Dependence of a solution of a linear ODE on parameter

Question

Is the following theorem known, or can be easily derived from known results?

Consider the differential equation $$w''-kz^{-1}w'=(\lambda+\phi(z))w,$$ where $k>0$ is fixed, $\lambda$ is a large (complex) parameter and $\phi$ is a complex valued function analytic on $[0,1]$. It is known that there exists a unique solution $w_0(z,\lambda)$ which is analytic at $z=0$ and $w_0(0,\lambda)=1$. It is also known that $f(\lambda)=w_0(1,\lambda)$ is an entire function.

Theorem (?). $f(\lambda)=(1+o(1))\exp\sqrt{\lambda},$ as $\lambda\to\infty$, $|\arg\lambda|\leq\pi-\epsilon$, where $\sqrt{\lambda}$ is the principal branch.

Remark. There is another solution, normalized by $w_1(z,\lambda)\sim z^{k+1},z\to 0$, and for this solution I know how to prove the result, and know the references, for example Olver, Asymptotics and special functions, Ch. 12.

More remarks: 1. When $\phi=0$ this reduces to Bessel's equation; the Theorem is true in this case but not trivial.

2. In the definition of $w_0$ the crucial word is ANALYTIC. There are infinitely many other solutions $w$ satisfying $w(0)=1$: adding to $w_0$ any multiple of $w_1$ does not change the value at $0$. And the conclusion of the Theorem does NOT hold for some of solutions satisfying $w(0)=1$. For example, when $\phi=0$ there is a solution with $w(0)=1$, for which $w(1,\lambda)$ decays exponentially for $\lambda>0$.

3. The most important case for me is when $\phi$ is even, which may help. But on my opinion, the problem is interesting in the general case as well.

Remark. Now I proved this, but the question remains whether this follows from some known, published results.

Answer 1

Consider the analytic vector field $$X(z,w,W)=z \partial_z+zW\partial_w+(kW+z(\lambda+\phi(z))w)\partial_W$$ whose orbits project to the graphs of solutions $z\mapsto w(z)$ (it is simply the companion vector field given by $W:=w'$ and scaled by $z$ to remove the pole).

The set $\{z=0\}\cap\{W=0\}$ is a line of singularities, and the special solutions $w_0,w_1$ pass through that line. The linearization of $X$ at $(0,0,0)$ is the matrix $$M_0=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & k \end{array}\right]$$ and at $(0,1,0)$ it is $$M_1=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ \lambda & 0 & k \end{array}\right].$$ If $k\neq 1$ the matrix $M_1$ is diagonalizable: in the new linear basis of $\mathbb C^3$ the linear part of $X$ has now the form $M_0$. Observe that one can require that only $W$ be affected by the diagonalization transform. In particular, $w_0$ is left untouched and so is its value at $1$.

If I understood correctly the $M_0$ case has been dealt with already, although it could happen that this part gets broken by the diagonalization (which does not preserve the companion structure). But I do not know the specifics of the OP argument.

In the case $k=1$ the argument does not work since $M_1$ is not diagonalizable anymore. Maybe something similar can be cooked up.

Answer 2

Maybe, something to adjust in view of the following results of Maple.

Order := 3; dsolve(((D@@2)(w))(z)-(D(w))(z)/z = (lambda+phi(z))*w(z), w(z), series, z = 0);

$$ w \left( z \right) ={\it \_C1}\,{z}^{2} \left( (1+ \left( {\frac { \lambda}{8}}+{\frac {\phi \left( 0 \right) }{8}} \right) {z}^{2}+O \left( {z}^{3} \right) ) \right) +{\it \_C2}\, \left( \ln \left( z \right) \left( ( \left( -\lambda-\phi \left( 0 \right) \right) {z}^ {2}+O \left( {z}^{3} \right) ) \right) +(-2+O \left( {z}^{3} \right) ) \right) $$

Order := 3; dsolve(((D@@2)(w))(z)-2*(D(w))(z)/z = (lambda+phi(z))*w(z), w(z), series, z = 0);

$$w \left( z \right) ={\it \_C1}\,{z}^{3} \left( (1+ \left( {\frac { \lambda}{10}}+{\frac {\phi \left( 0 \right) }{10}} \right) {z}^{2}+O \left( {z}^{3} \right) ) \right) +{\it \_C2}\, \left( (12+ \left( -6 \,\lambda-6\,\phi \left( 0 \right) \right) {z}^{2}+O \left( {z}^{3} \right) ) \right) $$