Existence and uniqueness of an Euler-type ODE with varying parameters

Question

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.

Answer 1

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.

Monotonicity

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 $[1,\infty)$.

Comparison

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 $f(1) \neq 0$.

Uniqueness

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.

Answer 2

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.