Return to Answer

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general real solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary real constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds whose values are real on $(1,\infty)$; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}>1. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

According to Sections 15.23 and 15.33 of Whittaker and Watson, 4th ed.,
\begin{equation*} P_s(\infty-)=\infty,\quad Q_s(\infty-)=0. \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_1=0$ and hence \begin{equation*} F(\infty-)= c_2 B, \end{equation*} which is in general nonzero.

Thus, in general no solution to your differential problem exists.

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general real solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary real constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds whose values are real on $(1,\infty)$; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}>1. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

According to Sections 15.23 and 15.33 of Whittaker and Watson, 4th ed.,
\begin{equation*} P_s(\infty-)=\infty,\quad Q_s(\infty-)=0,\quad Q_s>0\text{ on }(1,\infty). \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_1=0$ and hence \begin{equation*} F(\infty-)= c_2 B. \end{equation*} Also, if $a\in(0,1/2)$, then the inequality in (3) implies $B>0$. So, $F(\infty-)\ne0$ -- unless $c_2=0$ and hence $F=0$.

Thus, there is no nonzero solution to your differential problem.

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary complex constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

It appears that for $x>0$ we have \begin{equation*} \text{$\Im Q_s(x)=-\frac\pi2\,P_s(x)\to-\infty$ and $\Re Q_s(x)\to0$ as $x\to\infty$.} \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_2$ is real and $c_1=i\frac\pi2\,c_2$, so that \begin{equation*} F(r)= c_2 \left(i\frac\pi2\,P_s\left(\frac{r}{a}-1\right) +Q_s\left(\frac{r}{a}-1\right)+i\frac\pi2\, A+ B\right). \end{equation*} So, by (3), \begin{equation*} F(\infty-)= c_2 \left(i\frac\pi2\, A+ B\right), \end{equation*} which is in general nonzero.

Thus, in general no solution to your differential problem appears to exist. (I am using here "appears to exist" instead of "exists", because at this point I do not know how to verify the identity in (3). I know how to verify the limits in (3).)

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general real solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary real constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds whose values are real on $(1,\infty)$; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}>1. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

According to Sections 15.23 and 15.33 of Whittaker and Watson, 4th ed.,
\begin{equation*} P_s(\infty-)=\infty,\quad Q_s(\infty-)=0. \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_1=0$ and hence \begin{equation*} F(\infty-)= c_2 B, \end{equation*} which is in general nonzero.

Thus, in general no solution to your differential problem exists.

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary complex constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

It appears that for $x>0$ we have \begin{equation*} \text{$\Im Q_s(x)=-\frac\pi2\,P_s(x)\to-\infty$ and $\Re Q_s(x)\to0$ as $x\to\infty$.} \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_2$ is real and $c_1=i\frac\pi2\,c_2$, so that \begin{equation*} F(r)= c_2 \left(i\frac\pi2\,P_s\left(\frac{r}{a}-1\right) +Q_s\left(\frac{r}{a}-1\right)+i\frac\pi2\, A+ B\right). \end{equation*} So, by (3), \begin{equation*} F(\infty-)= c_2 \left(i\frac\pi2\, A+ B\right), \end{equation*} which is in general nonzero.

Thus, in general no solution to your differential problem appears to exist. (I am using here "appears to exist" instead of "exists", because at this point I do not know how to verify the identity in (3) and the second limit there. I know how to verify the first limit in (3).)

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary complex constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*}

It appears that for $x>0$ we have \begin{equation*} \text{$\Im Q_s(x)=-\frac\pi2\,P_s(x)\to-\infty$ and $\Re Q_s(x)\to0$ as $x\to\infty$.} \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_2$ is real and $c_1=i\frac\pi2\,c_2$, so that \begin{equation*} F(r)= c_2 \left(i\frac\pi2\,P_s\left(\frac{r}{a}-1\right) +Q_s\left(\frac{r}{a}-1\right)+i\frac\pi2\, A+ B\right). \end{equation*} So, by (3), \begin{equation*} F(\infty-)= c_2 \left(i\frac\pi2\, A+ B\right), \end{equation*} which is in general nonzero.

Thus, in general no solution to your differential problem appears to exist. (I am using here "appears to exist" instead of "exists", because at this point I do not know how to verify the identity in (3). I know how to verify the limits in (3).)