q'+(Q')^2+Q''. \end{gather*} So by \eqref{e3.17}-\eqref{e3.20}, we have \begin{gather} \varphi_1(z)=B_1e^P+2(A_0'+A_0Q')e^Q, \label{e3.21} \\ \varphi_2(z)=B_2e^P+A_0e^Q, \label{e3.22}\\ H_3=\frac{1}{\varphi_2(z)}H_4, \label{e3.23}\\ H_2=\frac{1}{\varphi_2(z)}[A_0^2e^{2Q}+H_5], \label{e3.24} \end{gather} where \begin{gather*} \begin{aligned} H_5&=[2A_0(A_1'+A_1P')+A_0B_2-2A_1(A_0'+A_0Q')]e^{P+Q}\\ &\quad +[2B_2(A_1'+A_1P')-A_1B_1]e^{2P}-[A_0''+2A_0'Q'+A_0(Q')^2\\ &\quad +A_0Q''-2(\frac{A_0'}{A_0}+Q')(A_0'+A_0Q')]e^Q\\ &\quad -[B_2(\frac{A_0''}{A_0}+2\frac{A_0'}{A_0}Q'+(Q')^2+Q'')-B_1(\frac{A_0'}{A_0}+Q')]e^P, \end{aligned} \\ H_4=A_1A_0e^{P+Q}+A_1B_2e^{2P}-2(A_0'+A_0Q')e^Q-B_1e^P,\\ B_1=A_1''+2A_1'P'+A_1(P')^2+A_1P''-\frac{A_1}{A_0}(A_0''+2A_0'Q'+A_0(Q')^2 +A_0Q''), \\ B_2=A_1'+A_1P'-A_1(\frac{A_0'}{A_0}+Q'). \end{gather*} Clearly, $B_1,B_2$ are meromorphic functions with \$\sigma(B_j)