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.



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


