Consider this ODE on $[1, \infty)$

$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - ({4a} + m(m+1))f(r) = -4af(1) $

with initial conditions

$\frac{a}{1-2a} f(1) + f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$

where $0\leq a < \frac{1}{2}$, $m$ is a positive integer, and $C \in \mathbb{R}$.

I want to ask if there exists a unique solution (at least for $a$ small enough).

If $a=0$, then this becomes the Euler equation:

$r^2f''(r) + 2r f'(r) - (m(m+1))f(r) = 0 $

$ f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$

which we know the unique solution is: $f(r) = \frac{-C}{\alpha r^{\alpha}}$ where $\alpha = \frac{1}{2} + \frac{\sqrt{1+4m(m+1)}}{2}$

Can I prove existence and/or uniqueness for the $a>0$ case using some kind of continuity method?

I know for instance that injectivity is a continuous property for elliptic operators, and one has the method of continuity to prove surjectivity of a 1-parameter family of elliptic operators. Is there something similar in this context?

Any help or references is appreciated.

$\textbf{EDIT} $: I wrote the equations incorrectly above. I apologize for that. I allowed the right side to decay to $0$ and so I believe it's possible to prove existence now. Here are the correct equations:

$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{4a^2}{r(r-2a)}f(1)+ \frac{4a(1-2a)}{(1-a)r(r-2a)} C $

with initial conditions

$f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$

There is an ODE that is somehow related to the above non-local differential equation.

$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{2a}{r(r-2a)} D $

with initial conditions

$\frac{2a}{1-2a}f(1) + \frac{2}{1-a}f'(1) = D, \qquad \lim_{r\to \infty} f(r) = 0$ where $D$ is any real number.


2 Answers 2


As Iosif said, in general the system you specified does not admit a solution. Here we will give a more pedestrian argument using only comparisons.


Claim: if a solution exists, and $f(1) > 0$, then the function is monotonically decreasing; if $f(1) < 0$, then the function is monotonically increasing.

Proof: we will focus on the positive case. The negative case is similar. Let $\zeta = \frac{4a}{4a + m(m+1)} \in (0,1)$ (if $a\in (0,\frac12)$). The second derivative test shows that $f$ cannot have a local maximum with $f(r) > \zeta f(1)$ or a local minimum with $f(r) < \zeta f(1)$. This immediately implies monotonicity in light of $f(1) > 0 = \lim f(r)$.

More details on monotonicity proof: Given $f(1) > f(\infty) = 0$. Suppose $f$ were not monotonic. Then there exists $r_m, r_M$ with $1 < r_m < r_M < \infty$ such that $f(r_m) < f(r_M)$.

I claim that $f(r_M) > 0$ and $f(r_m) < f(1)$. Suppose not: if $f(r_M) \leq 0$ then $f(r_m) < 0$ and $f$ would have a negative local minimum, in contradiction to the second derivative test. If $f(r_m) \geq f(1)$ then $f$ would have a local maximum with value $> f(1)$, against in contradiction to the second derivative test.

Therefore there must exist a local minimum $s_m\in (1,r_M)$ with $f(s_m) \leq f(r_m)$ and a local maximum $s_M\in (r_m,\infty)$ with $f(s_M) \geq f(r_M)$.

We have therefore established $$ \zeta f(1) \leq f(s_m) < f(s_M) \leq \zeta f(1) $$ which is a contradiction.

In particular, we must have $f' \leq 0$ on $[1,\infty)$.


We have then

$$ f'(1) - f'(r) = -\int_1^r f''(s) ~ds = \int_1^r \frac{4a}{s^2 - 2as} f(1) - \frac{4a + m(m+1)}{s^2 - 2a s} f(s) + \frac{2(s-a)}{s^2 - 2as} f'(s) ~ds $$

Since $f'$ is signed, we know that it is absolutely integrable on $[1,\infty)$. Since $f$ is monotonic (and hence bounded) the second integrand is also absolutely integrable. We conclude then that $\lim_{r\to\infty} f'(r)$ exists.

Since $\lim_{r\to\infty} f(r) = 0$, we must have also $\lim_{r\to\infty} f'(r) = 0$.

But now writing $f'(r) = - \int_r^\infty f''(s) ~ds$ using the above formula, we see that asymptotically $|f'(r)| \sim \frac1r$ (coming from the $4a f(1)$ term if it is non-zero; the other two terms can both be bounded by $O(1/r) f(r) = o(1/r)$). But this contradicts the integrability of $f'(r)$.

And hence we have proved:

Claim: no solution can exist with $f(1) \neq 0$.


When $f(1) = 0$, the same maximum principle argument shows that $f$ must be identically zero. This shows that

Theorem The only solution to your system is $f \equiv 0$, with $f(1) = f'(1) = 0$ and $C = 0$.

Final remark

Heuristically, if you want to look for asymptotically constant solutions to your equation, you probably want it to converge to $\zeta f(1)$ in the limit.

  • $\begingroup$ Very nice! I thought of such qualitative analysis, but was too lazy for that. $\endgroup$ Aug 23, 2021 at 23:31
  • 1
    $\begingroup$ @Laithy : It is seen from my answer that, if we replace $-4af(1)$ with a nonzero constant, then there is no solution vanishing at $\infty$. $\endgroup$ Aug 24, 2021 at 3:58
  • 1
    $\begingroup$ @Laithy: I added more details on the proof of monotonicity. $\endgroup$ Aug 24, 2021 at 14:27
  • 1
    $\begingroup$ Also: if you replace $-4a f(1)$ by $-4a p$ for any $p\in \mathbb{R}$, the same argument shows that $f$ is eventually monotonic. If $p > 0$, then the argument says no local max with $f(r) > \zeta p$ and no local min with $f(r) < \zeta p$. Such a condition rules out any solution with $r_1 < r_2 < r_3 < r_4$ such that $f(r_2) < \min(f(r_1), f(r_3))$ and $f(r_3) > \max(f(r_2), f(r_4))$. This leads to eventual monotonicity. Then the asymptotic argument can kick in and also rule out any solution that limits to $0$ at $\infty$. $\endgroup$ Aug 24, 2021 at 14:42
  • 1
    $\begingroup$ To be clear: the problem is with the asymptotic limit. For any $p, f_0, f_1\in \mathbb{R}$, you can always solve (globally on $[1,\infty)$) the linear inhomogeneous ODE $$ (r^2 - 2ra) f'' + 2(r-a) f' - (4a + m(m+1)) f = -4a p$$ with initial data $f(1) = f_0$ and $f'(1) = f_1$. The claim is that when $p\neq 0$, no solution has $\lim_{r\to\infty} f(r) = 0$. I suspect the only possible values for $\lim_{r\to\infty} f(r)$ are $\{\pm \infty, \zeta p\}$. $\endgroup$ Aug 24, 2021 at 14:51

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.

  • $\begingroup$ Thank you so much for your answer! I actually wrote the equations incorrectly. I should have allowed for the right side to decay to 0. I think now one can prove existence. But we cannot make use of Legendre polynomials anymore. Do you have any idea how to approach this (new) problem? I edited the original post. $\endgroup$
    – Laithy
    Aug 24, 2021 at 15:00
  • $\begingroup$ @Laithy : The amended equation can still be solved more or less explicitly, with the expression of the solution involving an integral of an integrand expressed in terms of Legendre functions. However, this expression of the solution seems to be much harder to analyze. It may be better to post the amended equation separately. $\endgroup$ Aug 24, 2021 at 15:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .